The Oxford Solid State BasicsSolutions to Exercises Steven H. Simon Oxford University CLARENDON PRESS . OXFORD 2015 iii These are the solutions to exercises from the Book The Oxford Solid State Basics by Steven H. Simon, published by Oxford University Press, 2013 edition. Please do everyone a favor and do not circulate these solutions. Do not post these solutions on your website. Do not put them on Russian websites. Do not copy them and hand them out to students. While there is no way for me to enforce these reasonable rules, be assured that I, being a professor at Hogwarts, am in possession of powerful hexes which I have used to protect the secrecy of these solutions. Those who attempt to circulate these solutions unlawfully will activate the hex and will suffer thirty years of bad luck, including spiders crawling into your underwear. Some of these solutions have been tested through use in several years of courses. Other solutions have not been completely tested. Errors or ambiguities that are discovered in the exercises will be listed on my web page. If you think you have found errors in the problems or the solutions please do let me know, and I will make sure to fix them in the next version. Doing so will undoubtedly improve your Karma. , Steven H Simon Oxford, United Kingdom January 2014 . and Thermal Expansion 49 9 Vibrations of a One-Dimensional Monatomic Chain 55 10 Vibrations of a One-Dimensional Diatomic Chain 71 11 Tight Binding Chain (Interlude and Preview) 81 12 Crystal Structure 95 13 Reciprocal Lattice.and Dia-Magnetism 159 . Einstein. Waves in Crystals 99 14 Wave Scattering by Crystals 111 15 Electrons in a Periodic Potential 125 16 Insulator. and Debye 3 3 Electrons in Metals: Drude Theory 15 4 More Electrons in Metals: Sommerfeld (Free Electron) Theory 21 5 The Periodic Table 35 6 What Holds Solids Together: Chemical Bonding 39 7 Types of Matter 47 8 One-Dimensional Model of Compressibility. Sound. Semiconductor.Contents 1 About Condensed Matter Physics 1 2 Specific Heat of Solids: Boltzmann. or Metal 135 17 Semiconductor Physics 139 18 Semiconductor Devices 149 19 Magnetic Properties of Atoms: Para. Brillouin Zone. and Ferri-Magnetism 167 21 Domains and Hysteresis 175 22 Mean Field Theory 179 23 Magnetism from Interactions: The Hubbard Model 191 .vi Contents 20 Spontaneous Magnetic Order: Ferro-. Antiferro-. About Condensed Matter Physics 1 There are no exercises for chapter 1. . . Show that the high temperature limit agrees with Using the partition function. pacity 3kB . Consider a three dimensional simple harmonic oscilla- (b) Quantum Einstein Solid: tor with mass m and spring constant k (i. in agreement with the law of (a) Classical Einstein (or “Boltzmann”) Solid: Dulong and Petit.7 for more on the same topic) (a) p2 k H= + x2 Z 2mZ 2 dp Z= dx e−βH(p. The Hamiltonian is given in the Calculate the quantum partition function usual way by p2 k X H= + x2 Z= e−βEj 2m 2 Calculate the classical partition function j Z Z dp where the sum over j is a sum over all eigenstates. Einstein. the law of Dulong of Petit. and Debye 2 (2. tors. Sketch the heat capacity as a function of tempera- Conclude that if you can consider a solid to consist ture.x) (2π~)3 Since.e. Z ∞ 2 p dy e−ay = π/a −∞ in three dimensions. Find an expression for the heat capacity.1) Einstein Solid ity should be 3N kB = 3R.Specific Heat of Solids: Boltzmann. we get h p p i3 Z = 1/(2π~) π/(β/2m) π/(βk/2)) = (~ωβ)−3 p with ω = k/m. From the partition function U = −(1/Z)∂Z/∂β = 3/β = 3kB T . calculate the heat ca.. Note: in this problem p and x are three dimensional vec. then the heat capac. of N atoms all in harmonic wells. Z= dx e−βH(p. (See also exercise 2. the mass Now consider the same Hamiltonian quantum mechan- is attracted to the origin with the same spring constant ically. in all three directions).x) (2π~)3 Explain the relationship with Bose statistics. 1. we have eigenenergies En = ~ω(n + 1/2) p with ω = k/m. The high temperature limit gives nB (x) → 1/(x+ x2 /2) = 1/x− 1/2 so that hEi → kB T .n2.75 where nB is the boson occupation factor 0. on horizontal.n3 = ~ω[(n1 + 1/2) + (n2 + 1/2) + (n3 + 1/2)] as large. (b) Quantum mechanically. .n3 = [Z1d ]3 n1. Einstein.n3≥0 and correspondingly 1 hEi = 3~ω(nB (β~ω) + ) 2 So the high temperature limit is hEi → 3kB T and the heat capacity C = ∂hEi/∂T = 3kB . we obtain 00 1 2 eβ~ω C = kB (β~ω)2 Fig. 2. More generally we obtain eβ~ω C = 3kB (β~ω)2 (eβ~ω − 1)2 Plotted this looks like Fig. In three dimensions. The partition function is then X Z1d = e−β~ω(n+1/2) n≥0 = e−β~ω/2 1/(1 − e−β~ω ) = 1/[2 sinh(β~ω/2)] The expectation of energy is then hEi = −(1/Z)∂Z/∂β = (~ω/2) coth(β~ω/2) 1 1 = ~ω(nB (β~ω) + ) 2 0. and X Z3d = e−βEn1 . More generally.n2 . 2. Units are kb on vertical axis and kb T /ω In 3D. and Debye Thus the heat capacity ∂U/∂T is 3kB .5 nB (x) = 1/(ex − 1) 0. for a 1d harmonic oscillator.1 Heat capacity in the Einstein (eβ~ω − 1)2 model (per atom) in one dimension. 4 Specific Heat of Solids: Boltzmann.25 (hence again the relationship with free bosons). the heat capacity per atom is three times En1 .n2 . We thus obtain Z ωcutof f E = dωg(ω)[nB (β~ω) + 1/2]~ω 0 where 12πω 2 9ω 2 g(ω) = N = N (2π)3 nv 3 ωd3 and we have replaced nL3 = N where n is the density of atoms. the low temperature limiting form is 3 12N kB π 4 T C= (2. Note that there is no dependence of g(ω) on the density n (it cancels). obtain the high and low tem. 5 (2.1) 5 TD and the high temperature limit is 3N kB . 1.0 8.2) Debye Theory I (b) The following table gives the heat capacity C for potassium iodide as a function of temperature. TD = ~ωDebye /kB . 5 . Here ωd3 = 6π 2 nv 3 is the Debye frequency.12 You may find the following integral to be useful 8 .6 × 10−4 perature limits of the heat capacity analytically.8 0 e −1 n=1 0 n=1 n 15 20 6. and dx x 2 = make an estimate of the Debye temperature. Derive the Debye heat capacity as a function of T (K) C(J K−1 mol−1 ) temperature (you will have to leave the final result in terms of an integral that cannot be done analytically). 0 (e − 1) 15 . (a)‡ State the assumptions of the Debye model of heat capacity of a solid.1 8. there . Here.3 By integrating by parts this can also be written as Z ∞ x 4 ex 4π 4 Discuss.59 Z ∞ X∞ Z ∞ X∞ 10 1. This shows that until the cutoff is imposed. The full derivation goes as follows. For a crystal containing N atoms. For oscillators with frequency ω(k) a system has a full energy Z E = L3 d3 k(2π)3 ~ω(k)[nB (β~ω(k)) + 1/2] One includes also a factor of 3 out front to account for the three different sound modes (two transverse and one longitudinal) and we cut off the integral at some cutoff frequency ωcutof f . (a) The key assumption of Debye theory is that the dispersion curve is linear (ω = vk) up to a cut-off frequency ωDebye determined by the requirement that the total number of vibrational modes is correct. 0. We use the assumption that ω = v|k| although it is not much harder to consider three different velocities for the three different modes. and ~ωd = kB TDebye defines the Debye temperature.5 × 10−7 From the final result. with reference to the Debye theory.1 x3 1 π4 dx x = x3 e−nx = 6 4 = 15 2. in the low T limit we should have (from Eq.1 2. .1 1. 2. the exponents can be expanded such that the Debye integral becomes Z ~ωd /kB T Z ~ωd /kB T 4 ex dxx x 2 = dxx2 = (1/3)(~ωd /kB T )3 0 (e − 1) 0 which then recovers the law of Dulong-Petit C = 3N kB (b) Given the heat capacity and the temperature. we can extend the integral out to infinity whereupon it just gives the constant 4π 4 /15 recovering the above claimed result Eq. In the low temperature limit. we find that the proper value of ωcutof f is exactly the Debye frequency ωd that we just defined.2 × 10−1 5.1. the T 3 scaling does indeed work). (Note that at low enough T . and Debye is actually no knowledge of the underlying lattice — only the overall volume and sound velocity.6 × 10−4 1. 2. Einstein.1) 1/3 12Rπ 4 T 3 TD = 5C The table of heat capacity looks like T (K) 0. thus we have Z ωcutof f 3N = dωg(ω) 0 performing this integral. We should choose the cutoff frequency such that we have the right number of modes in the system.8 6.0 5 8 10 15 20 −1 −1 C (J K mol ) 8. The fact that the T 3 fit is not perfect is a reflection of (a) that Debye theory is just an approximation (in particular that phonons have a nonlinear spectrum!) and (b) that one needs to be in the low T limit to obtain perfect T 3 scaling. The general heat Debye theory heat capacity will then be Z ωd kB 2 eβ~ω C = dhEi/dT = dωg(ω)(~ω) (kB T )2 0 (eβ~ω − 1)2 Defining x = ~ω/kB T we obtain 3 Z ~ωd /kB T T ex C = dhEi/dT = N kB 9 dxx4 x TDebye 0 (e − 1)2 This integral is known as the Debye integral.5 × 10−7 8.9 × 10−1 1. In the high temperature limit.3 1/3 12Rπ 4 T 3 5C (K) 132 131 127 119 121 132 135 So TDebye is about 130K.6 Specific Heat of Solids: Boltzmann. terms of a definite integral. but you will need to leave your result in terms of the gral that one cannot do analytically. Since phonons obey bose statistics we have Z |k|=kDebye 2 d k E = 2A ǫ n (βǫk ) 2 k B 0 (2π) Z |k|=kDebye 2 d k 1 = 2A 2 ~ck β~ck 0 (2π) e −1 Z |k|=kDebye 2π 1 = 2A k dk ~ck β~ck (2π)2 0 e −1 Z ΘDebye A dǫ ǫ 1 = ǫ π 0 ~c ~c eβǫ − 1 Let z = βǫ = ǫ/(kb T ) and we get Z ΘDebye /(kb T ) A(kb T )3 z 2 dz E= π~2 c2 0 ez − 1 For large T . Find K in perature. capacity of a two dimensional solid as a function of tem. 7 (2. So high T heat capacity should be C = 2kb N (Law of Dulong-Petit).3) Debye Theory II At high T . Θ/T is small so z is small. At low T . show that Cv = KT n Find n. State your assumptions. In 2d there should be 2N modes. You will need to leave your answer in terms of an inte. So ΘDebye = ~kDebye c with c the sound velocity. Riemann zeta function. show the heat capacity goes to a con- Use the Debye approximation to determine the heat stant and find that constant. so z 2 dz =z ez − 1 so we get Z ΘDebye /(kb T ) zdz = (ΘDebye /(kb T ))2 /2 0 so in this limit A(kb T )Θ2Debye 2 AkDebye kb T (4πN/A)kb T E= = =A = 2N kb T 2π~2 c2 2π 2π . So √ kDebye = 4πn with n = N/A the density. Z |k|=kDebye 2 2 d k 2(πkDebye 2N = 2A = 0 (2π)2 (2π)2 with A the area. If you are brave you can try to evaluate the integral. Assume longitudinal and transverse sound velocities are equal. Also good guesses are Noble gases where the Rubidium 56 K spring constant is very low (weak interaction between the atoms). One presumably wants a soft material of some Xenon 64 K sort – also possibly a heavy material.8 Specific Heat of Solids: Boltzmann.. Diamond is the obvious guess (and indeed Argon 92 K it does have the highest Debye temperature). For small T . but you should be Guess the element with the highest Debye temperature. the upper limit of the integral goes to infinity and we have Z A(kb T )3 ∞ z 2 dz E= π~2 c2 0 ez − 1 So Cv = KT 2 where Z ∞ 3Akb3 z 2 dz K= π~2 c2 0 ez − 1 To evaluate the integral we have Z Z −z ∞ ∞ z 2 dz z 2 dz X −nz ∞ = e 0 ez − 1 0 e n=0 ∞ Z ∞ X ∞ X = dzz 2 e−nz = 2/n3 = 2ζ(3) n=1 0 n=1 Thus we obtain 6Akb3 ζ(3) K= π~2 c2 (2. Material ΘDebye Largest Debye temperature should be the one with the highest speed of sound which is probably the hardest element (ie. Radon 64 K Soft and heavy metals like mercury are good guesses. and Debye which gives C = dE/dT = 2N kb as expected. (in fact mercury Mercury 69 K is liquid at room temperature and one has to go to low T to measure Potassium 91 K a Debye temperature). ΘDebye = 2230K. Many of these are gas or Some Low Debye Temperatures liquid at room T and a Debye temeperature can only be measured at low T . Also Cesium 32 K heavy soft group 1 metals are good guesses. highest spring con- Neon 75 K stant) and/or smallest mass.4) Debye Theory III The lowest? You might not guess the ones with the abso- Physicists should be good at making educated guesses: lutely highest or lowest temperatures. . Einstein. able to get close. The Krypton 64 K lowest is harder to guess. 9×10−7J/(mol − K4 ) Then using the formula 3 12N kB π 4 T C= 5 TD We obtain TD ≈ 2200K The reason that it does not match the Debye temperature given in the figure caption has to do with the comment in the caption. In the isotropic calculation we use 12πω 2 g(ω) = N (2π)3 nv 3 Recall the origin of these factors. suppose in the x ˆ. vy and vz respectively. Debye theory predicts the heat capacity at all possible temperatures. (2.3 of the main text) 4πω 2 g(ω) = 3L3 (2π)3 v 3 where the factor of 3 out front is for the three polarizations of the sound waves. 2.3 (main text) estimate the Debye temper. the sound ve- locity is independent of the direction the sound wave is locity is vx . listed in Table 2. 9 (2. capacity assuming that the longitudinal and transverse (b) Instead suppose the velocity is anisotropic.5) Debye Theory IV ature of diamond. The Debye temperature measured here is chosen to give a good fit at the lowest temperatures (where Debye theory can actually be exact). How does this change propagating.6) Debye Theory V* gitudinal velocity is vl . 2. Really we had (See Eq. Why does it not quite match the result From Fig. The derivation of the heat capacity follows the text (or exercise 2. the Debye result? (a) Suppose the transverse velocity is vt and the lon- (a) This is actually quite simple. How does this change the Debye In the text we derived the low temperature Debye heat result? State any assumptions you make. The only difference is in the density of states.2 (main text)? Extracting the slope from the figure gives C/T 3 ≈ 1. One could just as well have written it as 2 3 4πω 1 1 1 g(ω) = L + 3+ 3 (2π)3 v 3 v v . For ex- sound velocities are the same and also that the sound ve. ample.1). yˆ and zˆ direction. The Debye temperature quoted in the text is chosen so as to give a good fit over the full temperature range. if the three polar- izations have three different velocities. y.10 Specific Heat of Solids: Boltzmann. (b) If instead we have three different sound velocities in three different directions. Einstein. we would have q ˆ = v 2 kˆ2 + v 2 kˆ2 + v 2 kˆ2 v(k) x x y y z z Now. we cannot do the usual thing and convert to spherical polar coordinates directly. and Debye separating out the three different polarizations. If you consider a sound wave in direction kˆ (with kˆ = k/|k| a unit vector). Instead. following the usual derivation of Debye theory.3 Note that the high frequency cutoff is different for the two types of modes (but the k cutoff is the same for both modes). A reasonable guess would be as follows. the situation is more complicated (and here we neglect the differences between longitudinal and transverse modes). And now q ˆ ω(k) = v(k)|k| = vx2 kx2 + vy2 ky2 + vz2 kz2 Since the system is now not isotropic. Here we must make some assumption about the sound velocity in some arbitrary di- rection. Now. In an isotropic solid. defining v¯ such that 3 2 1 3 = 3+ 3 v¯ vt vl we obtain the low temperature capacity in the usual form 3 12N kB π 4 T C= 5 TD where now (kB TD )3 = 6π 2 n~3 v¯. we have 2 3 4πω 1 1 1 g(ω) = L + 3+ 3 (2π)3 v13 v2 v3 this is true since the density of states of the three different excitation modes simply add. the two transverse mode have the same velocity vt and the one longitudinal mode has velocity vl and we would have 4πω 2 2 1 g(ω) = L3 + (2π)3 vt3 vl3 The remainder of the derivation is unchanged. we rescale the axes first writing (with j = x. we start with Z L3 1 hEi = 3 dk x dky dkz ~ω(k) n B (β~ω(k)) + (2π)3 2 . Thus. z) Kj = kj vj So that q ˆ ω(K) = v(k)|k| = Kx2 + Ky2 + Kz2 . both in the bottom of a har.1 above.) this as a two particles in three dimensions. connected to (a) Analogous to exercise 2. sical partition function and show that the heat capacity monic well. and K is the spring constant holding perature if K ≫ k. Assume that the two particles (c)** How does the result change if the atoms are in- are distinguishable atoms.1 above. If you find this up of diatomic molecules.2) (2π)3 vx vy vz 0 2 . distinguishable? (a) We can write the partition function as Z Z dp1 dp2 Z= dx1 dx2 e−βH (2π~)3 (2π~)3 Considering the momentum integrals first. is again 3kB per particle (i. calculate the quan- H= + + x1 2 + x2 2 + (x1 − x2 )2 tum partition function and find an expression for the heat 2m1 2m2 2 2 2 Here k is the spring constant holding both particles in the capacity. We can (very crudely) model difficult.7) Diatomic Einstein Solid* (For this problem you may find it useful to transform to Having studied exercise 2.. the two particles together. Sketch the heat capacity as a function of tem- bottom of the well. we have Z 3/2 2 2πm dpe−βp /(2m) = β Then the spatial integrals are made simple by transforming Y = x1 − x2 y = (x1 + x2 )/2 So the spatial integrals are Z 3/2 3/2 −β (−ky2 −(k/4+K/2)Y 2 ) π π dYdye = kβ β(k/4 + K/2) . 11 and Z L3 1 hEi = 3 3 dKx dKy dKz ~ω(K) nB (β~ω(K)) + (2π) vx vy vz 2 . for simplicity you may assume that m1 = m2 .e. consider now a solid made relative and center-of-mass coordinates. calculate the clas- each other with a spring. p1 2 p2 2 k k K (b) Analogous to exercise 2. We can now use spherical symmetry to obtain Z ∞ 4πL3 2 1 hEi = 3 ω dω(~ω) nB (β~ω) + (2.1. (2. The rest of the derivation follows as usual to give the usual expression for heat capacity 3 12N kB π 4 T C= 5 TD where now (kB TD )3 = 6π 2 n~3 vx vy vz . 6kB total). The Hamiltonian is now written as Q2 q2 2 k K H= + + kY + + y2 2(m/2) 2(2m) 4 2 which comprises two independent three-dimensional harmonic oscillators with frequencies p ω1 = 4k/m p ω2 = (k/2 + K)/(2m) The heat capacity is then (analogous to 2.12 Specific Heat of Solids: Boltzmann. and Debye Putting these together we get a partition function 3/2 3/2 3/2 3/2 π π 2πm1 2πm2 Z= ∼ β −6 kβ β(k/4 + K/2) β β The energy is hEi = −∂ ln Z/∂β = 6/β So the heat capacity for the two particles is C = ∂hEi/∂T = 6kB (b) The case where the two masses are identical is fairly simple. Yk ] = i~δjk and [qj . Again. Einstein.4) 2 2 2m1 2m2 2 Note that r and z are canonically conjugate just like p and x. The potential can be viewed as a matrix which we can write as √ (k + K)/m1 −K/ m1 m2 √ −K/ m1 m2 (k + K)/m2 .1) eβ~ω1 eβ~ω2 C = 3kB (β~ω1 )2 + 3kB (β~ω 2 )2 (2. The general method is similar to the discussion outlined in problem *** below. these are canonical conjugates). First. rescale √ √ ri = pi mi and xi = zi / mi so that the Hamiltonian reads r1 2 r2 2 k k K √ √ 2 H= + + z1 2 + z2 2 + (z1 / m1 − z2 / m2 ) (2.3) (eβ~ω1 − 1)2 (eβ~ω2 − 1)2 The case of unequal masses is more tricky. we construct Y = x1 − x2 y = (x1 + x2 )/2 and correspondingly Q = (p1 − p2 )/2 q = (p1 + p2 ) Note that these two variables are constructed so that [Qj . yk ] = i~δjk and all other commutators are zero (in other words. the relative degree of freedom does not. Note that both of these have the correct Dulong-Petit high temperature limit of 3kb . 13 We can diagonalize this matrix to define two new decoupled degrees of freedom representing two independent harmonic oscillators. and ψn is symmetric or antisymmetric depending on whether n is even or odd. Due to the statistics. The fre- quencies of these oscillaltors squared are the eigenvalues of the above matrix. . Since the three dimesional harmonic motion wavefunction can be decomposed into three one-dimensional wavefunctions Ψ(r) = ψnx (x)ψny (y)ψnz (z).2 Heat capacity of two einstein Bose or Fermi statistics depending on the atom type. odd for fermions So when we write the partition function for this oscillator.nz ≥0 The sum can then be evaluated to give a partition function " 2 3 # 3 1 1 Zbose/fermi = e− 2 β~ω 2 ± 1 − e−β~ω 1 + e−β~ω which then can be differentiated to get the heat capacity. The center of oscillators. instead of X e−β~ω(nx +ny +nz +3/2) nx . we instead only include the terms of sum respecting the even/odd symmetry.ny . Ck_B p (k + K)(m1 + m2 ) + (k + K)2 (m1 − m2 )2 + 4K 2 m1 m2 5 ω1 = 2m1 m2 4 p (k + K)(m1 + m2 ) − (k + K)2 (m1 − m2 )2 + 4K 2 m1 m2 3 ω2 = 2m1 m2 2 1 And the heat capacity is given by formula 2.2. Here ω1 = 1 and ω2 = 10 mass degree of freedom (y above) has the same Einstein heat capacity as calculated before. I obtained 24e2ω/kb T 1 + 2e2ω/kb T + 5e4ω/kb T C = kb 2 1 + 2e2ω/kb T − 3e4ω/kb T (kb T )2 for the fermi case and 24e2ω/kb T 5 + 2e2ω/kb T + e4ω/kb T C = kb 2 −3 + 2e2ω/kb T + e4/kb T (kb T )2 for the bose case. 2. A plot is given in Fig.ny .nz ≥0 as usual. This restriction can be handled by writing X 1 X → (1 ± (−1)nx +ny +nz ) 2 nx +ny +nz =even/odd nx . we must have that nx + ny + nz = even for bosons. However. the relative wavefunction must obey Ψ(Y) = ±Ψ(−Y) with the ± depending on whether we have bosons or fermions. 2 4 6 8 10 T (d) If the two atoms are indistinguishable then they must obey either Fig. 2.3 using these two oscillator frequencies. Thus we obtain C ∼ 3kB N 1 − (β~ωD )2 /20 + . 0 where g(ω) = N 9ω 2 /ωD 3 .4 (main text).4 the ratio is 215/151 ≈ 1. The reason this does not match perfectly with our prediction is mainly because TDebye and TEinstein are likely fit over the full range of the heat capacities measured. 2 So κ = TEinstein /12.29TEinstein In the data from Fig 2. . . . and Debye (2. So Z ωD eβ~ω C = dωg(ω)kb (β~ω)2 β~ω 0 (e − 1)2 Z ωD ∼ kb dωg(ω) 1 − (β~ω)2 /12 + . It is not exactly the same though. 2. Why Einstein temperature. . the ratio would come out as we predicted here. . Expanding the heat capacity of a single Einstein oscillator eβ~ω C = kb (β~ω)2 (eβ~ω − 1)2 1 + β~ω + (β~ω)2 /2 ∼ kb (β~ω 2 ) [β~ω + (β~ω)2 /2 + (β~ω)3 /6]2 1 + β~ω + (β~ω)2 /2 ∼ kb [1 + (β~ω)/2 + (β~ω)2 /6]2 1 + β~ω + (β~ω)2 /2 ∼ kb [1 + (β~ω)/2 + (7/12)(β~ω)2 ] ∼ kb 1 + β~ω + (β~ω)2 /2 [1 − (β~ω)/2 + (5/12)(β~ω)2 ] ∼ kb 1 − (β~ω)2 /12 + . (The ratio you cal- culate should be close to the ratio stated in the caption For the Einstein model calculate κ in terms of the of the Figure. Compare your result to the values for C = N kB (1 − κ/T 2 + . We can handle the Debye case by realizing that the heat capacity is just an integration over Einstein oscillators. might it not be?).) silver given in Fig.42.14 Specific Heat of Solids: Boltzmann. Note that the integration is cut off so that the integral over 1 gives precisely 3N kb as it should. high temperature heat capacity is of the form From your results give an approximate ratio TEinstein /TDebye . . 2 2 2 So κ = TDebye /20. not just in the high temperature limit. . . Setting TEinstein /12 = TDebye /20 we would predict p TDebye = 5/3 TEinstein ≈ 1.8) Einstein Versus Debye* For the Debye model calculate κ in terms of the In both the Einstein model and the Debye model the Debye temperature. . Einstein. If they were fit pa- rameters for only the high temperature limit. the electron is generally not equal to the well-known value What practical difficulties would there be in mea- me = 9. induces an AC current j ∼ eiωt . the measured mass of atom (sodium has valence 1). meaning that τ → ∞. Use lation (in the presence of a magnetic field) to obtain an Drude theory to derive an expression efor the matrix e ρ for expression for the complex AC conductivity matrix σ (ω).) How might these difficulties be addressed). which we will explain in Part VI. (In fact. and you may assume is a matrix.) the long axis. You might again find it convenient to as- for the conductivity matrix σ. hard. (This exercise might look e (c) Define the Hall coefficient. If it does not scatter to momen- tum zero (with probability 1 − dt/τ ) it simply accelerates as dictated by its usual equations of motion dp/dt = F Thus dt hp(t + dt)i = 1 − (p(t) + Fdt) τ or dp p = F− (3. (a) We consider an electron with momentum p at time t and we ask what momentum it will have at time t + dt. as the cyclotron resonance) to measure the mass of the You may assume that there is one free electron per sodium electron. The density of sodium atoms is roughly Explain how could one use this divergence (known 1 gram/cm3 .1) Drude Theory of Transport in Metals (d) What properties of metals does Drude theory not explain well? (a)‡ Assume a scattering time τ and use Drude theory (e)* Consider now an applied AC field E ∼ eiωt which to derive an expression for the conductivity of a metal. This is a result of band structure suring the Hall voltage and resistivity of such a specimen. (You may assume B parallele e For simplicity in this case you may assume that the metal to the zˆ axis. There is a probability dt/τ that it will scatter to momentum zero. sume B parallel to the zˆ axis. a metal in a magnetic field. but if you think about it for a bit. and sodium has atomic mass of roughly 23. The under-tilde means that the quantity ρ is very clean. it isn’t really Estimate the magnitude of the Hall voltage for a much harder than what you did above!) specimen of sodium in the form of a rod of rectangular At what frequency is there a divergence in the con- cross-section 5mm by 5mm carrying a current of 1A down ductivity? What does this divergence mean? (When τ is its long axis in a magnetic field of 1T perpendicular to finite.1) dt τ where here the force F on the electron is just the Lorentz force F = −e(E + v × B) .1095 × 10−31 kg. the divergence is cut off.) Invert this matrix to obtain an expression e that E ⊥ B.Electrons in Metals: Drude Theory 3 (3. in real metals. in metals. Modify the above calcu- (b) Define the resistivity matrix ρ as E = ρj. and using p = mv and j = −nev. and they are all moving at velocity v then the electrical current is given by e2 τ n j = −env = E m or in other words. dp/dt = 0 so we have mv = p = −eτ E with m the mass of the electron and v its velocity.16 Drude Theory In absence of magnetic field dp p = −eE − dt τ In steady state. we obtain an equation for the steady state current j×B m 0 = −eE + + j n neτ or 1 m E= j×B+ 2 j ne ne τ We now define the 3 by 3 resistivity matrix ρ which relates the current vector to the electric field vector e E = ρj e We then obtain components of this matrix m ρxx = ρyy = ρzz = ne2 τ and if we imagine B oriented in the zˆ direction. then B ρxy = −ρyx = ne Inverting this equation we obtain a conductivity matrix σzz = ne2 τ /m σxx = σyy = ρxx /(ρ2xx + ρ2xy ) = σzz /[1 + (eBτ /m)2 ] σyx = −σxy = ρxy /(ρ2xx + ρ2xy ) = σzz (eBτ /m)/[1 + (eBτ /m)2 ] . If there is a density n of electrons in the metal.2) m (b) In both an electric and a magnetic field dp = −e(E + v × B) − p/τ dt Again setting this to zero in steady state. the conductivity of the metal is e2 τ n σ= (3. ex.e. We start with the equation of motion simplified by setting τ → ∞ dp = −e(E + v × B) dt ˆEx0 eiωt and B = B zˆ. thermoelectric coefficients. So the total Hall voltage is IB V = = 4. The cross section of the rod is L by L with L = 5mm. so the current density is j = I/L2 .6 × 1028 m−3 hence the same density of electrons assuming one free electron per atom. setting E = x mv˙ x = −eEx0 eiωt − evy B mv˙ y = evx B We can differentiate the first equation to give vx = −iωeEx0 eiωt − ev˙ y B m¨ then plug in the second equation to give v¨x = −iωe(Ex0 /m)eiωt − (eB/m)2 vx . (e) Drude theory at finite frequency. Drude theory also fails to explain why the sign of the Hall effect can be different in different samples. 17 with all other entries in the σ matrix being zero.8 × 10−8 Volts Lne Some of the problems with making this measurement might be: • This is a very small voltage: One needs a sensitive voltmeter • There may be contact resistance: Use a high impedance voltmeter • Contacts may not be perfectly aligned: Try varying (reversing) the magnetic field to pick out only the B dependent part (I. • Could have heating Tutors might also use this problem as an opportunity to discuss how useful lock-in amplifiers are (which most students do not appear to know). This results in incorrect predictions of. The Hall resistivity is ρxy = B/(ne) so the Hall voltage is jρxy L. (d) Drude theory fails to explain why the electrons do not carry heat capacity of 3/2kB per electron as a classical gas would. (c) The Hall coefficient is RH = ρyx /B which is −1/ne in Drude theory.. measure (V (B) − V (−B))/2). Drude theory does not explain why we should only count valence electrons. If sodium n= 1 gm /cm3 with atomic mass M = 23. this is then a density of atoms of n = NA × n/M = 2. Starting with dp = −e(E + v × B) dt Writing E = E0 eiωt and j = j0 eiωt and also j = n(−e)v = n(−e)p/m we then have iωmj0 /(n(−e)) = −e[E0 + j0 × B/(n(−e))] or iωm E0 = j0 − B × j0 /(n(−e)) ne2 So assuming B in the zˆ direction. The motion in the z-direction is unaffected by the magnetic field in the z direction so that we have σzz = ne2 /(iωm) and off-diagonal terms including z are zero. This divergent response is easy to detect experimentally as a strong absorbtion of the ac electric field at a particular frequency. and then invert last. and solve for E to obtain the finite frequency resistance matrix. (Then this obviously can be converted into a measurement of the mass). . we have a resistivity matrix iωm ne2 B/(−ne) 0 ρ = B/(ne) iωm ne2 0 e 0 0 iωm ne2 which we invert to get the same result as above.18 Drude Theory which is the equation of a driven harmonic oscillator. the calculation may look a bit nicer if you set v or equivalently j. We use the ansatz solution vx = vx0 eiωt so we obtain −ω 2 vx0 = −iω(eEx0 /m) − (eB/m)2 vx0 which we solve −iω(eEx /m) vx0 = (eB/m)2 − ω 2 and similarly −(eB/m)(eEx /m) vy0 = (eB/m)2 − ω 2 with the current being j = −env we obtain σxx = iω(ne2 /m)/[(eB/m)2 − ω 2 ] σyx = (eB/m)(ne2 /m)/[(eB/m)2 − ω 2 ] The cyclotron frequency eB/m is the natural oscillation frequency of a particle of charge −e of mass m in magnetic field B. Note. Lets try doing it that way also. 94 ameter.3 × 10−15 sec The second part should say room temperature and pressure. room temperature and compare your result to the Drude scattering times for electrons in Ag and Li metals. For a nitrogen molecule at room temperature. hvi is the average velocity (see Eq. Given that both Ag and Li are monovalent (i. Multiplying by Avagadro’s number give the density that we should use in the equation 1 τ= ≈ 2 × 10−10 sec nhviσ So electrons scatter much much much more often — this is not surprising considering how much higher their density is than that of the nitrogen gas.4 main text). (This should be the usual 22. So the velocity at 300 K is s 8kB T hvi = ≈ 475m/sec π28mp uncoincidentally being close to the speed of sound in air.4 moles per liter that people remember..e. and σ is the cross-section of the gas Ag 1.37nm.5 107. we can use d = . Note: the table should read 10−8 not 108 ! We use σ = ρ−1 = N e2 τ /m with m the free electron mass and where N here is the electron density which we calculate by Avagadro Number N =n 106 mol-weight in grams/cm3 Solving for τ we get τAg = 3. . have one free electron per atom). The weight of a Nitrogen molecule is about 28 times that of a proton (two nitrogen atoms of atomic weight 14). one can estimate the scat- The following table gives electrical resistivities ρ. calculate the Drude Calculate the scattering time for nitrogen gas at scattering times for electrons in these two metals.8 molecule—which is roughly πd2 with d the molecule di- Li 9. and atomic weights w for the metals silver and 1 lithium: τ = nhviσ ρ (Ωm) n (g/cm3 ) w where n is the gas density. The density is given by n = P/RT with R the gas constant. this gives .59 × 108 10. At P = 105 pascals and T = 300 K. 3.2) Scattering Times In the kinetic theory of gas.8 × 10−14 sec τLi = 8. tering time using the equation sities n.28 × 108 0. den. but we used 300 K instead of 273 and we approximated the pressure).025 mol/m3 . 19 (3.53 6. So the total conductivity is the sum of the two independent conductivities 2 ne τe ni τi σ = σe + σi = e + me mi And thus 1 ρ= ne τe ni τi e2 me + mi The thermal conductivity is similar – both pieces add 2 4kB T ne τe ni τi κ = κe + κi = + π me mi Note that the Weidemann-Franz law continues to hold here in the ratio of σ to κ. To simplify. The conductivity tensors are σj = ρ−1j and then the total conductivity tensor is σ = σe + σi . The Hall resistivity is more complicated. We define tensors ρe and ρi for the two separate species in terms of rj = mj /(nj qj2 τj ) and Rj = qj /nj with j = e or i. Finally this is inverted to give the tensor ρtotal = σ −1 . For a single species. scatter- ically poor. we need only keep track of conductivity in the x. there are no free electrons that would transport cur. ture. positive ions can move throughout the sample in Suppose free electrons have density ne and scattering response to applied electric fields. mass mi and charge +e. y plane (i.3) Ionic Conduction and Two Carrier Types roughly the same magnitude—such materials are known In certain materials. Using Drude theory. If we fix the electric field. it is mainly observable in materials where ing time τi . Suppose that the free ions have density ni . −e). However. we can think of this as a two dimensional problem). we have (See exercise 3. as mixed ion–electron conductors. There is a lot of algebra involved in this. resulting in what is time τe (and have the usual electron mass me and charge known as ionic conduction.20 Drude Theory (3. particularly at higher tempera. if we apply magentic field in the z direction. rent. Since this conduction is typ.e. (a) Calculate the electrical resistivity. I obtained B 2 (re Ri2 + ri Re2 ) + ri re (re + ri ) ρxx = B 2 (Re + Ri )2 + (re + ri )2 B B 2 Re Ri (Re + Ri ) + Ri re2 + Re ri2 ρxy = B 2 (Re + Ri )2 + (re + ri )2 . occasionally it can occur that a material (b) Calculate the thermal conductivity.1) r BR ρ= −BR r where r = m/(nq 2 τ ) and R = q/n with q the charge on the charge carrier. both species respond to the electric field in- dependently.. has both electrical conduction and ionic conduction of (c)* Calculate the Hall resistivity. dN/dEF can be written as 3N/2EF . merfeld) Theory of Metals (c) Estimate the value of EF for sodium [The density (a)‡ Explain what is meant by the Fermi energy.iii) Fermi surface is the surface in momentum space separating the filled and unfilled states at zero temperature.1) Fermi Surface in the Free Electron (Som.i) Fermi Energy EF is chemical potential at T = 0. This is annoying — why define a new quantity if it is just another name for the old quantity?! (a.ii) Fermi temperature TF = EF /kb with kb being Boltzmann’s con- stant. for example. Or the manifold of states having energy EF . You may assume that there is one free electron per sodium atom (sodium has valence (b)‡ Obtain an expression for the Fermi wavevector and one)] the Fermi energy for a gas of electrons (in 3D). (This is ill-defined for the case of a filled band – but we don’t do band theory until later in the course). and sodium has temperature and the Fermi surface of a metal. Ob- tain an expression for the density of states at the Fermi surface. Fermi of sodium atoms is roughly 1 gram/cm3 . atomic mass of roughly 23. (d) Now consider a two-dimensional Fermi gas. Note that it need not be a sphere. This result implies kF = (3π 2 N/V )1/3 . if there is a filled band the chemical potential is mid-gap. if the effective mass (defined later!) is anisotropic you get an ellipsoid instead. Show that the density of states at the Fermi sur. Note that some books define fermi energy to be chemical potential as a function of temperature.More Electrons in Metals: Sommerfeld (Free Electron) Theory 4 (4. (b) Z dk 2V 4πkF3 N = 2V 3 = k<kF (2π) (2π)3 3 Note the factor of 2 out front is for two species of spins. Note. (a. (a. face. and this differs from the conventional intuition that it is the highest filled state at zero tempera- ture. 1. Here is a short way to show that the density of states at the Fermi surface is 3N/2EF .2eV dividing by Boltzmann’s constant this gives about 37.) With 1 free electron per atom. . Thus we have (1gm/cm3 )(102 cm/m)3 (mole/23gm)(6. although we actually don’t care about its actual value).22 Sommerfeld Theory which gives us the Fermi energy ~2 (3π 2 N/V )2/3 EF = (4.02 × 1023 atoms/mole) = 2. Thus we obtain kF = (2πN/A)1/2 The Fermi energy is then ~2 (2πN/A) ~2 πN/A EF = = 2m m The density of states is then independent of energy and is given by dN/dE = Am/(~2 π) = N/EF One can ask the same question in d dimensions and use a similar scheme to do the calculation. this gives us the density N/V and we can plug this into Eq. 4. (See also part (c) of problem 3.1.1 yielding EF = 5 × 10−19 J = 3. 4.1) 2m with m the (effective) electron mass. We then have dEF /dN = (2/3)CN −1/3 = (2/3)EF /N which immediately gives us dN/dEF = (3/2)N/EF (c) Sodium had density of 1gm/cc and atomic mass 23.6 × 1028 atom/m3 .000 Kelvin. We write the density as EF = C(N )2/3 where C is a bunch of constants (given in Eq. (d) For a 2d Fermi gas we have Z dk 2A N = 2A = πk 2 k<kF (2π) 2 (2π)2 F where A is the (2d) area of the system. meaning there is one free electron per atom.. and vd is the average drift velocity. The current density is then j = ne2 /(τ m)E yielding the usual expression for the Drude conductivity ne2 τ σ= m Now if the typical velocity of an electron is on the order of the Fermi velocity vF . 300K is close to zero kelvin so we can ignore this temperature. (ii) estimate λ for copper at 300K and com- density of electrons.9 × 107 Ω−1 m−1 at 300K. This is about 0. i. vF = ~(3π 2 n)1/3 we obtain vF ≈ 1. The σ = ne2 λ/(mvF ). in an applied electric field E is vd = |σE/(ne)|. the mean free path is λ = τ vF . n is the density of the electrons.1. (c) Assuming that the free electron theory is applicable (a) As in the previous problem. Thus we can rewrite the Drude conductivity as ne2 λ σ= mvF (c. . (b) If an electric field E is applied. we have the Drude theory expression dp p = F− dt τ With F = −eE we then have the steady state momentum p = −eE/τ corresponding to the steady state drift velocity vd = −eE/(τ m).45 × 1028 m−3 . Thus we obtain |vd | = |σE/(ne)| Deriving the conductivity from the mean free path is an exercise in Drude theory. Thus pF = ~kF and the fermi velocity is vF = ~kF /m. and show that σ is given valent. (i) calculate the values of both vd and vF for ble: show that the speed vF of an electron at the Fermi copper at 300K in an electric field of 1 V m−1 surface of a metal is vF = m ~ (3π 2 n)1/3 where n is the and comment on their relative magnitudes. The electrical current is then given by j = −envd where −e is the electron charge. kF = (3π 2 n)1/3 with n = N/V . The in terms of the mean free path λ of the electrons by density of atoms in copper is n = 8. ment upon its value compared to the mean (b) Show that the mean drift speed vd of an electron spacing between the copper atoms. As in problem 3.5% of the speed of light. conductivity of copper is σ = 5. in 3D. then in the scattering time τ the distance traveled.6 × 106m/sec. 23 (4. a current density j = σE flows. Very fast.i) On the scale of the Fermi temperature.e.2) Velocities in the Free Electron Theory to copper: (a) Assuming that the free electron theory is applica. where σ You will need the following information: copper is mono- is the electrical conductivity. with σ the conductivity. Thus. I.e. it will gain an energy of order kb T . (d) The experimental specific heat of potassium metal (a)‡ Give a simple but approximate derivation of the at low temperatures has the form: Fermi gas prediction for heat capacity of the conduction electron in metals. (4.ii) Use here λ = vF σm/(ne2 ) plugging in numbers this gives λ ≈ 3. the electron energy (above the ground state) is roughly (kb T )2 D(EF ).1. . Here susceptibility is Explain the origin of each of the two terms in this χ = dM/dH = µ0 dM/dB at small H and is meant to expression. At temperature T .4 × 10−3 m/sec.08 mJ mol−1 K−2 and α = the Fermi gas prediction for magnetic susceptibility of 2. consider the magnetization of the electron spins only. Thus the heat capacity is approximately C = dE/dT ≈ 2kb (kb T )D(EF ) In a slightly more careful (but still not exact) treatment. (c. (a) Let the density of states at the Fermi surface be given by D(EF ) (and assume this is a nonzero quantity).6 mJ mol−1 K−4 .9 × 10−8m. Typically if an electron is excited. the conduction electron in metals. or about 400 angstoms – roughly 100 lattice spacings. Make an estimate of the Fermi energy for potassium (c) How are the results of (a) and (b) different from metal. electrons within an energy kb T of the Fermi surface can be excited above the Fermi energy.3) Physical Properties of the Free Electron that of a classical gas of electrons? Gas What other properties of metals may be different In both (a) and (b) you may always assume that the from the classical prediction? temperature is much less than the Fermi temperature.24 Sommerfeld Theory The drift velocity is vd = σE/(ne) which gives vd ≈ 4. 4.. one approxi- mates the smooth fermi function as a simple function as shown here in Fig. very slow. C = γ T + αT3 (b)‡ Give a simple (not approximate) derivation of where γ = 2. or exercise 4. 25 In this picture. Thus in the magnetic field the energy of an electron with spin up or down . An exact calculation (See Ashcroft+Mermin. the approximation to the fermi function is given by 0123 1 (E − µ)/kb T < −2 n(E) = 0 (E − µ)/kb T > 2 1/2 − (E − µ)/(4kb T ) otherwise In this approximation.2) This calculation is an exercise given as an additional problem.67(K/T )/kB the conventional Bohr magneton. Note that as discussed above. 4. one can calculate that the energy is given by Z 2/kb T 124536789 E(T ) = Constant + D dEE(1/2 − E/(4kb T )) −2/kb T = Constant + D(kb T )2 (4/3) Fig. we assume that the magnetic field couples only to the spins of the electron (we ignore orbital effects).1 The Fermi function (green) and a simple approximation to the which results in C = (8/3)kb (kb T )D(EF ). for a free fermi gas D(EF ) = 3N/2EF . (b) There are several ways that the electrons can respond to the mag- netic field.9. The Hamiltonian (neglecting the Lorentz force of the magnetic field) becomes p2 H= + gµB B · σ 2m where g = 2 is the g-factor of the electron B is the externally applied magnetic field and σ is the spin of the electron which takes eigenvalues ±1/2. Here µB ≈ . Thus up to constants of order one we have C ∼ kb N (kb T /EF ) which is very small since T ≪ TF . First. fermi function (blue).b below) of this result is C = (π 2 /3)kb (kb T )D(EF ) = N (π 2 /2)kb (kb T )/EF (4. to the Fermi wavevector). (assuming that the chemical potential does not change) we will have (g(EF )/2)µB B fewer spin ups electrons per unit volume. Thus. ↓) = − µb |B| 2m The spin magnetization of the system in the direction of the applied magnetic field will then be 1 dE M =− = −([# up spins] − [# down spins]) µB /V (4. for a single isolated spin 1/2 we can calculate the partition function (this calculation was done in stat mech class last year). Thus. the specific heat is given by the equipartition law C = 3kB N which is larger than the result above by roughly a factor of EF /kb T which could be a factor of 100 or more.26 Sommerfeld Theory (with up meaning it points the same way as the applied field) ~2 |k|2 ǫ(k. Near the Fermi level the density of states per unit volume for spin up electrons is g(EF )/2 and similarly the density of states per unit volume for spin down electrons is g(EF )/2. ↑) = + µb |B| 2m ~2 |k|2 ǫ(k.3) V dB So when the magnetic field is applied. so we will have (g(EF )/2)µB B more spin downs per unit volume. Z = eβgµB B/2 + e−βgµB B/2 The expectation of the moment (per spin) is then m = −d log Z/d(Bβ) = (gµB /2) tanh(βgµB B/2) For small B this is m = (gµB /2)2 (B/kb T ) . so our assumption that the chemical potential did not change is correct. using Eq. the spin downs will be less costly by the same amount. Similarly. 4. both the spin up and spin down states are filled up to the Fermi energy (i. the spin ups will be more costly by an energy µB B. When B is applied.3 the moment per unit volume is given by M = g(EF )µ2B B and hence the magnetic susceptibility χ = ∂M/∂H is given (at T = 0 by) χP auli = µ0 µ2B g(EF ) (c) For a classical monatomic gas. Note that the total number of electrons in the system did not change. Let us now calculate the Pauli paramagnetism of the free electron gas at T = 0. Similarly. so more of them will point down. it is lower energy for the spins to be pointing down. With zero magnetic field applied.e. (Recall that chemical potential is always adjusted so it gives the right total number of electrons in the system). 2 yields EF = (π 2 /2)R(T /Clinear ) =≈ 2 × 104 KkB ≈ 1. Define Fermi energy and Fermi temperature. if there is a filled band the chemical potential is mid-gap.. Cal- 2mL2 culate the Fermi energy as a function of the density for Find the numerical values of ad for d = 1. Another method would be to use the density of potassium and assume the valence is 1. Peltier Coefficinent. Using the above formula Eq. (4. 3 and calculate the density of states (b) Show also that the density of states at the Fermi at the Fermi energy in each case.. If you do this. Fermi Energy EF is chemical potential at T = 0. Average Electron Velocity. 2. one-dimensional organic conductor which has unit cell of Show that the Fermi energy is given by length 0. . 4. 2.8 nm. The T - linear term is the specific heat of free electrons. where each unit cell contributes one mobile electron. Nd g(EF ) = Why do metals held at room temperature feel cold 2Ld EF to the touch even though their Fermi temperatures are much higher than room temperature? Assuming the free electron model is applicable. ~2 EF = (N ad )2/d (c) Consider relativistic electrons where E = c|p|. Compressibil- ity. (d) The T 3 term is clearly from Debye phonon specific heat. 27 thus the total susceptibility for N spins (recall susceptibility is measured per unit volume) is dM/dH = µ0 dM/dB = N µ0 (gµB /2)2 (1/kb T ) which. The free electron model of a metal describes electrons in a metal as a noninteracting gas of fermions at some fixed density (usually chosen to be v electrons per unit cell of the metal where v is the valence).4) Another Review of Free Electron Theory energy is given by What is the free electron model of a metal. you get something much closer to the right answer.7eV The real value is roughly 2. Due to Pauli exclusion. es- (a) A d-dimensional sample with volume Ld contains timate the Fermi energy and Fermi temperature of a N electrons and can be described as a free electron model. and this differs from the conventional intuition that it is the highest filled state at zero tempera- ture. This is annoying — why define a new quantity if it is just another name for the old quantity?! Fermi temperature TF = EF /kb with kb being Boltzmann’s constant. and 3. for any T ≪ TF is much larger than the Pauli susceptibility calculated above (by a factor of approximately TF /T ). a metal can have a very high Fermi tempera- .1 eV. Note. Note that some books define fermi energy to be chemical potential as a function of temperature. Other properties that differ from the classical prediction include: Ther- mopower. electrons in d = 1. So EF = c~|kF | = c~(N 1/d /L)ad So here we have dEF /dN = (1/d)(EF /N ) and g(EF ) = (1/V )(dN/dEF ) = (N/V )(d/EF ) . so that a1 = π/2 and a2 = 2π and a3 = 3π 2 .e. if the material is in its ground state. (a) In any number of dimensions we can write Z d dd k N = 2L d |k|<kF (2π) with the 2 accounting for spin.2. as noted above EF = (~2 /(2m))(π/2)2 (N/L)2 where here N/L = 1/(..28 Sommerfeld Theory ture (high chemical potential) even if the material is at zero temperature — i. In 1. 3d we obtain 2kF /π 1d N/Ld = kF2 /(2π) 2d 3 kF /(3π 2 ) 3d So that (N/L)(π/2) 1d kF = (N 1/2 /L)(2π)1/2 2d (N 1/3 /L)(3π 2 )1/3 3d And in any dimension EF = ~2 kF2 /(2m). The integration is over a d-dimensional ball. Then dEF /dN = (2/d)(EF /N ) and g(EF ) = (1/V )(dN/dEF ) = (N/Ld )(d/(2EF )). this is because heat has flowed from the material to you. For a one dimensional system. We discuss physics of this type in Chapter 18. (b) We have EF = Cd /L2 N 2/d for some constant Cd .8nm) This gives me EF = 2. (c) Our above expression for kF still holds.4 × 10−20 J or TF = EF /kb = 1700K. If the material is in its ground state (despite having a high fermi temper- ature) it cannot lower its own energy and therefore cannot transfer heat to you. When you touch a material and it feels hot. Note: Having a high chemical potential DOES mean that the material might have a tendency to transfer electrons to another body with fewer electrons (although this might create a charge imbalance that then prevents further flow of electrons). the relationship of n to µ is to a good approximation independent of T so long as µ ≫ kb T .5) Chemical Potential of 2D Electrons long as T ≪ µ. Hint: first examine the density of states Show that for free electron gas in two dimensions. there is more than one eigenstate at the given energy. in 2d. all the higher energy eigenstates will be empty. 29 (4. the in two dimensions. Let us call the degeneracies of these levels g1 and g2 respectively. From the previous problem. chemical potential µ is independent of the temperature so The key here is to realize that the density of states is independent of energy.6) Chemical Potential at T = 0 at T = 0 the energy of the chemical potential is precisely Consider a system of N non-interacting electrons.e. To see this in more detail rewrite as Z ∞ Z ∞ 1 e−βx n/g = dx βx = dx −βx −µ e +1 −µ e +1 1 = ln(eβµ + 1) β Now for large βµ we can expand to get n/g = µ + O(e−βµ ) so. we have EF ∼ N . Show that This is a bit more difficult than it appears as most generally one should consider the possibility that the highest filled state at energy E1 or lowest unfilled state at energy E2 are degenerate – i. as claimed. Now. and g(EF ) ∼ N/EF so that the density of states is a constant independent of temperature. (4. At half way between the highest energy filled eigenstate and T = 0 the N lowest-energy eigenstates will be filled and the lowest-energy unfilled eigenstate. . the chemical potential is set by Z ∞ 1 n= dEg(E) β(E−µ) 0 e +1 where g here is now constant. The point here is that except for correc- tions exponentially small in βµ the value of the integral is independent of β. Let us focus on these two energies only and assume for now (to be justified in retrospect) that we can ignore any other states in the system with the intuition that any state with energy below E1 is completely filled and can be ignored any any state with energy above E2 is completely empty and can be ignored. Therefore the dependence of n on µ is to a very good approximation independent of temperature. given a fixed density of electrons. 30 Sommerfeld Theory We take our system (E1 with degeneracy g1 and E2 with degeneracy g2 ) and will fill it with exactly g1 electrons. (Were we to choose fewer, the lowest unfilled would move down to E1 were we to choose more the highest filled would move up to E2 ). Fermi occupation then gives us g1 g2 N = g1 = β(E −µ) + e 1 + 1 eβ(E2 −µ) + 1 Defining x = eβ(E1 −µ) and z = e−β(E2 −E1 ) this then becomes g1 g2 g1 = + x + 1 x/z + 1 Solving for x (and taking the positive root only) gives p (g2 − g1 )z + (g2 − g1 )2 z 2 + 4g1 g2 z x= 2g1 Since temperature is small, z becomes very small, so we can expand this expression to give √ x = g1 g2 z Or equivalently taking log of both sides 1 β(E1 − µ) = (ln(g1 g2 ) − β(E2 − E1 )) 2 Taking the limit of small temperature or large β, this becomes µ = (E1 + E2 )/2 + . . . (4.7) More Thermodynamics of Free Electrons 2.53×1022 cm−3 and that of potassium is 1.33×1022 cm−3 , (a) Show that the kinetic energy of a free electron gas and given that both of these metals are monovalent (i.e., in three dimensions is E = 35 EF N . have one free electron per atom), calculate the bulk modu- (b) Calculate the pressure P = −∂E/∂V , and then the lus associated with the electrons in these materials. Com- bulk modulus B = −V ∂P/∂V . pare your results to the measured values of 6.3 GPa and (c) Given that the density of atoms in sodium is 3.1 GPa respectively. (a) We begin with the fact that the density of state g(ǫ) ∼ ǫ1/2 . To see this note that ǫ ∼ k 2 so dǫ ∼ kdk and dN ∼ dk ∼ k 2 dk, so we have dN/dǫ ∼ k ∼ ǫ1/2 . We can thus write g(ǫ) = Cǫ1/2 for some constant C and Z EF 3/2 N = V dǫg(ǫ) = CV (2/3)EF 0 Z EF 5/2 E = V ǫdǫg(ǫ) = CV (2/5)EF 0 Thus dividing one by the other we get E/N = (3/5)EF 31 (b) From exercise 4.4.a or on dimensional grounds EF ∼ V −2/3 , so E ∼ V −2/3 so the pressure is ∂E 2E 2 N EF P =− = = ∂VN 3V 5 V Since P ∼ V −5/3 we then have the bulk modulus ∂P 5 10 E 2EF N B = −V = P = = ∂V 3 9 V 3 V (c) From exercise 4.1 or 4.4 we calculate the fermi energy EF = 5.0 × 10−19 J and EF = 3.3 × 10−19 for sodium and potassium respectively. Thus we obtain B = 8.5 GPa and B = 2.9 GPa respectively. Not too shabby!! (4.8) Heat Capacity of a Free Electron Gas* tions: In Exercise 4.3.a we approximated the heat capacity of Z ∞ x 2 ex π2 a free electron gas dx = = ζ(2)/2 −∞ (ex + 1) 2 3 (a*) Calculate an exact expression for the heat capacity of a 2d metal at low temperature. Note that for the 3d case you have to worry about the (b**) Calculate an exact expression for the heat capac- fact that the chemical potential will shift as a function of ity of a 3d metal at low temperature. temperature. Why does this not happen (at least for low The following integral may be useful for these calcula- T ) in the 2d case? (a) As pointed out in exercise 4.6, in two dimensions µ is indepdendent of temperature — a result of the fact that the density of states is a constant (the answer to the last part of this question). First we need to evaluate the value of the density of states. Using the result of 4.4, we obtain at the fermi energy (and therefore at all energies) m g= ~2 π We thus have µ = EF = n/g We can then write the total energy as Z ∞ Z ∞ Z ∞ g x µ E=V dǫ ǫ β(ǫ−µ) =Vg dx βx +Vg dx βx 0 e +1 −µ e +1 −µ e +1 where we have defined x = ǫ − µ. We recognize the second term here as simply being µN , and as discussed in 4.6, this is independent of temperature. To evaluate the first term, we integrate by parts to get ∞ Z ∞ gV x2 /2 . x2 −βeβx E − N µ = βx . − gV dx e + 1 −µ −µ 2 (eβx + 1)2 . . to within corrections that are of order of e−βµ we can throw away the first term and we can extend the integral in the second term all the way to −∞ (since the argument is peaked near x = 0 wiht only exponentially small tails). .5) V 0 Z EF π = dǫg(ǫ) + (µ − EF )g(EF ) + (kb T )2 g ′ (µ) + . ∞ ∞ 6 Using this equation to write an expression for the density Z ∞ N = g(ǫ)nF (ǫ)dǫ (4.2). this is exact – in other words. However. there is no correction term at any order in T . One could carry this expansion to higher order and pick up terms related to H ′′ etc as well.32 Sommerfeld Theory Assuming βµ is large. one must use the so-called Sommerfeld expansion. then we will go back to derive Sommerfeld’s formula. First we will quote the key formula and use it to derive our result for the heat capacity.6) V 6 .4) ∞ ∞ 6 The intuition behind this formula is that at low temperature the fermi function is a step function (given by the first term). Note that as we determinmed in 4. we can conclude Z ∞ Z EF π H(ǫ)nF (ǫ)dǫ = H(ǫ)dǫ + (µ − EF )H(EF ) + (kb T )2 H ′ (EF ) + . Scaling out the factor of β and using the integral given we obtain π2 E − N µ = V g(kb T )2 + O(e−βµ ) 6 Differentiating to obtan the heat capacity we obtain C/V = gkb (kb T )π 2 /3 which is precisely the result claimed above in exercise 4.3 (see Eq. if the argument H is constant. . Defining the Fermi function 1 nF (ǫ) = eβ(ǫ−µ) +1 The Sommerfeld formula is Z ∞ Z µ π H(ǫ)nF (ǫ)dǫ = H(ǫ)dǫ + (kb T )2 H ′ (µ) + O(T 4 ) (4.6 and above. ∞ 6 N (T = 0) n π o = + (µ − EF )g(EF ) + (kb T )2 g ′ (EF ) (4. 4. The finite slope of the fermi function where it is almost a step creates the small correction term which is determined by H ′ . there is no subleading T n with n > 1 term at all! (b) In the more general case where g(E) is not a constant. . This is quite a bit more complex. here note that up to exponentially small corrections. Expecting that the chemical potential will remain very close to the Fermi energy at low temperatures. then up to exponentially small terms. 8) ∞ 2 ∂ǫ . we have Z ∞ −∂nF (ǫ) I= K(ǫ) dǫ ∞ ∂ǫ there are no boundary terms (to accuracy of O(e−βµ ) because ∂nF /∂ǫ decays very rapidly away from the chemical potential. Thus we obtain E E(T = 0) π 2 = + (kb T )2 g(EF ) + . We can then write a similar expression for the energy density Z ∞ E = ǫg(ǫ)nF (ǫ)dǫ V 0 Z EF π = ǫg(ǫ)dǫ + EF (µ − EF )g(EF ) + (kb T )2 (ǫg(ǫ))′ǫ=EF + . ∞ 6 E(T = 0) n π o = + EF (µ − EF )g(EF ) + (kb T )2 g ′ (EF ) V 6 π2 2 + (kb T ) g(EF ) + .7) 6 Note that in the final equation the term in brackets is the same as the term in brackets from Eq. (4. Finally we return to prove the Sommerfeld formula 4. V V 6 Which we differentiate to obtain the heat capacity π2 C/V = kb (kb T )g(EF ) 3 as claimed. Note also that for g ′ = 0. thus we have Z ∞ 1 2 ′′ −∂nF (ǫ) I= K(µ) + (ǫ − µ) K (µ) + . µ = EF independent of temperature. dǫ (4. . The quantity we would like to evaluate is Z ∞ I = H(ǫ)nF (ǫ)dǫ ∞ Let us define a function Z ǫ K(ǫ) = dǫ′ H(ǫ′ ) −∞ So that. . the term in brackets must remain zero for all temperatures (accurate to order T 2 ). Note now that the function ∂nF /∂ǫ is a symmetric (even) function around the chemical potential. 4. such as in 2d. . 33 Since we are fixing the density N . Thus let if expand K in a taylor series around the chemical potential only the even terms will have a nonzero contribution. . . integrating by parts. . This confirms that µ = EF + O(T 2 ) and confirms our suspicion that it should be very close to EF at low temperature.4. . .6 which we have set to zero in order to keep the density constant as a function of temperature. 34 Sommerfeld Theory Z ∞ −∂nF (ǫ) =1 ∞ ∂ǫ the first term in the expansion of Eq.8. 4. 4. Thus we have obtained the first two terms of Eq.8 is just Z µ K(µ) = H(ǫ)dǫ −∞ as required. The next term gives a prefactor of K ′′ (µ) = H ′ (µ) and requires that we evaluate the integral Z ∞ Z 1 2 −∂nF (ǫ) 1 ∞ x2 eβx (ǫ − µ) dǫ = = (kb T )2 π 2 /6 ∞ 2 ∂ǫ 2 ∞ (eβx + 1)2 using the given integral. . By followin a similar procedure one can evaluate higher terms in the expansion and in particular we will find that the next term in the expansion must be proportional to T 4 . 5. 18.The Periodic Table 5 (5.e.1) Madelung’s Rule expected to be a noble gas. 118. This brings us to element 168. Note that the “exponents” add to 74. 41 47 48 49 Element 118 has a filled 7p shell. 5g. Or equivalently we write 31 37 38 [Xe] 4f14 5d4 .. 6f.) Assuming which has atomic number 74. For entertainment sake (and you can try to prove this) note that 61 67 68 69 6 6 the sequence of nobel gas element numbers 2. and for cases of the same n + l. 54. that Madelung’s rule continues to hold. l = 1 is called p. Note that the noble gases occur whenever a p-shell has just filled. Madulung’s rule fills orbitals according to the diagram Fig. So we have 01 Tungsten atomic number 74: 21 27 1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 5s2 4d10 5p6 6s2 4f14 5d4 . 7d and finally 8p. (No Use Madelung’s rule to deduce the atomic shell fill. 10. real chemistry tests have been performed on the ele- ing configuration of the element tungsten (symbol W) ment yet.1: from lowest n + l to highest. what should the Element 118 has recently been discovered. i. and is atomic number be for the next noble gas after this one? Angular momentum l orbitals (l = 0 is called s. as the nucleus decays very quickly. Madelung’s rule tells us that we then 51 57 58 59 5 have to fill 8s. 36. . etc) contain up to 2(2l + 1) electrons. fill the lower n first. is in group VIII. 86. . . . . . . . . . . . (18-10)/2 = 4. atoms (Madelung’s rule). (118-86)/2=16.e.e. text for estimating the effective nuclear charge: and Nsame is the total number of electrons in the nth principal shell (including the electron we are trying to remove from the atom.2) Effective Nuclear Charge and Ionization • (Approximation a) Energy (a) Let us approximate an electron in the nth shell Z = Znuc − Ninside (i. 168 has successive differences which are twice the perfect squares each occurring twice. Ninside is the number of electrons in shells inside of n (b) Consider the two approximations discussed in the (i.. 5.e. (86-54)/2 = 16. and so forth. (54-36)/2 Fig.. principal quantum number n) of an atom as being • (Approximation b) like an electron in the nth shell of a hydrogen atom with an effective nuclear charge Z. the energy required to pull the electron where Znuc is the actual nuclear charge (or atomic num- away from the atom) as a function of Z and n. (168-118)/2 = 25. electrons with principal quantum numbers n′ < n). Use your knowledge of Z = Znuc − Ninside − (Nsame − 1)/2 the hydrogen atom to calculate the ionization energy of this electron (i. hence the −1). (10-2)/2 = 4 .. . (5.1 Ordering of filling orbitals in = 9. (36-18)/2 = 9. ber). one can solve the Schroedinger equation in detail. Why is this? Neglecting fine structure. Use these approximations to calculate the ioniza. Notice with this second approximation one obtains a dip in the ionization energy for filled s- shells (such as atomic numbers 4 and 12) which is seen in the experiment. Your results should be qualitatively quite good. (i. the energy of an electron in the nth shell of hydrogen is −Ry En = n2 where Ry=13. although is weaker in reality. Ionization Energies 250 experiment "approx1. resulting in binding energy which is Z 2 as strong. In fact.txt" "approx2. Exact com- pared to the two proposed approxima. In this case one gets the following figure (in the two apperoximations discussed above).36 The Periodic Table Explain the reasoning behind these two approxima.3) by assuming that p-shells are “outside” of s-shells. Make a plot of your results and compare them to the mations begin to break down. For a hydrogenic atom with nuclear charge Z. However. So roughly one should be able to estimate the binding energy by setting these equal to each other.6 eV is the Rydberg constant. without doing this one can get it by a scaling argument as well. In other words. this is getting to be pretty decent. the Coulomb interaction is Z times as strong as in hydrogen. but quantitatively not so good I guess. actual ionization energies (you will have to look these up tions. One can do a bit better (See Fig. the simple approxi- 21. 5.txt" 200 150 100 Fig. 0 tions 0 5 10 15 20 25 The two approximations are plotted here with the exact ionization energies. Setting the the length scale to a. 5.2 Ionization Energy (eV) as a 50 function of atomic number. The bound state is a balancing of the kinetic with the potential energy. define a to be the effective Bohr radius) we have ~2 Ze2 KE = 2 = PE = ma 4πǫ0 a solving for a obtains a ∼ 1/Z and plugging back into P E or KE we determine that the kinetic energy should scale as Z 2 . To see this in detail. If you tion energies for the atoms with atomic number 1 through try this for higher atomic numbers. on a table). . a single electron in a p-shell sees a charge of Z = 1 since the entire s-shell is inside of it.e. Qualitatively they are OK. 37 Ionization Energies 140 experiment "approx3. Once you get to the transition metals.txt" 120 100 80 60 Fig. (5. 5.) portant. the fact that 3d is inside 4s makes the 4s electron less well bound than you might otherwise expect. Explain why the periodic table is so im. Thus the d electrons can fill preferentially over the s in some cases.3) Exceptions to Madelung’s Rule elements followed Madelung’s rule and the Aufbau prin- Although Madelung’s rule for the filling of electronic ciple? shells holds extremely well. chapter before attempting this exercise. Explain how the statement “3d is inside of 4s” tions to the rule. Here are a few of them: might help justify this exception in copper. Cu = [Ar] 4s1 3d10 Pd = [Kr] 5s0 4d10 Ag = [Kr] 5s1 4d10 Au = [Xe] 6s1 4f 14 5d10 What should the electron configurations be if these Madelung’s rule incorrectly predicts: Cu = [Ar] 4s2 3d9 Pd = [Kr] 5s24d8 Ag = [Kr] 5s2 4d9 Au = [Xe] 6s2 4f 14 5d9 For copper. atom was understood. the creator of the periodic table. Exact compared to the two proposed approx- 20 imations with the modification that p- 0 shells are declared to be outside of s- 0 5 10 15 20 25 shells. (If you do not already have some inating Mendeleev.3 Ionization Energy (eV) as 40 a function of atomic number. you may want to read the next a Nobel Prize. Remember that the periodic table (1869) was .txt" "approx4. they lose their s-electrons. for background in chemistry. (5. the d-shells really are not very easily described as being inside or outside of anything. there are a number of excep.4) Mendeleev’s Nobel Prize devised many years before the structure of the hydrogen Imagine writing a letter to the Nobel committee nom. And often when transition metals ionize. 38 The Periodic Table Dear Nobel Committee. Professor Steven H. Do I have to smack you upside the head? Do the right thing and give the prize to Mendeleev for God sake! Sincerely. Simon . in an eigenstate).What Holds Solids Together: Chemical Bonding 6 (6. terials that have these types of bonds. Explain why the force is attractive and pro- reference to location on the periodic table). We write p = χE with χ the polarizability. the atom will develop a polarization (i. The second atom then. separated a distance r in the xˆ direction. there is a dipole moment p = er where r is the vector from the electron to the proton. atoms. but the point of this exercise is to learn the (a) Qualitatively describe five different types of chem- information in the table!) ical bonds and why they occur. if an electric field is applied to the atom..1) Chemical Bonding (Yes. The potential energy between these two dipoles is −|p1 ||p2 | −p1 χE −|p1 |2 χ U= = = 4πǫ0 r3 (4πǫ0 )r3 (4πǫ0 r3 )2 . you can just copy the table out of the chapter summary.e. suppose this dipole moment is in the zˆ direction). With the electron “orbiting” (i. Now. portional to 1/R7 where R is the distance between two Describe some of the qualitative properties of ma. Suppose one atom momentarily has a dipole moment p1 (for definiteness. (b) van der Waals forces are from correlated dipole flucuations. (a) Just look at the table in the Chapter Summary of chapter 6. However. let us suppose we have two such atoms. Then the second atom will feel an electric field p1 E= 4πǫ0 r3 in the negative zˆ direction. it will be more likely for the electron to be found on one side of the nucleus than on the other).e. Describe which combinations of what types of (b) Describe qualitatively the phenomenon of van der atoms are expected to form which types of bonds (make Waals forces. . due to its polariz- ability. the average dipole moment is zero. If the electron is a given fixed position. develops a dipole moment p2 = χE whi ch in turn is attracted to the first atom. Let ǫ be the LCAO.6 below. is nonzero. R1 is the position of the first nucleus This is known as a linear combination of atomic orbitals. Note that while for a single isolated atom hpi = 0 the result is propor- tional instead to h|p|2 i ∼ h|r|2 i ∼ with r the position of an electron.40 Chemical Bonding corresponding to a force which is attractive and proportional to 1/r7 .1) and the hopping matrix element where φ is the vector of N coefficients φn .m = hn|H|mi Why can we write Vcross and t equivalently using either one of the expressions given on the right-hand side? with H the Hamiltonian of the full system we are consid. Show that the eigenvalues of our Schroedinger ering. or tight binding (it is used heavily in numerical energy of the atomic orbital around one nucleus in the simulation of molecules). In other words We would like to find the lowest-energy wavefunction we can construct in this form.1 are given by hψ|H|ψi E = ǫ + Vcross ± |t| E= hψ|ψi Argue (perhaps using Gauss’s law) that Vcross Show that minimizing this energy with respect to should roughly cancel the repulsion between nuclei. the more accurate our results will be.2) Covalent Bonding in Detail* the second eigenvalue of the effective Schroedinger equa- (a) Linear Combination of Atomic Orbitals: tion will be an approximation to the first excited state of In Section 6. In Exercise 6. single atomic orbital.2. . N . bitals in our basis. This calculation is done more carefully in problem 6. where V is the Coulomb interaction between the electron n and the nucleus.) Let us write a trial wavefunction for our ground state p2 H= + V (r − R1 ) + V (r − R2 ) = K + V1 + V2 as X 2m |Ψi = φn |ni. generally.5 the full Hamiltonian as we properly consider a non-orthonormal basis. that. (The more (K + V2 )|2i = ǫ|2i states we use in our basis. 6. Why? . i.2 we considered two atoms each with a the system.. More generally we may Let us return to the case where there are only two or- consider any set of wavefunctions |ni for n = 1. the best approxima. in the lower eigenstate the total energy is indeed (Caution: φn is generally complex! If you are not com. write everything in This approximation must fail when the atoms get terms of real and imaginary parts of each φn . and R2 is the position of the second nucleus. lower when the atoms are closer together. . (6. let us assume this basis is orthonormal two identical nuclei and a single electron which will be hn|mi = δn. absence of the other. so each φn gives the same eigenvalue equation. sufficiently close.e. To prove this.1. We called the orbital |1i around nu- (b) Two-orbital covalent bond cleus 1 and |2i around nucleus 2. one cannot assume that shared between them to form a covalent bond. This pertains to a case where we have For simplicity. We write the basis set of orbitals is orthonormal.m (More generally. 6. . and H is the t = −h1|V2 |2i = −h1|V1 |2i N by N matrix These are not typos! Hn. fortable with complex differentiation. Eq. . let us construct the energy equation Eq. (K + V1 )|1i = ǫ|1i tion to the actual ground-state wavefunction.) Similarly.) We claim that the ground state is Define also the cross-energy element given by the solution of the effective Schroedinger equa- tion Vcross = h1|V2 |1i = h2|V1 |2i Hφ = E φ (6. m φn Hnm φm φ 0 = = P − P P n 2 ∂φ∗n p |φp | 2 n |φn | 2 p |φp | X 0 = Hnm φm − Eφn m where we have used Eq.2 to identify E. When the nuclei are very close together. Since the charge distribution of the electron on atom 2 is roughly spherical we can use Gauss’s law to calculate its interaction energy with the nucleus of atom 1. and only sees the other nucleus. (One has to choose whether you want orthogonal orbitals or orbitals that are eigenstates of the single atom Hamiltonian.e..m E= = P 2 (6. Note that t can always be taken as real by simply making a gauge transform on the single particle wavefunction (i. redefining the phase of one of the wavefunctions to absorb the phase of t). 6. (b) The 2 by 2 matrix given by H in this basis is ǫ + Vcross t t∗ ǫ + Vcross which has eigenvalues E = ǫ + Vcross ± |t|. when the atoms get close together. Gauss’s law tells us that we can treat the entire spherical distribution of charge as if it is all at the center of the sphere. the assumption of orthogonality of grounds tate orbitals on different sites breaks down. 41 (a) writing X |ψi = φn |ni n P hψ|H|ψi φ∗n Hnm φm n. this argument no longer works. the nucleus does not see the electron charge at all. Note: When working with complex quantities we can simplify life by treating φn and φ∗n as independent variables.2) hψ|ψi n |φn | We can extremize by differentiating with respect to φ∗n .) . We thus have P P ∗ ∂E m Hnm φm n. If the nucleus of atom 1 is outside of this spherical distribution of charge of the electron. the charge of the electron on atom 2 exactly cancels the charge of the nucleus of atom 2. as the nucleus is then inside much of the distribution of the electron charge. When the two nuclei get close. If the electron charge distribution remains spherical. Also. In this way. Here Vcross is the potential felt by the electron on atom 2 due to the nucleus of atom 1. the nucleus of atom 2 will only see electron charge that is at smaller distances (inside) to the center of this spherical distribution. 4) Ionic Bond Energy Budget ergy.5 -5 -2. E1 − E2 ≫ t then conversely all of the weight of the wavefunction ends ing on site 1 (solid) or site 2 (dashed) up on the second atom. show that the bonding orbital becomes more localized on Here we have instead. of 4. the one with the lower energy). The electron affinity of a chlorine atom is about 3.1 it is shown how the weight of the as a function of E2 − E1 .4 0. When a single sodium atom bonds with a single chlo.2. the ground state moves more towards the lower energy site. For simplicity you may use the For Exercise 6. the two by two hamiltonian matrix ǫ1 t t∗ ǫ 2 where we have now absorbed Vcross into the values of ǫi . As the difference in these two energies increases covalent and ionic bonding. calculate the total energy released when a sodium The ionization energy of a sodium atom is about 5.1 and calculation can be used to describe a crossover between ǫ0.1 Probability (squared ampli- tude) of ground state wavefunction be.8 (X. suming that the cohesive energy is purely Coulomb en- (c) The cohesive energy is (with d the bond distance) e2 Ecoh = = 6.10eV 4πǫ0 d . The lower energy eigenstate is 1n p o Eground = (ǫ1 + ǫ2 ) + (ǫ1 − ǫ2 )2 + 4t2 2 Probability 1 with normalized eigenvector 0. (6. if Fig. 6.62 molecule.b consider now the case where the orthogonality assumption h1|2i = 0. ror. until the bond is completely ”ionic“ meaning that the electron is completely transferred from one atom to the other. Qualitatively account for the sign of your er- rine atom.14 atom and a chlorine atom come together to form a NaCl eV.26 eV. In Fig. the bond length is roughly 0. 6.236 nm.2 with p HE2 .6 0. 2t) ψ= √ 4t2 + X 2 0.5 When E2 − E1 ≫ t then X ≫ t and all of the wavefunction ends up on the first atom (i. Similarly. Compare your result to the experimental value eV.e.E1Lt X = E2 − E1 + (E2 − E1 )2 + 4t2 -7.5 2.3) LCAO and the Ionic–Covalent Crossover the lower-energy atom.. As.2 . one has an equal shar- ing of the wavefunction in the ground state (as in the prior problem). Explain how this atomic orbitals |1i and |2i have unequal energies ǫ0.42 Chemical Bonding (6.5 5 7. However. When the energies on the two sites are equal. as the energy difference is increased. wavefunction moves from towards the lower energy atom as a function of energy. 2.58eV which is slightly larger than the experimentally measured bonding en- ✲x ergy. 6. The upper tude) of ground state wavefunction be- curve includes a short range repulsion. Let us write This equation is known as a “generalized eigenvalue prob- N X lem” because of the S on the right-hand side. If the atomic orbitals are s-orbitals then we may assume also that t is real and Si. or ing on site 1 (solid) or site 2 (dashed) as a function of E2 − E1 .m φn Snm φm n.2 6= 0. (a) This is very similar to 6. Without loss of generality we may as- N by N overlap matrix S whose elements are sume hi|ii = 1 and S1. so overlaps Sij must be real and positive.2 is real.j = hi|ji positive (why?).m φn Snm φm X X 0 = Hnm φm − E Snm φm m m where we have used Eq.m φn Hnm φm Snm φm 0 = ∗ = P m ∗ − P ∗ P n. In that exercise we assumed that our basis of orbitals is orthonormal. 6.2. tially many atoms). In this Hφ = ESφ (6.2 we introduced the method of lin. P ∗ hψ|H|ψi n. Let us write the h1|2i = S1.2 Probability (squared ampli- In the Fig.m ∗ ∂φn n. In this case do not assume that S is diagonal. Fig. derive the (a)* In Exercise 6.m φn Hnm φm E= = P ∗ (6.62eV + 6.2.4 to identify E.2 the lower curve is the pure Coulomb energy. The reason for the discrepancy is that there must be a repulsive force in addition to the coulomb attractive force which reduces the mag- nitude of the cohesive (binding) energy. 6. new “Schroedinger equation” ear combination of atomic orbitals. there would be no minimum in the curve! (6.14eV + 3.4) hψ|ψi n.5) LCAO Done Right* Using a similar method as in Exercise 6.3 to derive the eigenenergies of the system. 6.3) exercise we will relax this assumption. (b) An s-orbital can be taken to be manifestly postive everywhere (no nodes). 43 Thus the total bonding energy is ✻ V (x) E = −5.m φn Snm φm n. Consider now many orbitals on each atom (and poten. |ψi = φi |ii (b)** Let us now return to the situation with only two i=1 atoms and only one orbital on each atom but such that for an arbitrary number N of orbitals.10eV = 4. Use Eq. Here the easiest thing .m φn Snm φm We can extremize by differentiating with respect to φ∗n to give P P ! P ∗ ∂E Hnm φm n. The repulsion must be there. with the same notation for H and φ as in Exercise 6. ll . zi are the components of ri .6) Van der Waals Bonding in Detail* atoms. the force is proportional to 1/R7 ). H1 = + Perturbing H0 with the interaction H1 . 1. Thus conclude that the leading correction to the 4πǫ0 |R − r1 | 4πǫ0 |R + r2 | ground-state energy is proportional to 1/R6 (and hence Here H0 is the Hamiltonian for two non-interacting hy. Show that for large R and small ri . ✒ where xi . mi and have energies En = −Ry/n2 with H = H0 + H1 Ry = me4 /(32π 2 ǫ20 ~2 ) = e2 /(8πǫ0 a0 ) the Rydberg (here l ≥ 0. The matrix we want to diagonalize is then −1 1 ǫ − St t − ǫS S H= 1 − S2 t − ǫS ǫ − St The eigenvalues are easily seen to be 1 E= ([ǫ − St] ± |t − ǫS|) 1 − S2 (6. let us assume that R is in of the van der Waals force between two hydrogen atoms. i. drogen atoms. n2 . let us write the Hamiltonian as ǫ t H= t∗ ǫ where we have absorbed Vcross into ǫ. 0. and H1 is the interaction between the Recalling second-order perturbation theory show . l2 . and 1 S S= S 1 with S = S12 .44 Chemical Bonding to do is to just apply S −1 to both sides of the equation to give the eigenvalue problem SHφ = Eφ Here. 0. Show that this ■ r1 is just the interaction between two dipoles. the x ˆ direction. show that 4πǫ0 |R| 4πǫ0 |R − r1 + r2 | e2 e2 to first order in H1 there is no change in the ground-state − − energy. The ground state of H0 is e2 e2 |1. 0.e. l. the First. H1 = (z1 z2 + y1 y2 − 2x1 x2 ) + O(1/R4 ) 4πǫ0 |R|3 - . let the positions of the two nuclei be separated by interaction Hamiltonian can be written as a vector R. Recall that the (assuming fixed positions of nuclei. 0i. m1 . using Born– eigenstates of hydrogen are written in the usual nota- Oppenheimer approximation) as tion as |n. |m| ≤ l and n ≥ l + 1). m2 i with energies H0 = + − − 2m 2m 4πǫ0 |r1 | 4πǫ0 |r2 | En1 . Thus the eigenstates p1 2 p2 2 e2 e2 of H0 are written as |n1 .n2 = −Ry(1/n21 + 1/n22 ). and let the vector from nucleus 1 to electron 1 be r1 and let the vector from nucleus 2 to electron 2 be e2 r2 as shown in the following figure. R r2 ✲ + (b) Perturbation Theory: + The eigenvalues of H0 can be given as the eigen- Let us now write the Hamiltonian for both atoms values of the two atoms separately. (a) Here we will do a much more precise calculation Without loss of generality. yi .. 0| H1 |n1 . 0. 0|x2 |1.n2 in the denominator with E2. 45 that we have a correction to the total energy given by 0. thus leaving us with only the contribution of the terms coming from the square bracketed terms of Eq.n2 n1 . (6. If we replace where a0 = 4πǫ0 ~2 /(me2 ) is the Bohr radius. (This last En1 . δE = Make these replacements.n2 by e−r/2a0 . l2 .) (a) In fact.0 − En1 . Thus the smallest En1 . 0. You will need the matrix element for a hydrogen atom (c)*Bounding the binding energy: First. (b) The expectation of x or y or z in the ground state of the hydrogen atom |100i is zero due to the fact that the state is spherically symmetric. 0. 0i = a20 zero if either n1 = 1 or n2 = 1.5) R R2 2 R2 2 R4 Applying this to all the terms in e2 e2 H1 = + 4πǫ0 |R| 4πǫ0 |R − r1 + r2 | e2 e2 − − 4πǫ0 |R − r1 | 4πǫ0 |R + r2 | We discover that the first two leading orders completely cancel. 1.n2 that appears in the denominator is E2. l2 6e2 a50 8e2 a50 6 ≤ |δE| ≤ m1 .2 then the |δE| we identity is easy to derive if you remember that the ground- calculate will be greater than than the |δE| in the exact state wavefunction of a hydrogen atom is proportional to calculation. . 6. if we replace En1 . n2 .2 .5. show that the numerator in this expression is h1. As a result taking the expectation of H1 in the ground state gives zero. m2 4πǫ0 R 4πǫ0 R6 Show that the force must be attractive. and perform the remain- X | < 1. then the |δE| will always be less than the δE of the exact calculation. m1 . . ll . n2 l1 . Derive the bound E0. Thus the leading correction to the energy occurs at second order in H1 . We then obtain 2 e2 r1 + r22 − |r1 − r2 |2 3(R · (r2 − r1 ))2 − (R · r1 )2 − (R · r2 )2 ) H1 = + 4πǫ0 R 2R2 R4 with R in the x ˆ direction. this is more or less the definition of dipole interaction! Let us start by deriving 1 1 = q |R − a| 2a·R R 1 + R2 + R a2 2 1 a·R 1 a2 3 (a · R)2 = 1− + − + + . this simplifies to e2 H1 = 3 r1 · r2 + 3 (x1 − x2 )2 − x21 − x22 4πǫ0 R which then simplifies to the desired answer. On the other hand. m2 i|2 ing sum by identifying a complete set. 0. 0. 0. Given then the form of the 2nd order perturbation theory. ll . l2 . 0. 0. we have E0.0 =-2Ry and E2. Thus the interaction is of the form −C/R6 for some positive constant C (to be calculated). 0| H12 |1. m1 .0 − E2. n2 . 1. m2 × hn1 . 0. 0. 0i (6.6) n1 . l2 m1 . since x or y or z has zero expectation in the ground state. 0. m1 . 0.7) n1 . 1. the upper bound is defined by setting En1 . n2 . 0.2 = −Ry/2 so E0. and us- ing the fact that the Rydberg is e2 /(8πǫ0 a0 ) we have 12e2 a50 W = I/Ry = 4πǫ0R6 so our inequality is W 2W < |δE| < 2 3 which is the required result. (c) As just above. m2 | H1 |1. ll . 0. 0. m1 . ll . this means H1 has zero matrix element unless n1 > 1 and n2 > 1 (for n = 1 we must have l = m = 0). so we do the sum over the set to obtain I = h1.2 whereas the lower bound is defined by setting En1 . 1. which we will show next).0 − E2. note that every term has an overall negative sign (numerator is positive. 0.2 Here using the excitation spectrum of the hydrogen atom. n2 l1 . denomi- nator is negative). 0. 0.46 Chemical Bonding and is proportional to 1/R6 (assuming it does not vanish. m2 X = h1. and hence the force is attractive.2 = −(3/2)Ry. 1.8) We notice the complete set in the middle here. 0| H1 |n1 . 0. l2 .n2 = E2. 0. n2 l1 . l2 m1 . . n2 . First let us consider the quantity X I= | < 1. m2 i (6. 1. Thus we obtain 2 e2 I= 6a40 4πǫ0 R3 Now. 0i 2 2 e = hy12 ihy22 i + hz12 ihz22 i + 4hx21 ihx22 i 4πǫ0 R with all expectations being in the ground state of the respective hydrogen atom (each bracket gives a20 ). 0| H1 |n1 .n2 = 0 so we obtain I I < |δE| < E0. l2 . m2 i|2 (6.0 E0. .Types of Matter 7 There are no exercises for chapter 7. . 8.1) Potentials Between Atoms small energies.e.1.One-Dimensional Model of Compressibility. and mum as comparing to Eq. we can truncate the series at the cubic As a model of thermal expansion. calculate the values of the coef- κ κ3 ficients x0 . By expanding Eq.4) dx3 x=x0 . . (8. The exponent 6 determines the long range behavior of the potential and is fixed by the form of the Van der Waals interaction. For in Exercise 8.3) dx2 x=x0 and d3 V √ = −756 2 ǫ/σ 3 = −κ3 ≈ 1069ǫ/σ 3 (8.2 around its minimum. Taking second and third derivatives at this position d2 V = 36 × 22/3 ǫ/σ 2 = κ ≈ 57ǫ/σ 2 (8. and κ3 for the Lennard-Jones potential in V (x) = (x − x0 )2 − (x − x0 )3 + . This potential can be expanded around its mini. atoms. . why is the exponent where x is the distance between the two neighboring necessarily chosen to be 6).1) 2 3! terms of the constants ǫ and σ. κ. we study the distance term. We will need these results where the minimum is at position x0 and κ3 > 0.3. x0 x What is the meaning of the exponent 6 in the sec- ond term of this expression (i. By setting dV /dx = 0 we find the minimum at x0 = 21/6 σ. (Note that we are defining the energy at the bot- between two nearest-neighbor atoms in an anharmonic tom of the well to be zero here.. 8.2) x x where ǫ and σ are constants that depend on the particular kB T ✻ ✲ atoms we are considering. Sound. and Thermal Expansion 8 (8.) potential that looks roughly like this A very accurate approximate form for interatomic po- ✻ tentials (particularly for inert atoms such as helium or argon) is given by the so-called Lennard-Jones potential V (x) σ 12 σ 6 V (x) = 4ǫ − +ǫ (8. .. 8. hxiβ = R βκ 2 h i dye− 2 y 1 + βκ6 3 y 3 + .2) Classical Model of Thermal Expansion Why is this expansion of the exponent and the ex- (i) In classical statistical mechanics.. with kB Boltzmann’s constant. we write the ex. since we have treated the cubic term perturbatively. dy e− 2 y Using Z 2 p dxe−ax = π/a as a generating function. 2κ2 Thus 1 dhxiT 1 κ3 kb = x0 dT x0 2κ2 In order for this calculation to be valid. we have Z Z p p 2 2 dxx4 e−ax = (d/da)2 dxe−ax = (d/da)2 π/a = (3/4) π/a5 gives κ3 (kb T ) hxiT = x0 + + . Sound. . this term actually must be small compared to the leading term. . so bad as an approximation!) R dx x e−βV (x) hxiβ = R dx e−βV (x) with −βV (x) − βκ 2 βκ3 e =e 2 (x−x0 ) 1+ (x − x0 )3 + . and hence show that the coefficient of thermal expan- dx x e−βV (x) sion is hxiβ = R 1 dL 1 dhxiβ 1 kB κ3 dx e−βV (x) α= ≈ = L dT x0 dT x0 2κ2 Although one cannot generally do such integrals for ar.. . .. one can expand the In what temperature range is the above expansion exponentials as valid? While this model of thermal expansion in a solid is βκ 2 βκ3 e−βV (x) = e− 2 (x−x0 ) 1 + (x − x0 )3 + .50 Compressibility. . why is it invalid for the 6 case of a many-atom chain? (Although actually it is not and let limits of integration go to ±∞. bitrary potential V (x) as in Eq. 6 Redefine y = (x − x0 ) so we have R βκ 2 h i dy (y + x0 ) e− 2 y 1 + βκ6 3 y 3 + . and Thermal Expansion (8. . . valid if there are only two atoms. tension of the limits of integration allowed? pectation of x as Use this expansion to derive hxiβ to lowest order in R κ3 .1. βκ3 R βκ 2 6 dy y 4 e− 2 y = x0 + R βκ 2 + . . In the next ex- of the longitudinal velocity is about 1600 m/sec. one must solve for the normal modes of the chain. 8.2 are given by ǫ = 10meV and σ = . κ and κ3 for the Lennard Jones case. 51 2 Roughly. if the leading term is most important.0012/K x0 2κ 48ǫ . the sound wave velocity in solid argon. However.2! (a) Sound The actual thermal expansion coefficient of argon is Given that the atomic weight of argon is 39. From our above problem on Lennard- Jones. estimate the thermal For argon. Note: You can do this Eq. we havep κ(x − x0 ) ∼ kb T which means that the typical deviation is |x − x0 | ∼ kb T /κ. we have κ|x − x0 |2 ≫ κ3 |x − x0 |3 or kb T ≫ κ3 (kb T /κ)3/2 or equivalently kb T ≪ κ3 /κ23 For a many atom chain.34nm. estimate approximately α = 2 × 10−3 /K at about 80K. You will part even if you couldn’t completely figure out Exercise need to use some of the results from Exercise 8.0399kg/NA . Then in order to have the leading term be larger than the cubic term.1. What we have done in this problem is more or less the thermal expansion of an Boltzmann model of a solid. Which gives v = 1320m/s Thermal expansion. we obtain 1 κ3 kb 7kb α= 2 = ≈ .2. 8. the Lennard-Jones constants ǫ and σ from expansion coefficient α of argon.3) Properties of Solid Argon Using the results of Exercise 8. Plugging in our above expression for x0 . Sound: From the text in one dimension we have q v = κx20 /m where x0 is the neighbor distance. ercise will use a more sophisticated quantum model to (b) Thermal Expansion understand why this is so. Then at finite temperature one should “occupy” the phonons thermally and then calculate the effect of the nonlinear terms in this state.9. The actual value at lower temperature α drops quickly. we have κx20 = 72ǫ and the mass m = . (8. 5? at low temperature? Using perturbation theory it can be shown that. Sound. and Thermal Expansion (8. Recall that one where |ni is the eigenstate of the Harmonic oscillator can define operators a and a† such that [a. (b) P P hn|x|nie−βEn nP n hn|(x0 + En κ3 /(2κ2 ))|nie−βEn hxiβ = −βEn = P −βE ne ne n hEiβ κ3 = x0 + 2κ2 where hEiβ is the energy expectation of a harmonic oscillator of fre- quency ω at temperature β = 1/(kb T ). and it matches that of Exercise 8. Since the well is asymmetric around the minimum. this expectation is 1 ~ω hEiβ = (nB (β~ω) + )~ω = coth(β~ω/2) 2 2 . we take it as given. For now. Hint: It is easiest to perform this calculation by using hn|x|ni = x0 + En κ3 /(2κ2 ) (8. δV is the perturbation (see Eq. one should use the ”reduced” mass µ = m1 m2 /(m1 + m2 ) which for two identical atoms is m/2. the expectation value of the position is slightly greater than x0 even in the ground state. 2mω (a) For a system of two atoms. 8. a† ] = 1 and whose energy is √ a† |ni0 = n + 1|n + 1i0 1 √ En = ~ω(n + ) + O(κ3 ) n≥0 a|ni0 = n|n − 1i0 . As derived above when we discussed Einstein model. turbed Hamiltonian H0 .6) raising and lowering (ladder) operators. 8. In terms of these operators. 8.52 Compressibility. the expectation of x deviates from x0 . when 6 is the classical calculation from Exercise 8.6.4) Quantum Model of Thermal Expansion (b)* Use Eq. . In (c) we will prove Eq. H0 = + (x − x0 )2 (8.1) In what range of temperatures is our perturbation κ3 expansion valid? δV = − (x − x0 )3 + . to (c)** Prove Eq.2 valid? where we will throw out quartic and higher terms. Why does the thermal expansion coefficient drop What value of m should be used in Eq. . Quantum mechanically the oscillator does not sit in its exact mini- mum. 2 p Note that these are kets and operators for the unper- with ω = κ/m. have the operator x − x0 given by Note that even when the oscillator is in its ground r state.5) 2m 2 Examine the high temperature limit and show that is the Hamiltonian for the free Harmonic oscillator. In light of the current quantum calculation.6 by using lowest-order perturbation lowest order in κ3 the following equation holds theory. Physically ~ why is this? x − x0 = (a + a† ). rather it has “quantum fluctuations” around this minimum .2. 8. 8. We write P −βEn H = H0 + δV n hn|x|nie hxiβ = P −βE ne n where p2 κ Derive the coefficient of thermal expansion.6 to calculate the quantum expectation (a) In quantum mechanics we write a Hamiltonian of x at any temperature. 53 with nB the boson occupation factor. Thus we obtain hxiβ = x0 + (κ3 ~ω/(4κ2 )) coth(β~ω/2) The coefficient of thermal expansion is then 1 dhxiβ κ3 dhEi α= = x0 dT 2x0 κ2 dT where C = dhEi/dT is exactly the specific heat of a harmonic oscillator (recall einstein model). Thus we obtain κ3 C κ3 2 eβ~ω α= = kb (β~ω) 2x0 κ2 2x0 κ2 (eβ~ω − 1)2 We can see that the thermal expansion drops as modes ”freeze” out, entirely analogous to the specific heat dropping at low temperature. In the high T limit, the specific heat C goes to kb , so we obtain α = (κ3 kb /(2x0 κ2 )) in agreement with the classical result. For the classical calculation to be valid, we must have kb T ≫ ~ω so that quantum mechanics can be replaced by classical mechanics. (c) First let us consider the unperturbed harmonic oscillator. Recall the raising and lowering operators r mω a = [(x − x0 ) + (i/mω)p] 2~ r mω a† = [(x − x0 ) − (i/mω)p] 2~ (8.7) So that the position is given by p x − x0 = ~/(2mω)(a + a† ) Write the eigenstates |ni0 to mean the eigenstates of the unperturbed oscillator with energy En = ~ω(n + 1/2). In lowest order perturbation theory, the perturbed eigenket is given by 3/2 X † 3 −κ3 ~ 0 hm|(a + a ) |ni0 |ni = |ni0 + |mi0 6 2mω m En − Em So that we have 2 X −2κ3 ~ 0 hn|a + a† |mi0 0 hm|(a + a† )3 |ni0 hn|x − x0 |ni = 6 2mω m En − Em There are two nonzero terms of the sum, the one where n = m + 1 which gives us a denominator ~ω and a numerator † † † † 0 hn|a |n − 1i0 0 hn − 1|a aa + aa a + aaa |ni0 √ √ √ √ = n n(n − 1) + n n + (n + 1) n = 3n2 54 Compressibility, Sound, and Thermal Expansion and also the term with n = m − 1 which gives a denominator −~ω and a numerator + 1i0 0 hn + 1|a† a† a + a† aa† + aa† a† |ni0 0 hn|a|n √ √ √ √ = n + 1 n n + 1 + (n + 1) n + 1 + (n + 2) n + 1 = 3(n + 1)2 Thus we get 2 2κ3 ~ 1 hn|x − x0 |ni = (6n + 3) = En κ3 /(2κ2 ) 6 2mω ~ω Vibrations of a One-Dimensional Monatomic Chain 9 (9.1) Classical Normal Modes to Quantum Note that since there are two initial degrees of freedom, Eigenstates there are two normal modes. In Section 9.3 we stated without proof that a classical Now consider the quantum-mechanical version of the normal mode becomes a quantum eigenstate. Here we same problem. The Hamiltonian is prove this fact for a simple diatomic molecule in a poten- tial well (see Exercise 2.7 for a more difficult case, and see p21 p2 H= + 2 + U (x1 , x2 ) also Exercise 9.7 where this principle is proven in more 2m 2m generally). Consider two particles, each of mass m in one dimen- Again transform into relative and center of mass co- sion, connected by a spring (K), at the bottom of a poten- ordinates. tial well (with spring constant k). We write the potential Define the corresponding momenta prel = (p1 − p2 )/2 energy as and pcm = (p1 + p2 ). (b) Show that [pα , xγ ] = −i~δα,γ where α and γ take k 2 K the values cm or rel. U= (x1 + x22 ) + (x1 − x2 )2 2 2 (c) In terms of these new coordinates show that the Write the classical equations of motion. Hamiltonian decouples into two independent harmonic Transform into relative xrel = (x1 − x2 ) and center oscillators with the same eigenfrequencies ωcm and ωrel . of mass xcm = (x1 + x2 )/2 coordinates. Conclude that the spectrum of this system is (a) Show that in these transformed coordinates, the 1 1 system decouples, thus showing that the two normal Enrel ,ncm = ~ωrel (nrel + ) + ~ωcm (ncm + ) 2 2 modes have frequencies p where ncm and nrel are non-negative integers. ωcm = k/m (d) At temperature T what is the expectation of the p ωrel = (k + 2K)/m. energy of this system? (a) The equations of motion are mx¨1 = −kx1 − K(x1 − x2 ) mx¨2 = −kx2 − K(x2 − x1 ) Taking the sum and difference of these two equations gives m¨ xcm = −kxcm m¨ xrel = −(k + 2K)xrel (c) We have p1 = pcm /2 + prel p2 = pcm /2 − prel x1 = xcm + xrel /2 x2 = xcm − xrel /2 So the Hamiltonian becomes 1 2 k K H = (p + p22 ) + (x21 + x22 ) + (x1 − x2 )2 2m 1 2 2 1 2 k K = (p /2 + 2p2rel ) + (2x2cm + x2rel /2) + x2rel 2m cm 2 2 which decouples into (note the total mass is 2m and the reduced mass is m/2) p2cm 2k 2 H1 = + x 2(2m) 2 cm p2rel (k/2 + K) 2 H2 = + xrel 2(m/2) 2 which are two independent harmonic oscilators with frequencies p ωcm = k/m p ωrel = (k + 2K)/m . x1 ] + [p1 . x2 ] − [p2 . xcm ] = ([p1 . x2 ]) = (−i~)(1 + 0 − 0 − 1) = 0 1 [pcm . x2 ] + [p2 . xj ] = −i~δij . xrel ] = ([p1 . xcm ] = ([p1 . x2 ]) 2 1 = (−i~)(1 + 0 + 0 + 1) = −i~ 2 One could alternately dmonstrate that prel = −i~∂/∂xrel etc using ja- cobians to confirm these commutations. x2 ]) 4 1 = (−i~)(1 + 0 − 0 − 1) = 0 4 [pcm . x1 ] − [p1 . Defining prel = (p1 − p2 )/2 and pcm = p1 + p2 and xrel = x1 − x2 and xcm = (x1 + x2 )/2 we obtain 1 [prel . xrel ] = ([p1 . x1 ] + [p2 . x1 ] − [p2 . x1 ] + [p1 .56 Vibrations of a One-Dimensional Monatomic Chain which we identify as harmonic oscillators of frequencies p ωcm = k/m p ωrel = (k + 2K)/m (b) We have [pi . x2 ] + [p2 . x2 ] − [p2 . x1 ] − [p1 . x1 ] − [p2 . x1 ] + [p2 . x2 ]) 2 1 = (−i~)(1 − 0 − 0 + 1) = −i~ 2 1 [prel . of modes per angular frequency. It is here to point out that coupled deg rees of freedom act just like a single simple harmonic oscillator once the degrees of free- dom are “rediagonalized”. and gral that you cannot do analytically. The fact . The purpose of this exercise is not just to do another quantum me- chanics problem. you can expand exponentials for high be in one dimension). or it may be occupied with mul- tiple phonons corresponding to a larger amplitude oscillation. (a)‡ Explain what is meant by “normal mode” and by (e) Find the expression for g(ω). A phonon is a quantum of vibration. (a) A normal mode is a periodic collective motion where all particles move at the same frequency. (b)‡ Derive the dispersion relation for the longitudinal (f) Write an expression for the heat capacity of this oscillations of a one-dimensional mass-and-spring crystal one-dimensional chain. spring constant κ (motion of the masses is restricted to (g)* However.) them as a function of k. By expanding to next relation is periodic in reciprocal space (k-space). Hence show that the dispersion Dulong–Petit in one dimension). the expectation of the energy of this system will correspondingly be hEi = ~ωrel (nB (β~ωrel ) + 1/2) + ~ωcm (nB (β~ωcm ) + 1/2) where nB (x) = 1/(ex − 1) is the usual Bose factor. (d)‡ Derive the phase and group velocities and sketch C/N = kB (1 − A/T 2 + . . What is the sound velocity? where ~2 κ Show that the sound velocity is also given by vs = A= 2 6mkB . Sketch g(ω). but it carries momentum as well.ncm = ~ωrel (nrel + ) + ~ωcm (ncm + ) 2 2 At temperature T .2) Normal Modes of a One-Dimensional 1/ βρ where ρ is the chain density and β is the com- Monatomic Chain pressibility. temperature to obtain a high-temperature approxima- (c)‡ Show that the mode with wavevector k has the tion. This is important motivation for treating phonons (coupled modes of springs) as individual harmonic oscillators. non-trivial order. show that How many different normal modes are there. A single mode may be occupied by a single phonon. lattice spacing a. √ (9. You will inevitably have an inte- with N identical atoms of mass m. 57 such that the spectrum can be written as 1 1 Enrel . the density of states “phonon”. The vibration does carry energy. [I do not like the definition ”a quantum of vibrational energy”. It should be obvious that the high-temperature same pattern of mass displacements as the mode with limit should give heat capacity C/N = kB (the law of wavevector k + 2π/a. so why specify energy only?] Each classical normal mode of vibration corresponds to a quantum mode of vibration which can be excited multiple times. . Explain briefly why phonons obey Bose statistics. 9. The dispersion is periodic in k → k + 2π/a. then k = 2πm/L but k is identified with k + 2π/a so that there are therefore exactly N = L/a different normal modes. Using the ansatz xn = Aeiωt−ikna we obtain −ω 2 meiωt−ikna = κeiωt (eik(n+1)a + eik(n−1)a − 2eikn ) ω2m = = 2κ(cos(ka) − 1) or p p ω= (2κ/m)(cos(ka) − 1) = 2 κ/m | sin(ka/2)| κ ω=2 √m Fig.1 Dispersion relation for vibrations of the one-dimensional ω monatomic harmonic chain.58 Vibrations of a One-Dimensional Monatomic Chain that the same state may be multiply occupied by phonons means that phonons must be bosons. k=−π/a 0 k=+π/a (c) e−i(k+2π/a)na = e−i(k+2π/a)na = e−ikna If you assume periodic boundary conditions. (b) The equation of motion for the nth particle along the chain is given by m¨ xn = κ(xn+1 − xn ) + κ(xn−1 − xn ) = κ(xn+1 + xn−1 − 2xn ) note that na is the equilibrium position of the nth particle. (d) p vphase = ω(k)/k = 2 κ/m | sin(ka/2)|/k and p q vgroup = dω(k)/dk = κ/m a cos(|k|a/2) = (a/2)ω0 1 − ω 2 /ω02 . 59 5676589 84 . 5676589 84 . 3 3 01 02 03 3 2 1 45676589 3. 01 02 03 3 2 1 45676589 3. there are N modes total. The density of the chain is ρ = m/a and the compressibility is β = −(1/L)dL/dF = 1/(κa). Thus we obtain v −2 = ρβ (e) Note first that (ω(k)/2)2 + (vgroup (k)/a)2 = κ/m (9. 9. then k is no longer the same as k + 2π/n.1. Note that the phase velocity is not periodic in the Brillouin zone! One can understand this if you think carefully about aliasing of waves. The density of states in k is therefore dN/dk = N a/(2π) = L/(2π) where L is the length of the system. However. when we make n non-integer.1) Density of states is uniform in k. Right: Phase velocity.2 The monatomic harmonic chain. the velocity should have negative sign. p where ω0 = 2 κ/m.2 The sound velocity is the velocity at small k. Thus we have Na g(ω) = dN/dω = (dN/dk)(dk/dω) = 2πvgroup N = p 2π κ/m cos(|k|a/2) 2N = p (9. we have k is the same as k + 2π/a. 9. Note velocities are signed quantities. However. The phase velocity is the velocity at which the peaks of waves move. to the left of the origin. Left: Group velocity.2) 2π (κ/m) − (ω(k)/2)2 where we have used Eq. This is p v = a κ/m . For sketches see figure 9. However. 0 0 03 03 Fig. but this only defines the value of the wave for integer n. If there are N sites in the system. the waves are only defined at the position of the masses along the chain. For integer n. . We write cos(kna) for the positions of the masses at some time. the ”peak” of this function may be between the integer values of n. 60 Vibrations of a One-Dimensional Monatomic Chain 856789 . 5 6 412 4 312 3 012 012 3 312 4 56789 . (f) The energy stored in the chain is given by Z U = dωg(ω)~ω(nB (ω) + 1/2) so the heat capacity is C = ∂U/∂T . The additional factor of 2 that appears up top is to account for the fact that for each value of ω > 0 there are actually two values of k with that ω. we use the high temperature expansion (expanding 1/(ex − 1) for small x) kB T 1 ~ω nB (ω) + 1/2 = + + . Note that the p DOS diverges at ω = 2 k/m where the group velocity goes to zero. ~ω 12 kB T So that we now have Z ∂U 1 1 ~ω C= = kB N − 2 dω~ωg(ω) ∂T T 12 kB So that the coefficient A defined in the problem has the values Z ~2 A= 2 dωω 2 g(ω) 12N kB Inserting our expression for g(ω) we obtain Z ωmax ~2 ω2 A= 2 dω p 12πkB 0 (κ/m) − (ω/2)2 . 9. (g) To recover the law of Dulong-Petit.3 The one dimensional harmonic chain. Note that we can drop the +1/2 since it has no derivative.. 5 6 Fig. (Note if you integrate over frequency you correctly get back N degrees of freedom).. Density of states g(ω). one takes the high temperature limit of nB (ω) = kB T /~ω so th at we have Z Z ∂ C= dωg(ω)(kB T ) = kB dωg(ω) = kB N ∂T To go further. (9.3) so r 2κ1 2κ2 ω= (cos(ka) − 1) + (cos(2ka) − 1) m m (b) To obtain the sound velocity. . Generalize this model to include springs not (a) Calculate the dispersion curve ω(k) for this model. Thus we obtain ~2 κ A= 2 6kB m as required.3) More Vibrations called κ1 and the spring constant between second neigh- Consider a one-dimensional spring and mass model of bors be called κ2 . Show the group neighbors. only between neighbors but also between second nearest (b) Determine the sound wave velocity.3 to given mω∂ω/∂k = −2aκ1 sin(ka) − 4aκ2 sin(2ka) At the zone boundary k = π/2 both terms on the right hand side are zero. 9. hence we have zero group velocity. (a) Use the same approach m¨ xn = κ1 (xn+1 − xn ) + κ1 (xn−1 − xn ) + κ2 (xn+2 − xn ) + κ1 (xn−2 − xn ) = κ1 (xn+1 + xn−1 − 2xn ) + κ2 (xn+2 + xn−2 − 2xn ) Using the same ansatz xn = Aeiωt−ikna we obtain −mω 2 = 2κ1 (cos(ka) − 1) + 2κ2 (cos(2ka) − 1) (9. a crystal. Let the mass of each atom be m. expand for small k to obtain r r ! 2κ1 (ka)2 2κ2 (2ka)2 κ1 + 4κ2 ω = + = a k m 2 m 2 m Thus the sound velocity is r κ1 + 4κ2 vs = a m The easiest way to examine ∂ω/∂k at the zone boundary is to differ- entiate Eq. 61 p Defining x = (ω/2) m/κ we obtain Z ~2 8κ 1 x2 A= 2 dx √ 12πkB m 0 1 − x2 The integral is evaluated to give π/4 (make the substitution x = sin θ). Let the spring constant between neighbors be velocity vanishes at the Brillouin zone boundary. one can always go back to the equations of motion m¨ xn = κ(xn+1 − xn ) + κ(xn−1 − xn ) = κ(xn+1 + xn−1 − 2xn ) Using the ansatz xn = Aeiωt−ikna we obtain −ω 2 meiωt−ikna = κeiωt (eik(n+1)a + eik(n−1)a − 2eikn ) ω2m = = 2κ(cos(ka) − 1) . which means x = π/2 + real. Multiplying by z we obtain the quadratic z 2 − 2izy − 1 = 0 Solving this equation gives p z = iy ± 1 − y2 Note that for y > 1 we have z = eix along the positive imaginary axis. but rather will decay as The usual form is Ω = ω/ωmax = ± sin(ka/2) with ± chosen so Ω is positive. For y = 1 we have x = π/2 whereas for y < 1 we obtain a complex valued z with unti magnitude. What happens upon the chain (say by a driving force) the “wave” will to this length as ω → ωmax ? not propagate along the chain. If one is concerned that this solution might not work. The complex part gets smaller and vanishes as Ω approaches 1 from above.6 below. Thus ka = ±2 sin−1 Ω When Ω = 1 then ka = π. When Ω > 1 then ka becomes complex with a real part of π.3). If a vibration with frequency ω > ωmax is forced the decay length of this evanescent wave. The complex part of k gives the inverse length scale of the wave’s decay (and it can decay left-going or right- going depending on the ± sign). 9. which is quite similar. posed (this is sometimes known as an “evanescent” wave). consider z − z −1 y = sin(x) = 2i where z = eix . 9. the length scale of decay grows longer and longer until at 1 it extends over the entire system and is a nondecaying wave.62 Vibrations of a One-Dimensional Monatomic Chain (9. See also the solution of 9.3 for a complex k to determine ωmax . thus a real x. To check these properties of the arcsin. there is a maximum possible frequency of oscillation With ω > ωmax solve Eq. So as Ω approachs 1 from above.4) Decaying Waves one moves away from the point where the oscillation is im- In the dispersion curve of the harmonic chain (Eq. 5) Reflection at an Interface* from the left. it should read T eiωt−ikR na n≥0 δxn = iωt−ikL na Ie + Reiωt+ikL na n<0 On both the left and the right. the ansatz form will then satisfy the equations of motion except near the boundary. κL κL κL κR κR κR m m m m m m m I T R A wave with amplitude I is incident on this interface Oops sorry about the typo.R and ω q ω=2 κL.R = sin−1 m/κL. Using double angle formula 4κ ω2 = sin2 (ka/2) m The only thing we would need to be careful about is choosing the sign of the square root. where it can be either transmitted with Consider a harmonic chain of equally spaced identical amplitude T or reflected with amplitude R. (9.R /m| sin(kL. as shown in this figure. we have equations of motion for δx−1 and δx0 which involve δx coordinates on both sides of the junction. Using the masses of mass m where left of the n = 0 mass the spring following ansatz form constant is κL but right of the n = 0 mass.R a/2)| or q 2 ω kL. T eiωt−ikL na n≥0 δxn = iωt−ikR na Ie + Reiωt+ikR na n < 0 a n=0 derive T /I and R/I given ω. κL . we must have the usual relations between kL. 63 Note that nowhere did we specify that k is real (!) so we can happily extend into the complex plane. the spring constant is κR . κR and m.R a 2 with these values of k. Near the boundary. We write ¨ −1 m δx = κL (δx−2 + δx0 − 2δx−1 ) m δx¨0 = κL (δx−1 − δx0 ) + κR (δx1 − δx0 ) . except Use an ansatz of the form for the mass at position n = 0 which instead has value M < m as shown in this figure: δxn = Aeiωt−q|n|a a n=0 with q real to solve for the frequency of this impurity mode.4. have the same value κ and masses have value m.5) and the first equation becomes simply I +R=T which is just extending the wavefunction from the left hand side until it hits position zero and matching it to the value of the wave from the right at this position.5 to give us −mω 2 /2 + κL (1 − e−ikL a ) = iκL sin(kL a) and similarly for the right hand wave so that we get 2 T /I = κR sin(kR a) 1+ κL sin(kL a) (9. κ κ κ κ κ κ m m m M m m m .64 Vibrations of a One-Dimensional Monatomic Chain Plugging in the ansatz form of the wavefunction and doing a bit of algebra gives us (−mω 2 + 2κL ) IeikL a + Re−ikL a = κL IeikL 2a + Re−ikL 2a + T (−mω 2 + κL + κR )T = κL IeikL a + Re−ikL a + κR T e−ikR a (9.4) The first equation simplifies quite a bit by rewriting the dispersion rela- tion as −mω 2 = 2κL (cos(kL a) − 1) So that −mω 2 + 2κL = κL eikL a + e−ikL a (9. Plugging this into the second equation of Eq. there can be a Consider a harmonic chain where all spring constants standing wave normal mode localized near the impurity.4 and solving to obtain κL 2i sin(kL a) T /I = −mω 2 + κL (1 − e−ikL a ) + κR (1 − e−ikR a ) In the denominator we can simplify further applying 9. Consider your result in the context of Exercise 9. 9.6) Impurity Phonon Mode* Along with traveling wave solutions. one can go a bit further analytically. writing z = eqa we have (M/m)((z + z −1 )/2 + 1) = z −1 + 1 which is a quadratic equation in z.6) or equivalently ω = ωmax | sin(ka/2)| or ka = 2 sin−1 (ω/ωmax) p with ω > ωmax = 2 κ/m we have decaying solutions of the form given by the usual where (as shown in Exercise 9. if M = m the only place the two sides are equal is at q = 0.4. since the left hand side grows with increasing q and the right shrinks with q it is clear that there can be no solution for M > m. However. 65 Another error here: q needs to have both real and imaginary parts. Further. So we have Aeiωt+iπn+qan n ≤ 0 δxn = Aeiωt−iπn−qan n ≥ 0 with q > 0 this solves all of the equations of motion except for the one at n = 0 ¨0 M δx = κ(δx−1 + δx1 − 2δx0 ) Plugging in our ansatz. For 0 < ω < ωmax the only solution to our equations of motion (except at the impu- rity site) is for the usual waves −mω 2 = 2κ(cos(ka) − 1) (9. In fact. for M < m there is one crossing of the two curves. which we can solve to given 2m z = eqa = −1 M . Then on the left we choose + and on the right we choose − so that the wave decays away from the impurity.6 and eliminating ω 2 we obtain M (cosh(qa) + 1) = m(e−qa + 1) Now. 9. which finds the q appropriate for the bound state. which means there is no bound state. This problem is fairly easy once you have solved 9. we obtain the three equations −M ω 2 = κ(−e−qa − e−qa − 2) Taking this equation with to Eq.4) ka = π ± iqa with q here real. 8) U= xa Va.b √ ma mb Hamiltonian is rewritten as Thus show that the solutions are X 1 (m) 2 1 2 (m) 2 H= [P ] + ωm [Y ] (9. Y (n) ] = −i~δn. a where ωm is the mth eigenvalue of the matrix S with Conclude that the quantum eigenfrequencies of the sys- (m) corresponding eigenvector sa . however.b yb b [P (m) .b. (Can you derive this result from the modes of the system. X (m) √ sider a situation in 1d. (i) First given the expression for U . (Here we con. the eigenvec- where xa is the deviation of the position of particle a from tors can be taken to be real. Construct the transformed its equilibrium position and V can be taken (without loss coordinates of generality) to be a symmetric matrix. Consider a set of N particles a = 1.12) ya(m) = e−iωm t s(m) m 2 2 . we will see that to go Y (m) = sa x a m a (9. since we require q > 0 has a solution only for M < m. prior two equations?) 2 Yet another typo.b xb b .b Since S is symmetric as well as hermitian.7) ma interacting via a potential a X (m) 1X [s(m) ∗ a ] [sb ] = δa. Finally to obtain the frequency.10) √ (i) Defining ya = ma xa .66 Vibrations of a One-Dimensional Monatomic Chain Or 2m qa = log −1 M which.) P (m) = s(m) a p a / ma (9. .9) to 3d we just need to keep track of three times as many a X √ coordinates. (ii) Recall the orthogonality relations for eigenvectors come Quantum Eigenstates∗ of hermitian matrices This proof generalizes the argument given in Exercise X (m) ∗ (n) 9.n (9. one just has to plub back into our dispersion relation to give −mω 2 = 2κ(cos(π + iqa) − 1) which gives a real frequency greater than ωmax (9. . a tions of motion may be written as X show that these coordinates have canonical commutations y¨a = − Sa.m (9. differentiate with resepect to xa to obtain the force on particle a ∂U X Fa = − =− Va.1. . These are the N normal tem are also ωm .b = √ Va.b xb ∂xa b The equation of motion is then X ma x¨a = − Va. (9.b xb m 2 a. N with masses [sa ] [sa ] = δm.11) where 1 1 and show that in terms of these new coordinates the Sa.7) General Proof That Normal Modes Be. show that the classical equa. in part (i) it should say −ωm is the mth eigenvaluer. (ii) Given X √ Y (m) = s(m) a xa ma a X (n) √ P (n) = s b p b / mb b we calculate X (n) p [P (n) . 67 √ Now defining xa = ya / ma we obtain X 1 1 X y¨a = − √ Va. xa ] ma /mb a.b Using the canonical commutations [pb .b.b a. Next let us look at the terms of the proposed Hamiltonian X1 XX 1 (m) √ [P (m) ]2 = s(m) a s b p a p b / ma mb m 2 m a.13) b which is solved by sa being the eigenvector of S and ω 2 its eigenvalue.b yb ma ma b b −iωt as required.13 to replace ω 2 with the matrix S giving us X1 XX 1 √ 2 ωm [Y (m) ]2 = s(m) (m) a Sb. Y (m) ] = −i~ s(m) (n) a sa = −i~δnm a where we have used the orthogonality of eigenstates. Using the ansatz ya = e sa we have X ω 2 sa = Sa.c sc xa xb ma mb m 2 m a. Y (m) ] = sb s(m) a [pb .b But here we can use Eq. 9.b sa (9. Now examining the second term of the proposed Hamiltonian X1 XX 1 √ 2 2 (m) ωm [Y (m) ]2 = s(m) a ωm sb xa xb ma mb m 2 m 2 a.b 2 X 1 = p2a a 2m a where we have used the orthogonality of s to collapse the m sum and generate δab .b xb 2 2 a.b √ yb = − Sa.c 2 X1 √ X1 = Sb.a xa xb ma mb = xa Va. xa ] = −i~δab we have X [P (n) .b . Now let us define . First of all. Further the P and Y satisfy canonical commutations for momenta and positon so each m is simply a harmonic oscillator with frequency ωm .8) Phonons in 2d* Consider a mass and spring model of a two-dimensional triangular lattice as shown in the figure (assume the lat- tice is extended infinitely in all directions). one assumes that the crystal starts in equilibrium with the springs unstretched. Let us let the edge have unit length for simplicity of notation. Calculate the dispersion curve ω(k). that if one of the masses is displaced in a direction perpendicular to one of its attaching springs.68 Vibrations of a One-Dimensional Monatomic Chain where we have used the orthogonality of s and taken the sum over m to give a δac . Ok.b The new form of the Hamiltonian in terms of P and Y has each m coordinate completely decoupled. the spring is not stretched at all (it is only rotated)! So in other words. Forces on the springs will be proportional to the amount of stretching. Probably it should have two stars. and it might be useful to read those chapters first and then return to try this exercise again. If we let the position of a sites in equilibrium be called rn and let the displacements from this equilibrium be called urn . Thus the given Hamiltonian is equivalent to the original Hamitlonian X 1 X1 H= p2a + xa Va. one could follow the usual procedure of writing 1 √ √ a† = √ P/ ω + i ωY 2~ to rewrite the Hamiltonian for each decoupled mode as H = ~ω(a† a + 1/2) (9.b xb a 2ma 2 a. In Chap- ters 12 and 13 we study crystals in two and three dimen- sions. this one is pretty hard if you have never seen such a thing before. to linear order. Assume that identical masses m are attached to each of their six neigh- bors by equal springs of equal length and spring con- stant κ. we need only keep track of the stretching in the direction parallel to the spring. The two- dimensional structure is more difficult to handle than the one-dimensional examples given in this chapter. Note however. If one wanted to derive the spectrum from the Hamiltonian. Fig. The force on a mass.5 k · ai Si = sin2 ( ) -3 -2 -1 1 2 3 2 These spectra are shown in figures 9.4 .6. Note that the spectrum is periodic in the Brillouin zone – which has the shape of a hexagon. 3/2).6 Cut along k = (kx. a2 = (−1/2. ky space.4 Contour plot of the transverse we get mode frequency of the triangular lattice phonons in kx .5 Contour plot of the longitu- 3 κ X α β ik·aj dinal mode frequency of the triangular Dαβ (k) = a a (e + e−ik·aj − 2) lattice phonons in kx . 9. Plugging in the values of aj and using some trig identities we gets √ √ κ 4 0 2 k · a1 1 3 2 k · a2 1 − 3 2 k · a3 D= sin ( )+ √ sin ( )+ √ sin ( ) m 0 0 2 3 3 2 − 3 3 2 For each k there will be two eigenvalues which correspond to the longi- tudinal and transverse phonon modes.9. 0). 9. X3 X β m¨uαrn = −κ aα j [ur +a + uβr −a − 2uβr ]aβj n j n p n 6 j=1 β 4 Using a wave ansatz 2 uα α iωt−ik·rn rn = U e 0 We obtain the two by two eigenvalue problem -2 X ω2U α = Dαβ (k)U β -4 -6 β -6 -4 -2 0 2 4 6 where the so-called dynamical matrix is given by Fig. and hence using -2 newton’s law we have -4 −∂U -6 uα m¨ α rn = Frn = -6 -4 -2 0 2 4 6 ∂uα rn where α is a basis direction (say. 3/2). a2 = (1/2. x or y). . 9. Carefully taking the derivative Fig. The eigenvalues of this becomes q 2κ 2 ω2 = S1 + S2 + S3 ± S12 + S22 + S32 − S1 S2 − S1 S3 − S2 S3 m 1. m j=1 j j Note that this matrix is symmetric in α and β. The total potential energy of the system can then be written as 3 κ XX 2 6 U= urn − urn +aj · aj 2 r j=1 4 n 2 where here we have included each spring once in the sum and we have dotted the displacement with its direction so as to only count stretching 0 of the spring and not rotation. 0) of the spectrum of the triangular lattice phonons. ky space.5 1 where 0. 69 three unit vectors ai pointing √ in the three directions √ of the lattice edges a1 = (1. . Acoustic waves ω ∼ k for small k. (d) Sketch the dispersion in both reduced and extended and motion of particles is in one dimension only). spring crystal where the unit cell is of length a and each p Show that the sound velocity is also given by vs = unit cell contains one atom of mass m1 and one atom of β −1 /ρ where ρ is the chain density and β is the com- mass m2 connected together by springs with spring con. how many modes are there there m1 m2 at each k? (e) What happens when m1 = m2 ? The following figure depicts a long wavelength acoustic wave: All atoms in the unit cell move in-phase with a slow spatial modulation. Determine the sound velocity and show that the nal oscillations of a one-dimensional diatomic mass-and. (b) Derive the dispersion relation for the longitudi. group velocity is zero at the zone boundary. as shown in the figure (all springs are the same.e. margin note 4 of this chapter!). zone scheme.1) Normal modes of a One-Dimensional Di.. stant κ. how many different normal modes are there? How many branches of excitations are there? I. κ κ in reduced zone scheme. pressibility. (c) Determine the frequencies of the acoustic and op- atomic Chain tical modes at k = 0 as well as at the Brillouin zone (a) What is the difference between an acoustic mode boundary. a If there are N unit cells.Vibrations of a One-Dimensional Diatomic Chain 10 (10. a acoustic κ κ m1 m2 . Describe the motion of the masses in each case (see Describe how particles move in each case. and an optical mode. (In general a long wavelength optical mode is any long wavelength mode where not all atoms in the unit cell are moving in phase).72 Vibrations of a One-Dimensional Diatomic Chain The following depicts a long wavelength optical wave: The two differ- ent types of atoms move out of phase.2) which is an eigenvalue problem from ω 2 . We can assume that the equilibrium position of xn is given by na and the equilibrium position of yn is given by na + d. Optical modes have ω nonzero as k → 0. with a slow spatial modulation. a optical κ κ m1 m2 (b) Let xn be the position of the nth particle of mass m1 and yn be the position of the nth particle of mass m2 . Thus we need to find the roots of the determinant . We write the equations of motion for the deviations from these equi- librium positions δxn and δyn . ¨n m1 δx = −κ(δxn − δyn−1 ) − κ(δxn − δyn ) ¨ m2 δy = −κ(δyn − δxn ) − κ(δyn − δxn+1 ) n Writing the ans¨ atze δxn = Ax eikan−iωt δyn = Ay eikan−iωt we obtain the equations −m1 ω 2 Ax eikna = −2κAx eikna + κAy (eikna + eik(n−1)a ) −m2 ω 2 Ay eikna = −2κAy eikna + κAx (eikna + eik(n+1)a ) or ω 2 Ax = 2(κ/m1 )Ax − (κ/m1 )(1 + e−ika )Ay (10. Note that the amplitude of motion of the different atoms in the cells is generally not the same.1) 2 ika ω Ay = 2(κ/m2 )Ay − (κ/m2 )(1 + e )Ax (10. . . 2(κ/m1 ) − ω 2 −(κ/m1 )(1 + e−ika ) . . . . −(κ/m2 )(1 + eika ) 2(κ/m2 ) − ω 2 . . For the optical mode at k = 0 we have Ax = −(m2 /m1 )Ay meaning that the two different masses move in opposite directions with the heavier mass moving with lower amplitude. so the two modes have energy r r 2κm1 2κm2 ω= and m1 m2 m1 m2 the greater of which is the optical mode.2 to find the relation between Ax and Ay . the lesser being the acoustic mode. For example. An example of a zone boundary mode is shown in the following figure a zone boundary κ κ m1 m2 . 73 which gives the equation κ2 0 = ω 4 − ω 2 (2κ(1/m1 + 1/m2 )) + 4 − (1 + eika )(1 + e−ika ) m1 m2 2 2(m 1 + m 2 )κ κ 0 = ω4 − ω2 + (2 − 2 cos(ka))) m1 m2 m1 m2 with the solution (skipping a few steps) q 2 κ 2 2 ω = m1 + m2 ± m1 + m2 + 2m1 m2 cos(ka) m1 m2 q κ = m1 + m2 ± (m1 + m2 )2 − 4m1 m2 sin2 (ka/2) m1 m2 (c) At k = 0.1 and 10. the higher mass particles move and the lower mass particle stays fixed. cos(ka) = 1. for the lower frequency mode. 10. For the acoustic mode at k = 0 we obtain Ax = Ay which means the two masses move in phae with the same amplitude. every other higher mass particle moves in the opposite di- rection (thus compressing symmetrically around the fixed particle. Since we are at the zone boundary. To find the motions corresponding to these modes we need to plug our frequencies back into Eqs. the acoustic mode has zero energy. At the zone boundary the two modes cor- respond to one of the masses staying still and the other mass moving. whereas the optical mode has energy s 2κ(m1 + m2 ) ω= m1 m2 At the zone boundary cos(ka) = −1. since the group velocity is dω/dk and since dω/d cos(ka) is nonsingular.74 Vibrations of a One-Dimensional Diatomic Chain To find the sound velocity. The density of the chain is ρ = (m1 + m2 )/a. expand the cos around k = 0. one obtains the acoustic mode velocity ω = vk with r κ v=a 2(m1 + m2 ) We check that v −2 = ρβ. Near the zone boundary. the group velocity must be zero by using the chain rule since d cos(ka)/dk = a sin(ka) = 0 at the zone boundary (k = π/a). the compressibility of two springs in series is κ/2 so the compressibility of the chain is β = −(1/L)dL/dF = 2/(κa). 5676589 8 85 84 . 1 2 3 45676589 . 01 02 03 3 2 1 5676589 8 85 84 . 1 2 3 01 02 03 3 2 1 45676589 . . there are 2N modes. Both pictures use m1 /m2 = . 10. Top Reduced Zone Scheme.4. therefore 2N atoms. Bottom Extended Zone Scheme. If there are N unit cells. Fig.1 Diatomic Chain. There are 2 modes per k in the reduced zone scheme. therefore two branches. whereas for ω between the zone boundary acoustic and optical model frequencies. if this chain is driven at Consider the alternating diatomic chain dispersion as a frequency ω for which there are no propagating wave discussed in the text Eq.4. 10. From the text Eq.6.6 for a complex k. the compressibility. it to vs = β −1 /ρ where ρ is the chain density and β is ferent spring constants κ1 and κ2 and lattice constant a. instead.2) Decaying Waves wave modes. As in Exercise 9. or evanescent. Consider a general diatomic chain as shown in Fig. Whereas the arcos of a number less than -1 is π + imaginary with the imaginary part growing from zero as the argument decreases from -1.6 we can rearrange to obtain 1 ka = cos−1 (mω 2 − κ1 − κ2 )2 − κ21 − κ22 2κ1 κ2 The argument on the right hand size is greater than 1 for ω larger than the q = 0 optical mode freuquency. we simply want to analytically extend k to complex numbers. the right hand side is less than −1. ω− (k = π/a) and ω+ (k = π/a) there are no propagating As in problem 9. then there will be a decaying. find the length ing wave modes. and similarly for frequencies between scale of this decaying wave.6 and shown in Fig. the gap at the Brillouin zone boundary vanishes and the two branches become the single branch of the monatomic chain (this is most easily described in extended zone scheme).4. just a bit more algebra to keep track of ¨n m1 δx = −κ1 (δxn − δyn−1 ) − κ2 (δxn − δyn ) ¨ m2 δy = −κ2 (δyn − δxn ) − κ1 (δyn − δxn+1 ) n Writing the ans¨ atze δxn = Ax eikan−iωt δyn = Ay eikan−iωt . 10. By solving 10. (10. (a) This is the same approach as the prior cases. 10. (10. 10. The length scale of decay is always given by L = a/q with q the imaginary part of k. In this limit. 75 (e) When m1 = m2 the unit cell is now of size a/2 so the Brillouin zone is doubled in size. modes.3) General Diatomic Chain* (a) Calculate the dispersion relation for this system. wave For frequencies above ω+ (k = 0) there are no propagat.1 (b) Calculate p the acoustic mode velocity and compare with two different masses m1 and m2 as well as two dif. The arccos of a number greater than one is pure imaginary and grows from 0 as the argument increases from unity. The density of the chain is ρ = (m1 +m2 )/a. the spring constant of the two springs in series is κ1 κ2 /(κ1 κ2 ) so the compressibility of the chain p is β = −(1/L)dL/dF = (κ1 κ2 )/(κ1 + κ2 )/.76 Vibrations of a One-Dimensional Diatomic Chain we obtain the equations −m1 ω 2 Ax eikna = −(κ1 + κ2 )eikna Ax + (κ2 eikna + κ1 eik(n−1)a )Ay 2 ikna −m2 ω Ay e = −(κ1 + κ2 )eikna Ay + (κ2 eikna + κ1 eik(n+1)a )Ax or κ1 + κ2 κ2 κ1 ω 2 Ax = Ax + + e−ika Ay m1 m m1 1 κ1 + κ2 κ2 κ1 ω 2 Ay = Ay + + eika Ax m2 m2 m2 which is an eigenvalue equation for ω 2 . Plugging this into vs = β −1 /ρ gives the same result. The secular equation is then 0 = (κ1 + κ2 − m1 ω 2 )(κ1 + κ2 − m2 ω 2 ) − κ21 − κ22 − 2κ1 κ2 cos(ka) or (κ1 + κ2 )(m1 + m2 ) 2 4κ1 κ2 ω4 − ω − sin2 (ka/2) = 0 m1 m2 m1 m2 with the solution (skipping a few steps) (κ1 + κ2 )(m1 + m2 ) ω2 = 2m1 m2 s 1 (κ1 + κ2 )2 (m1 + m2 )2 16κ1 κ2 ± 2 − sin2 (ka/2) 2 (m1 m2 ) m1 m2 (b) Simply taylor expanding the dispersion gives us r κ1 κ2 ω= ak (κ1 + κ2 )(m1 + m2 ) or s a2 κ1 κ2 vs = (κ1 + κ2 )(m1 + m2 ) Now the hydrodynamic approach. . with spring constant κ′ connecting equivalent masses in sider the diatomic chain from Exercise 10. . Determine the dispersion relation for to the spring constant κ between neighboring masses. In addition adjacent unit cells. What happens if κ′ ≫ κ? pose that there is also a next nearest-neighbor coupling We write the equations of motion ¨n m1 δx = −κ(δxn − δyn−1 ) − κ(δxn − δyn ) − κ′ (δxn − δxn−1 ) − κ′ (δxn − δxn+1 ) ¨ m2 δy = −κ(δyn − δxn ) − κ(δyn − δxn+1 ) − κ′ (δyn − δyn−1 ) − κ′ (δyn − δyn+1 ) n Writing the usual ans¨ atze δxn = Ax eikan−iωt δyn = Ay eikan−iωt we obtain the equations −m1 ω 2 Ax eikna = −2(κ + κ′ )Ax eikna + κAy (eikna + eik(n−1)a ) + κ′ Ax (eik(n+1)a + eik(n−1)a ) −m2 ω 2 Ay eikna = −2(κ + κ′ )Ay eikna + κAx (eikna + eik(n+1)a ) + κ′ Ay (eik(n+1)a + eik(n−1)a ) or m1 ω 2 Ax = 2 [(κ + κ′ ) − κ′ cos(ka)] Ax − κ(1 + e−ika )Ay m2 ω 2 Ay = 2 [(κ + κ′ ) − κ′ cos(ka)] Ay − κ(1 + eika )Ax which is an eigenvalue problem for ω 2 .1. 77 (10. sup. Using some trig identies and some algebra the secular equation can be reduced to 0 = m1 m2 ω 4 − (m1 + m2 ) 2κ + 4κ′ sin2 (ka/2) ω 2 + [4κ2 + 16κκ′ ] sin2 (ka/2) + 16κ′2 sin4 (ka/2) with the solution 1 ω2 = (m1 + m2 )(2κ + 4κ′ sin2 (ka/2)) 2m1 m2 q ± (m1 + m2 )2 (2κ + 4κ′ sin2 (ka/2))2 − 4m1 m2 ([4κ2 + 16κκ′ ] sin2 (ka/2) + 16κ′2 sin4 (ka/2)) 1 = (m1 + m2 )(κ + 2κ′ sin2 (ka/2)) m1 m2 q ± 4m1 m2 κ2 cos2 (ka/2) + 4(m1 − m2 )2 κ′2 sin4 (ka/2) + κκ′ sin2 (ka/2) Note that when κ2 = 0 we recover the diatomic chain from Exercise 10. this system.4) Second Neighbor Diatomic Chain* Con. When κ′ ≫ κ we get two decoupled monatomic chains (as should be expected!).1. 78 Vibrations of a One-Dimensional Diatomic Chain (10. (a) At k = 0 how many optical modes are there? Cal- With three particles in the unit cell. m2 . m3 with positions x. Declaring the unit cell to be one set of m1 . (b)* If all the masses are the same and κ1 = κ2 de- a termine the frequencies of all three modes at the zone boundary k = π/a. Hint: You will get a Consider a mass-and-spring model with three different cubic equation. You will have a cubic equation. z our equations of motion are ¨n m1 δx = −κ3 (δxn − δzn−1 ) − κ1 (δxn − δyn ) ¨ m2 δy = −κ1 (δyn − δxn ) − κ2 (δyn − δzn ) n ¨n m3 δz = −κ2 (δzn − δyn ) − κ3 (δzn − δxn+1 ) Plugging in the usual wave ansatz gives us the secular determinant . and m1 = m2 determine the frequencies of all three modes at As usual. in one dimesion there is one acoustic mode and two optical modes. you already know one of the masses and three different springs per unit cell as shown roots since it is the energy of the acoustic mode at k = 0 in this diagram.5) Triatomic Chain* culate the energies of these modes. guess one of the roots. Again you should be able to dimension. assume that the masses move only in one the zone boundary k = π/a. However. but you should be able to guess one root which corresponds κ3 κ1 κ2 κ3 κ1 to a particularly simple normal mode. m3 m1 m2 m3 m1 m2 (c)* If all three spring constants are the same. y. . . κ 3 + κ 1 − m1 ω 2 κ1 κ3 e−ika . 0 = . κ1 κ 1 + κ 2 − m2 ω 2 κ2 . . . κ3 eika κ2 κ 2 + κ 3 − m3 ω 2 . and setting ka = π we obtain 0 = m3 ω 6 − 2m2 (2κ1 + κ3 ) ω 4 + (2κ1 κ3 + κ21 )(3m) ω 2 − κ21 κ3 (4) (10. This we could have guessed in advance since we know the acoustic mode should come down to zero frequency at k = 0.3) At k = 0. which multiplied out gives the secular polynomial 0 = m1 m2 m3 ω 6 − [κ1 (m1 + m2 )m3 + κ3 m2 (m1 + m3 ) + κ2 m1 (m2 + m3 )] ω 4 + [(κ2 κ3 + κ1 κ2 + κ3 κ1 )(m1 + m2 + m3 )] ω 2 − [κ1 κ2 κ3 (2 − 2 cos(ka))] (10. (b) Setting all the masses the same and κ1 = κ2 .4) . Once we remove the obvious factor of ω 2 the remaining polynomial is just quadratic and we can obtain the two roots in the obvious way. the final term vanishes and the secular polynomial clearly has ω = 0 a root. this frequncy solves Eq. We can thus factor out this root to obtain a quadratic which we then solve to get the two other roots at the zone boundary p 2 3κ1 + 2κ3 ± 9κ21 − 4κ1 κ3 + 4κ23 ω = 2m (c) Setting all of the spring constants the same and m1 = m2 and setting ka = π we obtain 0 = m21 m3 ω 6 − [2κm1 (2m3 + m1 )] ω 4 + (2m1 + m3 )(3κ2 ) ω 2 − 4κ3 (10. 10. but since we are at the zone boundary every other κ3 spring moves in the opposite direction.5) Very similar reasoning suggests a mode at this wavevector with frequency ω 2 = κ/m1 (which is indeed a solution) which we can then factor out to give the two additional zone boundary modes p ! 3m + 2m ± 9m 2 − 4m m + 4m2 3 1 1 3 ω2 = κ 3 1 2m1 m3 .4. However. with some intuition we realize that there should be a mode at the zone boundary where the κ3 spring does not compress at all – the two masses connected to κ3 moving in unison. Thus we expect a frequency of ω 2 = κ1 /m. Indeed. 79 Here we have a difficult cubic. The two masses moving in unison have mass 2m and have a restoring force of 2κ1 from the two springs on the two sides. . In other words.3). Thus there are exactly N different values of k within the Brillouin zone. N (you may assume What is the effective mass of the electron near the periodic boundary conditions. If we started with a Hilbert space of N dimensions (described by states |ni). then we expect that there should be exactly . Thus k must be quantized k = 2πm/(N a). If there are N sites in the system.Tight Binding Chain (Interlude and Preview) 11 (11.2. What is the heat capacity if each atom is divalent? Write a general wavefunction as X Ψ= φn |ni n The schroedinger equation (See problem 6. for t > 0 the minimum occurs at k = 0 whereas for t < 0 the minimum is at the zone boundary..e. (Hint: Use the effective Schroedinger equations of Exer. suppose hn|H|mi = ǫ for n = m and tron) what is the density of states at the Fermi surface? hn|H|mi = −t for n = m ± 1. to that of Exercise 9. i. Give an approximation of the heat capacity of the Derive and sketch the dispersion curve for electrons. This is to be expected. Note however.1) Monatomic Tight Binding Chain cise 6. Let the distance between How many different eigenstates are there in this atoms be called a. system? bital on atom n as |ni for n = 1 . bottom of this band? thonormality of orbitals. the total length of the system (as- sumed periodic) is N a. .) trons hopping between atoms. Suppose What is the density of states? there is an on-site energy ǫ and a hopping matrix element If each atom is monovalent (it donates a single elec- −t. and you may assume or. . hn|mi = δnm ). and here let us label the atomic or. The resulting equation should look very similar Consider a one-dimensional tight binding model of elec.2a) is Eφn = ǫφn − t(φn+1 + φn−1 ) write the ansatz φn = eikna and we obtain E(k) = ǫ − 2t cos(ka) (I won’t sketch this because it is rather trivial). system (see Exercise 4.2a. then the band would be com- pletely full. Here g(EF )V = N 1 π |t| . If there were two electrons per atom. (Note if you integrate over frequency you correctly get back 2N degrees of freedom. and there would be no freedom in the system at all. as in exercise 9. If each atom is monovalent. (And one gets twice as many eigenstates if you include spin). Expanding around the minimum one obtains E = const + |t|a2 k 2 which we set equal to a free particle dispersion ~2 k 2 /(2m∗ ) = |t|a2 k 2 yielding an effective mass m∗ = ~2 /(2|t|a2 ) The density of states in k space is uniform as usual dN/dk = L/(2π) = N a/(2π) (I have not included spin here yet) so the density of states in energy dN/dE = (dN/dk)(dk/dE) Now using q dE/dk = 2ta sin(ka) = 2|t|a 1 − ((E − ǫ)/(2t))2 we obtain N 2∗2 dN/dE = q 2π 2 2|t| 1 − ((E − ǫ)/(2t)) and. The spin susceptibility would be zero too.82 Tight Binding Chain (Interlude and Preview) N eigenstates.2 there is an extra factor of 2 being that one has two possible k states for each energy E and still another factor of 2 to include the 2 spin states. now with the factor of 2 for spin included). Thus the density of states is N 1 dN/dE = π |t| The heat capacity is given in terms of the density of states as 2 CV = γ˜ kB g(EF )V T with γ˜ = π 2 /6 (the exact number is from a more complicated calculation that the students are not responsible for). . the band is exactly half filled (since two electrons can go into each orbital). The heat capacity would be zero. I.2) Diatomic Tight Binding Chain studying that exercise again. The unit cell a is now the distance from an A atom to the next A atom. If you are stuck. hn|H|ni is ǫA for n being bottom of the lower band? on a site of type A and is ǫB for n being on a site of type If each atom (of either type) is monovalent.) We now generalize the calculation of the previous exer. Let φA n be the amplitude of the wavefunction on the n th site of B th type A and φn be the amplitude on the n site of type B.e. (All hopping matrix elements −t are still identical to system a metal or an insulator? each other. as follows: What happens if ǫA = ǫB ? What happens in the “atomic” limit when t be- −A − B − A − B − A − B− comes very small. where we solve for the roots of the determinant . Suppose that the onsite energy of type A is different from What is the effective mass of an electron near the the onsite energy of type B. (This is ex. is the B. The Schroedinger equation becomes EφA n = ǫA φA B B n − t(φn + φn−1 ) EφB n = ǫB φB A A n − t(φn + φn+1 ) Using the ans¨ atz φA n = Aeikna φB n = Beikna gives EA = ǫA A − t(1 + e−ika )B EB = ǫB B − t(1 + eika )A again giving a two by two eigenvalue problem. 83 (11.) *Given the results of this exercise. try lator. explain why LiF Calculate the new dispersion relation. Sketch this dispersion relation in both the reduced cise to a one-dimensional diatomic solid which might look and extended zone schemes. (which has very ionic bonds) is an extremely good insu- tremely similar to Exercise 10.1. . . ǫA − E −t(1 + e−ika ) . . . . −t(1 + eika ) ǫB − E . yielding the secular equation 0 = E 2 − E(ǫA + ǫB ) + (ǫA ǫB − t2 (2 + 2 cos(ka))) with the solutions 1 p E± (k) = ǫA + ǫB ± (ǫA − ǫB )2 + 4t2 (2 + 2 cos(ka)) 2 Note that in the limit that ǫA = ǫB we recover E = ǫ ± 2|t| cos(ka/2) . 84 Tight Binding Chain (Interlude and Preview) 5676589 8 2 3 . 01 02 03 3 2 1 45676589 . 5676589 8 2 3 . 01 02 03 3 2 1 45676589 . The system is therefore an insulator. there are now two electrons per unit cell. Left: Reduced Zone scheme. Expanding around the minimum gives 2t2 (ka)2 E = constant + p (ǫA − ǫB )2 + 16t2 which we set equal to ~2 k 2 /(2m∗ ) to yield p ∗ ~2 (ǫA − ǫB )2 + 16t2 m = 4t2 a2 If each atom is monovalent. (See the figure for sketches of the dispersion) In the limit of small t the bands become very flat at energies ǫA and ǫB . If ǫA = ǫB this becomes a monotonic chain. the gap closes and it becomes a metal. Right: Extended Zone scheme. For LiF we can expect a much lower energy for electrons on F than on Li (F high high electron affinity. and this fills exactly the lower band (leaving the upper band empty). but for the change in the definition of the size of the unit cell. In both pictures we have chosen ǫA = 2. 11. So .56 which matches the solution of part (a) above.56 and ǫB = 1. Fig.1 Diatomic Tight Binding Problem in 1D. Li has low ionization energy). (11. n = m±1. Similarly we imagine a hopping amplitude quantities ǫA B atomic − ǫatomic .m−1 ) Snm = Aδnm + Bδn. Again we assume an on-site en. let us no longer assume that or. A and B hav.3) Tight Binding Chain Done Right Why is this last assumption (the |n − m| > 1 case) Let us reconsider the one-dimensional tight binding reasonable? model as in Exercise 11. Calculate and sketch the dispersion of the two re- ine hopping amplitudes tAA which allows an electron on sulting bands. if we look a the eigenvectors in the lower band.1. the energy shift of the atomic orbital due make a one-dimensional chain of such atoms and let us to neighboring atoms. The generalized Schoedinger equation is X X Hnm φm = E Snm φm m m where here Hnm = ǫ0 δnm − t(δn. Thus the free electron is transferred almost completely from the higher to the lower energy atom.5. However. What happens in this limit is that the band are extremely far apart – thus a very good insulator. Use the effective Schroedinger equation from Ex- bitals are orthonormal.m+1 + B ∗ δn. let us assume hn|mi = A ercise 6. for |n − m| > 1. we obtain E(A + Beik + B ∗ e−ik ) = ǫ0 − 2t cos(ka) or ǫ0 − 2t cos(ka) E= A + Beik + B ∗ e−ik (11. derive a condition on the boring atom. We imag. Now suppose we sume that V0 .4) Two Orbitals per Atom tBB that allows an electron on orbital B of a given atom (a) Consider an atom with two orbitals.m−1 (note we did not promise that B is real!). (We as- ing eigenenergies ǫA B atomic and ǫatomic . Further. assume that these orbitals remain orthogonal. Plugging in our wave ansatz φm = eimka into the Schroedinger equation. to hop to orbital B on the neighboring atom. Go back suppose hn|H|mi = ǫ for n = m and hn|H|mi = −t for and look at that exercise. One can assume that orbitals far apart have no overlap since the tails of wavefunctions decay exponentially. orbital A of a given atom to hop to orbital A on the neigh. Instead. If the atom is diatomic. we will find that they are almost completely on the lower energy atoms. is very similar to what we did in Exercise 6.5 to derive the dispersion relation for this one- for n = m and hn|mi = B for n = m + 1 with hn|mi = 0 dimensional tight binding chain. Treating the possible non-orthogonality of orbitals here ergy ǫ and a hopping matrix element −t. as well as tAA and tBB which . is zero). now.m+1 + δn. 85 we can set ǫA ≪ ǫB . In other words. the dispersion as in Eqs. then it will be a metal if the bands overlap and an 11. Here.2. 86 Tight Binding Chain (Interlude and Preview) determines whether the system is a metal or an insulator. tAA = 1. Note if we are tricky we can summarize the two above conditions as one condition which. ǫA = 0.2) -2 A sketch is given in Fig.2). we have a matrix schroedinger equation EφA n = ǫA φA A ∗ A B ∗ B n − tAA φn+1 − tAA φn−1 − tAB φn+1 − tAB φn−1 EφB n = ǫB φB B ∗ B A ∗ A n − tBB φn+1 − tBB φn−1 − tAB φn+1 − tAB φn−1 Using the usual ans¨ atz φA n = Aeikna φB n = Beikna we obtain EA = ǫA − tAA eika − t∗AA e−ika A + −tAB eika − t∗AB e−ika B EB = ǫB − tBB eika − t∗BB e−ika B + −tAB eika − t∗AB e−ika A which is an eigenvalue equation for E.1 and atom is divalent. Defining TAA = tAA eika + t∗AA e−ika = 2Re[tAA eika ] (11.4) TAB = tAB eika + t∗AB e−ika = 2Re[tAB eika ] (11. If the Fig. tBB = 2. if satisfied. 11.1 has a band which extends from ǫA − 2|tAA | to ǫA + 2|tA A| (and similar for the other band – this holds even if t’s are complex). insulator (or semiconductor) if the bands do not overlap. tells us we have an insulator |ǫA − ǫB | > 2 (|tAA | − |tBB |) (b) When there is hopping tAB between an A orbital on one site and a B orbital on a neighboring site. A band with ǫB = 4. So we 4 get two independent dispersions (we assume t’s real here) 3 2 E = ǫA − 2tAA cos(ka) (11.5) . A to hop to orbital B on the neighboring atom (and vice (b)* Now suppose that there is in addition a hopping versa). What is the dispersion relation now? term tAB which allows an electron on one atom in orbital 5 (a) The two orbitals are completely decoupled from each other. 11.1) 1 and -3 -2 -1 1 2 3 -1 E = ǫB − 2tBB cos(ka) (11. It should say ”’if the atom is divalent” not diatomic (doh!). In order for the bands to not overlap we must have either ǫA + 2|tAA | < ǫB − 2|tBB | or ǫB + 2|tBB | < ǫA − 2|tAA | If neither of these is satisfied the bands must overlap and we have a metal. 11. Horizontal axis is ka.3) TBB = tBB e + t∗BB e−ika ika ika = 2Re[tBB e ] (11. 11.2 Dispersion in the two orib- tal tight binding model (Eqs. such that it has an atomic orbital energy which differs by (b) Consider instead a continuum one-dimensional ∆ from all the other atomic orbital energies. and find the nian given in Eq.m = ǫ0 δn. Now consider the situation where eigenstate energy.m ) + ∆δn. show that there Consider the one-dimensional tight binding Hamilto. Examining the schroedinger equation at position zero. 2m∗ x (a) Using an ansatz Similarly show that there is a localized eigenstate for any negative ∆ and find its energy.m δn.m + δn−1.2. Compare your result to φn = Ae−qa|n| that of part (a).7) these give solutions with q real for (ǫ0 − E)/2t > 1. 87 We can write the secular equation as 0 = E 2 + E (TAA + TBB − ǫA − ǫB ) 2 + (ǫA − TAA ) (ǫB − TBB ) − TAB (11. we can patch together two of these evanescent waves at the impurity. we have Eφ0 = (ǫ0 + ∆)φ0 − t(φ1 + φ−1 ) which gives us E − ǫ0 − ∆ = 2te−qa Then plugging in the value of E in terms of q from Eq.7. (a) Consider first without the impurity. In this case Hamiltonian with a delta-function potential the Hamiltonian becomes ~2 2 H=− ∂ + (a∆)δ(x). for energies below the bottom of the band. This exercise is very similar to Exercise one of the atoms in the chain (atom n = 0) is an impurity 9. These are analogous to the evanescent waves discussed in Exercises 9. or in other words. Now. Hn.6) which we solve as a quadratic equation to give q 1 2 2 E= ǫA + ǫB − TAA − TBB ± (ǫA − ǫB − TAA + TBB ) − 4TAB 2 (11. 11. and a the lattice constant. is a localized eigenstate for any negative ∆. 11. we obtain −∆ = 2 sinh(qa) recalling that ∆ was negative we have a solution for any value of ∆ < 0 with qa = sinh−1 (|∆|/(2t)) .4.5) Electronic Impurity State* with q real. Propose exponentially decay- ing or growing solutions φn = Ae±qa|n| plugging these in we obtain E = ǫ0 − 2t cosh(qa) (11.0 .6.4 and 10.m − t(δn+1. (11. Use the schroedinger equation at position n = −1 Eφ−1 = ǫ0 φ−1 − t(φ0 + φ−2 ) and at position zero Eφ0 = (ǫ0 + ∆)φ0 − t(φ1 + φ−1 ) .5). (b) For the continuum problem we have similarly a decaying waves of the form e−qx with energies −~2 q 2 E= 2m∗ Then we need to patch together solutions at zero. T e−ikna n≥0 vious exercise representing a single impurity in a chain. we use the value of the effective mass m∗ = ~2 /(2ta2 ) to obtain q = |∆|/(2t) which matches our above result for small ∆. we have ∂ψ− − ∂ψ+ = 2q = (a|∆|)(2m∗ )/~2 or q = (a|∆|m∗ )/~2 with energy E = −a2 m∗ ∆2 /(2~2 ) To compare to the above tight binding problem.6) Reflection from an Impurity* wavefunction Consider the tight binding Hamiltonian from the pre. 11. φn = e−ikna + Re+ikna n<0 Here the intent is to see how this impurity scatters a plane wave incoming from the left with unit amplitude to determine the transmission T and reflection R as a (this is somewhat similar to Exercise 9. Correspondingly the energy is E = −∆2 /(4t) which also matches the above result for small ∆ so long as we choose the bottom of the band to be called zero energy. Use an ansatz function of k.7 we obtain the bound state energy p E = ǫ0 − 2t [∆/(2t)]2 + 1 and note that for ∆ = 0 this gives us the energy at the bottom of the band.88 Tight Binding Chain (Interlude and Preview) Plugging back into Eq. J q = K(TL − TR ) ing to the right is q q Z ∞ where J q = jL − jR and TL − TR is assumed to be small.7) Transport in One Dimension* conductance is routinely measured in disorder free one- (a) Consider the one-dimensional tight binding chain dimensional electronic systems. whereas the particles moving towards the right will be (b) Now suppose that the chemical potentials at both filled up to chemical potential µL . purity is placed in this chain between the two reservoirs voir at chemical potential µR and the left end of the to create some backscattering. confirm the Wiedemann– for left moving current. The particles moving to. with v(k) = (1/~)dǫ(k)/dk the group velocity and nF the hT −∞ ∂E Fermi occupation factor. µR (ii) Define the thermal conductance K to be (i) Argue that the total current of all the particles mov. With R = T − 1. reservoirs are the same. defined as K π 2 kB2 = Jtotal = GV TG 3e 2 where Jtotal = jL − jR and eV = µL − µR . (iii) In the context of Exercise 11. Landauer formula and is a pillar of nano-scale electron- wards the left will be filled up to chemical potential µR . as shown in the bot. dk Derive that the thermal conductance can be rewritten as jR = v(k)nF (β(E(k) − µL )) Z 0 π −2 ∞ ∂ K= dE(E − µ)2 nF (β(E − µ)). the following figure (i) Argue that the heat current j q of all the particles moving to the right is Z ∞ µL q dk jR = v(k) (E(k) − µ) nF (βL (E(k) − µ)) µL µR 0 π and a similar equation holds for left-moving heat current. and also diagrammed schematically in TR respectively. and show (Note that this is a relationship between conductance and G = 2e2 /h with h Planck’s constant. we have complete transmission T = 1 and R = 0. and an analogous equation holds Evaluating this expression.4.2. we have |R|2 + |T |2 = 1 which indicates current conservation. (11. The former equation simply tells us that we must have 1 + R = T (To see this recall that the relationship between E and k is fixed by the plane wave away from positon zero. So this first equation tells us that position φ0 must just be the usual continuation of this wave-front with no changes. 11. ics. Franz ratio for clean one-dimensional systems (ii) Calculate the conductance G of this wire. discussed in this chapter at (or near) zero temperature.) The second equation then gives us E T = (ǫ0 + ∆)T − t(T eika + eika + Re−ika ) Using 1 + R = T as well as using E = ǫ0 + 2t cos(ka) this becomes 1 T = ∆ 2it sin(ka) +1 Note that when ∆ = 0. imagine that an im- Suppose the right end of this chain is attached to a reser. 89 and we plug in the ansatz form. Argue that the conduc- chain is attached to a reservoir at chemical potential µL tance is reduced to G = 2e2 |T |2 /h. This is known as the and let us assume µL > µR . This “quantum” of thermal conductance rather than between conductivity . but the temperatures are TL and tom of Fig. The . one integrates over all possible k. the occupancy of that k times the group velocity. identity using the techniques analogous to those men- gral you will want to use tioned in footnote 20 of Chapter 2. Once should use k near kF to calculate T .) In evaluating the above inte. it is delocalized over the entire system. To see why this is missing we have to think carefully about the definition of current – which counts the number of particles crossing a given point in some unit of time. ω = Ek /~ To find the total right going current. but the 2 is cancelled by a factor of 2 out front for two spins. chapter. As we raise the temperature. (ii) We can write the total current as Z ∞ Z −∞ e dE(k) dE(k) J = − nF (β(Ek − µR ))dk − nF (β(Ek − µL ))dk π 0 dk dk Z ∞ Z ∞ 0 e = − dEnF (β(E − µR )) − dEnF (β(E − µL )) ~π 0 0 e = − [µL − µR ] ~π where in the last step we have assumed that µ ≫ kB T so that the occupancy at E = 0 is unity — in which case the integral of the Fermi function simply gives the chemical potential (one can see this by starting at T = 0 where the Fermi function is just a step function. so even though the fermi function gets smaller below µ and larger above µ the integral is unchanged). There should be an additional factor of −e out front to turn this into an electrical current. If one has a particle in a k eigenstate. the change in the fermi function is symmetric around the chemical potential. Thus the ”probability“ that with velocity v it crosses that given point in a unit of time is given by v/L. The integral over k also usually comes with a factor of volume (or length in this case) out front. as well as the fact that Z ∞ the Riemann zeta function takes the value ζ(2) = π 2 /6 ∂ 1 π2 dx x2 x +1 =− which you can prove analogous to the appendix of that −∞ ∂x e 3 . If you are very adventurous. you can prove this nasty (a)(i) Genereally the velocity of a wavepacket is the group velocity v = dω/dk.90 Tight Binding Chain (Interlude and Preview) and thermal conductivity. T is a function of wavevector and hence energy. The resulting current is then reduced by the probability that each particle is transmitted. to obtain the current. (iii) If the amplitude of transmission through the impurity is T then the probability of transmission is |T |2 . Here. Thus we obtain G = 2e2 |T |2 /h Note that as we discovered in exercise 11. Plugging in the expression for voltage e2 J =− V π~ or G = 2e2 /h. The integral over k usually has 2π downstairs.6. For completeness.8) where we have made the approximation that βR − βL is small to replace the finite difference with a derivative with respect to β. we evaluate the integral. The difference from this T = 0 energy is the excess heat added. Here energy is measured with respect to the chemical potential – this is an appropriate definition since even at T = 0 the addition of an electron carries the chemical potential worth of energy. Note that the formula also extends the lower limit of integration to −∞. Here each electron is moving with velocity v. First. Z ∞ −1 q dnF (β(E − µ)) J = (βR − βL ) dE(E − µ)2 ~πβ 0 dE then using TL − TR βR − βL ≈ kB T 2 gives the desired result. 91 reason for this is that all of the net current is coming from near the Fermi surface. note that the inte- . This is allowed since dnF /dE is strongly peaked around µ. Finally scaling out some factors of β we obtain Z ∞ −2T d 1 2π 2 kB 2 T K= dxx2 = h −∞ dx ex + 1 3h Thus dividing by the expresion for G we obtain K π 2 kB 2 T = 2 GT 3he which is the Weideman-Franz law. (b) (i) The argument here is quite similar. but is carrying energy E − µ. (ii) We take the difference of right moving and left moving energy currents (the sign gives current moving from left to right) Z ∞ Z −∞ −1 dE(k) dE(k) Jq = (Ek − µ)nF (βR (Ek − µ))dk − (Ek − µ)nF (βL (Ek − µL ))dk π dk dk Z0 ∞ Z ∞ 0 −1 = dE(E − µ)nF (βR (E − µ)) − dE(E − µ)nF (βL (E − µ)) ~π 0 0 Z ∞ −1 dnF (β(E − µ)) = (βR − βL ) dE(E − µ) ~π 0 dβ (11. This then becomes. Well below the fermi surface the currents of left and right movers exactly cancel. Suppose the lower band is filled and the Consider a chain made up of all the same type of atom. chain with the valence as in the first part of this problem. First. It can be writen as ∞ X x= an sin(nx) n=1 with Z π 1 an = x sin(nx) = −2(−1)n /n π −π The calculate Z π dx x2 = 2π 3 /3 −π but also caculate the same quantity in terms of its fourier transforms (using percival’s theorem) Z π Z π ∞ X X∞ 4 dx x2 = dx = |an |2 sin2 (nx) = 2 π −π −π n=1 n=1 n setting these two expressions equal to each other gives X∞ 1 π2 = n=1 n2 6 (11. Calculate the total ground-state energy of nates as long-short-long-short as follows the filled lower band. Show that making a distortion whereas the longer bonds (marked with −) have hopping whereby alternating springs are shorter/longer by δx matrix element tlong = t(1 − ǫ). consider the function x in the range [−π. π] as a fourier series. the shorter bonds (marked with atoms connected together with identical springs with =) will have hopping matrix element tshort = t(1 + ǫ) spring constant κ.92 Tight Binding Chain (Interlude and Preview) grand is symmetric around zero so we can write Z ∞ Z ∞ Z ∞ 2 d 1 2 d 1 e−x I = dxx = 2 dxx = −4 dxx −∞ dx ex + 1 0 dx ex + 1 0 1 + e−x Z ∞ X ∞ ∞ X (−1)n = 4 dxx (−ex )n = 4 0 n2 " ∞ n=1 ∞ n=1 # X 1 X 1 2 = −4 2 − 2 2 = 4(1 − )ζ(2) = 2ζ(2) n=1 n n=1. This is known as a Peierls distortion. (The onsite en. Calculate the tight. ergy ǫ is the same on every atom). Conclude that for a binding energy spectrum of this chain. costs energy proportional to (δx)2 . in this case?). and show it decreases linearly with increasing ǫ.8) Peierls Distortion* linear order in ǫ.evens n 4 The method of calculating ζ(2) proceeds similar to the appendix of Chapter 1. −A = A − A = A − A = A− Now consider a chain of equally spaced identical A In a tight binding model. . upper band is empty (what is the valence of each atom but in such a way that the spacing between atoms alter. Expand your result to a distortion of this sort will occur spontaneously. However. Set the onsite energy epsilon0 to zero for simplicity and assume t is real (without loss of generality). 93 OK. all we need to do is to make an estimate of how this changes to lowest order in ǫ. but this is very difficult analytically (we could do it numerically though!). Ideally. we need Z dk Etot = 2L E(k) 2π which is some horrid ellipic integral. We then have the tight-binding schroedinger equation EφA n = −t(1 + ǫ)φB B n − t(1 − ǫ)φn−1 EφB n = −t(1 + ǫ)φA A n − t(1 − ǫ)φn+1 Using the usual ansatz this gives us EA = −t(1 + ǫ)B − t(1 − ǫ)Be−ika EB = −t(1 + ǫ)A − t(1 − ǫ)Aeika which is an eigenvalue problem with solutions p E(k) = ± |2t[cos(ka/2) + iǫ sin(ka/2)]| = ±|2t| ǫ2 + (1 − ǫ2 ) cos2 (ka/2) To find the total energy of a filled lower band. since the cosine term is small the effect of epsilon is most pronounced where the energy reduction in the lower band is −|2tǫ|. we just evaluate this horrid integral and we are done. For the region further from the zone boundary. Call the left side the A site and the right site the B site. we have roughly Z cutoff hp i δE ≈ |2tL| dκ ǫ2 + (κa)2 /4 − κa/2 κa=ǫ Z cutoff ǫ2 ≈ |2tL| dκ ≈ |2tL|ǫ2 log(ǫ/cutoff) κa=ǫ κ . Define κa = π − ka to be the distance to the zone boundary (focusing for now on the zone boundary at ka = π). which we are not interested in. For small κ we see that the two term in the square root become roughly equal when κa ≈ ǫ And when κa ≫ ǫ the cosine term dominates. Let us then split the integration roughly at this intermediate point κa ≈ ǫ. Near the zone boundary. For smaller κ. the energy reduction can be approximated as −|2tǫ| for each value of κ and integrating then a range of κ that is ǫ large we get an energy reduction of ∼ L|t|ǫ2 . Consider the unit cell to be a single unit like this A = A−. I messed up this problem a bit. so bear with me. First we need to evalulate the spectrum. As we move away from the zone boundary the cosine term becomes more important. Calculate the dispersion relation for this tight binding model. What does the dispersion relation look like near the bottom of the band? (The two-dimensional structure is more diffi- t2 cult to handle than the one-dimensional examples given in this chapter.m − t1 (φn+1. (ii) The energy of stretching springs is always quadratic in the stretch- ing (hookes’ law). and assuming both t1 and t2 real E = ǫ0 − 2t1 cos(kx ax ) − 2t2 cos(ky ay ) Assuming both t1 and t2 positive.m = ǫ0 φn.) Consider a rectangular lattice in two dimensions as shown in the figure.m+1 + φn. In Chapters 12 and 13 we study crys. Let the wavefunctions on the sites be φn.m = eikx nax +ky may where ax and ay are the lattice distances in the two directions. . we can expand to gets 1 E ≈ (ǫ0 − 2t1 − 2t2 ) + t1 (kx ax )2 + t2 (ky ay )2 2 which has equal-E contours which are ellipses. hence proportional to ǫ2 . Thus near the bottom of the band we have an ellipsoidal bowl. and where the hopping matrix element is hn|H|mi = t1 if sites n and m are neighbors in the horizontal direction and is = t2 if n and m are neighbors in the vertical direction. the minimum occurs at kx = ky = 0. The Schroedinger equation is then Eφn. Now imagine a tight binding model where there is one orbital at each lattice site.m accordingly. For small ǫ the electronic energy saving ǫ2 log ǫ always wins. and it might be useful to read those chapters first and then return to try this This is a lot easier than it looks! Let us label the sites (m.94 Tight Binding Chain (Interlude and Preview) here the cutoff is some arbitrary momentum much further from the zone boundary (it will not matter where we choose the cutoff!).n).m−1 ) Using an ansatz φn.9) Tight Binding in 2d* exercise again. t1 tals in two and three dimensions. Thus we see that the energy saving is proportional to ǫ2 log ǫ. so the system always distorts sponta- neously! (11.m ) − t2 (φn.m + φn−1. Near this minimum. 563 nm. . twelve nearest neighbors. one can define the fcc lattice as being vectors of the form a2 (n. m. m. 0] → d= a + b2 2 1p 2 [±1. Here the two entries that are 1 could be either ±1 which gives four possibilities. ±1. b.2815 nm whereas the Na-Na distance is a 2/2 = . If the lattice tance from a sodium atom to the nearest other sodium constant is a = 0.398 nm. l) where n. each the same distance from the and c which are all different. ±1. Further the 0 can be in one of three places. (b) In units of the three different (unequal) conventional unit cell lattice constants. 1.21. 1. What are the distances to initial point. Thus we have √ 12 possibilities. ±1] → d= a + c2 2 1p 2 [0. ±1] → d= b + c2 2 There are four nearest neighbors corresponding to the smaller two of a. 0] are of the form a2 [1. 0. the 12 points are still of the form 12 [1.1) Crystal Structure of NaCl Consider the sodium atom to the nearest chlorine? What is the dis- NaCl crystal structure shown in Fig. What is this distance if the conventional these twelve neighboring points now? How many nearest unit cell has lattice constant a? neighbors are there? (b)∗ Now stretch the side lengths of the fcc lattice (a) Given the primitive lattice vectors. 0. The distance from [0. 0] to any of these points is a 2/2 (analogous to previous problem). l are either all even or two odd and one even. 0] and permu- tations. The distances to these 12 points are then 1p 2 [±1. c. 0. (12. 12. 0]. Among this sets of possible lattice points. the closest ones to [0. what is the distance from a atom? Super easy:√ The Na-Cl distance is a/2 = . b.2) Neighbors in the Face-Centered Lattice. such that you obtain a face-centered orthorhombic lattice (a) Show that each lattice point in an fcc lattice has where the conventional unit cell has sides of length a.Crystal Structure 12 (12. e.541 nm. (12.0] and S at position [ 14 . 12.383 nm. 1 2 a Zn= S= Fig.8. 14 .96 Crystal Structure 1 (12. (b) The volume of a conventional unit cell is Vcell = a3 . Thus the packing fraction is Vsphere /Vcell = π/6 ≈ . 4 4 neighbor Zn–Zn. (a) The lattice is FCC (otherwise known as ”cubic F”) (b) The basis can be described as Zn at position [0. Choose the radius of the spheres to be such that neighbor. and S–S distances.. nearest neighbor 2 2 2 √ Zn-S is a 1/4 + 1/4 + 1/4 = 0.22 shows a plan view of a structure of cubic ZnS (zincblende) looking down the z axis. Each conven- √ tional unit cell contains two lattice points which are a distance a 3/2 . but less standard notation [ 14 . −1 4 ] all in units of the lattice constant. (a) Calculate the packing fraction for a simple cubic ing spheres just touch (see for example.0. volume).3) Crystal Structure 2 The diagram of Fig. The radius of this sphere is a/2 so its volume is Vsphere = 4π/3(a/2)3 . (a) The volume of a conventional unit cell is Vcell = a3 . packing fraction is the fraction of the volume of all of (b) Calculate the packing fraction for a bcc lattice. The numbers attached to some atoms represent the 1 3 heights of the atoms above the z = 0 plane expressed as a 4 4 fraction of the cube edge a. 12.52. 34 ] (or equivalently. space which is enclosed by the union of all the spheres (c) Calculate the packing fraction for an fcc lattice. √ (c) (just a bit of geometry:)pnearest neighbor Zn-Zn is a/ 2 = 0. a 1 1 2 2 (a) What is the Bravais lattice type? (b) Describe the basis. 3 1 (c) Given that a = 0. The lattice.383 nm. Fig. nearest neighbor S-S is a/ 2 = 0.23 Plan view of conventional unit cell of zincblende. 14 . 12.234 nm. the ratio of the volume of the spheres to the total Consider a lattice with a sphere at each lattice point. Unlabeled atoms are at z = 0 and z = a. Each cell cor- responds to a single sphere.4) Packing Fractions (i. Zn–S. calculate the nearest. Thus we have a packing fraction 4Vsphere /Vcell = π/(3 2) ≈ . [1/2. 3/4. Then there is one in the center. 1/4]. [1/4. Thus we have a packing fraction 2Vsphere /Vcell = π 3/8 ≈ . 0. 0]. Thus√ the 3radius of the each sphere is a 3/4 so has volume Vsphere√= 4π(a 3/4) /3. 1/2. Fig. 12. 1/2.74 (12. 1/2]. 97 √ apart from each other.5) Fluorine Beta Phase three-dimensional form along with a plan view. [3/4. 1/2] .68 (c) The volume of a conventional unit cell is Vcell = a3 . Unlabeled atoms are at height 0 and 1 in units of the lattice constant. 12. 0.24 shows What is the lattice and the basis for this crystal? the cubic conventional unit cell for beta phase fluorine in 1 1 2 2 Fig. 0]. Each conven- √ tional unit cell contains four lattice points which are a distance√a 2/2 apart from each other. Each atom on the face counts 1/2. Fluorine can crystalize into a so-called beta-phase at How many atoms are in this conventional unit cell? temperatures between 45 and 55 Kelvin. 0. Lines are drawn for clarity Top: Three-dimensional view. [1/2. 1 in the corner (8 times 1/8). [0. Bottom: Plan view. 3/4]. All atoms in the picture are fluorine. [1/2. 1/2. 1/4. Thus √ the radius of the each sphere is a 2/4 so has volume Vsphere =√4π(a 2/4)3 /3. [0. 1/2].23 A conventional unit cell for fluorine beta phase. 1 3 1 1 3 4 and 4 2 4 and 4 1 1 2 2 There are 8 atoms in the conventional unit cell. The basis is [0. 0]. The lattice is simple cubic. . we must therefore have ei2πRi ·Sj = 1 for all i and j which is easy to check is true. For the orthorhombic face centered system one would similarly obtain a reciprocal lattice which is an orthorhombic body centered systems with basis vectors 4π/ai . 1/2) in units of the reciprocal cubic lattice constant 4π/a. Let us start with the direct lattice of the BCC .1) Reciprocal Lattice stant for the conventional unit cell of the reciprocal bcc Show that the reciprocal lattice of a fcc (face-centered lattice? cubic) lattice is a bcc (body-centered cubic) lattice. given the bcc basis there is no fifth point we can add to the fcc basis.Reciprocal Lattice. we can show that given the fcc basis there is no third point that can be added to the bcc basis which would still have the same property. 0. we propose a basis for the reciprocal lattice which is S1 = (0. 1/2. and R2 = [1/2. show that the reciprocal lattice of a bcc with conventional lattice constants a1 . If an fcc lattice has conventional the reciprocal lattice now? unit cell with lattice constant a. What it lattice is an fcc lattice. 1/2] and R4 = [0. Further. For these to be reciprocal. Brillouin Zone. 1/2] in units of the lattice constant a. 0. Waves in Crystals 13 (13. In reciprocal space. a3 . 1/2. 0. Consider now an orthorhombic face-centered lattice respondingly. 1/2. what is the lattice con- Brief Solution Say we have an fcc lattice in real space. Cor. This can be written as cubic with a basis R1 = [0. Similarly. a2 . 0] and R3 = [1/2. 0) and S2 = (1/2. 0]. Detailed Solution The more straightforward way to find the reciprocal lattice of a direct lattice is by construction. 1/2]a we then construct the primitive lattice vectors of the reciprocal lattice via 2πbf aj × ak bi = a1 · a2 × a3 with i. 0. 1/2. 1/2.100 Reciprocal Lattice. l/2 integer or two half-odd integer and one integer. Waves in Crystals lattice which has primitive lattice vectors a1 = [1. l are even or two are odd and one is even. j. We assemble them together with integer coefficients to make b′ . k/2. 0]a a2 = [0. the selection rule is that h + k + l is even. . This means that either all h. Brillouin Zone. 0]a a3 = [1/2. This is precisely the condition we would use to define an FCC lattice. NOTE about making this transformation from b into b′ . Thus if we write 4π h k l G(hkl) = x ˆ + yˆ + zˆ a 2 2 2 we must have either all h/2. 0. −1/2) a 4π b3 = (0. 1. 0) a which shows us tht the fcc lattice has lattice constant 4π/a. This gives 4π b1 = (1/2. 0. k cyclic. For example. However we must also be able to reassemble the b′ with integer coefficients to get back the b in order for this to be an allowed transformation from one set of primitive lattice vectors to another. 1/2. if the direct lattice is BCC. We are guaranteed that that b are primitive lattice vectors. Slick Solution Once one has read chapter 14 on scattering and selection rules. 1/2) a 4π b′3 = b1 + b2 + b3 = (1/2. The prefactor 4π/a then gives us the lattice constant. 1) a we can then reassemble these vectors to give the standard primitive lattice vectors for FCC 4π b′1 = b1 + b3 = (1/2. k. 0. 1/2) a 4π b′2 = b2 + b3 = (0. −1/2) a 4π b2 = (0. 1/2. we simply note that the selection rules on miller indices tell us the form of the reciprocal lattice. the selection rule is h. k. l all even or all odd. if the direct space lattice is FCC. For the FCC to be a family of lattice planes. The moral of this story is that scattering occurs when (h. Copy this ily of lattice planes. k/2. 101 Similarly in reverse. So the question is not correctly written as it stands. This is precisely the condition for defining a BCC lattice. k. This means (h/2. l/2) all integer or all half-odd integer. l) repre- sents a reciprocal lattice vector. Is the same true in general for an orthorhombic What is the generalization of this formula for an crystal? orthorhombic crystal? .3) Directions and Spacings of Crystal Show that the spacing d of the (hkl) set of planes Planes in a cubic crystal with lattice parameter a is ‡Explain briefly what is meant by the terms “crys- tal planes” and “Miller indices”. The selection rules tell you when this is so. it must have half the plane spacing so that it is called (420) and it cuts though every lattice point. [210] y x (210) planes NOTE: As drawn here this is a family of planes (it is a family of lattice planes for the corresponding simple cubic). (13. (13.3.2) Lattice Planes figure and indicate the [210] direction and the (210) fam- Consider the crystal shown in Exercise 12. a d= √ Show that the general direction [hkl] in a cubic h2 + k2 + l2 crystal is normal to the planes with Miller indices (hkl). and we can resolve it parallel to n by taking the dot product with the unit normal vector nˆ . Note: In cases where the axes are not orthogonal. These intersection points (kx . By convention note that negative numbers are denoted as an integer with a bar on top. take any vector that connects two ad- jacent planes and find the component in the direction of the normal to the plane. the vector a/h connects two adjacent planes. 1. −1). assumed to be orthog- onal.102 Reciprocal Lattice. The unit normal in this direction is 1 h k l nˆ= p a + 2b + 2c (h/a)2 + (k/b)2 + (l/c)2 a2 b c To find the inter-planar spacing. bi = 2π/ai . ¯1) to denote (1. Much more succinctly one could also note that for orthogonal axes. b/k and c/l.1) (h/a) + (k/b)2 + (l/c)2 2 with the case a = b = c appropriate for cubic crystals. For example. ay and az are the length of the three basis vectors. c respectively. Thus we obtain 1 dhkl = nˆ · a/h = p (13. 0) and (0. Let the basis vectors of the lattice be a. In terms of lattice planes. 0) and (0. To construct a vector normal to this plane. and the family of lattice planes (hkl) is normal to the cor- responding reciprocal lattice vector. Let the lengths of these three lattice vectors be a. kz ) have the property that they are in the ratios 1 1 1 h:k:l= : : kx ky kz The actual values of h : k : l are then the smallest integer values with these ratios. the Miller indices (hkl) specifies the reciprocal lattice vector (2π)(hˆ x/ax +kˆ y/ay +lˆz/az where ax . 0. Miller indices are a set of three integers which specify a set of parallel planes (or equivalently specify a vector in reciprocal space). b. Waves in Crystals A crystal plane is a plane which intersects at least three non-colinear (and therefore an infinite number of) points of the lattice. 0. ky . one can determine the lattice vectors by picking any given plane from the family. and finding its intersection points with the three coordinate axes. The plane (hkl) can be defined as the plane connecting the points a/h. 1. take any two (noncolinear) vectors in this plane and take their cross product n = (a/h − b/k) × (a/h − c/l) abc h k l = a + b + c hkl a2 b2 c2 This is only parallel to the vector [hkl] in the case of the cubic crystal a = b = c. b and c. If the axes of a lattice are mutually orthogonal . ex (1. this formula does not work. Brillouin Zone. . 6) with h. It is trivial to then see that eiK·R = 1 (13. 103 (13. 13. l integers. a2 and a3 then primitive lattice vectors for show that the length of the reciprocal lattice vector the reciprocal lattice can be taken as G = hb1 + kb2 + lb3 is equal to 2π/d. (b) If we take a2 × a3 b1 = 2π a1 · (a2 × a3 ) a3 × a1 b2 = 2π (13. However. b2 . k.4) ‡Reciprocal Lattice What is the proper formula in 2d? (a) Define the term Reciprocal Lattice. (c) Define tetragonal and orthorhombic lattices. where d is the a2 × a3 spacing of the (hkl) planes (see question 13) b1 = 2π (13. For an (b) Show that if a lattice in 3d has primitive lattice orthorhombic lattice. If these are our basis vectors for the reciprocal lattice.5) a1 · (a2 × a3 ) a1 × a2 b3 = 2π a1 · (a2 × a3 ) Then we can show the key formula bi · aj = 2πδij . What it proves is that vector K of the form of Eq. and b3 can indeed be taken as the basis. the reciprocal lattice is defined by the set of points in k space such that eik·R = 1 for all points R in the lattice . Note that this set of points forms a lattice of values of k. vectors a1 . . we then have a general reciprocal lattice point given by Ghkl = hb1 + kb2 + lb3 (13. Hence.7) for any lattice vector R and any reciprocal lattice vector K.4) a1 · (a2 × a3 ) (a) Given a lattice of points (in 3d it would look like R[uvw] = ua1 + va2 + wa3 with [uvw] integers).6 are in the reciprocal lattice.3) a1 · (a2 × a3 ) a1 × a2 b3 = 2π (13. show that |bj | = 2π/|aj |.2) a1 · (a2 × a3 ) a3 × a1 b2 = 2π (13. We need to show that there are no other vectors in the reciprocal lattice as well so that b1 . this does not quite prove the desired statement. 104 Reciprocal Lattice, Brillouin Zone, Waves in Crystals To show this, consider an arbitrary vector K of the form of Eq. 13.6 but do not require that h, k, l are integers. Given an arbitrary Ruvw of the real space lattice, in order that Eq. 13.7 is satisfied, we must have uk + vh + wl = integer for this to be true for arbitrary u, v, w which are integers, we can conclude that k, h, l are integers. We can derive the 2d analogous formula by setting a3 = zˆ the unit vector normal to the plane, more conveniently written as a2 × ˆz b1 = 2π ˆz · (a1 × a2 ) ˆz × a1 b2 = 2π ˆz · (a1 × a2 ) (c) A tetragonal lattice is a lattice (in 3d) where all three basis vectors are normal to each other, and two of them are the same length but the third is a different length. An orthorhombic lattice is a lattice (in 3d) where all three basis vectors are normal to each other and all three have different lengths. (Note: crystals may have orthorhombic or tetragonal symmetry even if lattice constants in the three directions are all equal. The symmetry of a crystal has to do with whether it looks the same under various types of rotations. An orthorhombic crystal does not look the same under the three 90 degree rotations – this may be the case even if all three lattice constants are the same). For an orthorhombic crystal, without loss of generality, let us write a1 = a1 x ˆ and a2 = a2 y ˆ and a3 = a3 ˆz. We then just use the above formula Eq. 13.5 to obtain a2 a3 b1 = 2π yˆ × ˆz = 2πˆ x/a1 a1 a2 a3 a2 a3 b2 = 2π ˆz × x ˆ = 2πˆy/a2 a1 a2 a3 a2 a3 b3 = 2π xˆ×y ˆ = 2πˆz/a3 a1 a2 a3 So |bi | = 2π/|ai |. Thus the length of G = hb1 + kb2 + lb3 is given by p p |Ghkl | = h2 |b1 |2 + k 2 |b2 |2 + l2 |b3 |2 = (2πh/a1 )2 + (2πk/a2 )2 + (2πl/a3 )2 = 2π/dhk as given in Eq. 13.1. (13.5) More Reciprocal Lattice Label the reciprocal lattice points for indices in the A two-dimensional rectangular crystal has a unit cell range 0 ≤ h ≤ 3 and 0 ≤ k ≤ 3. with sides a1 = 0.468 nm and a2 = 0.342 nm. (b) Draw the first and second Brillouin zones using the (a) Draw to scale a diagram of the reciprocal lattice. Wigner–Seitz construction. The reciprocal lattice vectors are b1 = 2πˆ x/|a1 |, and b2 = 2πˆ y/|a2 |. Their magnitudes are |b1 | = 13.4nm−1 |b2 | = 18.4nm−1 105 (a) A diagram of the reciprocal lattice is given in the figure. 12 32 42 012 032 042 42 32 12 042 032 012 Fig. 13.1 Figures for problem 13. First Two Brillouin Zones. Red is the 1st Zone. Blue is the 2nd zone. To find the Brillouin zones, one first constructs perpendicular bisectors between the origin and any given lattice point (shown as dotted lines on the plot). Then starting at the origin, the region one can get to without crossing a dotted line is the first zone. Crossing only one dotted line gets one to the second zone. etc. (13.6) Brillouin Zones (b) Consider a triangular lattice in two dimensions (a) Consider a cubic lattice with lattice constant a. (primitive lattice vectors given by Eqs. 12.3). Find the Describe the first Brillouin zone. Given an arbitrary first Brillouin zone. Given an arbitrary wavevector k (in wavevector k, write an expression for an equivalent two dimensions), write an expression for an equivalent wavevector within the first Brillouin zone (there are sev- wavevector within the first Brillouin zone (again there eral possible expressions you can write). are several possible expressions you can write). (a) The reciprocal lattice of a cubic lattice with lattice constant a is a cubic lattice with lattice constant 2π/a. The first Brillouin zone is a cube centered around the origin in reciprocal space (0,0,0), with side length 2π/a. It spans kx , ky , kz ∈ [−π/a, π/a]. Given an arbitrary k = (kx , ky , kz ) we can write the equivalent wavevector as (kx′ , ky′ , kz )′ where kj′ = kj − [[(kj + π/a)/(2π/a)]] ∗ 2π/a (13.8) with j = x, y, z and [[ ]] is the floor function – meaning the greatest integer less than its argument. The point of Eq. 13.8 is that it takes k in units of the reciprocal lattice constant (2π/a) and returns a number be- tween -1/2 and +1/2 of the reciprocal lattice constant by appropriately 106 Reciprocal Lattice, Brillouin Zone, Waves in Crystals adding or subtracting integer units of 2π/a. (b) Start with the primitive lattice vectors a1 = aˆ x √ a2 = (a/2)ˆ x + (a 3/2)ˆ y We need to find vectors b1 , b2 such that ai · bj = 2πδi,j . One way to do this is to use the usual 3d formula a2 × a3 b1 = 2π a1 · (a2 × a3 ) a3 × a1 b2 = 2π a1 · (a2 × a3 ) a1 × a2 b3 = 2π a1 · (a2 × a3 ) √ z. The denominator is a2 3/2. Thus we have then assigning a3 = ˆ √ √ √ b1 = 2π[(a 3/2)ˆ y]/(a2 3/2) = (2π/a)[ˆ x − (a/2)ˆ x − ( 3/3)ˆ y] 2 √ √ b2 = 2πˆ ya/(a 3/2) = (2π/a)ˆ y(2 3/3) These form the primitive lattice vectors of a triangular lattice in k-space. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice — which is a hexagon centered around zero wavevector (see fig 12.6 of the book). Two write a formula that translates any k into the first Brillouin zone, we first write any k point in terms of the primitive lattice vectors k = α1 b1 + α2 b2 We are going to need to solve for αi in terms of k. To do this we write k · b1 = α1 |b1 |2 + α2 b1 · b2 k · b2 = α1 b1 · b2 + α2 |b2 |2 and we will want to solve for the αs. In this particular case, things are quite easy since |b1 |2 = |b2 |2 = −2b1 · b2 = (2π/a)2 (4/3). So this becomes βk · b1 = α1 − α2 /2 βk · b2 = −α1 /2 + α2 where we have defined 3 a2 β= 4 (2π)2 Solving this system of equations we obtain α1 = (β/3)(2k · b1 + k · b2 ) α2 = (β/3)(k · b1 + 2k · b2 ) 1/2]. It is simplest to assume periodic boundary conditions.) We are given a cubic lattice with lattice constant a and an overall size L on a size. Once we have α′ ’s we construct k′ = α′1 b1 + α′2 b2 which is now within the first Brillouin zone. Once hopping is turned back on. One can also consider the same problem with hard wall boundary conditions. although it is worth think- translation vectors (primitive lattice vectors) have length ing about whether this still holds for hard-wall boundary a. ky . The first Brillouin zone extends from −π/a ≤ kx . If one thinks in a tight-binding or atomic orbital picture. (One may assume tive cubic lattice whose mutually orthogonal fundamental periodic boundary conditions. . so the total number of states remains the same as using periodic boundary conditions. This agreement is not coincidental. but the total number of states stays constant. k = π/L in each direction. To do this we write α′j = αj − [[αj + 1/2]] again with [[ ]] being the floor function. so one is not really talking about pure plane waves. kz ≤ π/a. although it is less convenient. This has volume in k-space of (2π/a)3 . In this case the eigenstates are not of the form eikx but are rather sin(kx) where k > 0. kz ≤ π/a. Dividing this by the volume occupied by one eigenstate gives a total of (L/a)3 states in the first Brillouin zone. (13. Thus (considering all three directions) there is one eigenstate per volume (2π/L)3 in k-space.7) Number of States in the Brillouin Zone within the first Brillouin zone equals the number of prim- A specimen in the form of a cube of side L has a primi. ky . there should be exactly one orbital per atom per band when hopping is turned off. itive unit cells forming the specimen. Since eikx (x+L) = eikx x we must have kx = 2πn/L for n an integer. each atomic orbital spreads into a band that fills the Brillouin zone. 107 Now to obtain α within the first brillouin zone we need to shift both α’s by integers until they lie in the range [−1/2. In this case the analogue of the Brillouin zone goes from 0 ≤ kx . The number of atoms in this sample are N = (L/a)3 . but the density of k states is doubled. Show that the number of different allowed k-states conditions as well. π/a2 ]. ing a zone boundary.8 and 11. Show (a) In Exercises 9. you should do so now). Show that the dispersion curve sites. 9. However. π/a1 ] and ky ∈ [−π/a2 . Note that the dispersion must be symmetric as k approaches the zone boundary. Brillouin Zone.8.8 we have a triangular lattice.6. describe the Brillouin zone. Thus the dispersion is always flat approaching the zone boundary.8 that the derived freuquency spectrum is q 2 2κ 2 2 2 ω = S1 + S2 + S3 ± S1 + S2 + S3 − S1 S2 − S1 S3 − S2 S3 m where k · ai Si = sin2 ( ) 2 where ai are three independent vectors along the lattice. so long as the dispersion does not have a cusp at the zone boundary. Thus. (a) [First see solutions for 9. is always flat crossing a zone boundary. In Exercise 9.108 Reciprocal Lattice. In terms of primitive lattice vectors here we can take a1 and a2 to be primitive lattice vector.8 which clearly show the hexagonal periodicity of the dispersion curve. and then a3 = a2 − a1 . it must have zero derivative. Note that the dispersion must be symmetric around α = 1 since (by reflection symmetry of the problem) the frequency at k is the same as the frequency at k′ = −αbi /2 + k⊥ .b also). Recall from Eq. Note also the figures shown in the solution of 9. elements −t from each site to each of the nearest-neighbor odic in the Brillouin zone. Show that near k = 0 the dispersion is parabolic. Show that the dispersion curve is always flat cross- already solved those exercises. Imagine approaching the zone boundary via k = αbi /2 + k⊥ and taking α to one.8 and 11. Given the derived dispersion E = ǫ0 − 2t1 cos(kx a1 ) − 2t2 cos(ky a2 ) at the zone boundary. . In Exercise 11. Waves in Crystals (13. for a rectangular lattice with lattice vectors of length a1 and a2 the reciprocal lattice is rectangular with lattice vectors of length b1 = (2π)/a1 and b2 = (2π)/a2 .9. and its derivative is zero. whose reciprocal lattice is also triangular and has a Wigner-Seitz cell which is hexagonal (see exer- cise 13.8) Calculating Dispersions in d > 1* In Exercise 9.9 describe the Brillouin zone (you (b) Consider a tight binding model on a three- may assume perpendicular lattice vectors with length a1 dimensional fcc lattice where there are hopping matrix and a2 ). one can translate this by a reciprocal lattice vector to get (2 − α)bi /2 + k⊥ which is the same point reflected around the zone boundary. The Brillouin zone is hence a rectangle extending between kx ∈ [−π/a1 . Show that the tight-binding dispersion is peri.9] In Exercise 11.9 we discussed dispersion that the phonon dispersion is periodic in the Brillouin relations of systems in two dimensions (if you have not zone. Let the zone boundary wavevector be bi /2 + k⊥ where k⊥ · bi = 0. the cos is at its maximum. Determine the energy spectrum E(k) of this model. if the conventional unit cell has lattice constant a. (b) For an fcc lattice. 109 The frequency is an analytic function in k so long as the argument of the square-root is nonzero. j ∈ 1. For those who would prefer to see this statment proven a bit more directly. So we can write k⊥ = β(b1 + b2 /2) for some value of β. Thus we can conclude that the dispersion is analytic and symmetric around the zone boundary and therefore must have zero slope.. but it is perfectly rigorous. we take the following approach. Here S2 is independent of α so we are not concerned with that piece. Using sin2 (x/2) = (1 − cos(x))/2 we have that S1 = [1 − cos((β + α)π)]/2 S2 = [1 − cos((β − α)π)]/2 The cos is periodic in α → α ± 2.e. By rewriting the argument as 1 (S1 − S2)2 + (S2 − S3)2 + (S3 − S1)2 2 it is clear the argument must be nonzero except at k = 0. In particular we are concerned with the dependence of the dispersion on α near α = 1.6. We then have (using ai · bj = 2πδij for i. thus having zero slope. 2) k · a1 = (β + α)π k · a2 = β2π k · a3 = (β − α)π Now let us examine the terms in the dispersion. Thus we have the dispersion symmetric and analytic around the zone boundary.b) are given by √ √ √ b1 = 2π[(a 3/2)ˆ x − (a/2)ˆ y]/(a2 3/2) = (2π/a)[ˆ x − ( 3/3)ˆy] √ √ b2 = 2πˆ ya/(a2 3/2) = (2π/a)ˆ y(2 3/3) For simplicity let us imagine approachig the zone boundary k = αb2 /2 + k⊥ (the other zone boundaries will be the same by symmetry). Let us start with the primitive lattice vectors a1 = aˆ x √ a2 = (a/2)ˆ x + (a 3/2)ˆ y Recall that the reciprocal lattice vectors (see 13. α → 2 − α) turns S1 into S2 . the vectors to the 12 nearest neighbors of a lattice point are given by . Here the direction orthogonal to b2 is b1 + b2 /2. hence leaves the freqeuncy unchanged. and reflection of α around 1 (i. This may seem like a bit of a cheat. Again all we need to do is to show that the dispersion is symmetric and analytic around α = 1. Using the usual ansatz ψ(r) = Aeik·r we obtain X h i E = −t eik·[α. ±1.α. which is parabolic as claimed. . . ±1.β.0.β=±1 or E = −4t [cos(kx a/2) cos(ky a/2) + cos(kx a/2) cos(kz a/2) + cos(ky a/2) cos(kz a/2)] Expanding around k = 0 to second order we obtain E = −12t + t kx2 + ky2 + kz2 (a2 ) + .β]a/2 α=±1.0]a/2 + eik·[α. . 0]a/2 [±1.110 Reciprocal Lattice. ±1]a/2 [0.2) [±1.β]a/2 + eik·[0. Waves in Crystals (see exercise 12. ±1]a/2 Thus we have the schroedinger equation (setting the onsite energy ǫ0 to zero for simplicity) X Eψ(r) = −t ψ(r + u) u where the sum is over u being these 12 vectors. 0. Brillouin Zone. 5 (a two. and hence dimensional rectangular crystal having a unit cell with construct on your drawing the “scattering triangle” cor- sides a1 = 0. k′ and ∆k). A collimated responding to the Laue condition ∆k = G for diffraction beam of monochromatic X-rays with wavelength 0.468 nm and a2 = 0. A diagram of the reciprocal lattice is given in the figure. k′ of the incident and reflected X-ray beams. The scatter- 42 32 12 012 12 32 42 012 Fig. (a) Draw to scale a diagram of the reciprocal lattice. ing triangle is the triangle such that |k| = |k′ | and k + G = |k′ | (Note for fixed G there are two such possible triangles).342 nm).Wave Scattering by Crystals 14 (14.1) Reciprocal Lattice and X-ray Scattering (b) Calculate the magnitude of the wavevectors k and Consider the lattice described in Exercise 13. 14. is used to examine the crystal.1 A picture of the reciprocal Lattice with the scattering triangle for the (210) reciprocal wavevector.166 nm from the (210) planes (the scattering triangle includes k. . thus we obtain 2 I002 (fBa + fT i + 3fO ) = ≈ 15.4 I001 (fB a − fT i − fO )2 In reality the form factor depends on the scattering vector. You may assume that the atomic form factor is propor- Sketch the unit cell. but the multiplicity for all (00l) are the same!). where I(hkl) is the O [ 12 . 14. 21 . ZO = 8. 12 . etc. and the variation is different from each atom. If we are interested in (00l) we set h = k = 0 and obtain S(00l) = fBa + (−1)l fT i + [1 + 2(−1)l ]fO The Bragg peak intensity is proportional to the square of the structure factor (times a multiplicity factor. 12 ] Calculate the ratio I(002) /I(001) .) 01 23 4 Fig. and fd it the corresponding form factor (which we take to be proportional to Zd ). Ti [ 12 .112 Wave Scattering by Crystals (14.0. [0. (ZBa = 56. 0]. and neglect its dependence Show that the X-ray structure factor for the (00l) on the scattering vector. 21 . 21 ] intensity of the X-ray diffraction from the (hkl) planes. yd . tional to atomic number (Z).2) ‡ X-ray scattering II Bragg reflections is given by BaTiO3 has a primitive cubic lattice and a basis with h i atoms having fractional coordinates S(hkl) = fBa + (−1)l fT i + 1 + 2(−1)l fO Ba [0. zd ) are the positions of atom d in the unit cell.0] where fBa is the atomic form factor for Ba. ZTi = 22. [ 12 . so this is just an approximation. 12 ].2 Unit cell of BaTiO3 The X-ray structure factor is given by X X S(hkl) = fd eik(hkl) ·Rd = fd e2πi(hxd +kyd +lzd ) d d where Rd = (xd . . 0. ) what atoms are in the primitive unit cell.1/2).0) and (1/2. density of palladium is 12023 kg m−3 ? (Atomic mass of (c) Show that these selection rules hold independent of palladium = 106.0. Note that these selection rules hold for an orthorhombic lattice as well as for cubic (nowhere did we use the lattice constants in the three . wavelength 0. whereas the (110) planes of a crystal with a face-centred cubic lattice contain only half the lattice points.4.0. the lattice constant is the distance between one lattice point and the nearest neighbor lattice point. Peaks in the scattered X- (a) Explain what is meant by “Lattice Constant” for a ray pattern are observed at angles of 42.1/2. Calculate the lattice constant and the nearest- centered cubic lattice. 72. (b) The (110) planes of a body-centred cubic lattice contain all the lattice points. (b) Explain why X-ray diffraction may be observed in Identify the lattice type. Analogously we obtain f cc S(hkl) ∼ 1 + (−1)h+k + (−1)h+l + (−1)k+l This is only nonzero when h. The remaining fcc lattice points lie on a set of planes half-way in between the (110) planes. The fcc lattice is a cubic lattice with with basis (0.3) ‡ X-ray scattering and Systematic Ab. first order from the (110) planes of a crystal with a body.3◦ from the direction of the incident beam.1/2. (For one suggestion.2◦ .0) and (1/2. cubic crystal structure.4◦ .1/2) and (0. so long as the (e) How could you improve the precision with which lattice is bcc or fcc respectively. For a cubic lattice.10.1/2). we must have khkl = (2π/a)(hˆ x + kˆy + lˆz) (or in other words we consider reciprocal lattice vectors (h. If you assume there is only a single atom in the ba- Derive the general selection rules for which planes sis does this distance agree with the known data that the are observed in bcc and fcc lattices. 87. but not from the (110) planes of a neighbor distance. 113 (14. So we obtain bcc S(hkl) ∼ 1 + eik(hkl) ·[(1/2)ˆx+(1/2)ˆy+(1/2)ˆz] = 1 + (−1)h+k+l This is only nonzero when h + k + l = even.162 nm is incident upon a powdered sample sences of the cubic metal palladium. l are either all even or all odd.) Generally we sum over lattice points Ri in a unit cell to get the struc- ture factor X S(hkl) ∼ eiRi ·k(hkl) Ri The bcc lattice is a cubic lattice with a basis (0. k. and so X-rays reflected from these planes interfere destructively with X-rays reflected from the (110) planes. (d) A collimated beam of monochromatic X-rays of see Exercise 14. k. and 92. 49. Let us see this more analytically now: We view both the bcc and fcc lattices as being a cubic lattice with a basis.2◦ .1/2. crystal with a face-centered cubic lattice.0) and (1/2.) (a) For a cubic lattice.0.3◦ . l). the lattice constant is determined. 3) that for a cubic lattice (a2 /d2 ) = h2 + k 2 + l2 . (2. These are all even or all odd. The cube root of this gives 3.66 . These atoms may have scattering form factors fj as well. (1.3◦ . (c) This is the general principle that you multiply a lattice by a basis. so no higher accuracy estimate can be made from this data).872 × 10−29 m3 .1. m3 . .0. and (2.389 nm we see that the values of 1/d2 are in ratios 3:4:8:11:12. Recall (see problem 13. then the structure will also vanish for any lattice having an fcc lattice with any basis. we treat the Bravais lattice as a cubic lattice with a basis Ri (i=1. So the observed peaks are of the types (1.1) i. As above.0).2◦ .389 nm 72.67 . An equivalent calculation is to calculate the density .1064 kg = 1.022 × 1023 atoms Then for an fcc there are 4 atoms per unit cell. So if the fcc structure factor vanishes.0).1).2◦ .j j i The factor inside the brackets is exactly the structure factor for the Bra- vais lattice as calculated above. so the volume of the unit cell is 5.2).4◦ .2.1. . which is characteristic of the fcc lattice. So positions of all of the atoms (2m or 4m of them) in the full cubic unit cell can be written as Rij = Ri + rj Writing the structure factor " # X X X ik(hkl) Rij ik(hkl) Ri S(hkl) = fj e = fj e eik(hkl) rj (14... (2. m) with respect to the lattice points. 4 for fcc).114 Wave Scattering by Crystals directions!). All measured scattering points give the same estimate of the lattice constant a to three digits (which is the same number of digits that the measurement contains.389 nm 49.4682 × 10−29 m3 /atom 12023 kg 6. Suppose we have a Bravais lattice (either fcc or bcc) and then we have a basis of m atoms which are at positions rj (j = 1. We then examine the ratio’s between these distances. . .2.117 nm 3.33 .887 nm which is in good agreement.195 nm 1.3◦ .389 nm 87.389 nm 92.224 nm 1 .3). .00 .137 nm 2. √ 2θ d = λ/(2 sin θ) (dmax /d)2 aestimate = d h2 + k 2 + l2 42.2 for bcc and i = 1 .112 nm 4. (d) Note that the angle given is 2θ (the full deflection angle)! We are given 2θ then we use Bragg’s law λ = 2d sin θ to get the distance between lattice planes. ) (zincblende) structure or NaCl (sodium chloride) struc.4) ‡ Neutron Scattering Write down expressions for the structure factors (a) X-ray diffraction from sodium hydride (NaH) estab. .374 × 105 nm. error is minimized if one can work as close as possible to scattering angle of 90 degrees (See exercise 14. (b) How does one produce monochromatic neutrons for ture. 115 based on the lattice constant 3 4 ∗ . 41 . 21 ]. From Bragg’s law d ∼ 1/ sin θ so 1 ∂d = cot θ d ∂θ So if there is an error in measurement of θ one ends up with a fractional error in d given by δ log d = (cot θ)(δθ) As a result. What are the main differences between neutrons tensity of the Bragg peak indexed as (111) was found to and X-rays? be much larger than the intensity of the peak indexed as Explain why (inelastic) neutron scattering is well (200). 41 ] or by [ 21 . (e) to improve the precision of the measurement one would need to ob- tain more “digits” of resolution in the measurement of the angle.022 × 1023 atoms .363 × 105 nm. Why is it difficult to locate the positions of the H Hence.10). but X-rays are not. respectively. and the charge of H is very small. (14. (ii) the zinc blende (ZnS) structure. (Nuclear scattering length of Na The H atoms were thought to be displaced from the Na = 0. In addition some experimental issues can be listed. [This problem is based on a classic experiment by Shull et al. Temperature control is necessary since lattice constants do change as a function of temperature due to thermal expansion. 12 . A higher bright- ness source will frequently reduce the noise and make the signal easier to analyze.1064 kg 1 = 12006 kg/m3 6. There are several things that will help this. the scattering length should be in units of 10−5 nm not 105 nm. suited for observing phonons. deduce which of the two structure mod- atoms using X-rays? els is correct for NaH. It is difficult to see H atoms with X-rays since the form factor (am- plitude) of scattering is proportional to the charge of the nucleus (the atomic number).389 × 10−9 m in good agreement. At least as important is the calibration of the device that measures the angles! Defining the wavelength λ more precisely becomes necessary at some point. nuclear scattering length of H = atoms either by [ 14 . S(hkl) for neutron diffraction assuming NaH has lished that the Na atoms are arranged on a face-centered (i) the sodium chloride (NaCl) structure cubic lattice. To distinguish these models a neutron use in neutron diffraction experiments? powder diffraction measurement was performed. Better columnation will make measurements of angle more pre- cise. The in. to form the ZnS −0. who later won a Nobel prize for his work on neutron scattering] Note. It doesn’t change the problem much though. 1/2.1/4. (1) time of flight (a ”chopper”) can select a particular velocity (2) Bragg scattering of a given angle off of a crystal can select a particular wave- length. 14. The multiplicity of (111) is 8 (since each factor of 1 could have been ¯1).116 Wave Scattering by Crystals From Eq.1/2]. we have a basis of Na at [0.0] and H at [1/4. So we obtain S(hkl) = (1 + eiπ(h+k) + eiπ(k+l) + eiπ(h+l) )(bN a + bH eiπ(h+k+l) ) For the case of the ZnS structure we have a basis of Na at [0. whereas the multiplicity of (200) is 6 since the 2 could be of either sign. and any one of the 3 entries could contain the 2. Note that the first factor in brackets is the scattering from the fcc lattice which always gives (See problem 14) f cc Shkl = (1 + eiπ(h+k) + eiπ(k+l) + eiπ(h+l) ) For the case of the NaCl structure. Thus we obtain in the NaCl case a ratio of scattering intensities I(111) (bN a − bH )2 8 = = 6000 I(200) (bN a + bH )2 6 whereas in the ZnS case we have I(111) (b2 + b2H ) 8 = Na = 0. The intensity of scattering varies with the atomic form .1 above we have ! X X S(hkl) = eik(hkl) Ri bj eik(hkl) rj i j where j are the elements of the basis and i are the 4-elements of the fcc lattice basis when fcc is viewed as a cubic lattice with a basis.0.0.1/4] so we obtain S(hkl) = (1 + eiπ(h+k) + eiπ(k+l) + eiπ(h+l) )(bN a + bH ei(π/2)(h+k+l) ) The powder scattering intensity is proportional to the structure factor squared times a multiplicity factor that counts how many symmetry re- lated vectors give the same scattering pattern.0] and H at [1/2.67 I(200) (bN a − bH )2 6 Thus we conclude that NaH has NaCl structure. The main things to know about the differences between X-rays and neutrons are: X-rays are scattered by electrons. Here I have replaced fj by bj which is the conventional notation for the scattering length (which is analogous to the form factor in X-rays) in neutron scattering. and so X-rays typically penetrate only a few microns into a sample. (b) Two typical schemes for producing monochromatic neutrons. The electromagnetic interaction is relatively strong. which is independent of ∆k. The diffraction theory is the same as for X-rays. 117 factor f (∆k). and so interact weakly with matter.83◦ 50. Although with modern tools measurement of phonons using X-rays is now possible. it really is a good approximation because the scattering is essen- tially off of a point nucleus).68◦ 41. are electrically neutral. and can easily distinguish atoms with similar atomic number.154 nm). The incident X-ray ra.64◦ 33. The following scat. They penetrate typically a few cm into materi- als.7 g/cm3 . which approximately scales with the number of electrons in the atom. Neutrons can thus easily see small atoms. (14. Neutrons. If you use 2θ by mistake . Neutrons may scatter from nuclei (via the strong nuclear force) or unpaired electrons (via the magnetic dipole-dipole interaction). (Note.00◦ 39. on the other hand. we typically make the approximation that f is independent of ∆k. The reason inelastic scattering of neutrons is more useful than X-rays for studying phonons is because the X-ray velocity (velocity of light) is huge.5) And More X-ray Scattering 19. The scattering length b varies irregularly from nucleus to nucleus so that the scattering from light elements is sim- ilar in strength to that from heavy elements.42◦ A sample of aluminum powder is put in an Debye– Given also that the atomic weight of Al is 27. Magnetic neutron diffraction can also be used to determine magnetic structures (in which case there is a form factor similar to that for X- rays). density is 2. to make calculations easier. How far off are you? What causes that the wavelength is λ = . but it is not really such a good approximation. the error? tering angles were observed: Note: The angles listed are θ not 2θ here. This makes it very difficult to conserve both momentum and energy in any inelastic process involving light since a very small mo- mentum change corresponds to a huge energy change.05◦ 59. and the Scherrer X-ray diffraction device. use this information to calculate diation is from Cu–Ka X-ray transition (this just means Avagadro′ s number.48◦ 22. except that for scattering from nuclei the atomic form factor f (∆k) is replaced by the nuclear scat- tering length b. In the neutron case.35◦ 57. Probably the biggest possible error is that the density changes as a function of temperature.35◦ . If the density is measured at one tem- perature. The corresponding plane spacings are p d = a/ h2 + k 2 + l2 .0918 nm 6.7 g/cm3 we predict an atomic density of NA × 105 m−3 with NA = 6.24nm.1000 nm 5.64◦ .48◦ .3999 nm 22. so we would predict an atomic density of 4/(4nm)3 = 6.28 × 1028m−3 .1414 nm 2.2309 nm 1 . (Or in other words.3999 nm 39.00◦ . First. There are four atoms per unit cell.28 × 1023 ). So we are off by abour 4%. it is actually about 26. the measurements could even be done above room temperature!).022 × 1023 being Avagadro’s number. Where does the error come from? There are a few possible places.6) More Neutron Scattering What is the minimum energy of the neutrons? At what The conventional unit cell dimension for a particular temperature would such neutrons be dominant if the dis- bcc solid is . our slighly inaccurate prediction of Avagadro’s number is 6. Thus I suspect the data given above was measured at low temperature. Two orders of diffraction are observed.404 nm at room tempera- ture (which would match the stated density much more closely). tribution is Maxwell–Boltzmann.982 (depending a bit on isotopic abun- dances.4000 nm 41.4000 nm 59.00 . In fact the linear thermal expansion coefficient of aluminum is about (22 × 10( − 6))K−1 which would give a density different between T = 0 and room temperature of a few percent (and indeed. On the other hand.66 . the lowest two order diffraction peaks are the (110) and the (200).33 .2000 nm 1. For bcc.67 .4000 nm 57.68◦ . From the final column we conclude the conventional lattice constant is about .66 .4000 nm From the third column we see that we have an fcc lattice.4001 nm 33.33 . But this is an error of much less than an percent.1155 nm 4.05◦ . the atomic mass is not quite 27.4000 nm 50. In the literature X-ray measurements of the lattice constant of alumium give numbers of about . hence the disagreement.1206 nm 3. (14. with atomic weight of 27 and density 2. but this is typical). then there can be a substantial disagreement.0894 nm 6.33 .83◦ . but the lattice constant is measured at another temperature.42◦ .118 Wave Scattering by Crystals you will discover that the ratios of d values do not make sense! √ θ d = λ/(2 sin θ) (dmax /d)2 aestimate = d h2 + k 2 + l2 19.4000 nm. 14). The corresponding energy of neutrons is ~2 k 2 ~2 (2π)2 E= = = 2.α )+l(Rz. for the basis (i.a +ux. Thus we have X S(hkl) = fα e2πi(h(Rx. 119 √ Or d = a/ 2 and d = a/2.α ) e2πi(hRx.a +kRy.e. Using Bragg’s law.a ) α a which is simply the product of basis and lattice structure factors basis lattice S(hkl) = S(hkl) S(hlk) . We split this sum into two X X S(hkl) = fα e2πi(hux. all of the positons rj of atoms j in the unit cell can be written as the sum of a lattice point Ra and a basis vector uα .a +uy. d = λ/(2 sin θ) the wavelength must be at least as small as 2d in order to see a particular peak.28 × 10−21 J 2mn 2mn λ2 and E/kB = 165K (14.α +luz..α )) α.a Note that the atom type depends only on the basis vector not on the lattice point. Thus we can write rj = Ra + uα and the sum over j becomes a sum over a and α. scribed with a lattice and a basis) is the product of the We have the definition of the structure factor X S(hkl) = fj e2πi(hxj +kyj +lzj ) atoms j in (conventional) unit cell When we have a lattice and a basis.24nm.α )+k(Ry. prove Eq.α )+kuy.a +uz. 14.a +lRz.7) Lattice and Basis structure factor for the lattice times the structure factor Prove that the structure factor for any crystal (de. Or in other words λ ≤ a = . 0]. Thus peaks and the center are O and the lighter col. [0. [ 14 .3 Conventional unit cell of the fCu vanish unless h. Indeed. k. 3/4. Let us examine the last three terms. 1/4. 1/2. 1/2]. 3/4]. 14 . [0. 1/4]. 0]. [ 43 .3. The darker atoms at the corners the Cu alone form an fcc lattice (perhaps less obvious!). Cu [ 14 . Calculate the structure factor for this crystal. [1/2. this tells us that the oxygens alone form a bcc lattice. 0. [3/4. 1/2. only on fCu .5. What is the lattice (a) the conventional unit cell of Cu2 O is shown in Figure 14. 1/2]. 14.5 the basis of Flourine beta is [0. Note that 0 l = odd ik + i−k = 2 l = 0 mod 4 −2 l = 2 mod 4 . 12 ] cise 12. (b) From 12. [ 12 . The structure factor is S(hkl) = fO (1 + eiπ(h+k+l) ) h π π π π i +fCu ei 2 (h+k+l) + ei 2 (h+3k+3l) + ei 2 (3h+k+3l) + ei 2 (3h+3k+l) Note that sqaure bracketed expression can be simplified to π h i ei 2 (h+k+l) 1 + eiπ(k+l) + eiπ(h+l) + eiπ(k+l) Note that the coefficient of f0 vanishes unless h + k + l is even. which tells us that Cu2 O. [1/2. 14 ] . 0. 34 . 34 ] . 21 . [1/4. (b) Consider fluorine beta phase as described in exer- O [000] . 34 ] . 1/2] The structure factor is then 1 + eiπk+iπl/2 + eiπk+iπ3l/2 + eiπl+iπh/2 + eiπl+iπ3h/2 +eiπh+iπk/2 + eiπh+iπ3k/2 + eiπ(h+k+l) which can be simplified to 1 + (−1)h+k+l + (−1)k [il + i−l ] + (−1)l [ih + i−h ] + (−1)h [ik + i−k ] Note that the first two terms cancel if h + k + l is odd and otherwise give 2. 34 . 41 .8) Cuprous Oxide and Fluorine Beta type? Show that certain diffraction peaks depend only (a) The compound Cu2 O has a cubic conventional unit on the Cu form factor fCu and other reflections depend cell with the basis: only on the O form factor fO . l are all even or all odd. The coefficient of Fig. [1/2. such as (110) depend only on fO wheras peaks such as (111) depend ored atoms are Cu. 0]. 41 ] What are the selection rules? Sketch the conventional unit cell.120 Wave Scattering by Crystals (14. [ 43 . 1/2. 0. 14. (a) Start with 14.α radius. Derive Eq. of hydrogen. 14.5 Z XZ iG·x S(G) = dx e V (x) = dx eiG·x Vα (x − R − yα ) unitcell R. Now tron in a hydrogen atom to calculate the X-ray form factor use the definition of the structure factor Eq. Write the positions of sum over integrals of individual unit cells. 121 Thus we obtain the following rules for (hkl) where the structure factor does not vanish one even two odd if the even one is 2 mod 4 one odd two even if the two even indices are not the same mod 4 all even with either one or all three indices being 2 mod 4 (14.9) Form Factors rive an expression of the form of Eq. (b) Assuming the potential is V0 inside a radius a and zero outside we have Z radius=a Z a Z iG·x fα (G) = V0 dx e = 2πV0 r dr sin θdθeirGcosθ 2 0 0 .8 and hence derive (a) Assume that the scattering potential can be written expression 14. α indexes the particles in the (c)* Use your knowledge of the wavefunction of an elec- basis and yα is the position of atom α in the basis. 14. so we have XZ XZ iG·x S(G) = dx e Vα (x − yα ) = dx eiG·(x+yα ) Vα (x) α α X iG·yα = e fα (G) α with Z fα (G) = dx eiG·x Vα (x) as required.α unitcell Now we note that the integral dx over a unit cell summed over all possible unit cells indexed by R is equivalent to a single integral over all of space. (Hint: Use the fact as the sum over the contributions of the scattering from that an integral over all space can be decomposed into a each of the atoms in the system. where R are lattice points.5 and de.10. assume the scattering potential from an V (x) = Vα (x − R − yα ) atom is constant inside a radius a and is zero outside that R.α unitcell we then using eiG·R = 1 we have XZ S(G) = dx eiG·(x−R) Vα (x − R − yα ) R. 14.9).9 for the form factor.) the atoms in terms of a lattice plus a basis so that (b) Given the equation for the form factor you just de- X rived (Eq. (c) The normalized wavefunction for an electron in the ground state of a hydrogen atom is 1 2 −r/a0 ψ=√ 3/2 e 4π a0 with a0 the Bohr radius. What is the fractional error δa/a in the result- a of some crystal using X-rays. So setting V0 4πa3 /3 = Z makes the two equations match. error is minimized if one can work as close as possible to scattering angle of 90 degrees. We then have (using similar calculation as above) Z Z ∞ 2 sin(rG) f (G) = K dx eiG·x |ψ(r)|2 = K2π r2 dr |ψ(r)|2 0 rG So we want Z 2K ∞ 2 2 sin(rG) −2r/a0 f (G) = r dr e a30 0 rG The integration is not too difficult and gives the result 16K f (G) = 2 (a20 G2 + 4) (14.e.3. it looks identical except for the prefactor. Suppose a diffraction peak ing measurement of the lattice constant? How might this is observed at a scattering angle of 2θ. (e) From Bragg’s law d ∼ 1/ sin θ so 1 ∂d = cot θ d ∂θ So if there is an error in measurement of θ one ends up with a fractional error in d given by δ log d = (cot θ)(δθ) As a result. Now comparing this to 14. The scattering potential is proportional to the electron density |ψ|2 . suppose error be reduced? Why could it not be reduced to zero? See 14. However.10) Error Analysis that the value of θ is measured only within some uncer- Imagine you are trying to measure the lattice constant tainty δθ. Note however. that the charge of atom is spread out over a volume 4πa3 /3. Although at 90 degrees one could have .122 Wave Scattering by Crystals where θ is the angle between x and G.10. Let us call the constant of proportionality K. So we have Z a Z 1 Z a 2 irGz 2 sin(rG) fα (G) = 2πV0 r dr dze = 2πV0 r2 dr 0 −1 0 rG sin(x) − x cos(x) = 4πV0 a3 x3 where x = Ga. . One has only made the error in measurement zero to lowest order. this does not mean the error is actually zero. 123 this expression be exactly zero. One also has to worry about higher derivative term 1 ∂2d = 1 6= 0 d ∂θ2 at 90 degrees. However. . 2) or equivalently √ ψ = (Aeikx + Bei(k+G)x )/ L To maintain normalization we can insist that |A|2 + |B|2 = 1. Give a qualitative explanation of why these two and VG∗ = V−G assures that the potential V (x) is real. (b)We have our (variational) trial wavefunction given by |ψi = A|ki + B|k + Gi (15.1) ‡Nearly Free Electron Model that the eigenenergies at this wavevector are Consider an electron in a weak periodic potential in one dimension V (x) = V (x + a). ence from k to the zone boundary wavevector. Thus. use the above form of the wavefunction to show wavevector. Taking k and k + G both on a Brillouin zone boundary we have k = nπ/a and k + . Give a sketch (don’t do a full calculation) of the ary (such as k near π/a) the electron wavefunction should energy as a function of k in both the extended and the be taken to be reduced zone schemes. the scat- tering perturbation is weak. (b) For an electron of mass m with k exactly at a zone Calculate the effective mass of an electron at this boundary. In the nearly free electron picture. (a) Explain why for k near to a Brillouin zone bound. states are separated in energy by 2|VG |. the ψ = Aeikx + Bei(k+G)x (15. but not exactly at. there is an energy denominator which suppresses mixing of k-vectors which have greatly different unperturbed energies. so that we can treat the scattered wave in perturbation theory. In this case.1) zone boundary.Electrons in a Periodic Potential 15 (15. Write the periodic potential ~2 k2 as E= + V0 ± |VG | X iGx 2m V (x) = e VG G where G is chosen so |k| = |k + G|. (a) A periodic lattice can only scatter a wave by a reciprocal lattice vector (Bragg diffraction). Degenerate perturbation theory tells us that we should first diagonalize within the degenerate space spanned by only these two eigenstates. Give an expression for the energy E(k) where G is a reciprocal lattice vector such that |k| is close correct to order (δk)2 where δk is the wavevector differ- to |k + G|. where the sum is over the reciprocal lattice G = 2πn/a. the only mixing that can occur is between two states with similar energies that are separated by a reciprocal lattice vector. (c) *Now consider k close to. . If we simply calculate the expectation value of H in the trial state given by Eq. then the higher energy state is the ψ = cos(2nπr/a) which puts the maximum amplitude of the wavefunction exactly at the maxima of the potential. and we must have G = −2nπ/a the reciprocal lattice vector. the eigenstate is the trial wavefunction which minimizes the total energy while preserving the normalization. Assuming V2nπ/a > 0. Approach 1: Diagonalize H within the 2d degenerate space hk|H|ki = ~2 (nπ/a)2 /(2m) + V0 hk + G|H|k + Gi = ~2 (nπ/a)2 /(2m) + V0 hk|H|k + Gi = V2nπ/a hk + G|H|ki = V−2nπ/a Diagonalizing this two by two matrix 2 ~ (π/a)2 /(2m) + V0 V2nπ/a V−2nπ/a ~2 (π/a)2 /(2m) + V0 gives eigenstates ~2 (nπ/a)2 E= + V0 ± |V2nπ/a | 2m ∗ where we have used VG = V−G . One way to do this is to write A = cos(θ) and B = eiχ sin(θ) which is the most general form we can write while still preserving |A|2 + |B|2 = 1 (we can arbitrarily choose A to be real. Approach 2: Variational. since that only introduces an irrelevant overall phase). It gives minimum energy states for ψ = sin(2nπr/a) as above. Interpretation: If we have considered only the V2nπ/a and V−2nπ/a Fourier modes of the potential then we have V = 2V2nπ/a cos(2nπr/a). where here we have chosen the nth zone boundary. The eigenstates are correspondingly |ψ± i = |ki + |k + Gi which are (proportional to) the functions sin(2nπx/a) and cos(2nπx/a). In terms of these parameters we have hψ|H|ψi = ~2 (nπ/a)2 /2m + V0 + 2Re[V2nπ/a eiχ sin(θ) cos(θ)] for V2nπ/a > 0 this is minimized for χ = π and θ = π/4 (or equiv- alently χ = 0 and θ = 3π/4). Similarly. In the case of V2nπ/a < 0 the sin is the higher energy wavefunction. the lower energy wavefunction is the sin(2nπr/a) which has the minimum amplitude of the wavefunction at the maximum of the potential.2 we obtain hψ|H|ψi = ~2 (nπ/a)2 /(2m) + V0 + A∗ B V2nπ/a ∗ + B ∗ A V2nπ/a Using the variational principle. The Hamiltonian H in question is the usual Kinetic term plus V (x).126 Electrons in a Periodic Potential G = −nπ/a. 15. The effective mass is then obtained by setting 1 1 ~2 (nπ/a)2 = 1 ± 2m∗ 2m m|V2nπ/a | . Note that for this expansion to remain valid we must have the bracketed term in the square root two equations up small compared to the |V2nπ/a |2 term. only here we need to consider k not on the zone boundary. 15.1 for a sketch of the bands in the extended zone scheme.1 Diagram of the dispersion in a nearly free electron model. Letting k = nπ/a + δk and k + G = −nπ/a + δk we have hk|H|ki = ~2 (δk + nπ/a)2 /(2m) + V0 hk + G|H|k + Gi = ~2 (δk − nπ/a)2 /(2m) + V0 hk|H|k + Gi = V2nπ/a hk + G|H|ki = V−2nπ/a which we now need to diagonalize. Note that gaps open up at all zone boundaries (c) This calculation is entirely analogous to that above. Top: extended zone scheme. 127 See figure 15. We obtain s 2 2 ~2 [(δk)2 + (nπ/a)2 ] ~ 2(δk)nπ/a E± = + V0 ± + |V2nπ/a |2 2m 2m expanding the square-root we obtain ~2 (nπ/a)2 ~2 (δk)2 ~2 (nπ/a)2 E± = + V0 ± |V2nπ/a | + 1± 2m 2m m|V2nπ/a | which is a quadratic correction as we move away from the Brillouin zone. Bottom: reduced zone scheme. 2nd Zone 2nd Zone 1st Brillouin Zone Extended Zone Scheme −2π/a −π/a 0 π/a 2π/a Reduced Zone Scheme −2π/a −π/a 0 π/a 2π/a Fig. 128 Electrons in a Periodic Potential or equivalently . . . . . m . ∗ . m =. . ~2 (nπ/a)2 . . 1 ± m|V2nπ/a | . is the ground-state wavefunction of an electron bound to . with the + being for the upper band.3) Tight Binding Bloch Wavefunctions a nucleus on site R. consider a tight-binding wave ansatz of the form X ik·R X ik·R ψ(r) = e ϕ(r − R). show it is a modified plane wave). we then directly obtain X ρ(x) = eiG·x ρG G (15. In real space this ansatz can be Analogous to the wavefunction introduced in Chapter expressed as 11.. ρ(x) = ρG eiG·x G where the sum is over points G in the reciprocal lattice. Letting the delta function act and cancelling some factors. and |Ri Bloch’s theorem (i. so let us take a sum over all N lattice vectors in the system Z 1 X V X dk ρ(x) = ρ(x + R) = ρk eik·(x+R) N N (2π)3 ZR X R V dk ik·x = ρk e eik·R N (2π)3 R The sum gives X (2π)3 X 3 eik·R = δ (k − G) v R G where the sum is over all G which are reciprocal lattice vectors and v is the volume of the unit cell. (15.2) Periodic Functions Show that ρ can be written as Consider a lattice of points {R} and a function ρ(x) X which has the periodicity of the lattice ρ(x) = ρ(x + R).e. |ψi = e |Ri R R Show that this wavefunction is of the form required by where the sum is over the points R of a lattice. Generally we can always write ρ(x) in terms of its Fourier transform Z dk ρ(x) = V ρk eik·x (2π)3 Now by periodicity we know that ρ(x) = ρ(x + R) for any lattice vector R. 1! The fourier components V10 couple (π/a. 129 Start by writing the function ϕ in its fourier representation Z dq iq·r ϕ(r) = V e ϕq (2π)3 so that X Z dq iq·(r−R) ψ(r) = eik·R V e ϕq (2π)3 R The sum over R gives X (2π)3 X 3 ei(k−q)·R = δ (k − q − G) v R G and we allow the delta function to act. Let us label the points |1i = (+π/a. lattice with lattice constant a. giving V X −iG·r ψ(r) = eik·r e ϕk−G v G The sum over G is a function periodic in r → r + R (since eiG·R = 1) hence this is in Bloch form.4) *Nearly Free Electrons in Two Dimen. 0). The energies of the states are E = ~2 (π/a)2 /(2m) ± |V10 | (b) The second part of the problem is more complicated. all four points (±π/a. (a) Use the nearly free electron model to find the energies sions Consider the nearly free electron model for a square of states at wavevector G = (π/a. −π/a) |3i = (−π/a. Therefore we must treat all of these states in degenerate perturbation theory. Make use of adding together rows or + 4V11 [cos(2πx/a) cos(2πy/a)] columns of the determinant before trying to evaluate it!) Note. +π/a) . (b) Calculate the energies of the states at wavevec- tential is given by tor G = (π/a. Suppose the periodic po. π/a) as G since they are not reciprocal lattice vectors! (a) The first part of this problem is no different from the one di- mensional problem posed in 15. +π/a) |2i = (+π/a. π/a). (Hint: You should write down a 4 by 4 secular determinant. which looks difficult. but actu- V (x. we should not call the points (π/a. 0). 0) and (−π/a. Here. 0) and (π/a. (15. −π/a) |4i = (−π/a. y) = 2V10 [cos(2πx/a) + cos(2πy/a)] ally factors nicely. ±π/a) all have the same energy and are coupled together by the scattering potential. To find the eigenenergies we would thus like to solve the determinant equation .130 Electrons in a Periodic Potential P Writing a general wavefunction within this space as i φi |ii the Hamil- tonian matrix within this reduced Hilbert space is ǫ V10 V11 V10 V10 ǫ V10 V11 V11 V10 ǫ V10 V10 V11 V10 ǫ where ǫ = ~2 (π/a)2 /m. . . ǫ−E V10 V11 V10 . . . . V10 ǫ − E V10 V11 . 0 = . . . . . V11 V10 ǫ−E V10 . . V10 V11 V10 ǫ−E . So we can subtract row 3 from row 1 and subtract row 4 from row 2. Adding and subtracting rows and columns leaves the determinant un- changed. The result is . Then add column 3 to column 1 and add column 4 to column 2. . . 0 0 V11 − ǫ + E 0 . . . . 0 0 0 V11 − ǫ + E . 0=. . . V + ǫ − E 2V ǫ − E V . . 11 10 10 . . 2V10 V11 + ǫ − E V10 ǫ−E . decaying (evanes- valid? Here.5) Decaying Waves cent) waves still exist. method of minors can then evaluate the deter- minant easily to given 0 = (V11 − ǫ + E)2 (V11 + ǫ − E)2 − 4V10 2 Thus giving solutions E = ǫ + V11 (two solutions) E = ǫ − V11 ± 2|V10 | (15. a peri- odic potential opens a band gap such that there are no ψ(x) = eikx−κx plane-wave eigenstates between energies ǫ0 (G/2) − |VG | with 0 < κ ≪ k and κ real.e.. Find κ as a function of energy and ǫ0 (G/2) + |VG | with G a reciprocal lattice vector. in one dimension. Assume the form As we saw in this chapter. for k = G/2. absorb κ into k) k = −G/2 + iκ . at these forbidden energies. we can think of the wavevector k as taking an imaginary part (i. Because of all the zeros. For what range of VG and E is your result However. which can be solved to then give only one (possibly) positive solution r ~2 κ2 ~2 (G/2)2 ~2 (G/2)2 =− − E + 4E + |VG |2 2m 2m 2m In order for this solution to be valid.3) Thus. Thus we must have |VG | ≥ |~2 Gκ/(2m)| In Eq. 15.8). We can write this condition as 2 2 ~ (G/2)2 ~2 (G/2)2 +E ≤ 4E + |VG |2 2m 2m Or equivalently .3 we have a quadratic equation for κ2 (there are no lone factors of κ). we must have the right hand side be positive. 131 so that ~2 k 2 ~2 ((G/2)2 − iGκ − κ2 ) ǫ0 (k) = 2m = = ǫR − iǫI 2m ~2 (k+G)2 ~2 ((G/2)2 + iGκ − κ2 ) ǫ0 (k + G) = 2m = = ǫR − iǫI 2m where we have defined ~2 ((G/2)2 − κ2 ) ǫR = 2m ~2 (Gκ) ǫI = 2m As worked out in the text (Eq. we have q E = ǫR ± |VG |2 − ǫ2I Note here that for |ǫI | > |VG | the energy becomes imaginary and the solution is not valid. 15. the characteristic equation is 0 = (ǫ0 (k) − E) (ǫ0 (k + G) − E) − |VG |2 = (ǫR − iǫI − E)(ǫR + iǫI − E) − |VG |2 = (ǫR − E)2 + ǫ2I − |VG |2 (15. 2 . . ~ (G/2)2 . . − E . ≤ |VG | . 2m . 6) Kronig–Penney Model* function comb” potential in one dimension Consider electrons of mass m in a so-called “delta. or in other words. X V (x) = aU δ(x − na) n . that the energy is inside the gap! (15. nuity in slope at x = 0 derive ways of a plane wave form eiqE x with √ qE A − eika k[A cos(qE a) − B sin(qE a)] = 2maU B/~2 qE = 2mE/~. This then implies ψ(x + a) = eika ψ(x) as required. and uE (x) = uE (x + a) defines u outside of this interval. determine the size of the gap at the zone B = e−ika [A sin(qE a) + B cos(qE a)]. Between delta functions. Oops another error in this problem. result agrees with the predictions of the nearly free elec- rive tron model (i. Probably too many martinis. Actually this one is really bad. but the right-hand side is not. As a result. . Solve these two equations to obtain Using Bloch’s theorem conclude that we can write an mU a eigenstate with energy E as cos(ka) = cos(qE a) + sin(qE a) ~2 qE ψ(x) = eikx uE (x) The left-hand side of this equation is always between −1 and 1. we must have ψ of the form A sin(qE x) + B cos(qE x). The Schroedinger equation evaluated near x = a ~2 2 ∂ ψ(x) = aU δ(x − a)ψ(x) 2m x is equivalent to ~2 ∂x ψ(a+ ) − ∂x ψ(a− ) = aU ψ(a) 2m . (c) For small values of the potential U show that this (b) Using continuity of the wavefunction at x = 0 de. (a) Bloch’s theorem tells us ψ(x) = eikx u(x) with u periodic in the unit cell. it should be qE not k. an eigenstate of energy E is al. and using the Schroedinger equation to fix the disconti- between delta functions. (b) Using ψ(a+ ) = eika ψ(0) = Beika and ψ(a− ) = A sin(qE a) + B cos(qE a) gives the required result immediately.. Conclude that there where uE (x) is a periodic function defined as must be values of E for which there are no solutions of the uE (x) = A sin(qE x) + B cos(qE x) 0<x<a Schroedinger equation—hence concluding there are gaps in the spectrum. Then in the second to last equation there is an extra random factor of k in front of the brackets. boundary). there is no potential so the solution must be plane waves e±iqE x . AND the exponent is e−ika .132 Electrons in a Periodic Potential (a) Argue using the Schroedinger equation that in. The first part of the problem should ask you to show that ψ(x + a) = eika ψ(x) and it should be ψ that has the form A sin(qx) + B cos(qx) not u.e. Since the Hamiltonian is time reversal invariant we can choose the wavefunction to be real. so we expect an overall energy shift of U (from V0 ) and then a gap of magnitude 2|U | opening at the zone boundary as well. we have Z 1 1X VG = dxV (x)eiGx = Ua = U L L n Note this is true also at V0 = U . . 133 Using ψ(a) = Beika + ∂x ψ(a ) = eika qE A ∂x ψ(a− ) = qE [A cos(qE a) − B sin(qE a)] gives the desired result which now reads (after removing errors!) A − e−ika [A cos(qE a) − B sin(qE a)] = 2maU B/(qE ~2 ) The two equations (with errors removed) have the required solution. (c) First we note that for G = π/a. Setting k = π/a and qE = k + δ we have mU a −1 = cos(±π + δa) + sin(π + δ) ~2 (π/a + δ) Expanding for small δ and solving to lowest order we get two solutions 2mU a δ=0 or ~2 π Substituting back into E = ~2 (π/a + δ)2 /(2m) we obtain E = E0 or E0 + 2U giving the gap of 2|U | at the zone boundary as expected (and the mid- point of the gap shifted up by U due to V0 as well). . (Try to think up several (b) calcium. (c) C. which have similar cubic) unit cell. and atoms much closer together) therefore it is an insulator with a large band gap. Semiconductor. So calcium. one needs to con- sider single-electron orbitals. Before you bring the atoms together to make a solid. The higher orbitals in an atom are typically closer together in energy (as they are for hydrogen). Why is diamond transparent? (a) The conventional bcc unit cell contain 2 atoms. but the primitive unit cell contains only one atom. the periodic potential is extremely strong (no inner shell electrons to screen it.5 eV). Hence they are semiconductors. For C the valence electrons are in the 2p shell whereas for Si and Ge. structures.1 eV and . In a more chemical language. This could therefore constitute 2 completely filled bands.67 eV respectively). results in a half-filled 1st BZ – and thus gives a metal. as a monovalent atom. The fcc unit cell has 4 atoms. which is divalent. so the gap is smaller. the valence electrons are 3p and 4p respectively. whereas silicon and germanium. 4 electrons per atom can therefore form 4 covalent bonds with each of the 4 neighbors in the diamond lattice . Hence we expect smaller gaps for Si and Ge. could be either a metal or an insulator — depending on the strength of the periodic potential. is a metal. One argument is that for carbon. are semiconductors. is a metal. In the case of C (diamond) this is indeed an insulator with a large band gap (5.Si. or Metal 16 (16. The fact that it is a metal tells us that the potential is not strong enough to make it an insulator. is an electrical insula- (a) sodium. Another argument is given by thinking in the tight binding picture. and Ge are group IV elements and are therefore 4-valent. the band gap is smaller (1.1) Metals and Insulators (c) diamond. but the primitive unit cell has 1 atom. There are various explanations for this effect.Insulator. Thus sodium. which has four atoms in a fcc (conventional possible reasons!) cubic) unit cell. which has two atoms in a bcc (conventional tor. (b) The conventional fcc unit cell contains 4 atoms. For Si and Ge. but the primitive unit cell contains only a single atom. the potential is less strong. For Si and Ge. and therefore 4 electrons. which has eight atoms in a fcc (conven- Explain the following: tional cubic) unit cell with a basis. hence they are semiconductors. Write an expression for the dispersion of the elec- (b) Now suppose the lattice is not square. (b) If ax > ay one expects the hopping magnitude to be smaller in the x direction since the atoms are further apart (although this is not holy. Thus light cannot be absorbed (or reflected) by any electronic transition. ky ) = −2t (cos(kx a) + cos(ky a)) A contour plot of this energy is given in the left of Fig. may not be isotropic).136 Insulator. (16. 16. We then expect a dispersion of the form ǫ(kx . with ty > tx .2) Fermi Surface Shapes ax > ay . describe ax > ay ?) the shape of the Fermi surface. where is the shape of the Fermi surface now? (a) The dispersion of the tight binding model is given by ǫ(kx . If these atoms are monovalent.1 Left: Dispersion of a 2D tight binding model on a square lattice. a value −tx in the x-direction and a value −ty in the y- dimensional) square lattice where each atom has a single direction. If we are considering a monovalent unit cell. tronic states ǫ(k). (Why does this inequality match atomic orbital. This then gives a Fermi surface in the shape of a diamond. as shown in the right of Fig. 16. ky ) = −2tx cos(kx ax ) − 2ty cos(ky ay ) . as the orbitals. 3 3 2 2 1 1 0 0 -1 -1 -2 -2 -3 -3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Fig. In this case. or Metal structure. such as px orbitals. making the material an insulator.1. Right: the Fermi surface for one electron per unit cell. The electrons are all therefore tied up in covalent bonds. then the Brillouin zone is half filled. what ax and ay in the x and y directions respectively. Semiconductor. angular instead with primitive lattice vectors of length Suppose again that the atoms are monovalent. 16. but is rect. Diamond is transparent because it is an insulator with band gap that is larger than any visible light frequency. imagine that the hoppings have (a) Consider a tight binding model of atoms on a (two.1. (16. A contour plot of the energy is given in the left of Fig. This then gives a Fermi surface in the shape of . In the absence the first Brillouin zone remain empty? This is some nasty geometry. But it is shown in the right of Fig.2. then the Brillouin zone is half filled. let us choose ty = 2tx but ax = ay = a for simplicity.1).0. so the primitive unit cell (or the Bril- louin zone in this case) has volume 21 (4π/a)3 in k-space. that forms Brillouin zone boundary? What fraction of the states in an fcc lattice (with a single atom basis). 16. i’m not sure what to call it.3) More Fermi Surface Shapes* of a periodic potential. Right: the corresponding Fermi surface for one electron per unit cell. The BCC lattice has two lattice points per conventional unit cell. Now since we have a divalent unit cell in real space. If we are considering a monovalent unit cell. 3 3 2 2 1 1 0 0 -1 -1 -2 -2 -3 -3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Fig. For a BCC lattice. such as Ca or Sr..2 Left: Dispersion of a 2D tight binding model on a square lattice with anisotropic hoppings. Thus for electrons in the absence of a periodic potential we have 3 4πkF3 1 4π = ≈ 992/a3 3 2 a or kF = (24π 2 )1/3 /a ≈ 6. well.187/a We would like to know if this hits the Brillouin zone boundary or not. we should have enough electrons to fill exactly the volume of the Brillouin zone.2..0] is the point . 16. would the Fermi surface touch the Consider a divalent atom. the nearest neighbor of the point [0. recall that the reciprocal lattice of an FCC lattice with lattice constant a is a BCC lattice with lattice constant 4π/a (See excercise 13. First. 16. 137 As an example. 745/a.1/2] in units of the lattice √constant.441/6. We can check this by noting that the angle subtended by this spherical cap is only θ = cos−1 (5.441/a. Note that the center of this spherical cap is in the L direction (in the language of Brillouin zones. We should check that the spherical cap remains on the L-face.4/a3 3 However.6 of the book).187/a where the Brillouin zone boundary is of radius 5. note that we have 8 such spherical caps in all of the 8 equivalent L directions. the distance to this perpendicular bisector (i. thus telling us that the Fermi surface hits the Brillouin zone boundary (it is obvious that it must hit the Brillouin zone boundary since the volume of the fermi surface must equal the volume of the Brillouin zone and they are not the same shape!). The perpendicular bisector to this point is thus a distance 3/4. which is much smaller than the angle to say the K point.187/a − 5. Semiconductor. The height of the spherical cap is h = 6.e. The Fermi surface thus goes into the 2nd Brillouin zone as a spherical cap of radius 6. Thus the caps in the second Brillouin zone account for roughly 8% of the filled states. see fig 13.441/a which is less than the Fermi wavevector.187) ≈ 28. A well known geometric formula gives us that the volume of a cap is πh2 V = (3r − h) = 10.138 Insulator. or Metal [1/2. to the Brillouin zone boundary) is then √ dk = π 3/a ≈ 5. .441/a = 0. For our reciprocal lattice with lattice constant 4π/a.. thus giving a total volume of V = 83/a3 in the 2nd Brillouin zone.425 degrees.1/2. compared to the total volume of the fermi surface which is 992/a3 . since a filled band carries no (crystal) momentum. Note that it is typical to define the wavevector of a hole to be negative . and for electrons crystal momentum is always ~k. It is always true that the velocity of an electron in a state is ∇k Ek /~ where Ek is the electron energy. (this matches well to the intuition that p = mv with a positive effective mass for holes). calculate the moved from a state k = 2 × 108 m−1 x ˆ. where x ˆ is the unit current density and its sign.1 × 10−26 kg-m/s which is also in the negative x ˆ direction. This mass is positive in the usual convention. Getting the momentum and velocity right are tricky. what is meant by a hole the sign of!) and why is it useful? (i) the effective mass (b) An electron near the top of the valence band in a (ii) the energy semiconductor has energy (iii) the momentum (iv) the velocity. E = −10−37 |k|2 If there is a density p = 105 m−3 of such holes all where E is in Joules and k is in m−1 .e the velocity is in the negative xˆ) direction. or about 0. So m∗ = 5 × 10−32 kg or . it is equivalent to describe the dynamics of just the (few) holes. Thus we obtain hole momentum −~k = −2. (a) A hole is the absence of an electron in an otherwise filled valence band. The energy is E = (10−37 Joule·meter2 ))k 2 = 4×10−21 J. With p the density of such holes.05 the mass of the electron. calculate (and give (a) In semiconductor physics. For a hole.Semiconductor Physics 17 (17. First.8 × 105 m/s (i. For momentum.025 eV.1) Holes vector in the x-direction. (b) Effective mass ~2 k 2 /(2m∗) = (10−37 Joule · meter2 )k 2 . note that the velocity of an eigenstate is the same whether or not the state is filled with an electron. An electron is re. the total current density is pev = −6 × 10−9 Amp/m2 also in the negative xˆ direction (noting that the charge of the hole is positive). This energy is positive (it takes energy to ”push” the hole down into the fermi sea. like pushing a balloon under water). Thus the hole velocity here is negative v = −~k/m∗ = −3. having almost exactly this same momentum. This is useful since instead of describing the dynamics of all the (many) electrons in the band. the removal of an electron leaves the band with net momentum −~k which we assign as the momentum of the hole. 1 shows the relationship be- (a) Assume that the band-gap energy Eg is much tween charge-carrier concentration for a certain n-doped greater than the temperature kB T .1 Figure for Exercise 17. the density of states in the conduction band and the den.) Estimate the conduction electron concentration for intrinsic (undoped) silicon at room temperature.75 eV respectively. Describe in detail an experimental method by sity of states in the valence band (through their effective which these data could have been measured. (17. and suggest masses). but we are just making a rough estimates here.140 Semiconductor Physics of the wavevector of the missing electron. You may need to use the integral 0 dx x1/2 e−x = √ π/2. possible sources of experimental error. (a) The density of states per unit volume of free electron with disper- sion E = ~2 k 2 /(2m) is given by (including spin) √ m3/2 g(E) = 2E 3 2 ~ π So if the dispersion near the valence band edge and conduction band edges are Ee (k) = Ec + ~2 k 2 /(2me ) Eh = Ev − ~2 k 2 /(2mh ) we obtain density of states for conduction electrons and valence holes given by p 3/2 me ge (E > Ec ) = 2(E − Ec ) 3 2 ~ π p 3/2 m gh (E < Ev ) = 2(Ev − E) 3h 2 ~ π . conductors (c) The graph in Fig.1 eV and 0. roughly the same. (b) The band gaps of silicon and germanium are 1. You may assume the ef- fective masses for silicon and germanium are isotropic. and furthermore the effec- tive masses are both rather anisotropic. (Actually the effective masses are not quite the same. Derive expressions for n for p andR for the product ∞ np. germanium at room temperature.2) Law of Mass Action and Doping of Semi. 17. Make a rough estimate of the maximum concen- tration of ionized impurities that will still allow for this “intrinsic” behavior. and are roughly . Show that in a pure semiconductor. Estimate the conduction electron concentration for Fig. and on the band-gap energy.5 of the bare electron mass for both electrons and holes. 17. semiconductor at a fixed T . the product of the number of Estimate the band gap for the semiconductor and electrons (n) and the number of holes (p) depends only on the concentration of donor ions. 2) 4 π~2 Obtaining √ 3 1 2 mh me kb T np = e−β(Ec −Ev ) 16 π~2 which depends only on the band gap. one obtains 3/2 1 2me kb T n= e−β(Ec −µ) (17. that when we multiply np the variable µ completely vanishes. Assuming that µ is well below the conduction band (by at least energy kb T ). thus we obtain. and we are assuming mh = me = m/2 as well so we have 3/2 1 2(m/2)kb T n= e−β(Eg /2) 4 π~2 .1) 4 π~2 and similarly 3/2 1 2mh kb T p= e−β(µ−Ev ) (17. A bit more manipulation obtains √ 3/2 Z ∞ eβµ 2me p n= 3 2 (E − Ec ) e−βE dE ~ π Ec redefining variables y = E − Ec and performing the integral. 141 At fixed chemical potential µ and temperature β the number density of electrons in the conduction band is Z ∞ n= ge (E)nF (β(E − µ))dE Ec where nF (x) = 1/(ex + 1) is the Fermi occupation factor. T and the effective masses. the number of holes in valence band is given by Z Ev p= gh (E)(1 − nF (β(E − µ))dE −∞ Assuming that µ is well above the valence band (by at least energy kb T ) then x = β(E − µ) is very negative and we can replace the fermi factor 1 − nF (x) by ex resulting in Z Ev p= gh (E)eβ(E−µ) dE −∞ It is then clear immediately. (b) In the undoped case n = p. Z ∞ n= ge (E)e−β(E−µ) dE Ec Similarly. then x = β(E − µ) is very positive and it is acceptable to replace nF (x) by the Boltzmann factor e−x . (c) The concentration of donor atoms is simply the saturation concen- tration at low T (which I estimate from the figure to be about 2 × 1019 m−3 ). we need to measure the slope of the curve at high temperature.26 × 1015 m−3 for Silicon and n = 4. one needs to align contacts exactly parallel to each other in a hall bar. Several possible sources of error can occur here. To measure hall resistivity. then you measure some (nontriv- ial) combination of the Hall resistivities weighted by their concentrations and by their mobilities (in a very nontrivial way).54 × 1018 m−3 for Germanium. It might be useful to also mention to the students that Si and Ge both have valley degeneracies (i.9). Re ρ2e + Rh ρ2h RH = (ρe + ρh )2 (See exercise 17. To estimate the gap. then you no longer have intrinsic behavior. Note that at very low temperature one could get carrier freeze-out where the density drops again. One has to also make sure that the voltages being measured are due to the sample and not the contacts/wires/amplifiers etc. Plugging in numbers gives n = 5. when there are both electrons and holes present. multiple equi-energy minima in the conduction band). without mixing in longitudinal resisitivity. The concentration is most likely measured by a Hall effect measure- ment.142 Semiconductor Physics with Eg the gap value. First. These may add an additional factor to the law of mass action. There are other more obvious sources of experimental error such as heating when one runs current through a sample to measure it – thus it requires measuring small voltages accurately. One only gets an accurate absolute measurement of the electron concentration to the extent that the electron resistivity is much lower than the hole resistivity. Or equivalently n ∼ e−1500K/T We then set 1500K = Eg /(2kb ) and obtain a band gap of about . Extracting a slope log n/(m3 ) = −1500K/T + Constant.e. . When the doping level gets to on the order of the expected intrinsic level.26 eV. For simplicity let us assume me = mh .1 and Eq. the chemical potential is essentially right at the conduction (donor) or valence (acceptor) band and moves towards mid- gap as the temperature is increased. (Incidentally.3) Chemical Potential (c) A direct-gap semiconductor is doped to produce a (a) Show that the chemical potential in an intrinsic density of 1023 electrons/m3 . 17.. the thermally excited electrons are irrelevant compared to those that are there from doping). If the doner density is much higher than nint then n ≈ ndopant (i. 17. 17.0 eV. perature if the semiconductor is doped with (i) donors (ii) Hint: use the result of Exercise 17. When the intrinsic density exceeds .2. Then looking at Eq.a. so the doping can be thought of as going directly into the conduction band.25 and 0.2 for n and p respectively. (a) In an intrinsic semiconductor n = p so we can set n =1 p Referrring back to the previous problem we can insert the expressions Eq.1 fixing n to be ndopant and solving for µ we have ndopant µ ≈ Ec − kB T log 1 2me k T 3/2 b 4 π~2 Similarly for acceptor impurities pacceptor µ ≈ Ev + kB T log 1 2m k T 3/2 h b 4 π~2 So at low tempearture. so the intrinsic behavior is then simply that the chemical potential is fixed as a func- tion of temperature. 143 (17. and perature. Calculate the hole density semiconductor lies in the middle of the gap at low tem. this is why it is never a good idea to say that the fermi energy is the energy of the highest filled state.4 electron masses respectively. acceptors. Almost all of the nasty prefactors cancel and we obtains 3/2 me e−β(Ec −µ) 3/2 =1 mh e−β(µ−Ev ) We can solve this trivially to obtain Ec + Ev 3 µ= + (kB T ) log(mh /me ) 2 4 So at low temperature the chemical potential lies mid-gap. at room temperature given that the gap is 1. lence band are 0.e. the effective mass of carriers in the conduction and va- (b) Explain how the chemical potential varies with tem. (b) Let us assume for a moment we are well above the freezeout tem- perature. There may be a very large difference between the highest filled state and the chemical potential!). 4) Energy Density where n is the density of these electrons and ǫc is the Show that the energy density of electrons in the valence energy of the bottom of the conduction band. One can. at zero temperature the chemical potential is midway between the top of the impurity band and the bottom of the conduction band. since the real valence band is very far away compared to the temperature so we can ignore it. which is roughly to have the chemical potential mid-gap with a small slope dependent on the ratio of masses. then one expects to have µ given by the intrinsic expression from part a. I obtained nintrinsic = 1016 m−3 Then since ndopant ≫ nintrinsic we can set p = n2intrinsic /n (from the law of mass action) to obtain p = 109 m−3 (17. Now the story is a bit more complicated if one wants to think about the very low temperature regime where there is carrier freezeout. (c) We first calculate the undoped intrinsic carrier concentration using the law of mass action with n = p = nintrinsic . the chemical potential moves up towards the conduction band. As is usually the case. As the temperature is increased.2) Z ∞ E/V = Ege (E)nF (β(E − µ))dE Ec where nF (x) = 1/(ex + 1) is the Fermi occupation factor. as the temperature is raised.144 Semiconductor Physics the dopant density. do the calculation more rigorously writing the total energy density as (compare problem 17. At T =293 Kelvin. band of a semiconductor is 3 (ǫc + kB T ) n 2 In short this is just a matter of realizing that the electrons in the conduction band are essentially classical (are activated with Boltzmann factors not fermi factors). It is then only a matter of figuring out how the chemical potential moves between the filled impurity band and the empty conduction band. then an extra 32 kB T for the translational degrees of freedom as usual in equipartition theorem. At zero temperature all of the electrons are bound to their dopant nuclei and one can think of this as being a filled “impurity band” playing the role of a filled valence band. One obtains ǫc for each particle excited. of course. As in 17.2 it . and the nearby conduction band is empty. so classical stat mech applies and one can ap- ply the equipartition theorem. Since the density of states in the conduction band is larger than the those in the sparse impurity band. 145 is acceptable to replace nF (x) by the Boltzmann factor to get Z ∞ E/V = Ege (E)e−β(E−µ) dE Ec The same manipulation obtains √ 3/2 Z ∞ eβ(µ−Ec ) 2me p E/V = 3 2 [(E − Ec ) + Ec ] (E − Ec ) e−β(E−Ec ) dE ~ π Ec Note the second term in brackets. Explain the ducting crystal? significance of the distinction between a direct and an Optical absorbtion can occur when a photon is absorbed while exciting an electron out of the valence band into the conduction band. In the absorbtion process energy and momentum must both be con- served. (iii) band gap may be measured optically. whixh we recognize as being precisely ∂/∂β of the prior integral from 17. This requires a minimum of the gap energy (Small amounts of absorbtion can occur below the gap for impure semiconductors if there are impurity or defect states within the gap – also very weak nonlinear processes can allow multiple photons to be absorbed while exciting a single electron).ii) Sign of majority carrier and carrier concentration (assuming there is only one type of carrier) are both easily measured with Hall effect. the Ec term (compare 17. type is dominant) (iii) band gap (iv) effective mass (v) erties of a semiconductor sample: (i) sign of the majority mobility of the majority carrier. (17. (iv) Effective mass is measured with cyclotron resonance (v) mobility is easily measured via resistivity once concentration of carriers is known. (17.2) gives the same integral as the above calcualtion of n in 17. To evaluate the remaining term we redefine variables y = E − Ec and performing the integral. Since photons carry very little momentum given a certain energy .2 so we obtain nEc . (i. Since the prior integral was proportional to T ( 3/2) we obtain (3/2)(kB T ) times the prior integral and thus a total of (3/2)(kB T )n proving the result.6) More Semiconductors indirect semiconductor. Or by carrier concentration (es- sentially conductance) as a function of temperature.2.5) Semiconductors carrier (ii) carrier concentration (assume that one carrier Describe experiments to determine the following prop. What region of the optical spec- Outline the absorption properties of a semiconductor trum would be interesting to study for a typical semicon- and how these are related to the band gap. For what with n and p the densities of electrons in the conduction value of n − p is this conductivity achieved? Actually this is easy. Semiconductor gaps tend to be in the optical. This means that direct gap absorbtion (where the momentum of the electron does not change) is highly favored over indirect gap absorbtion. Using law of mass action np = n2intrinsic . (17. explaining why the energy levels are very close to ergy of electron states introduced by donor atoms into an the conduction band edge. dent of doping. Write down an expression for this Outline a model with which you could estimate the en. hence the bound states remain very close the conduction band edge.5 eV).7) Yet More Semiconductors n-type semiconductor. achieved is ity µe . energy. or roughly 3 eV to . the Rydberg becomes replaced by an effective Rydberg R∗ = R0 (m∗ /m)(1/ǫ2r ) This gives us a hydrogenic binding energy that can be extremely small for typical semiconductors. and ǫ0 is multiplied by the dielectric constant ǫr of the semiconductor.146 Semiconductor Physics (being that c is very large) one should think of this absorbtion as not tranferring any momentum to the system. indepen- Suppose holes in a particular semiconductor have mo. or infra-red range (some- where from 400 nm to 3 micron. One can consider a simple hydrogenic schroedinger equation with an attractive proton being the ionized donor and the single electron. (17. The total conductivity of the semiconductor will √ σ = 2e nintrinic µe µh be σ = e (n µe + p µh ) with nintrinsic the intrinsic carrier density.8) Maximum Conductivity* band and holes in the valence band. Only a very few wide gap semiconductors reach the optical blue range and UV. Indirect gap absorbtion can occur. but it must be assisted by a phonon or some other process that can account for the necessary momentum. Show that. As a result. The main differences are that the mass of the electron is replaced by the band electron mass. Thus we write nintrinsic σ = e n µe + µh n Now set dσ/dn = 0 to maximize and solve to obtain p n = nintrinsic µh/µe Which correspondingly results in p p = nintrinsic µe/µh . the maximum conductivity that can be bility µh and electrons in this semiconductor have mobil. We define tensors ρe and ρh for the two species with re = nµn and rh = pµh and Re = e/n and Rh = −e/p.1) r BR ρ= −BR r where r = nµ and R = q/n with q the charge on the charge carrier and n the carrier density. For a single species.9) Hall Effect with Both n.3c. I obtained B 2 (re Rh2 + rh Re2 ) + rh re (re + rh ) ρxx = B 2 (Re + Rh )2 + (re + rh )2 B B 2 Re Rh (Re + Rh ) + Rh re2 + Re rh2 ρxy = B 2 (Re + Rh )2 + (re + rh )2 . See also exercise 3. There is a lot of algebra involved in this. The conductivity tensors are σj = ρ−1j and then the total conductivity tensor is σ = σe + σi . 147 plugging into the original expression for σ with a tiny bit of algebra we obtain √ σ = 2e nintrinic µe µh as required.and p-Dopants* electrons in the conduction band with mobility µn . we have (See exercise 3. Use Suppose a semiconductor has a density p of holes in Drude theory to calculate the Hall resistivity of this sam- the valence band with mobility µh and a density n of ple. and we also obtain p p n − p = nintrinsic µh/µe − µe/µh (17. Finally this is inverted to give the tensor ρtotal = σ −1 . . 1) Semiconductor Quantum Well the electron. If one puts donor impurities outside of the well (on both sides. (a) A quantum well is formed from a layer of GaAs Sketch the shape of the potential for the electrons of thickness L nm.Semiconductor Devices 18 (18.45 me with me the mass of (b) *What might this structure be useful for? (a) This is a particle in a box problem. This is known as modulation doping. but the ionized dopants remain behind. One uses these structures heavily for fundamental physics studies of clean (unperturbed) electrons. (b) This type of quantum well device is useful to precisely design a band gap for example for a laser where one wants to fix the emission wavelength. gap of the quantum well is to be 0. 18.2). Both electron and holes are particles in a box of length L. Thus the lowest lying electron in the well is π 2 ~2 Ee = Ec + L 2m∗e with Ec the bulk conduction band minimum. surrounded by layers of Ga1−x Alx As and holes. (see Fig. It is useful because one can obtain extremely high mobility electrons within the quantum well since there are no ionized dopants in the well to scatter off of. You may assume that the band gap of the What approximate value of L is required if the band Ga1−x Alx As is substantially larger than that of GaAs. Similarly the highest lying hole state in the well is π 2 ~2 Eh = Ev − L 2m∗h Thus the difference in energy is ~2 π 2 1 1 Ee − Eh = Ebulkgap + − 2 L m∗e m∗h Setting Ee − Eh − Ebulkgap to 1eV and solving for L gives 8 nm.068 me whereas of GaAs bulk material? the hole effective mass is 0.1 eV larger than that The electron effective mass in GaAs is 0. say) the donated electrons can reduce their energies by falling into the well. . In our quantum well we must make a few minor changes. Thus for an electron in the conduction band of the quantum well we have ~2 π 2 a2 ~2 k2 E = Ec + + 2m∗e L2 2m∗e where Ec is the bulk conduction band bottom. . this quantum wire make a very good laser? (a) For a 2D electron gas for electrons with mass m. is the transverse quantum number of the particle in the well.) Describe the density of states for gas.2) Density of States for Quantum Wells (b) Consider a so-called “quantum wire” which is a (a) Consider a quantum well as described in the previ. First of all. Secondly the energy of the particle in the quantum well also includes its particle-in-a-box energy for its motion transverse to the 2D quantum well. 3. Why might box states. one-dimensional wire of GaAs embedded in surrounding ous exercise. Z kF dk N = 2A 0 (2π)2 with A the area and the factor of 2 out front for spin and k a two dimensional vector. Calculate the density of states for electrons AlGaAs. We then write more precisely that m g(E) = Θ(E) π~2 where Θ is the step function which has value 1 for nonnegative argument and value 0 for negative argument. and a = 1. we should use the effective mass rather than the actual mass of the electron. Fixing the transverse quantum number a. the density of states would be m∗e ~2 π 2 a2 g(E) = Θ E − Ec − π~2 2m∗e L2 . we quickly cal- cualte the density of states. .150 Semiconductor Devices (18. . electrons or holes within the quantum wire. This can be converted to kF2 n = N/A = 2π When the energy of an electron is given by ~2 k2 E= 2m we have k 2 = 2mE/~2 and we then have a density of states per unit volume of dn m g= = dE π~2 This is the correct answer for any E ≥ 0 and the density of states is zero for any E < 0. but don’t forget that there are several particle-in-a. Hint: It is a 2D electron square with side 30nm. (You can consider the wire cross-section to be a and holes in the quantum well. 2. a >0 2m∗h L2 2m∗h L2 1 2 (18. In general the energy of an electron in the wire is then given by ~2 π 2 a21 + a22 ~2 k 2 E = Ec + + 2m∗e L2 2m∗e where a1 and a2 are the mode indices (integers greater than zero) in the two transverse directions. Z kF dk N = 2L −kF 2π with L the Length of the system and the factor of 2 out front for spin. Stating your assumptions.a >0 2m∗e L2 2m∗e L2 1 2 and similarly in the valence band p ∗ −1/2 2mh X ~2 π 2 a21 + a22 ~2 π 2 a21 + a22 g(E) = Ev − − E Θ Ev − − E π~ a . Adding the density of states associated with all of these modes we obtain √ ∗ X −1/2 2me ~2 π 2 a21 + a22 ~2 π 2 a21 + a22 g(E) = E − Ec − Θ E − Ec − π~ a . This can be converted to 2kF n = N/L = π √ using k = 2mE/~ we then obtain a density of states √ dn 2m −1/2 g(E) = = E Θ(E) dE π~ Now in the seminconductor quantum wire we must consider the trans- verse modes. 151 Now accounting for the fact that there may be many transverse modes we have m∗e X ~2 π 2 a2 g(E) = Θ E − Ec − π~2 a>0 2m∗e L2 and this expression remains true up to an energy where the transverse modes spill out of the box. show that the Explain the origin of the depletion layer in an abrupt total width w of the depletion layer of a p-n junction is: p-n junction and discuss how the junction causes rectifca- w = wn + w p .3) p-n Junction* tion to occur. Analogously the density of holes in the valuence band in the quantum well is m∗ X ~2 π 2 a2 g(E) = h2 Θ Ev − − E π~ a>0 2m∗h L2 (b) First we determine the density of states for a one dimensional electron gas. while φ0 is the difference in and why might it be useful to use a diode as a capacitor potential across the p-n junction with no applied voltage. 1/2 2ǫr ǫ0 NA φ0 wn = eND (NA + ND ) and similarly 1/2 2ǫr ǫ0 ND φ0 wa = eNA (NA + ND ) . we immediately obtain −eND 2 φ(x) = x + CD x x<0 2ǫ0 ǫr eNA 2 φ(x) = x + CA x x>0 2ǫ0 ǫr where CD and CD are constants to be fixed here. Let the depletion widths be wn and wp respectively.2) Plugging Eq. We must solve the Poisson equation ∂x2 φ = ρ/(ǫ0 ǫr ) where ρ is the local charge density (which is constant in each region). and the doping densitities be ND and NA respectively. 18.1) 2ǫ0 ǫr We also note that the total charge in the two depletion regions must sum to zero (since the depletion occurs by annihilation of opposite charges) so we have wa NA = wn ND (18. This fixes the constants so that we now have −eND 2 φ(x) = x + 2wn x x<0 2ǫ0 ǫr eNA 2 φ(x) = x − 2wa x x>0 2ǫ0 ǫr The total potential drop across both regions is then e φ0 = φ(−wn ) − φ(wa ) = ND wn2 + NA wa2 (18.4 of the book for example). where we have an n doped region to the left (negative x) and a p doped region to the right (positive x). 18. permittivity and NA and ND are the acceptor and donor What is the differential capacitance of the diode densities per unit volume. We have additional boundary conditions that the electric field must go to zero at the edge of the depletion region. Setting φ(x = 0) = 0 for simplicity.152 Semiconductor Devices where 1/2 form of φ given the presence of the ion charges in the 2ǫr ǫ0 NA φ0 depletion region. in an electronic circuit? You will have to use Poisson’s equation to calculate the Let us set the position x to be zero.1 and solving for wn yields the desired expression.2 into Eq. so we have ∂x φ(x = −wn ) = ∂x (x = wa ) = 0. wn = eND (NA + ND ) Calculate the total depletion charge and infer how and a similar expression for wp Here ǫr is the relative this changes when an additional voltage V is applied. Within the depletion widths there is a net charge built up (see Fig 18. As in the p-n junction. The total depletion charge per unit cross sectional area is then 1/2 2ǫr ǫ0 ND NA (φ0 + eV ) q = wn ND = wa NA = e(NA + ND ) The differential capacitance per unit cross sectional area is 1/2 ∂q ǫr ǫ0 ND NA C= = (φ0 + eV )−1/2 ∂V 2e(NA + ND ) This provides a useful circuit element as it allows one to control a ca- pacitance by applying a DC voltage. Before the two semiconductors are brought together. The blue lines indicate the conduction band minima.2. This situation is shown in figure 18. nothing interesting happens in the valence band — so we only draw the conduction band.1 Before electrons are allowed to flow between the two seminconductors. The dotted line is the fermi energy (slighly below the conduc- tion band assuming that electrons are bound to dopants). It should say. there are electrons on the left at higher energy than empty states on the right. When equilibrium is established the electro- chemical potential is as shown in figure 18. it simply shifts φ0 to φ0 + eV . 153 When an additional voltage is added. The electrons are back . how this structure might result in a two-dimensional elec- conductor with minimum conduction band energy ǫc1 and tron gas at the interface between the two semiconductors. 18.4) Single Heterojunction* band energy ǫc2 where ǫc1 < ǫc2 . there are electrons in the conduction band on the left (or in dopant orbitals slightly below the conduction band as shown in the figure). electrons want to flow between the two semi- conductors in order to lower their energies. ǫc1 > ǫc2 . but empty states on the right at lower energy because the conduction band energy is lower. Describe qualitatively Consider an abrupt junction between an n-doped semi. (18. but in so doing charge builds up (a postive charge is left on the left and negative charged electrons accumulate on the right. an undoped semiconductor with minimum conduction Sketch the electrostatic potential as a function of position. In this problem.1 Fig. The situation is quite similar to the p-n junction. (18.154 Semiconductor Devices attracted to the positive charges they left behind. 18. Fig.5) Diode Circuit current) signal into a DC (direct current) signal. 18.2 Electrochemical potential of a semiconductor heterostructure. V = V0 sin(ωt). 18. thus accumulating in a roughly triangular well that forms near the interface. Fig. If the source voltage is sinusoidal.3 A half-wave rectifier. If the confine- ment is sufficiently strong. this becomes a particle in a box problem and the electrons in the well may become restricted to a single transverse wavefunction — thus becoming a strictly two dimensional electron gas. There is net postive charge on the left where electrons have left their donor atoms behind and net negative charge on the right where electrons have accumulated without positively charged nuclei. Design a circuit using diodes (and any other simple *Can you use this device to design a radio reciever? circuit elements you need) to convert an AC (alternating For the purpose of this problem we will make the crude assumption that a diode (arrow in a circuit diagram) is an ideal circuit element that allows current flow one way but no current flow the other way.3 is known as “half-wave rectifier”. The circuit shown in Fig. the voltage across the load . e. the voltage . 155 resistor is then V = V0 sin(ωt)Θ(sin(ωt)) where Θ is the Heaviside step function.4 A half-wave rectifier with a smoothing capac- itor. In order to smooth the nonnegative (but fluctuating) voltage into a smooth DC voltage one inserts a capac- itor into the circuit to act as a “battery” as shown in Fig.4. 18.5 A full-wave rectifier. One can do a bit better in circuit design by using more diodes to construct a “Full wave rectifier” as shown in Figure 18. Fig. Now instead of giving a true DC output. the current goes to zero instead of going negative. In this case. Fig. 18.. this gives an output that is nonnegative. in the absence of the smoothing capacitor. I.5. One should choose a capacitor such that Rload C ≫ 1/ω in which case the non-dc component of the resultant signal is of magnitude VAC ≈ V0 /(ωRC) . 18. This effectively damps the high frequency components of the signal leaving only the low frequency DC component. Here E1 is the “earphone” or output load of the circuit. we must first think about how radios encode information. If one wants to do a slightly bet- ter job.6 A radio circuit. This scheme has the advan- tage that the resulting DC voltage ends up being of higher magnitude than that of the half-wave. 18.156 Semiconductor Devices across the load is given by V = |V0 sin(ωt)| a smoothing capacitor can be used as above. In this figure the right hand side (D1.C3) attached directly to an incoming antenna and an outgoing earphone E1. In this case a signal A(t) (for simplicity let us assume the signal is everywhere positive) is multiplied by a high frequency carrier wave sin(ωt) and the product A(t) sin(ωt) is transmitted and then re- ceived by an antenna. Note that the capacitors are made tunable here so that the resonance frequencies can be modified. To convert the received signal back into A(t) one simply puts it through a rectifier. The middle of the circuit L2. which is inductively coupled to the incoming signal from the antenna. In order to design a radio. one wants to recieve only signals where the carrier frequency is very close to some given frequency ω0 . one needs to build a resonant (LC) circuit which will respond only to frequencies near its resonance. Fig. and the fluctuating DC component ends up being smaller.C3.E1) is just a half-wave rectifier as discussed above. A very crude radio could be made by with only the rectifier (D1. Many other designs are possible as well. The simplest encoding system is AM or amplitude modulation.6.C2 is the resonant LC circuit. To do this. The incoming antenna is on the far left. Here E1 is the “earphone” or output load of the circuit. . 18. A sample 1-diode radio circuit (often known as a “crystal” radio after the fact that the diode was often made of a small crystal) is shown in Fig. e.) which can act off by itself). 157 (18. 18. To put the circuit back in the “Off” state. when it is turned Design a circuit made of one n-MOSFET and one p. and when it is turned off it stays MOSFET (and some voltage sources etc. then this activates the P-MOSFET. allowing current to flow to the gate of the N-MOSFET.. When the switch (S1) is closed no current flows.7. 18. . Thus the circuit is latched in the “On” state. on it stays on by itself. so both transistors are “off” meaning they prevent current flow (and therefore prevent voltage from being transmitted). allowing current to flow to the gate of the P-MOSFET even when the ON voltage is removed. The reason for this is that there is no voltage on either of the two gates. Fig. one must disconnect the switch S1 momentarily.7 A CMOS latch circuit. which then activates it. However.6) CMOS Circuit* and can act a single bit memory (i. when a voltage is momentarily applied to the ON input. as a latch—meaning that it is stable in two possible states A simple (and very rough) CMOS latch circuit (also known as an SCR or silicon-controlled rectified) is shown in Fig. . 4s. If an atom has a net moment J 6= 0. 3s. Pauli para- magnetism in a metal is much weaker than Curie paramagnetism and can be roughly the same size as diamagnetic effects. paramagnetic atoms have net moment J 6= 0 which can be re-aligned. then it is paramagnetic. (iv) Dysprosium (Dy). 3d. 2p. and the atom is diamagnetic due to Larmor Diamagnetism. Both para and dia terms are weak and either one can win in this case. However (more advanced answer): it is also possible to have J = 0 while having L and S nonzero. with s.1) ‡ Atomic Physics and Magnetism termine L. 5s. . In this case. and J for the following isolated atoms: (a) Explain qualitatively why some atoms are param. 3p. S. . 4p. atomic number = 40 (b) Use Hund’s rules and the Aufbau principle to de.p. This is almost all other atoms.6. 2s. atomic number = 66 (a) To a first approximation.10 electrons respectively. from unfilled shells. (b) Shells are filled in the order 1s.Magnetic Properties of Atoms: Para. atomic number = 23 electronic structure of these materials. If an atom has completely filled shells (say. (i) 16 S: [Ne]3s2 3p4 (ii) 23 V: [Ar]4s2 3d3 (iii) 40 Zr: [Kr]5s24d2 . (i) Sulfur (S) atomic number = 16 agnetic and others are diamagnetic with reference to the (ii) Vanadium (V). then J = L = S = 0. the atom can either be para- magnetic or diamagnetic.and Dia-Magnetism 19 (19. One can also have Landau dia- magnetism (beyond the scope of the course) which is the diamagnetic response of the orbital motion of the conduction electrons. the noble gases). (Known as Van Vleck Paramagnetism). 4d . This can occur for atoms that are one electron short of half-filled shells. whereas the typical diamagnetic atoms have no moment. In metals one can have Pauli paramagnetism associated with re-orientation of the spins of the conduction electron.d shells containing 2. (iii) Zirconium (Zr). (19. Note that none of these atoms violate the Madelung rule which dic- tates the filling order or violates Hund’s rules when the atoms are iso- lated. Calculate the Land´e g-factor has eleven f electrons on it. is purely spin. (a) For 11 electrons in an f-shell. and since the shell is more than half full J = L + S = 2. Since the shell is less than half filled J = L − S = 3/2. L = 0 and S = 7/2 and J = S. these fill lz = 2. Using the formula for the Lande g factor (with g = 2) we have 1 1 S(S + 1) − L(L + 1) g˜ = (g + 1) + (g − 1) = 6/5 2 2 J(J + 1) This is almost exactly right. 1. so J = L = S = 0. atom. 2. (v) Dysprosium has 10 electrons in an f-shell. Since the shell is more than half filled J = L + S = 8.160 Magnetic Properties of Atoms: Para. Er typically is in a +3 state. 1 with spin up and lz = 1 with spin down. [Ar] con- tains 18. 0. these fill lz = 2. J is given by the sum J = 6 + 3/2 = 15/2. one electron tron from each atom forms a delocalized band so each Eu from each atom forms a delocalized band so each Er atom atom has seven f electrons. (iv) Xenon is a noble gas. these fill all the spin up state (7 of them) and lz = 3. so g˜ = g = 2 (it also comes out of the above formula as . these fill lz = −1. There are 7 electrons before losing one. Since the shell is less than half filled J = L − S = 2. Thus L = 1 and S = 1. Since the shell is more than half filled.2) More Atomic Physics (b) In solid europium (atomic number =63). For 7 electrons in an f-shell. (ii) Vanadium has 3 electrons in a d-shell. Calculate the Land´e g-factor for the seven electrons (the localized moment) on the Eu for the eleven electrons (the localized moment) on the Er atom.and Dia-Magnetism (iv) 54 Xe: [Xe] (All filled shells) (v) 66 Dy: [Xe]6s2 4f 10 Where filled shell configurations [Ne] contains 10 electrons. (Violations do sometimes occur but these atoms work as they are supposed to). using Hund’s first rule we obtain S = 3/2 and Hund’s second rule we have L = 6. (b) The counting here is messed up. 1 is from the f orbital. (iii) Zirconium has 2 electrons in a d-shell. 0 with spin up giving L = 3 and S = 3/2. one elec- (a) In solid erbium (atomic number=68). Thus (i) Sulfer has 4 electrons in a p-shell. [Kr] contains 36 and [Xe] contains 54. 2 of those are from the core s-orbitals. 1 with spin up giving L = 3 and S = 1. meaning all shells are filled. 1 for spin down giving L = 6 and S = 2. and J as mentum l (l = 0 means an s-shell. l) = x x=l−n+1 We can do this sum to obtain 1 L(n. and we get L = 3 = S and J = 0. n electrons. l) = 1 2 (n − 2l)(4l + 2 − n) 2l + 1 ≤ n ≤ 4l + 2 ‡ (19. (19.). the man- . l) = n(2l + 1 − n) 2 For 2l + 1 ≤ n ≤ 4l + 2 we can consider only the electrons in addition to the L = 0 half-filled shell. with an l shell having 2l + 1 orbital states and two The shell has 2l +1 orbital states and 2 spin states per orbital. we should have 6 electrons when one is lost. However. L. l) = L(n − (2l + 1). 1 2 n(2l + 1 − n) 0 ≤ n ≤ 2l + 1 L(n. etc.4) Para and Diamagnetism (a) Determine L. 161 well). Suppose this shell is filled with Suppose an atomic shell of an atom has angular mo. l) So we have in all. Determine the paramagnetic contribution to the vapor at 2000K with vapor pressure 105 Pa. l) = 1 2 (n − 2l − 1)(4l + 2 − n) 2l + 1 ≤ n ≤ 4l + 2 And thus we have (using Hund’s third rule) J = L − S for less than half filled and J = L + S for more than half filled so 1 2 n(2l − n) 0 ≤ n < 2l + 1 J(n. S. and J for an isolated manganese Manganese (Mn. l = 1 means a p-shell a function of l and n based on Hund’s rules. Derive a general formula for S.3) Hund’s Rules* spin states per orbital. You can (Curie) susceptibility of this gas at 2000K. (b) In addition to the Curie susceptibility. atomic number=25) forms an atomic atom. Hund’s first rule tells us that n 2 0 ≤ n ≤ 2l + 1 S(n. l) = 4l+2−n 2 2l + 1 ≤ n ≤ 4l + 2 The second rule tells us that for n ≤ 2l + 1 we have x=l X L(n. This is a van-vleck ion. In fact. consider this vapor to be an ideal gas. so L(n. so in fact the result is not what is predicted here. it is probably simplest to expand the partition function directly 5/2 X 1 Z ≈ 1 − (βgµB Bm) + (βgµB Bm)2 + . meaning a half filled d-shell. the density of Mn atoms is n = P/(kB T ). . ˚ngstrom. Since this is purely spin moment the g-factor is 2.162 Magnetic Properties of Atoms: Para. We will use this below. 2 m=−5/2 35 = 6+ (βgµB B)2 4 Calculating the moment 35 moment = −∂(kB T log Z)/∂B = β(gµB )2 B 12 Yielding a susceptibility of the gas 35 35 χ = ∂M/∂H = nµ0 β(gµB )2 = [P/(kB T )2 ] µ0 µ2B = 1. One can write in circular gauge 1 A= r×B 2 and the final term ends up being e2 e2 |B|2 (r⊥ )2 = |B|2 r2 8m 6m . and has zero expectation if there is a filled shell such that hLi = 0.and Dia-Magnetism ganese atom will also have some diamagnetic susceptibil.6 × 10−7 12 3 (b) Diamagnetism is understood from the schroedinger equation and expanding for small magnetic field (p + eA)2 p2 1 e2 A2 H= = + (p · A + Ap) + 2m 2m 2m 2m The middle term is a L · B. This then has L = 0 and S = 5/2. (a) Atomic Mn has orbital configuration 25 Mn: [Ar]4s2 3d5 . . Now we need to determine the paramagnetic susceptibility of a spin S = 5/2. Let us write the partition function 5/2 X Z= e−βgµB Bm m=−5/2 Since we will be concerned with small B. Determine the Larmor Make sure you know the derivations of all the formulas diamagnetism of the gas at 2000K. atomic radius of an Mn atom is one A ity due to its filled core orbitals. The final term is the source of diamagnetism. You may assume the you use! Using the ideal gas law. For an fcc crystal. the diamagnetism of the underlying silicon atoms? Si has (b) The wavefunction of an electron bound to an im.188 nm. At low temperature it the electron effective mass m∗ = 0.4 = 1. The radius of the avrgon atom is r√= . Since there are 20 core electrons per atom. 163 where we have used the spherical symmetry of the atom so that x2 +y 2 = (2/3)r2 . the nearest neighbor distance is r = a/(2 2) We then have four atoms √ per conventional unit cell for an overall atomic density of 4/a3 = 1/(4 2r3 ).09. Again using the result of the prior problem (and using m∗ rather than m) ρe2 2 χ = −µ0 hr i ≈ −10−11 6m∗ . which would correspond to hr2 i ≈ r/3.5) Diamagnetism the impurity contribution to the diamagnetic susceptibil- (a) Argon is a noble gas with atomic number 18 and ity of a Si crystal containing 1020 m−3 donors given that atomic radius of about . Measurement p gives us a suscpetability about 10−5 . (b) For a hydrogenic orbital (recall ψ ∼ e−r/a0 ) it is easy to calcualte that hr2 i = 3a20 . so we have (see 19. and density 2. We thus have the susceptibility ρe2 2 χ = −µ0 dE 2 /dB 2 = −µ0 hr i 6m where here ρ is the density of electrons. Estimate the magnetic susceptibility permittivity is ǫr = 12. How does this value compare to of solid argon.188 nm. atomic number 14. Using 1 angstrom for r we thus obtain χ = −4. we have ρ = 20n with n the gas density (using 25 here would also be sensible). Estimate g/cm3 .6nm.33 purity in n-type silicon is hydrogenic in form. atomic weight 28. We need only calculate the Bohr radius of an impurity in a silicon atom.4 for derivation) ρe2 2 e2 3 e2 χ = −µ0 hr i = −µ0 18 √ hr2 i = −µ0 √ = −1 × 10−4 6m (4 2r3 )6m 4 2 mr This number is actually a bit too big because the average hr2 i is smaller than this number since most of the electrons are further inside the atom than the full atomic radius.3 × 10−9 which is much much smaller than the paramagnetic contribution. Recalling that the expression for the Bohr radius is 4πǫ0 ~2 a0 = me2 we see that the Bohr radius in silicon should be rescaled by aSi ∗ 0 = a0 ǫr m/m = a0 ∗ 12/. (a) First we establish the density of Argon.4me and the relative forms an fcc crystal. even at 2000K! ‡(19. The number of electrons is 18 times the number of atoms. If we didn’t know the crystal structure.17 Angstrom.33g mol 6. (a) Calculate the magnetization as a function of B and “Monatomic atoms” ? What was I smoking when I wrote this? It should say monovalent. (Note: I think the table value is almost exactly this. Doh! I should have also specified the g-factor (as usual we can take it to be 2).09g mol m3 m3 Now we need to determine the atomic radius of silicon from the data given. The gas has density this gas due to the spins. ) ‡(19. We then have ρe2 2 χ = −µ0 hr i ≈ −4 × 10−6 6m This is much much greater than the diamagnetism of the few impurities. (a) Same calculation as we have done a million times. Determine the susceptibility. we might guess roughly that 1 r = n−1/3 2 which would be exact for a simple cubic lattice.164 Magnetic Properties of Atoms: Para. Consider a gas of monatomic atoms with spin S = 1/2 (b) Calculate the contribution to the specific heat of (and L = 0) in a magnetic field B. The nearest √ neighbor distance is from [0. function of µB B/kB T .02 × 1023 atoms (100cm)3 atoms n= = 5 × 1028 cm3 28. so the density is n = 8/a3 with a the lattice constant so a = 2n−1/3 . we know that Si is diamond structure with 8 atoms per conventional unit cell. a/4] or a distance of a 3/4. 1 1 Z = e+βgµB 2 B + e−βgµB B 2 F = −kB T log Z 1 1 m = −∂F/∂B = gµB tanh(βgµB B) 2 2 M = mn 1 χ = lim µ0 ∂M/∂B = µ0 n( gµB )2 β B→0 2 (b) Similarly. however. First let us calculate the density of atoms in the system 2.6) Paramagnetism T . 0] to [a/4. this is fortuitous as the table value is usually given in cgs which differs from the SI version by a factor of 4 pi. √0. so the atomic radius is a 3/8 = 41 n−1/3 = 1.and Dia-Magnetism Finally we would like to compare this to the diamagnetism of the pure silicon. a/4. However. Sketch this contribution as a n. we have 1 1 U = ∂ log Z/∂β = gµB B tanh(βgµB B) 2 2 1 1 2 2 1 C = ∂U/∂T = 2 (gµB B) sech (βgµB B) kB T 2 2 . Again the error is in the estimate of r2 . 7) Spin J Paramagnet* (b) Show that the susceptibility is given by Given the Hamiltonian for a system of non-interacting spin-J atoms g µB )2 J(J + 1) nµ0 (˜ χ= 3 kB T H = g˜µB B · J where n is the density of spins. −J + 2 . 165 (19. . J So the canonical partition function of the single spin is J X 2J X Z = e−βgµB Bm = eβ˜gµb BJ e−β˜gµB p m=−J p=0 1 − e−β˜gµB B(2J+1) eβ˜gµB B(2J+1)/2 − e−β˜gµB B(2J+1)/2 = eβ˜gµb BJ −β˜g µ B = 1−e B eβ˜gµB B/2 − e−β˜gµB B/2 sinh(β˜ g µB B(2J + 1)/2) = sinh(β˜ g µB B/2) we construct the free energy per spin F/N = −kB T log Z and then get the magnetic moment per spin ∂(F/N ) m=− = (˜ g µB /2) [(2J + 1) coth(β˜ g µB B(2J + 1)/2) − coth(β˜ g µB B/2)] ∂B and the total magnetizaton is then the moment per spin times the density of spins n. . . we want ∂M χ = lim µ0 B→0 ∂B To take this limit we take the argument of the coth to be small and we use the expansion 1 x lim coth x = + x→0 x 3 Note that the two 1/x terms cancel when subtracted leaving M ∼ n(˜ gµB /2) (2J + 1)2 β˜ gµB B/6 − β˜g µB B/6 So we have g µB )2 J(J + 1) nµ0 (˜ χ= 3 kB T as required. pression for part a!) (a) The eigenenergies of a single spin are Em = g˜µB Bm m = −J. (You can do this part (a)* Determine the magnetization as a function of B and of the exercise without having a complete closed-form ex- T. J − 1 . −J + 1 . . so M = n(˜ g µB /2) [(2J + 1) coth(β˜ g µB B(2J + 1)/2) − coth(β˜ g µB B/2)] (b) To obtain the susceptibility. Thus we propose G[J] = aJ 3 + bJ 2 + cJ Then we can also write the difference of two successive sums as just the new ad G[J + 1] − G[J] = 2(J + 1)2 which we multiply out to give a(3J 2 + 3J + 1) + b(2J + 1) + c = 2(J + 1)2 matching coefficients then gives the values of a. Further. . Finally we turn to derive Eqn. J X J X 1 Z = e−β˜gµB Bm = 1 − β˜ g µB B)2 m2 + . To do this. it must have a maximum power of J 3 . b.1. and the coefficient of the J 0 term must be zero since G[0] = 0.] 6 Using the free energy per spin is F/N = −kB T log Z.1) 3 m=−J I will derive this below. The third term is the only hard one. I claim that J X 1 G[J] = = (2J + 1)(J + 1)J (19. So we have 1 Z = (2J + 1)[1 + (β˜ g µB B)2 J(J + 1) + . The result of this sum must be some polynomial in J. g µB Bm + (β˜ 2 m=−J m=−J The first term in the sum gives (2J + 1) and the second term gives 0 by symmetry. since B is small we have 1 F/N = −kB T log(2J + 1) − g µB B)2 J(J + 1) (˜ 6kB T And the magnetic moment per site is ∂(F/N ) g µB )2 BJ(J + 1) (˜ m=− = ∂B 3kB T which multiplying by the density of spins n to give the magnetization. approximating it as an integral.166 Magnetic Properties of Atoms: Para. 19. note that we only concerned with small B so we can expand the partition function directly for small B. . . . but for now let us assume it is correct. 19.1.and Dia-Magnetism Note. c which proves Eq. . gives the same result as above. the question claims that part (b) may be achieved without com- pleting part (a). (b) Assuming an antiferromagnetic configuration. it generates other spin configurations. Although the intuition of alternating spins on alter- (a) For J > 0. for the case of a ferromagnet. Consider such a state of the system. Show that this is an eigenstate of the Hamiltonian (which is much more advanced. The system is in an energy eigenstate with energy E = −N gµB |B| − N zJS 2 /2 with z the number of neighbors of each site (=6 for a cubic lattice). . it becomes reasonable for sys- tuition tells us that the ground state of this Hamiltonian tems with large spins S.1) Ferromagnetic vs Antiferromagnetic (b) For J < 0. H=− J Si · Sj + gµB B · Si (20.1 and find its energy. 20. For smaller spins (like spin 1/2) should simply have all spins aligned. Consider such a one needs to consider so-called “quantum fluctuations” state. and Ferri-Magnetism 20 (20.ji Hamiltonian. the key here is to note that there are terms Si+ Sj− which do not vanish (where S + is ap- plied to a down spin.1) 2 i Show that the state in question is not an eigenstate of the hi. i. and for this exercise set B = 0. and S − is applied to an up spin). Hence this is not an eigenstate. this is 1X X not precisely true. one might expect that (at least for B = 0) the Consider the Heisenberg Hamiltonian state where spins on alternating sites point in opposite directions might be an eigenstate. This means that when the hamiltonian is applied to the proposed antiferromagnetic ground states. here). Antiferro-.Spontaneous Magnetic Order: Ferro-. students often need to be reminded of this! Maybe it is worth giving this as a hint!) (a) If each spin is aligned in the zˆ direction (it has Sz = 1). nating sites is not perfect..e. in. the case of an antiferromagnet on a cu- States bic lattice. so we will not do that Eq. It is useful here to recall that 1 Si · Sj = (Si+ Sj− + Si− Sj+ ) + Siz Sjz 2 (Indeed. Unfortunately. then the energy is −gµb B per spin and for each bond we have energy −JSi · Sj = −JSiz Sjz = −JS 2 since S + on the spins all give zero. but rather as being classical variables. and now contains one spin of each type A. Antiferro-. C.1. we can perturb around this . what does the (a) If the system consists of only three spins arranged ground state look like? Probably here I should have stated explicitly that all three spins have the same |S| (might be interesting to consider a case where they don’t all have the same spin!). Show that one obtains For the spin-S ferromagnet particularly for large S. particularly if S is large.1 with Si being a spin S all Si = zˆS (Assuming B is in the −ˆ z direction so the operator). we i~ ˆ H] = [O. (b) For a triangular lattice. dt j First recall the Heisenberg equations of motion for any operator where the sum is over sites j that neighbor i. C. B. and Ferri-Magnetism (20. or C in such a way that all triangles contain one site of type A. and one of type C. we must have S1 + S2 + S3 = 0 which implies that the spins are at 120 degree angles from each other – but can lie in any plane. Then (a) Derive equations of motion for the spins in the to consider excited states. In the ground state. (b) For an infinite triangular lattice. So choose three directions all at 120 degree angles from each other in any given plane.3) Spin Waves* Hamiltonian Eq.168 Spontaneous Magnetic Order: Ferro-. with J < 0. One can think of this as being now a crystal whose unit cell has three times the area of the original unit cell. each triangle must have three spins each at 120 degree angles from its neighbors. ˆ dO In the ferromagnetic case.2) Frustration in a triangle (as in Fig. (20. B.2) the ground state. 20. 20. our “classical” intuition is fairly good and we can use simple ! dSi X approximations to examine the excitation spectrum above ~ = Si × J Sj − gµb B (20. dt can treat the spins as not being operators. one of type B. Now we must assign each site on the triangular lattice one of the three values A. B. show that the ground state Consider the Heisenberg Hamiltonian as in Exercise 20 has each spin oriented 120◦ from its neighbor. and treat the spins as classical vectors. ground state has spins aligned in the zˆ direction). 20. (a) The Hamiltonian is H = J(S1 · S2 + S1 · S3 + S2 · S3 ) Since Si · Si = S 2 is a constant we can write H = (J/2)(S1 + S2 + S3 )2 + constant To minimize the energy. we can set with H the Hamiltonian (Eq.2). Call these three directions A. gµb (By Siy + Bz Siz ) − J (Sj Si + Sjz Siz )] dt j with the sum over j being over neighbors of i. . Expand the equations of motion (Eq. Siz = S − O((δS) /S) 2 What does this mean? Six = δSix Show that the dispersion curve for “spin-waves” of Siy = δSiy a ferromagnet is given by ~ω = |F (k)| where where we can assume δS x and δS y are small compared to S. δSix = Ax eiωt−ik·r How might these spin waves be detected in an ex- periment? δSiy = Ay eiωt−ik·r (c) Assume the external magnetic field is zero. We then have the Heisenberg equations dSix i~ = [Six . the spectrum you just derived.2) for small F (k) = gµb |B| perturbation to obtain equations of motion that are linear + JS(6 − 2[cos(kx a) + cos(ky a) + cos(kz a)]) in δSx and δSy (b) Further assume wavelike solutions where we assume a cubic lattice. Thus we conclude dSi X ~ = Si × J Sj − gµb B dt j Writing Si = zˆS + δSi we obtain dδSi X ~ = JS zˆ × δSj + δSi × (JZS zˆ + gµb |B|) dt j where Z is the number of neighbors of a site. (a) We need the angular momentum algebra [S x . H] dt The only terms that this does not commute with are those containing Siy and Siz . Given This ansatz should look very familiar from our prior con. 169 solution by writing Plug this form into your derived equations of motion. S y ] = iS z and cyclic permutations of this. show that the specific heat sideration of phonons. Thus we have dSix X y y i~ = [Six . Thus we obtain dSix X i~ = i (gµb )(By Siz − Bz Siy ) − J (Sjy Siz − Sjz Siy ) dt j and similar for the other two components of the spin. Show that Six and Siy are out of phase by π/2. 20. due to spin wave excitations is proportional to T 3/2 . . This is like scattering from phonons except that one uses (spin-polarized) neutrons in order to couple to the spin. . We rewrite this as Z ∞ V U = dEE 3/2 nB (E) 4π 2 (JSa2 )3/2 0 Z ∞ V (kb T )5/2 x5/2 = dx 4π 2 (JSa2 )3/2 0 ex − 1 which we differentiate to get the heat capacity Z ∞ 5 V (kb T )3/2 x5/2 C = dU/dT = kb dx 2 4π 2 (JSa2 )3/2 0 ex − 1 (one does not really need to carry all of the constants to see how it scales!).. and Ferri-Magnetism (b) Plugging in the wave ansatz (for a cubic lattice) we obtain i~ωAx = F (k)Ay i~ωAy = −F (k)Ax where F (k) = JS(6 − 2[cos(kx a) + cos(ky a) + cos(kz a)]) + gµb |B| This system of equations can be solved immediately to give dispersion ~ω = |F (k)| Spin waves are typically detected by inelastic neutron scattering. Antiferro-. we have Z Z ∞ dk V U =V 3 E(k)nB (E(k)) ≈ 2 dkk 2 E(k)nB (E(k)) BZ (2π) 2π 0 with nB the usual Bose factor and the approximation accurate for low temperatures. For those who are interested though. the energy spectrum is quadratic E = JSa2 k 2 We will also need k = (E/JSa2 )1/2 and dk = dE/(2(EJSa2 )1/2 ) Calculating the total energy stored in spin waves.170 Spontaneous Magnetic Order: Ferro-.1267 . (c) For small k = |k|. the integral can be eval- uated in the usual way Z ∞ ∞ Z ∞ X X∞ Z ∞ x5/2 5/2 −nx 1 dx x = dxx e = 7/2 dyy 5/2 e−y = Γ(7/2)ζ(7/2) 0 e − 1 n=1 0 n=1 n 0 √ where Γ(7/2) = 15 π/8 and ζ(7/2) ≈ 1. and hence small energy. . Note the obvious typo. when two spin 1/2’s are added they can form either a spin-0 singlet or a spin- 1 triplet (three Sz states). Now adding a spin-1/2 the spin-0 gives spin-1/2 and adding spin-1/2 to the spin-1 gives either spin-1/2 or spin-3/2. Thus the Hamiltonian has eigenstates 3J/4 for the singlet (one eigensate) and −J/4 for the triplet (three eigenstates). it should read 2S1 ·S2 = (S1 +S2 )2 −S1 2 −S2 2 . and spin 2 in one way. S2 = S(S + 1) = 15/4. (c) Now consider four spins forming a tetrahedron. 20. so the energy is −3J/4 (four eigenstates). The spin- 0 singlets (two eigenstates) have energy (−J/2)[0 − 4(3/4)] = 3J/2. Again assuming these spins are only two spins. (S1 + S2 )2 − S1 2 − S1 2 . so that S = 1/2. (a) H = −(J/2)[(S1 + S2 )2 − S21 − S22 ] Since (S)2 = S(S + 1) for spin-1/2 we have (S)2 = 3/4. We have 2 spin 0’s + 3 states for spin 1 in 3 ways + 5 states in spin 2. Again we should add up the total number of eigenstates to check that it is 24 = 16. So we have 2+9+5 = 16. When adding three spin- 1/2s. 171 (20. In the case that the three spins add to spin-1/2. calculate the en. (To see this. spin 1 in 3 ways. Hint: Write 2S1 · S2 = spectrum of the system.4) Small Heisenberg Models (b) Now consider three spins forming a triangle (as (a) Consider a Heisenberg model containing a chain of shown in Fig. we obtain energy 3J/4 (four eigenstates) whereas if the three spins add to spin-3/2. Further. Note that counting the total nbumber of eigenstates we should get 8 since each spin-1/2 has two possible Sz states. which gives a total of 8 possible Sz states. Hint: Use the same trick as in part (a)! H = −JS1 · S2 . think about adding the first two spin-1/2 to get spin-0 or spin- 1. Supposing these spins have S = 1/2. calculate the spectrum of the system. calculate the ergy spectrum of this system. (b) Similarly H = −(J/2)[(S1 + S2 + S3 )2 − S21 − S22 − S23 ] Here. So (S1 + S2 )2 takes the values 0 for the singlet or S(S + 1) = 2 for the S = 1 triplet. Each of the two possible spin-1/2’s can take two possible Sz states and the spin-3/2 can take 4 possible Sz states. Again assuming these spins are S = 1/2. we can obtain spin-1/2 in two ways and spin-3/2 in one way.2). again for spin 1/2 we have (S)2 = 3/4. The three spin-1 triplets (9 eigenstates) have energy (−J/2)[2 − 4(3/4)] = J/2 and the spin-2 fiveplets (5 eigenstates) have energy (−J/2)[2 × 3 − 4(3/4)] = −3J/2 . (c) Same story H = −(J/2)[(S1 + S2 + S3 + S4 )2 − S21 − S22 − S23 − S24 ] The sum of the four spins can give spin 0 in two ways. (b) The probability that a given R is in the +1 state is eβJ 1 P (R = +1) = βJ −βJ = −2βJ e +e e +1 Having R in the +1 state tells us that two consecutive spins are pointing in the same direction. and hence S = 1..σN (a) The first spin can be either in the spin-up or spin down state. Antiferro-. . σ1 .172 Spontaneous Magnetic Order: Ferro-. . To see this. rewrite P (M ) = exp[−(M − 1) log(e−2βJ + 1)] . so we leave σ1 as a variable to be summed over.. in zero magnetic field as Show that the free energy has no cusp or disconti- N−1 nuity at any temperature. and Ferri-Magnetism (20. We write the Hamiltonian for a chain of N spins evaluate the partition function. The partition function pointing in the same direction. calculate an expres- sion for the probability that M consecutive spins will be where each σi takes the value ±1.. small? You may assume N ≫ M .σ2 . The free energy is thus F = −kB T log Z = −kB T N log 2 − kb T N log cosh(βJ) which is a completely continuous function with no cusps or discontinu- ities at finite β. This prob- ability decays exponentially with M . The Hamitonian in terms of the R variables is N X −1 H = −J Ri i=1 So the partition function is Z = 2(e−βJ + eβJ )N −1 = 2(2 cosh(βJ))N −1 with the factor of 2 out front being the sum over the first spin.5) One-Dimensional Ising Model with B = 0 Using the transformation Ri = σi σi+1 rewrite the parti- (a) Consider the one-dimensional Ising model with spin tion function as a sum over the R variables. How does this probability can be written as decay with M for large M ? What happens as T becomes X Z= e−βH . N − 1 where each R can take the values ±1. i=1 (b) *At a given temperature T . and hence conclude that there X H = −J σi σi+1 is no phase transition in the one-dimensional Ising model. Looking at M − 1 consecutive values of R. the probability that all of these are +1 is then M−1 1 P (M ) = e−2βJ + 1 which would put M consecutive spins in the same direction. The remaining spins are defined by Ri for i = 1. . and the decay length becomes long.3) where T is a 2 by 2 matrix. so the decay length is e2βJ lattice sites and is exponentially long. We write the Hamiltonian (Eq. derive the magnetization.. given that the M th spin is in a particular state. can be written as where each σi takes the value ±1 and we have defined F/N ≈ −kB T log λ+ h = gµB B for simplicity of notation. Let us define a partial partition function for the first M where λ+ is the larger of the two eigenvalues of the matrix spins (the first M terms in the Hamiltonian sum Eq. Actually. ??) for a chain of Z(M. X PM and show that the susceptibility per spin is given by Z(M.−) e e−β(J−h) T = = e−βH(−. 20. σM ) = e−β i=1 Hi σ1 .. −1)..σM −1 Z(M − 1.+) −β(−J+h) e e−βH(+.−) e−β(J+h) e−β(−J−h) (b) The full partition function is then Z = vT N −1 v T where v = (1. (T is i=1 known as a “transfer matrix”). (a) The transfer matrix takes the form −βH(+.. (c) Assuming these two eigenvalues are not degenerate. (d) From this free energy. 173 so the decay length is 1/ log(e−2βJ + 1).3) T. it can still be brought to diagonal form via a Jordon decomposition T = JΛJ −1 where Λ is a diagonal matrix of the eigenvalues of T and J is a matrix made of the eigenvectors of T (these are no longer orthonormal!). 1). we then have for large N that Z ∼ λN+ −1 .. I. σM ) = TσM .σM −1 χ ∝ βe2βJ so that the full partition function is Z = Z(N. in the large N Hi = −Jσi σi−1 + hσi for i > 1 limit. (b) Write the full partition function in terms of the where matrix T raised to the (N − 1)th power. Although T is not hermitian. +1) + Z(N.6) One-Dimensional Ising Model with B 6= 0* (a) Show that these partial partition functions satisfy a recursion relation Consider the one-dimensional Ising model with spin X S = 1. In the low tempearture limit P (R + 1) becomes close to unity. H1 = hσ1 (c) Show that the free energy per spin.e. we can see that for very large β we have log(e−2βJ +1) ≈ e−2βJ .+) e−βH(−. which matches the Curie form at high T . and find the matrix T . (20. σM −1 ) N spins in magnetic field B as σM −1 N X H= Hi (20. In the end we will obtain ∂ 2 log λ+ χ = g 2 µ2B µ0 kb T = g 2 µ2B µ0 βe−2βJ ∂h2 as claimed. we want the susceptibility per spin χ = µ0 ∂m/∂B which one evaluates at B = 0. the value of the larger eigenvalue is e−β(h+J) p λ+ = 1 + e2βh + 1 − 2e2βh + e4βh + 4e4β(h+J) 2 Just to check that this agrees with the previous problem. . It is much easier to make expansions for small h since this is all we will need. Thus the free energy per site is F/N ∼ −kB T log λ+ For the record. and Ferri-Magnetism where λ+ is the larger of the two eigenvalues. note that setting h = 0 we correctly obtain λ+ = e−βJ + eβJ (d) Probably we want to derive the magnetic moment per spin which is ∂(F/N ) ∂ log λ+ m=− = gµB kb T ∂B ∂h The result is a bit of a mess.174 Spontaneous Magnetic Order: Ferro-. Antiferro-. Then once we have this. 1) Domain Walls and Geometry (d) Suppose the spins in this material are arranged in Suppose a ferromagnet is made up of a density ρ of a cubic lattice. In zero external paring this energy to the magnetic energy. calculate the magnetic energy of walls? this ferromagnet. This puts two magnetic charges of the same . neighbors is denoted J and the anisotropy energy is very (a) Suppose a piece of this material forms a long cir. a common magnetic material. For magnetite. The analogy here is thus A/(dµ0 ).Domains and Hysteresis 21 (21. Estimate where might it go to minimize the magnetic energy in the the width of the domain wall and its energy per unit area. on the other hand is analogous to the energy of a capacitor.33 × 10−11 J/m and the (c) If a domain wall is introduced into the material. anisotropy energy is κS 2 /a3 = 1.. How much energy does the domain wall cost? Com- cular rod of radius r and length L ≫ r.e. the r ≫ L.35×104 J/m3 . so qm correctly has dimension of a charge. (Hint: a volume of aligned magnetic (e) Note that factors of the lattice constant a are often dipoles is equivalent to a density of magnetic monopoles introduced in quoting exchange and anisotropy energies on its surface. use! (a) Following the hint. what should magnetic field. The energy stored in a capacitor is q 2 /(2C) where electrically the capacitance is ǫ0 A/d with A the area and d the spacing. let us consider a domain wall that cuts L in half (i. Thus the energy is µ0 (µB ρ(πr2 ))2 4π L (b) This problem. two different geometries. large. Thus the total energy stored is 2 µ0 (µB ρ)2 (πr2 )L E = qm µ0 d/(2A) = 2 (c) First. the magnetic energy between two monopoles ′ of charge qm and qm separated by distance r is given by ′ µ0 qm qm E= 4π r Here we have charge of magnitude qm = µB Aρ where A = πr2 is the area of the end. a plane parallel to the A surface). recall that a dipole is a charge times a length. and the exchange energy between nearest spins each with moment µB . What is the magnetic energy now? exchange energy is JS 2 /a = 1. Estimate how much magnetic Make sure you know the derivation of any formulas you energy is saved by the introduction of the domain wall. if all of the moments are aligned along the we conclude about which samples should have domain L-direction of the rod. Check that this has the right dimensions.) to make them into energies per unit length or unit vol- (b) Suppose now the material is shaped such that ume. 176 Domains and Hysteresis sign right next to each other and is therefore energetically unfavorable.1. for large enough L. here we will not be able to consider the system “dipolar” until the size of the domains d is smaller than L. the energy will again drop as in the (a) case proportional to d6 /L3 (d) Since the lattice spacing will be a = ρ−1/3 . This completely kills the leading energy cost and now we have instead some sort of dipolar interaction between the two ends — with energy typically on the much smaller order µ0 (µB ρ(πr2 /2))2 r2 E= 4π L3 Similarly with the capacitor configuration of part (b). Now let us consider the (b) case. E/V = E0 − |M ||B| cos θ − κ′ |M |2 (cos θ)2 (d) Can you make an estimate (in terms of actual num- show that for |B| < Bcrit there is a local energy minimum bers) of how big this activation barrier would be for a fer- where the magnetization points opposite the applied field. we can convert “charge” energy to “dipole” energy by placing a domain wall in the diraction along the axis (in the direction of L). Now consider a domain wall perpendicular to this plane — i. The energy per unit area is then Jκ = 4 × 107 Joules per a2 area. the width of a domain wall should be J/κ√which here is 3100 lattice spacings. In this case we have an energy . The number of domains is πr2 /d2 and the total aread of the walls will be πr2 L/d and their total domain wall en- ergy will be πr2 LJρ2/3 /d. romagnetic crystal of magnetite that is a sphere of radius and find Bcrit . instead of having a single charge at each end of the rod. that runs along the L axis. (a) Given that the energy of a crystallite in a magnetic (c) For small B. the energy of intro- ducing a domain wall of area A is AJρ2/3 = 2LrJρ2/3 . 1 nm ? You may use the parameters given in Exercise (b)* In part (a) we have assumed B is aligned with the 31. In the (a) case. Thus.e. Once we do this. the magnetic energy is not very large to begin with. Let us consider introducing a domain wall network spaced by d. there will be some happy medium which optimizes the total energy. (21.2) Critical Field for Crystallite can occur if these directions are not aligned. The function E(z) is an upside-down parabola with maxium at z = −B/(2κ′ |M |) < 0. so introducing an energy cost proportional to L is not a very good idea..e (you may need to estimate some other parameters anisotropy direction of the magnetization. However. (a) Let cos θ = z. In the (a) case we can view this as. The absolute minium is always at θ = 0. now we have two opposite charges along each end of the rod (each with charge ±qm /2). Describe what as well). As a function of d. one at θ = 0 and the other at θ = π. as discussed in part (c) then magnetic energy drops strongly when d < L. If this parameter is z > −1 then there are two minima. roughly how large (in energy per unit field is given by volume) is the activation barrier for the system to get from the local minimum to the global minimum. p(e) As mentioned in the text. (b) Let us consider B again pointing in the zˆ direction and the anisotropy direction oriented at an angle φ from zˆ. we should instead write κS 2 (cos θi )2 and then sum over all spins i. These differ only by a constant. First write b = 2(1/a) Jκ so that we have θ(x) = 2 tan−1 (ebx ) . write θ = θ0 + δθ and expand to linear order in δθ. (c) Consider small B.1. then the minimum will always be close to 0. (21. and replace the tion r −1 √ κ sum over sites with an integral.1 of the en. In this case. To is minimized when be more precise. but in the former case. 177 functional E/V = E0 − |M ||B| cos θ − κ′ |M |2 [cos(θ − φ)]2 If y = κ|M |/(2B) is small. it is still possible to solve the problem so long as the spin twists slowly so that we can replace (b) Verify that this differential equation has the solu- the finite difference δθ with a derivative. p (b) This is an exercise in nasty algebra. (d) The activation barrier is just the total anisotropy energy. so we have a total energy of about 10−22 J ≈ 10Kelvin activation energy. we can write θ(x) = 2 tan exp 2(x/a) J the energy of the domain wall as p Z ( 2 ) thus demonstrating the same L ∼ J/κ scaling. then there may be a metastable min- imum as well.35 × 104 J/m3 and the size is (4/3)π × 1nm3 ≈ 10−26 m3 . then the metastable minima always exist near to φ and φ + π. Isolating the terms inside the integral linear in δθ. (a) This is a trivial exercise in calculus of variations. dx JS 2 a2 dθ(x) (c) Show that the total E= 2 − κS [cos θ(x)] 2 √ energy √ of the domain wall be- a 2 dx comes Etot /(A/a2 ) = 2 2S 2 Jκ. This is 1. although for y close to 1/2 then there is only a metastable minimum if φ is close to zero. If this parameter is greater than 1/2. Although this makes things (Ja2 /κ)d2 θ/dx2 − sin(2θ) = 0 more complicated. (a) Using calculus of variations show that this energy ergy of the anisotropy (κ) term is annoyingly crude. the background energy is zero whereas as written the background energy is finite. and then integrating by parts to remove the derivatives gives 2 2 d2 θ(x) 2 −JS a + 2κS cos θ(x) sin θ(x) δθ(x) dx2 Setting this variation to zero immediately gives the desired√result. In this case. So the activation barrier is κ|M |2 . To clarify it. so z is close to zero. If y is large.3) Exact Domain Wall Solution* with a the lattice constant. The approximation used in Section 21. It is better to write the first equation as + sin2 instead of − cos2 . Emin = E0 − κ|M |2 and Emax = E0 . 21.178 Domains and Hysteresis From this we obtain on the left of the equation ebx b dθ/dx = 2b 2bx = (21.2 along with the expression for b imme- diately confirms the desired solution. . (c) Starting with Z ( 2 ) dx JS 2 a2 dθ(x) 2 2 E= + κS [sin θ(x)] a 2 dx b we have dθ/dx = cosh(bx) and 2ebx sin θ = 2 cos θ/2 sin θ/2 = = 1/ cosh(bx) 1 + e2bx This then naturally gives Z ( 2 ) dx JS 2 a2 b 2 1 E= + κS a 2 cosh(bx) cosh2 (bx) Plugging in the value of b this simplifies to Z Z dx 1 2κS 2 1 E = 2κS 2 = dx a cosh2 bx ab cosh x √ plug in the value of b to get a prefactor of S 2 2κJ and the integral of sech2 is tanh thus giving a factor of 2.2) cosh2 (bx) On the right of the equation we need to evaluate θ θ θ θ sin 2θ = 2 sin θ cos θ = 4 sin cos (cos2 − sin2 ) (21. yielding the desired result.3) 2 2 2 2 We have θ/2 = tan−1 (ebx ) so ebx = tan 2θ so we have 1 + e2bx = sec2 θ 2 so θ 1 cos = √ 2 1 + e2bx θ eb x sin = √ 2 1 + e2bx Plugging these results into the prior expression for sin 2θ the gives ebx 1 − e2bx −2 sinh(bx) sin 2θ = 4 = 2bx 1+e 1+e 2bx cosh2 (bx) Comparing this result to Eq.1) 1+e cosh(bx) b2 sinh(bx) d2 θ/dx2 = − (21. J X X (b) Calculate hSi i in terms of the temperature and the H=− Si · Sj + gµB B Si (22. J > 0. We then have the usual calculation Zi = exp(βgµb Bi. there are solutions of the electron charge is negative. and net write down an effective Hamiltonian for this spin. which we will assume is in the zˆ direction for simplicity. Each site i is assumed to have Write the susceptibility in terms of this critical tem- a spin Si of spin S = 1/2. dM/dH = µ0 dM/dB in this approximation. with the sum indicated with hi. 0. (a) The effective Hamiltonian for one spin-1/2 is X Hi = Si · −J hSj i + gµb B = +Si · gµb Bi. ing all other variables Sj with j 6= i as expectations hSj i tonian on the cubic lattice: rather than operators. summing over i and j being neighboring sites of the cu. and B is the externally applied magnetic field.1) ‡ Weiss Mean Field Theory of a Ferromag. (a) Focus your attention on one particular spin Si . treat- Consider the spin-1/2 ferromagnetic Heisenberg Hamil.Mean Field Theory 22 (22. Thus we can treat these quantities as scalars rather than vectors. therefore the magnetic mo. ment opposes the spin direction. final term has a + sign out front is from the fact that the below the critical temperature. The fact that the (e) Show graphically that in zero external field (B = 0). spins is counted only once. (b)Assuming B aligned with zˆ tells us that hSk i should be aligned with −ˆz if nonzero. +1). (Note the factor of 1/2 out front is now missing).ef f j where X gµb Bi. Here µB is the conventional perature.ef f = gµb B − J hSj i j with the sums being over sites j neighboring site i.ef f (1/2)) (22. ji means hSi and the density of spins.ji equation. If one were to assume (f) Repeat parts (a)–(d) but now assuming there is an that these were nuclear spins the sign would be reversed S = 1 spin on each site (meaning that Sz takes the values (and the magnitude would be much smaller due to the −1.2) 1 hSiz i = − tanh[βg µ˜b Bef f (1/2)] 2 . Bohr magneton defined to be positive.ef f (1/2)) + exp(−βgµb Bi. find the susceptibility χ = bic lattice. larger nuclear mass). (c) At high temperature. self-consistency equation with M 6= 0. (d) Find the critical temperature in this approxima- The factor of 1/2 out front is included so that each pair of tion. Write the magnetization M = |M| in terms of Here.1) fixed variables hSj i to obtain a mean-field self-consistency 2 i hi. ef f ) (22. the right side is a tanh.2 is replaced by Zi = exp(βgµb Bi. however at T < Tc they intersect as well at a finite value of hSi. (e) Graphically we plot the right and left sides of Eq. these intersect only at hSi = 0. (f) The procedure is the same for S = 1. the partition function Eq.3 as shown roughly in Fig. Note that the point at which the curves are tangent is the critical temperature. 22.3) 2 where z is the number of neighboring spins for each site (which is z = 6 for a cubic lattice). The left side of the equation is a straight line. At high T we expand for small x we obtain the equation −2x −2 hS z i = = β(gµb B − JzhS z i) 3 3 which then gives us (−2/3)βgµb B hS z i = 1 − 23 βJz .180 Mean Field Theory which gives us the self consistency equation 1 hSz i = − tanh[β(gµb B − JzhSz i)(1/2)] (22. 22. Thus for small field we can expand the tanh for small argument to obtain.1. 1 hSz i = − [β(gµb B − JzhSz i)(1/2)] 2 or (−1/4)β(gµb B) hSz i = 1 − βJz/4 The moment per site is thus (1/4)(gµb )2 B m = −gµb hSz i = kb T − Jz/4 giving the susceptibility (µ0 /4)(gµb )2 N (1/4)(gµb )2 N χ= = (22. (d) the critical temperature is the point where the susceptibility di- verges.ef f ) + 1 + exp(−βgµb Bi. or kb Tc = Jz/4. The horizontal axis is −hSi.4) kb T − Jz/4 kb (T − Tc ) refWeissMeanFerro where N is the density of spins. 22.5) leading to −2 sinh(x) hSiz i = 2 cosh(x) + 1 with x = βgµb Bef f = β(gµb B −JzhSiz i). In this case. At high enough T . (c) At high T the magnetization is zero in the absence of a field. We will show in the next problem that below Tc the solution with hSi = 0 is unstable. 22. calculate the entropy 2 2 of the system in the large N limit. expectation of the Hamiltonian. 1 1 + ( − s) log( − s) termined by s). Let act result. Thus the probability will have a tendency to have the same spin because that of a spin being an up spin is P↑ = 1/2 + s whereas the will lower their energies. we can = N kB T ( + s) log( + s) 2 2 count the number of configurations (microstates) which have the given number of spin-ups and spin-downs (de. 181 1 2 01 02 2 1 02 01 Fig. For simplicity we will Use this result to calculate an approximation to the again assume spin 1/2 variables on each site. F = E − T S (a) Consider first a case where sites do not interact 1 1 with each other. sites that are next to each other the average spin value be hSiz i = s. ing the Weiss mean-field equations. +gµB Bz N s − JN zs2 /2 (b) Assuming all sites have independent probabilities where z is the number of neighbors each spin has. Using S = kB ln Ω. (22. but we have ignored this effect probability of a spin being a down spin is P↓ = 1/2 − s.3 or a susceptibility of (2/3)µ0 (gµb )2 N χ= kb T − 23 Jz and a critical temperature of kb Tc = (2/3)Jz (where N is the density of spins). The total number of up spins or down spins is then N P↑ (c) Putting together the results of (a) and (b) above.1 graphical solution of Eq. and N P↓ respectively where there are N total lattice sites derive the approximation to the free energy in the system. Note: This is not an ex- Assume there are N lattice sites in the system.2) Bragg-Williams Approximation same direction and the probability that two neighboring This exercise provides a different approach to obtain. and we P↑ and P↓ of pointing up and down respectively. In the micro-canonical ensemble. calculate have assumed the external field B to be in the zˆ direc- the probability that two neighboring sites will point in the tion. sites will point in opposite directions. as in reality. (Again we assume the spin is electron spin so that . here. 22. . (a) The number of configurations is N N! Ω= = N↑ (N (1/2 + s))!(N (1/2 − s))! giving the entropy (using Stirling’s approximation) 1 1 1 1 S = kb ln Ω = −kb N ( + s) log( + s) + ( − s) log( − s) 2 2 2 2 (b) The expected energy per bond is −J(1/2)2 times the probability of having like spins on neighboring sites plus J(1/2)2 times the probability of having unlike spins. At non-zero B? three solutions of the mean field equations. maximum of the free energy rather than a minimum. We thus have E/bond = −(J/4)[P↑ P↑ +P↓ P↓ ]+2(J/4)P↑ P↓ = −(J/4)(P↑ −P↓ )2 = −Js2 If there are N z/2 bonds in the whole system (with z being the number of neighbors of each site) we thus obtain a total energy E = −JN zs2 /2 We must add to this the coupling to the external field which is simply N gµB s. Adding these terms together F = E − T S gives us the desired result. show that the s = 0 solution is actually a (d) Extremize this expression with respect to the vari. able s to obtain the Weiss mean field equations.) spect to s. Sketch F (s) both above and below the critical tem- Below the critical temperature note that there are perature for B = 0. (c) 1 dF 1 1 = gµb B − sJz + kb T log( + s) − log( − s) N ds 2 2 Setting this to zero we obtain 1 1 β(−gµb B + sJz) = log( + s) − log( − s) 2 2 defining the left to be x. we exponentiate both sides to get 1/2 + s ex = 1/2 − s which we rearrange to 1 ex − 1 1 1 s= x = tanh(x/2) = tanh(β(−gµb B + sJz)/2) 2e +1 2 2 which is the same self-consistency condition.182 Mean Field Theory the energy of a spin interacting with the external field is By examining the second derivative of F with re- +gµb B · S. 183 (d) Look at the second derivative 1 d2 F 1 1 = −Jz + kb T + N ds2 1/2 + s 1/2 − s Now examine this at s = 0. we obtain . 1 d2 F . . = −Jz + 4kb T N ds2 . 9 01263 01268 01267 0123 0124 124 123 5 9 01264 01263 01268 01267 0123 0124 124 123 5 . The free energy as a function of s is plotted in Fig.m=0 Note that this changes sign exactly at the critical temperature! Thus below Tc the s = 0 solution becomes a maximum of the free energy rather than a minimum as it is as above Tc .2. 22. Using the result of Exercise 19. Left: in zero applied B. (It may be useful to re-solve Exercise 19. 22. Right: In gµb B = 0. remember how this is done. at.7 use mean field the.3) Spin S Mean Field Theory number z and nearest-neighbor exchange coupling Jex . having coordination . and below Tc .02Tc (22.7 if you don’t ory to calculate the critical temperature for a spin S fer.) romagnet with a given g-factor g. Fig.2 Free energy (vertical axis) as a function of s (horizontal axis). tion from Exercise 22. the magnet (c)* In fact this exponential behavior is not observed is fully polarized. the effective field seen from neighboring spins is gµB Bef f = Jex zhSi Thus we have at the crtical point S(S + 1) hSi = βJex zhSi 3 which has the solution for the critical temperature S(S + 1) kb Tc = Jex z 3 (22. At zero temperature. Using some results from becomes exponentially small as T goes to zero. using the results of 22. the self consistency equation is 1 Sz = tanh(βJzSz /2) 2 For large argument x we have 1 − e−2x tanh x = ≈ 1 − 2e−2x 1 + e−2x So at low temperature we have 1 Sz ≈ − e−βJzSz 2 Since at low temperature. 22.7.1.4) Low-Temperature Mean Field Theory your result conveniently in terms of the result of Exercise Consider the S = 1/2 ferromagnet mean field calcula.184 Mean Field Theory From Exercise 19. Sz is very close to 1/2. but are not ature limit. determine (roughly) the low-temperature (b)* Now consider a spin S ferromagnet. this becomes 1 Sz ≈ − e−βJz/2 2 .3. which are explored in Exercise 20. the magnetization in the low T limit.1. in a field B the resulting hSi per site (for small perturbation B) is given by S(S + 1) hSi = βgµB B 3 And in mean field theory. that exercise. Show that the deviation from fully polarized included in mean field theory. Determine behavior of the magnetization of a ferromagnet. You can express (a) In absence of external field. experimentally! The reason for this has to do with spin- (a) Calculate the magnetization in the very low temper.3. waves. The low energy excitations excite Sz by a single step at activation energy JzS.5) Mean Field Theory for the Antiferromag. if all the neighbors are aligned. The total num- ber of spins is V /(a3 ) and the number of excitations (the number of lost steps) at finite T is Z dk V nB (~ωk ) (2π)3 with nB the bose factor. so neighboring spins want to point (a) Write the mean field Hamiltonian for a single site in opposite directions (compared to a ferromagnet where on sublattice A and the mean field Hamiltonian for a sin- J > 0 and neighboring spins want to point in the same gle site on sublattice B.) . Thus. Recall that J < 0. Let sA = hS z iA be the average value of the spins mean field) approximation for the spin-1/2 antiferromag. whereas B sites have lattice coordinates with i + j + k (c) Let B = 0. for each spin the reduction in magneti- zation is Z ∞ 3/2 Z ∞ 4π 1 1 kb T x2 a3 3 dkk 2 βJS 2 a2 k 2 = 2 2 dx x (2π) 0 e −1 2π JS 0 e −1 and the final integral gives 2ζ(3) ≈ 2. sublattice A and sublattice B respectively (i. the ordered ground state of this sA = tanh(β[JZsB − gµB B]/2) 2 Hamiltonian will have alternating spins pointing up and 1 down respectively. sB = tanh(β[JZsA − gµB B]/2) 2 ing sites. (b) Probably the easiest way to do this one is to use the final comment of part (a). (Hint: Use symmetry to simplify! Try plotting sA versus sB . 185 Another way to see this is to note that.4. Thus the energy for flipping a given spin in this effective field is Jz/2. direction). the effective field is gµBef f = JzS. thus we obtain Sz = S − e−βJzS I have no idea what the comment about expressing the answer in terms of 22. 1 At mean field level.e. and sB = hS z iB be the average value of netic model on a three-dimensional cubic lattice. now we have J < 0. Let us call the sublattices of alternat. on sublattice A. k) with i + j + k odd with β = 1/(kB T ). In mean field theory the interaction between neighbor- net ing spins is replaced by an interaction with an average For this Exercise we use the molecular field (Weiss spin.3 the low energy spectrum of spin waves (in absence of external field) is given by ~ω = JSa2 |k|2 Each excitation reduces the magnetization by one step. tions to a single self-consistency equation. If all the spins are aligned to extremize Sz . 22. except that oriented in the ±ˆz direction). Reduce the two self-consistency equa- even). A sites have lattice coordinates (i. then gµBef f = Jz 21 . The the spins on sublattice B (we assume that these are also full Hamiltonian is exactly that of Eq. For simplicity let us assume that the external (b) Derive the mean-field self-consistency equations field points in the zˆ direction. j. (22. So the activated deviation from fully aligned is e−βJz/2 .1.3! (c) As noted in problem 20. . it was this type of temperature Tc below which the system is antiferromag. Going into the anti- ferromagnetic phase there is a new unit cell of lattice constant 2a. |sB (T )|). Thus the problem is reduced to 1 1 sA = − tanh(βJZsA /2) = tanh(β|J|ZsA /2) 2 2 which is identical to the relation we obtained in the ferromagnetic case. Derive the critical a ferromagnet (Exercise 22). the magnetic sus. (d) As in the ferromagnet kb Tc = z|J|/4. s(T ) = |sA (T )| = (why the factor of 1/2 out front?).B are small near the critical point and Compare your result with the analogous result for expand the self-consistency equations. (g)* For T < Tc show that mentally? (f) In this mean-field approximation.B = Si · −J hSj A i + gµb B j (b) solving as above gives the desired self-consistency equations. sA.B become non-zero).186 Mean Field Theory (d) Assume sA. terms of Tc . This follows very much problem 22 (a) X Hi. (e) This staggered moment is most easily observed with neutron scat- tering where the scattering will be spin-dependent. (Since tanh is an odd function these are the only possible solutions for J < 0. thus one sees new diffraction peaks at k = n2π/(2a) (with n odd) which are not present above Tc . (N/4)µ0 (gµb )2 (1 − (2s)2 ) ceptibility can be written as χ= kB T + kB Tc (1 − (2s)2 ) ∂(sA + sB ) χ = −(N/2)gµ0 µB lim B→0 ∂B with s the staggered moment (ie.e.A = Si · −J hSj B i + gµb B j X Hi. Give a sketch of the susceptibility at all T . . (c) For B = 0 we have 1 sA = tanh(βJZsB /2) 2 1 sB = tanh(βJZsA /2) 2 These are solved by 1 1 sA = −sB = tanh(βJZsB /2) = − tanh(βJZsA /2) 2 2 Recall that J < 0. Derive this susceptibility for T > Tc and write it in Compare this low T result with that of part f. which one can check graphically by plotting sA versus sB ). In fact. measurement that first suggested the existence of antifer- netic (i. romagnets! (e) How does one detect antiferromagnetism experi. Expanding the self consistency equations for high T we obtain sA = β[JZsB − gµb B]/4) sB = β[JZsA − gµb B]/4) One can solve this system of two equations in two unknowns to obtain −βgµb B sA = sB = 4 − βJz Thus giving (1/4)µ0 (gµb )2 N χ= kb (T + Tc ) compare to the T − Tc factor in Eq. 22.4. (g) This required implicit differentiation. Differentiating our two self- consistency equations gives . 187 (f)The factor of 1/2 is because we have only N/2 A sites or B sites. . ∂sA . . βJZ ∂sB . . βgµb = − sech2 (βJZsB /4) ∂B . B=0 4 ∂B . B=0 4 . . ∂sB . . βJZ ∂sA . . βgµb = − sech2 (βJZsA /4) ∂B . B=0 4 ∂B . we add both equations together to get .B=0 4 Noting that sA = −sB = s so both sech terms are the same. . ∂(sA + sB ) . . βJZ ∂(sA + sB ) . . βgµb . = . We thus have . we then have sech2 (βJZs/4) = 1 − tanh2 (βJZm/4) = 1 − (2s)2 where we have used the self-consistency equation to replace the tanh by s. − sech2 (βJZs/4) ∂B h=0 4 ∂B B=0 2 Making note that sech2 + tanh2 = 1. . ∂(sA + sB ) . . βJZ ∂(sA + sB ) . . βgµb . = . − (1 − (2s)2 ) ∂B B=0 4 ∂B B=0 2 This then can be rearranged into . ∂(sA + sB ) . . . At Tc there is a cusp. Above Tc . Below Tc . This behavior is illustrated in the figure 22.3 . and the susceptibility drops rapidly. we have s = 0 and we have the same behavior as part f above. 1 − βJZ(1 − (2s)2 )/4 = −(βgµb /2)(1 − (2s)2 ) ∂B B=0 or (N/4)µ0 (gµb )2 (1 − (2s)2 ) χ= kb T + kb Tc (1 − (2s)2 ) with s the staggered moment (with N the density of spins). the factor 1 − (2s)2 goes quickly down to zero. 188 Mean Field Theory 789 . 8. 9. 9 99. . 9 . 012 016 . 9 9 ! 015 014 013 00 012 3 . . 3912 . 6) Correction to Mean Field* a block of two connected spins σi and σi′ where the neigh- Consider the spin-1/2 Ising ferromagnet on a cubic lat. 22. (22. spin hσi. (↓. The critical temperature you calculate should (it will be a transcendental equation). ↓). estimate of the critical temperature higher or lower than To improve on mean field theory. we can instead treat that calculated in the more simple mean-field model? We have to consider four configurations in our partition function: (↑. ↑). Use this improved mean field the- bors on each side will be considered to have an average ory to write a new equation for the critical temperature spin hσi. averaged neighbors. bors outside of this block are assumed to have the average tice in d dimensions. ↑). (↓. 4 412 5 Fig. Is this improved be kB Tc = Jz/4.3 Susceptibility of an Antiferromagnet. ↓) Correspondingly. and then setting B to zero we obtain e−βJ/4 sinh(β(Jhσi(z − 1)) m = hσi = 2e−βJ/4 cosh(β(Jhσi(z − 1)) + 2eβJ/4 . the partition function is Z = e−β(Jhσi(z−1)+gµB B)−βJ/4 + 2eβJ/4 + eβ(Jhσi(z−1)+gµB B)−βJ/4 From this we calculate the magnetization per site (note that this parti- tion function represents two sites). (↑. When we consider mean field theory. Each of the spins in the block has 2d − 1 such we treat exactly a single spin σi and the z = 2d neigh. so this expression always gives a lower prediction for Tc than our prior mean field theory. 189 Expanding for small hσi we find equality (which finds Tc ) when J(z − 1) e−βJ/4 kB T = 4 cosh(βJ/4) The final factor in brackets is always less than unity. . . How does this alter the result of part (d)? (a. neighbors and the lattice constant a. n↓ = (n/2)(1 − α) (23. M = −µb nαB.i) The tight binding spectrum is E = E0 − 2t [cos(kx a) + cos(ky a) + cos(kz a)] which we expand to get E = Emin + ta2 |k|2 . density of spin-down electrons be n↓ . We can write these (d) Adding together the kinetic energy calculated in as part b with the interaction energy calculated in part c.1) Itinerant Ferromagnetism calculate the total energy of the system per unit (a. fixing the total density of model on a square lattice with a Hubbard interaction. where the total net magnetization of the system is given Explain why this calculation is only an approxima- by tion.1 and 23. Argue that this remains true to all orders in α (a. one can view the X i i electrons as being free electrons with this effective mass HHubbard = U N↑ N↓ m∗ . model on a cubic lattice. (e) Consider now a two-dimensional tight binding Using the result of part (a). 23.i) Review 1: For a three-dimensional tight binding volume as a function of α. calculate the magnetization as a function of U . calculate the effective mass in Expand your result to fourth order in α.Magnetism from Interactions: The Hubbard Model 23 (23.1) become non-zero. with densities n↑ and n↓ as given by Eqns.2) For values of U not too much bigger than this value. terms of the hopping matrix element t between nearest Show that α = 0 gives the lowest possible energy. electrons in the system n. For a system of spinless electrons show that the total i energy per unit volume (at zero temperature) is given by with U ≥ 0 where Nσi is the number of electrons of spin E/V = nEmin + Cn 5/3 σ on site i. given that the up and down electrons form Fermi seas Calculate the constant C.2. determine the value of U for which it is favorable for α to n↑ = (n/2)(1 + α) (23. Calculate the expectation value of this interaction term where Emin is the energy of the bottom of the band.ii) Review 2: Assuming the density n of electrons (c) Now consider adding a Hubbard interaction term in this tight binding band is very low. (b) Let the density of spin-up electrons be n↑ and the Write this energy in terms of α. so there is no factor of 2 out front) Z kf Z kf 3 V V 2 V 4πkf N= dk = 4πk dk = (2π)3 0 (2π)3 0 (2π)3 3 or kf = (6π 2 n)1/3 The total energy is given by Z kf 2 2 Z kf 2 2 V ~ k V ~ k E − Emin N = dk = 4πk 2 dk (2π)3 0 2m∗ (2π)3 0 2m∗ V ~2 4πkf5 V ~2 5 = = k 3 (2π) 2m ∗ 5 20π 2 m∗ f Thus we have E/V = Emin n + Cn5/3 with 1 ~2 2 5/3 ta2 C= (6π ) = (6π 2 )5/3 20π 2 m∗ 10π 2 (b) Note that n remains fixed. . We thus have ta2 |k|2 = ~2 |k|2 /(2m∗ ) or ~2 m∗ = 2ta2 (a. the expectation of the hubbard interaction per unit volume is Ehubbard /V = (U/a3 )(n↑ a3 )(n↓ a3 ) = U a3 n↑ n↓ = U a3 (n/2)2 (1 + α)(1 − α) = U a3 (n/2)2 (1 − α2 ) . note that the odd terms of the expansion cancel leaving only even terms E/V − Emin n = 2C(n/2)5/3 1 + (1/2!)(5/3)(2/3)α2+ (1/4!)(5/3)(2/3)(−1/3)(−4/3)α4 + . (c) The expected number of electrons per unit site is na3 and simi- larly n↑ a3 and n↓ a3 are the expected number of spin up and spin down electrons per site. . Note that successive terms of the expansion always have positive coeffi- cient. .192 Magnetism from Interactions: The Hubbard Model where Emin = E0 − 6t is the bottom of the band. So we have h i 5/3 5/3 E/V − Emin n = +C n↑ + n↓ h i = C(n/2)5/3 (1 + α)5/3 + (1 − α)5/3 Taylor expanding here. Thus. = 2C(n/2)5/3 1 + (5/9)α2 + (5/243)α4 + .ii) As we have calculated many times (note we are considering spin polarized electrons here. . For a ferromagnet. Consider a tight binding model with hopping t and a (b) On a square lattice. there must always be one electron per site. strong Hubbard interaction and large U . . so it is higher energy. the transition is discontinuous going suddenly from unpolarized spins to fully polarized spins. i state and the antiferromagnetic state. . (23. at an energy cost of U . If the hopping is t at 2nd order in perturbation theory. Thus. 193 (d) We thus obtain a total energy given by h i h i Etotal /V = constant+α2 2C(5/9)(n/2)5/3 − U a3 (n/2)2 +α4 2C(n/2)5/3 (5/243) +. per site. we have a total energy of the form Etotal = const + α2 (C˜ − KU ˜ ) Once U becomes large enough that α = 0 is not the lowest energy solution. 23. (b) For the antiferromagnet. with one electron per site.3. . of −2zt2 /U where z is the number of neighbors. we have the kinetic term given by ˜ + α)2 + (1 − α2 )] = 2C[1 E = C[(1 ˜ + α2 ] and there is no quartic term. so E/V ∼ n2 . Minimizing with respect to α gives s −2C(5/9)(n/2)5/3 + U a3 (n/2)2 α= 4C(n/2)5/3 (5/243) and M = −µb nα. we can use the quartic form of the energy given in Eq. (e) In the 2d case. As a result. the key here is that N ∼ kf2 and ǫ ∼ EF2 .3) And taking the limit of small α we see that α = 0 is the solution only for U ≤ 2C(5/9)(n/2)−1/3 For U not too much bigger than this. if the interaction term (a) If U is strong enough. no excursion is allowed by the Pauli principle. This makes a traffic jam of electrons where no one can move (so long as there is no doping). this gives a reduction in the ground state energy.2) Antiferromagnetism in the Hubbard U is very strong. then α immediately goes to its maximum possible value 1. Thus. explain qualitatively why the system Model must be an insulator. use second-order perturbation theory to de- X i i termine the energy difference between the ferromagnetic HHubbard = U N↑ N↓ . Which one is lower energy? (a) If there is one electron per site. (23. each spin can make a virtual excursion to each of the neighboring sites.
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