Quantum Theory IJens Kortus
[email protected] TU Bergakademie Freiberg Some recommendations : ● A.C. Phillips, Introduction to Quantum Mechanics ●Sakurai & Napolitano, Modern Quantum Mechanics ● Feynman, Lectures ● Greiner, Quantum Mechanics: An Introduction ● Konishi & Paffuti: Quantum Mechanics: A New Introduction ● many more Warning! This script may contain errors. Warning! It is strongly recommended to check all derivations and Formulas themselves. . Please send any corrections and comments to
[email protected] Pictures: If not stated explicitly otherwise, the pictures are from http://de.wikipedia.org/ or have been created by the author. 2 Motivation Why do we need quantum mechanics: ● stability of atoms (positive and negative charges) ● System of elements ● Chemical bonding, molecules ● Solid state ● spectroscopy (colors -> emission + absorption, laser) Applications ● NMR, ESR, Squids ● medicine (nuclear spin tomography, laser-scalpel) ● chemistry Dreams (but in principle already doable) ● Atomic construction (STM -> move single atoms) ● Theoretical predictions of materials properties ● Computer aided molecular design (pharmacy, dyes) 3 almaden.com/vis/stm/atomo.IBM Almaden STM Image Gallery Xenon on nickel (110) Iron on Copper (111) Carbon Monoxide on Platinum (111) Kanji characters for atom http://www.ibm.html 4 . Calculation of the energy of the electromagnetic field in the oven ➔ Due to reflecting boundary conditions there will be standing electromagnetic waves ➔ Each wave has energy kT (kT/2 electric.1. Some experimental observations 1.1 Planck radiation law (black body radiation) Thermal radiation (e. the colors depend only on temperature but not on the material. iron shows different colors with increasing temperature) IR (heat) → red → yellow → white Interestingly.g. kT/2 magnetic) . Oven T Black body radiation: because there is no radiationat T = 0 Which frequencies and intensities will occur? Rayleigh-Jeans law: ● classical electrodynamics (dipole radiation) and statistics (equal partition) At thermal equilibrium each degree of freedom contains an energy of kT/2. k=Boltzmann constant ➔ Need to count the number of such waves in a frequency interval [v + dv] ● 5 . n1 = 2aλ cos α n2= 2aλ cosβ n3 = 2aλ cos γ using cos2 α+cos 2 β+cos2 γ=1 n21 +n22 +n23 =(2 aλ )2 =(2 a νc )2 λ ν=c ν= 2ac √ (n21 +n22 +n23 ) Frequencies between 0 und v are within a sphere with radius 2av/c.Cube with length a ● standing electric waves have minima at the walls ● standing magnetic waves have maxima at the walls y λ/2 α x cos=n1 / 2 y cos =n 2 / 2 z cos =n 3 / 2 x Standing waves are created. 1 4 π 2a ν 3 N ( ν)= ( ) 8 3 c 2 d N ( ν)=4 π a 3 ν3 d ν c spectral energy density ων d ν=2kT d N ( ν)∼T ν2 d ν 6 . if they can fit an integer number times there wavelength/2 into the box. so that in case of a>>λ. the number of frequencies is equal to 1/8 of the volume of such sphere. there are only positive integer n. However. Rayleigh-Jeans law: ω ~ T ν2 2T T Radio IR light UV Roentgen Only correct in case of small frequencies or large wave length (ν=c/λ) -> UV catastrophe: intensity diverges for small λ -> ∞ Classical physics is giving a wrong result here! Wien law: 3 − ~ e T =h/ k Only correct for large frequencies or small wave length → we need a mix of both radiations laws ● 7 . ):∼ν2 kT Interpolation between Rayleigh-Jeans and Wien law. April 1858 in Kiel † 4.Max Karl Ernst Ludwig Planck * 23. Oktober 1947 in Göttingen Nobelpreis 1918 Planck radiation law (1900) ν ω∼ 3 hν kT e −1 h ν /kT 3 −h ν kT ν large(e ≫1):∼ν e x ν small(e =1+ x+ ..626 10-34 Js ћ=h/2π 8 . Interpretation: Atoms behave like harmonic oscillators and can only have discrete energies E=hv (n+ ½) = ћω (n+½) The oscillators can only accept or emit energies equal to an integer n times a constant.. h = Planck quantum = 6. Light has wave length between 400-700nm.de/beispiele/rahmen.. This corresponds to energies of hv = 2 . http://www.webgeo..k_304. Surface temperature of our sun is 5800K.1 9 . 4 eV.php?string=1. + Some α-particles were scattered back to the source. According to classical electrodynamics: Any accelerated charge will emit EM-radiation.1. which results in a loss of energy. 1937 Nobelpreis Chemie 1908 Scattering of α-particles (He2+) at atoms in a gold foil. This would result in a radiation decay of atoms during a time span of about 10 -8 – 10-10 s. so that the electron will fall into the core. 10 . According to classical pictures electrons move around the core like planets around the sun. However. movement along a circle is an accelerated motion. 1871 – October 19. This means that there should be heavy positively charged centers in the gold.2 Stability of atoms Ernest Rutherford 1st Baron Rutherford of Nelson August 30. Explanation by Einstein 1905 Albert Einstein * 14. ● hv corresponds to the binding energy of the electron in the min metal. 1) There is a minimal frequency for each metal below no photo-electrons can be found. 4. 1879 Ulm † 18. 1955 Princeton Nobelpreis Physik 1921 Light = particles with energy E=hv (photon) ● Each photon kicks an electron from the metal.1. 4) Number of electrons proportional to intensity. ● Electron number is proportional to the number of photons (intensity).3 Photoelectric effect Metal plate in vacuum radiated with UV-light. ● Kinetic energy is proportional to h(v-v ) min 11 . 2) Very fast process (<10-9s) 3) kinetic energy of the electrons for v>vmin proportional to the frequency v of the light.3. 4 Particle-wave dualism a) Wave character of light Interference and diffraction (Huygens 1678) * 14. Intensities can not be superposed. fields can be superposed. Picture is not sum of single pictures 2 2 2 2 I = Er⋅Er =∣Er∣ =∣E1 E2∣ ≠∣E1∣ ∣E2∣ b) Particle character of light -> see photo-electric effect 12 . Er = E1 E2 The intensity is the energy density ~ |E|2. April 1629 in Den Haag † 8. Juli 1695 Double slit Diffraction at single slit Light is an electromagnetic field.Christiaan Huygens 1. . polarization.12Å (Röntgen radiation). de-Broglie Wellenlänge Phase velocity can not be measured Indivisibleness of particles: there will always be complete electrons in contrast to light (reflected and diffracted ray).. Normandie † 19. Applications: electron microscopy. Quantum teleportation . . The de Broglie wave length of moving particles with momentum p is λ= h p e. direction. Neutron scattering 13 Quantum optics. August 1892 in Dieppe. electrons with 10 keV kinetic energy have λ=0. wave length easy to measure matter-waves: direction in direction of momentum.g. Quantum communication. März 1987 in Louveciennes Nobelpreis Physik 1929 All particles (atoms. Quantum computer.Waves of matter: de Broglie 1924 (Dissertation) Louis-Victor Pierre Raymond de Broglie * 15. Waves: Amplitude and phase. solids) can have wave properties. molecules. Univ.html 14 . Harrison.upscale.Double slit experiment with classical particles only 1 open only 2 open both open Double slit experiment with waves How about electrons? Bilder von David M.utoronto. of Physics.utoronto.upscale. of Toronto http://www. Dept.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html http://www.ca/GeneralInterest/Harrison/DoubleSlit/Flash/Histogram . The Jönsson experiment has been selected in September 2002 by the journal 'Physics World' as the most beautiful experiment of all times.Wave character of electrons 1927 Electron diffraction using a Ni-crystal (Davisson & Germer) 1961 Double slit experiment with electrons Claus Jönsson. http://physicsweb.org/articles/world/15/9/2 15 . Tübingen. 454 (1961) Electrons show particle or wave properties similar to light. Zeitschrift für Physik 161. . 680-682. 14.Wave-particle dualism for molecules: C60 16 "Wave-particle duality of C60" Markus Arndt et al. Nature 401.October 1999 . therefore: ∫w(r.t)|2=ψ*ψ is the probability density to find a particle at location r at time t.2.t) d3r =|ψ(r.t)|2=ψ*ψ A single particle has to be somewhere in a given volume.t = ∫∣ r .t) is a complex scalar. Probability density w(r. The wavefunction ψ(r.t)|2d3r=∫ψ*ψ d3r=1 Normalization: If the above condition is not fulfilled for ψ(r.t).1 Wavefunction Based on the particle-wave duality we can describe the state of any physical system with help of a wavefunction ψ(r.t ∣2 d 3 r .t) =|ψ(r. The square |ψ(r. The probability to find the particle at time t in a volume element d 3r located at r is given by w(r. because there was no polarization found In contrast to electromagnetic waves.t)|2d3r=ψ*ψ d3r. Wave equation for free particles 2.t) then we can always normalize to 1.t) d3r =∫|ψ(r. t ∣2 17 r . by: ∣ r . 18 . so that the alternate possibilities can not be realized.Wavefunction ψ(r.t)|2 = large -> many photons |ψ(r. but only a probability. For many equal photons (bosons) the intensity is given by the square of the wavefunction resulting from the superposition of the single photons |ψ(r. One possible interpretation of the wavefunction of matter is a statistical interpretation. If the experimental setup is changed.t) Probabilities are positive definite numbers. the interference will vanish too. Only the square has a real physical interpretation.t)|2=0 no photons If there are alternative possibilities which will result in the same result then there will be interference phenomena observed in QM. There is no certainty to find a particle at a given location and time. The wavefunction (also called probability amplitude) ψ(r. A single measurement will always detect a single particle.t) itself can not be measured and no real physical interpretation. however the distribution in space and time is given by the square of the wavefunction. Schrödingers equation can not be derived directly from first principles. Therefore we try to 'guess' Schrödingers equation based on the wave-particle duality in analogy to light waves.Motivation Schrödingers equation Goal: To find the equation which gives as solution the wavefunction for particles. Wave-particle duality well proofed experimentally ● All particles with constant momentum have a wavelength (de Broglie 1924) ● QM as a general theory should contain the macroscopic correct classical mechanic in certain limits (Hamilton-Jacobi theory) ● 19 . because Maxwell eq.t) -> kλ = 2π k = 2π/λ ω=c k -> 2π v = c 2π/λ -> v λ = c Interference: E will be added (super position).t+τ)=E(x.t)=E(x.t = E0 e i k⋅r − t E will be a solution if ω=c k Wave vector k points in the direction of motion. e.t) -> ωτ = 2π ω = 2π/τ = 2π v Wavelength λ: E(x+λ. t)=E0 cos(k x -ωt) Period τ : E(x. t c ∂t Ansatz: plane waves ∂ = ∂ 2 x 2 ∂ ∂ 2 y 2 ∂ ∂ 2 z 2 r . k || x-Axis E(x. t = 2 2 E r .g. are linear differential equations Intensity: ~ Energy-current density (Poynting vector) ~ |Re E|2 2∣2 =∣Re E1∣2 ∣Re E 2∣2 2 Re E1⋅Re E 2 ∣Re E1 Re E Intensity can not be superposed! 20 .2.2 Light waves (classical electrodynamics) Wave equation from the Maxwell equations in case of the electric field: E r 1 ∂2 . Light m0=0 photons have no rest mass but momentum -> E2 = p2 c2 E = |p|c p=|p| E = hv = ћω = ћ c k = p c E = ћω p = ћk -> p= ћk ( p = ћk ) Energy photon momentum photon because k=|k|=2π/λ p= ћk = ћ 2π/λ = h / λ Therefore any particle with fixed momentum has a well defined wave length.Remark on theory of special relativity ED is a relativistic theory (c= speed of light). Transition to classical mechanics in case of small p (c>>|v|) E= m c p c =m0 c 2 0 4 2 2 2 p 2 c2 1 p 2 c 2 1 p 2 2 2 1 2 4 ~m0 c 1 =m0 c 2 m20 c 4 2 m0 m0 c 1 x~1 12 x . so that a relativistic wave equation results into quantum electrodynamics. E=mc2 E2 = m2c4 = p2 c2 + m02c4 Energy and mass increase with momentum (velocity). de Broglie 1924 21 ... 2.bosons) 2 2 2 1 ∂ m0 c − 2 = 2 c ∂ t2 ℏ 22 .t = 0 e i k⋅r − t Due to the experimental known wave-particle duality we additionally require the Energy and momentum relations from de Broglie: E = ћω p = ћk energy momentum Using the relativistic equation now with m0≠0: E2 = m2c4 = p2 c2 + m02c4.3 Relativistic equation matter waves (Klein-Gordon equation) Similar to light we are searching for an equation where the solution for free particles will be a wavefunction ψ(r.t) given by a plane wave r . we obtain: ℏ 2 2 =ℏ 2 c 2 k 2 m 20 c 4 2 2∂ 2 2 2 4 −ℏ =−ℏ c m0 c 2 ∂t Klein-Gordon equation 1926 ∂ =−i ∂t ∂ =i k j ∂xj (only correct for spin less particles . r .t −ℏ iℏ = r .t U r r . p = ћk 2 2 ℏ k 2m ∂ −ℏ 2 iℏ = ∂t 2m ℏ = ∂ =−i ∂t ∂ =i k j ∂ xj In case of a particle in external potential U(r) one can generalize this using the classical Hamilton-function H = p2/2m + U(r).2. Januar 1961 in Wien Nobelpreis Physik 1933 23 . which corresponds to the total energy of the system. E = ћω . 2 ∂ r .4 Non-relativistic equation for matter (Schrödinger 1926) We start from classical mechanic and require energy and momentum relations for light due to the wave-particle duality.t = 0 e i k⋅r − t Classical kinetic energy E = p2/2m. August 1887 in Wien-Erdberg † 4.t ∂t 2m Schrödinger equation Erwin Rudolf Josef Alexander Schrödinger * 12. 24 .5 Wave packets and coherence length A Gauss wave packet is a wave modulated by a Gauss-function (Multiplication of the wavefunction by a Gauss-function).2. e.g. This corresponds to a coherence length lc=c*τ ~ 3m of the wave packet. but are rather given by Wave packets. ψ ψ*ψ Coherence length = Length of the wave In reality wave have mo infinite extension like plane waves. Interference will only occur if the coherence length l c is large compared to the dimension of the measurement apparatus. realistic life times for radiation of a photon from an atom are about τ~10-8s. t) w(⃗r . t )∣2 ψ* (⃗ r . t )= = ∫∣ψ(⃗r . t U r r .t )d3 ⃗r The wavefunction is the solution of the Schrödinger equation. Schrödinger equation and operators The state of a QM system will be described by its wavefunction ψ(r.3.t)=wavefunction=state function Probability density to find an electron at location r and time t is defined by the square of wavefunktion. t ) ψ(⃗ r . Wavefunction. ∣ψ(⃗r . t )∣2 d 3 ⃗r ∫ ψ* (⃗r .t ∂t 2m H = Hamilton-operator (Operator: Function -> Function) Build from the Hamilton-function as defined in classical mechanics. 2 ∂ r . t ) ψ(⃗r . 25 .t −ℏ iℏ = H = r . t −ℏ 2 iℏ =H = r . t )=ψ0 e i(⃗ k⋅⃗ r −ωt ) with (ℏ ⃗ k )2 ℏ ω= 2m 2 2 −ℏ 2 ℏ k 2 i ℏ−i = −k ℏ = 2m 2m Normalization of the wavefunction ∫ * d3 r=∫ 0* e−i k⋅r − t 0 e i k⋅r − t d3 r =∫ *0 0 d3 r= 0* 0∫ d3 r = *0 0 V =1 1 V We get a plane wave with fixed energy E = ћω and momentum p = ћk wave vector k || p and group velocity ∣ 0∣= v g k = d dk = ℏk =v m Where do we find the particle? |ψ(r.Example: free particle in large volume V free = no acting forces -> potential energy U(r)=0 Schrödinger equation: ∂ r . 26 .t ∂t 2m Solution will be plane waves: ψ(⃗r .t)|2 = 1/V The probability is equal in all points of space. t x i ∂x i ∂x −ℏ x p x − p x x r . t = p x x r . t = x 27 .t i ∂x ℏ ∂ ℏ ∂ r . t = x p x r . . t = x r .3. t = r .1 Operators and measurements Momentum operator: classical momentum p will be replaced by QM operator pop ℏ ∂ ℏ ℏ ∂ ∂ . t i x p x r . t p x x r . = grad= ∇ i ∂x ∂ y ∂z i i ℏ ℏ ∂2 ∂2 ∂2 2 2 2 2 pop = pop⋅pop = ∇ ⋅ ∇ =−ℏ ∇⋅∇=−ℏ =−ℏ i i ∂ x 2 ∂ y 2 ∂ z2 pop = Position (space) operator: classical position r will be replaced by QM position operator rop r -> rop = just multiplication with value r Function depending on coordinates U(r) -> Uop(r) multiplication with value of function Index op will not be used further! The order of operators can not be permuted in the general case! ℏ ∂ r . t = r . Eigenvalues of operators correspond to physical values of measurements.t = 0 e i k⋅r − t p =ℏ k nach de Broglie B) Hamilton operator free particle p2op −ℏ 2 H op = = 2m 2m 2 pop ℏ 2 k2 H op = = =E 2m 2m Eigenvalues of the Hamilton operator are just possible values of the energy. Eigenvalue equation from linear algebra a 11 a 21 a 12 x 1 = x 1 a 22 x 2 x2 A x= x QM: Operator function = Number function A) momentum measurement for plane waves: ℏ pop = grad = p i r . C) general case: Operator A corresponds to a certain physical measurement Eigenvalue equation A ψ=a ψ Eigenvalue a=possible value of measurement 28 . which is not true in QM. iℏ ∂ r . .t). = grad = ∇ i ∂x ∂ y ∂z i i ∂ E i ℏ ∂t As a results we obtain a linear partial differential equation for the wavefunction ψ(r.t)|2d3r=∫ψ*ψ d3r=1 Remark: The choice of classical generalized coordinates may in some cases not unique.t) will be a solution: Therefore normalization needed: ∫|ψ(r. The order of mixed position and momentum coordinates does not matter in classical mechanics.3.t =H ∂t Schrödinger equation Linerar equation: if ψ(r. The freedom of choice vanishes if we use always Cartesian coordinates together with an additional replacement rule: p q -> (qp+pq)/2 29 .2 Formal quantization In the classical Hamilton function H=p2/2m+U(r) all classical values like space and momentum are replaced by there corresponding QM operators p ℏ ∂ ℏ ℏ ∂ ∂ . then also constant*ψ(r.t) will be a solution. t) one obtains the stationary Schrödinger equation H Φ(r) = E Φ(r).t )+ U (⃗r ) ψ(⃗ r .t) = e-iωt Φ(r) 30 . t )=f (t )ϕ( ⃗ r) Ansatz: Separation of variables i ℏ ϕ(⃗ r) ∂ f (t ) =H f (t ) ϕ(⃗r ) ∂t Dividing by ψ(r. t ) −ℏ iℏ =H ψ= Δ ψ(⃗ r . even if the wavefunction is still time dependent ψ(r. The time part equals to f(t)=e-iωt with E = ћω. * w( ⃗ r . t )=ψ ( ⃗ r .3 Stationary (time independent) states In case of an Hamilton operator which does not depend on time.3. t )=e iωt * ϕ ( ⃗r )e −i ω t * ϕ( ⃗r )=ϕ ( ⃗ r )ϕ( ⃗ r )=w( ⃗ r) The probability density is time independent. The solution Φ(r) are called stationary states which have fixed energy. t ) ψ( ⃗ r . t) ∂t 2m ψ( ⃗r . 2 ∂ ψ(⃗r . it is possible to derive from the full time-dependent Schrödinger equation a stationary Schrödinger equation. 4 probability current * ∂ ∂ * * ∂ = ∂t ∂t ∂t Schrödinger equation and the complex conjugated one 2 * 2 ∂ r . t =0 ∂t The change in time of the probability w(r. t ∇ * r .t −i ℏ = r . t − r .t ∂t 2m ∂t 2m ℏ ∂ * = * − * ] [ ∂t i 2m Definition of the probability current density: ∇ =grad j r . or mass conservation in continuum mechanics) 31 .t = ℏ [ * r .t) in a given volume equals to the probability current through the surface of the volume. t ∇ r . t ] i 2m This allows to formulate a continuity equation ∂ w r . t −ℏ * iℏ = r .t div j r .3.t −ℏ ∂ r . (similar to charge conservation in ED. 4. Some very basics of probability calculus Result of a single experiment can not be predicted in the general case. However,it is possible to make statistical predictions for many experiments. Def.: Physical property : A (energy, momentum, ...) Measurement value of A: a (possible values from measuring A) probability w(a), to measure the value a N (a) N →∞ N N =number of measurements N (a)=number of mesurements with result a 0≤w (a)≤1 ∑ w(a)=1 w(a)= lim a Physical properties are often continuous (e.g. space) and not discrete. w(a) becomes a probability density w(a)da= probability to find the value a within the interval [a-da/2,a+da/2] ∑ ∫ ∑ w a ∫ w a da 32 4.1 Averages (expectation values) Average of the property A is also called expectation value 〈 A 〉= A =∑ a w a a Average of a function f(A) 〈 f A〉=∑ f a w a a 〈 f A 〉=∫ f a w a da a Which experiment is more accurate? 2,5 2,3 2,2 2,1 2 1,9 1,8 1,7 1,6 1,5 1,4 2 1,5 1 0,5 0 0 2 4 6 Messungen 8 10 12 0 2 4 6 Messungen 8 10 12 33 Average of the distance from the expectation value is not good: 〈 A−〈 A 〉〉=〈 A 〉−〈 A 〉=0 4.2 average quadratic deviation = uncertainty 2 A= 〈 A−〈 A 〉 〉 2 A =〈 A 2−2 A 〈 A 〉〈 A 〉2 〉=〈 A 2 〉−2 〈 A 〉 〈 A 〉〈 A 〉 2=〈 A2 〉−〈 A 〉2 A= 〈 A2 〉−〈 A 〉 2 Most physical properties have ∆A≠0. -> measurements of the same property deliver different results (Heisenberg uncertainty principle: ∆x∆p≥ћ/2) There are cases, e.g. stationary cases, where the uncertainty may vanish (∆A=0). The aim of QM is often to calculate the possible eigenvalues (measurement values) a of a physical property A together with their probabilities w(a). 34 2 2 −ℏ ∂ x U x x =E x 2m ∂ x2 Finite energies are only possible if ψ(x)=0 outside the box. t iℏ = H x . t −ℏ ∂ x .: 2 ∂ x . t = U x x . t ∂t 2m ∂ x2 We search for stationary states (solutions for fixed energy) Hψ=Eψ. Boundary conditions for the solution of Schrödinger equation between 0 ≤ x ≤ L will be ψ(0)=ψ(L)=0.5. There exist an infinite number of solutions ψn fulfilling the boundary condition with different energies En. Simple Examples 5.1 Particle in 1D-box with infinite walls U U=0 =∞ 0 L for 0 ≤ x ≤ L outside 2 1D Schrödinger eq. 2 −ℏ 2 ∂ n x = En x 2 2m ∂x n x = a n sin k n x n x = b n cos k n x 35 . because sin(nπ)=0 -> kn= nπ/L n x =an sin n x L n=1. 3. an from normalization L L 0 2 L [ L 1=∫ *n n dx =∫ a2n sin 2 0 a n= ] L n 2n 2n L 1 1 L x dx =a2n ∫ 1−cos x dx =a2n x − sin x =a2n L 2 2n 2 L L 0 2 0 2 2 2 cos 2 =cos −sin =1−2sin 1 2 sin = 1−cos2 2 ψ ψ*ψ x/L x/L 36 . . 4. because cos(0)=1 ψ(L)=0 -> knL= nπ...The boundary conditions select from the general solution compatible ones ψ(0)=0 -> all bn=0. 2. The uncertainty of the energy is the state ψn vanishes ∆En=0.Energies can be obtained from putting ψn(x) in the Schrödinger equation: 2 2 2 2 −ℏ ∂ n x ℏ n n E n x = = a sin x n 2 2m 2m L L ∂x 2 2 ℏ 2 k 2n ℏ n E n= = 2m L 2m Due to the boundary condition not all energies are allowed anymore. practical energies appear continuous L small: only discrete energy differences allowed (-> sharp spectral lines) 37 . This is a qualitatively new results of QM. A n=an n A2 n = A A n = A a n =a 2n n Distance between two energy levels 2 2 2 2 ℏ ℏ 2 2 E n1 −E n = n1 −n = 2n1 2m L2 2m L 2 L large: very small energy differences. The discrete energy levels correspond to the measurable energy eigenvalues of the system. Molekülphysik und Quantenchemie.259 38 . S.Absorptions spectra of aromatic ring molecules 2 2 2 2 ℏ ℏ 2 2 E n1 −E n = n1 −n = 2n1 2m L 2 2m L 2 E = hv = ћω = ћ c k = h c /λ Absorption spectra of aromatic molecules Show with increasing molecular size a shift to larger wave length (smaller energies). Electrons in π-bonds are only weakly bonded and can be approximately described by particles in a box of a length given by the size of the molecule. Haken&Wolf. L3 =0 Ansatz: Separation of variables ψ(x. z =0 x . z = x . z = E x . y . y . y .0 . z 2 2 2m ∂ x ∂y ∂z L3 L1 L2 Similar boundary condition as before: 0. y . y . L 2. z = 1 2 3 n 2 sin 1 x L1 L1 n 2 sin 2 y L2 L2 n 2 sin 3 z L3 L3 2 2 2 ℏ 2 2 n 1 n2 n3 En n n = 2 2 2 2m L1 L 2 L 3 1 2 3 Sommerfeld model of metals: explanation of electric conductivity using QM 39 . y .0 = x .2 Particle in 3D-box with infinite walls −ℏ 2 ∂ 2 ∂2 ∂2 2 x .5. z = L1. z = 0 x .z)= ψ(x) ψ(y) ψ(z) n n n x . y .y. order differential eq. Scattering U0 kinetic energy smaller than U0 U0 kinetic energy larger than U0 QM: What are the states with fixed energy? I II 0 (general solution for 2. -> see Mathematics) III L ℏ2 d 2 I: − x =E 1 x 2m dx 2 1 ℏ2 d 2 II: − x U 0 2 x = E 2 x 2m dx 2 2 2 ℏ d2 III: − x =E 3 x 2m dx 2 3 1 x = A 1 e ik1 x −i k 1 x B1 e 2 x = A2 e i k2 x B2 e −i k 2 x 3 x = A3 ei k x B 3 e−i k 3 3 x ℏ2 2 E= k 2m 1 ℏ2 2 E= k U 0 2m 2 2 ℏ 2 E= k 2m 3 Energy given and constant in all parts (energy conservation) k 1 =k 3 = 1 2m E ℏ k 2= 1 2m E −U 0 ℏ k2 for E<U0 imaginary k2=iκ 40 .3 Tunnel effect classical Mechanic Reflexion.5. -> see Mathematics) III L ℏ2 d 2 I: − x =E 1 x 2m dx 2 1 ℏ2 d 2 II: − x U 0 2 x = E 2 x 2m dx 2 2 2 ℏ d2 III: − x =E 3 x 2m dx 2 3 1 x = A 1 e ik1 x −i k 1 x B1 e 2 x = A2 e i k2 x B2 e −i k 2 x 3 x = A3 ei k x B 3 e−i k 3 3 x ℏ2 2 E= k 2m 1 ℏ2 2 E= k U 0 2m 2 2 ℏ 2 E= k 2m 3 Energy given and constant in all parts (energy conservation) k 1 =k 3 = 1 2m E ℏ k 2= 1 2m E −U 0 ℏ k2 for E<U0 imaginary k2=iκ 41 . Scattering U0 kinetic energy smaller than U0 U0 kinetic energy larger than U0 QM: What are the states with fixed energy? I II 0 (general solution for 2.3 Tunnel effect classical Mechanic Reflexion. order differential eq.5. Boundary condition: wavefunction should be continuous d d 1 0 = 0 dx dx 2 d d 2 L = L dx dx 3 1 0 = 2 0 2 L = 3 L I II 0 III L 4 equations but 6 unknowns. one additional condition due to normalization -> one of the unknowns is not determined -> use experimental setup to define one e. A2 e i k 2 A1 B1 = A 2 B 2 L −i k L ik B2 e = A3 e 2 1 L A1 k 1 − B 1 k 1= A2 k 2 − B 2 k 2 A 2 k 2 e i k L− B 2 k 2 e − i k L = A 3 k 1 e i k 2 2 1 L 42 . particles come only from part I but not from III → B 3=0 (no back scattering in III) I II III A3 corresponds to classical forbidden contribution 0 L Using the conditions together with the general solution we obtain a linear system of equations to solve for the unknown constants.g. k2 real numbers → solution have wave character b) E<U0 (classical reflexion) 2 ∣ ∣ A3 admission coefficient D= A1 2 ∣ ∣ B1 reflexion coefficient R= A1 Particle should not vanish in the barrier (II). A3 4 k 1 k 2 e−i k L = A 1 k 1k 22 e−ik L −k 1 −k 2 2 eik 2 1 2 2 L A3 4 k 21 2 D= = 2 2 2 2 2 2 A1 k 1 sinh L4 k 1 ∣ ∣ 43 . therefore D+ R = 1.a) E> U0 → k1. The bigger the mass the smaller is the admission coefficient D.k 1 =k 3 = 1 2m E ℏ sinh L= 2 ∣ ∣ = 1 2m U 0− E ℏ 1 L − L 1 L e −e ≈ e 2 2 2 2 falls L≫1 1 sinh 2 L≈ e 2 L 4 2 2 A3 4 k 1 16 k 1 −2 L D= = 2 ≈ e 2 2 2 2 2 2 2 2 A1 k 1 sinh L4 k 1 k 1 Typical for solids (metals) U0-E~5eV.1 L=2Å : e-2κL ~ 0.01 L=10Å : e-2κL ~ 10-10 The admission decreases exponentially with the width L of the barrier. In case of electron : L=1Å : e-2κL ~ 0. 44 . Is the tunnel effect important? Qualitative understanding of the chemical bond Due to the tunneling process the total energy of the system can be lowered -> this is one reason for binding energies benzen: Collective tunneling of all double bonds -> all C-C bonds have the same length Radioactive α-decay 1/r α-particles are able to tunnel out of the nuclear box 45 . Januar 1862 Königsberg † 14. the is λψ (λ= complex number) also element of the set. David Hilbert * 23.1 Hilbert space space = mathematical construct: vector space a) The linear complex space is the set of mathematical objects having the following properties 1.6. ψ-ψ=0. If ψ is an element of the set. Hilbert space and linear operators (mathematical foundation QM) 6. It exist one element 0 (zero element) with ψ+0=ψ. then ψ±φ is also an element of the set. 2. Februar 1943 Göttingen 46 . 3. ψ and φ are elements of the set. then is ψ±φ also a solution function identical zero is solution of Schrödinger equation λψ(r. Set of all wavefunction (state functions) which are solution of Schrödinger equation ψ(r.t) and φ(r.Examples: I.g. matrices) 47 .0)=a zero vector (0. then is a±b=(ax±bx.t) are solutions of Schrödinger equation.t) is also a solution III. Set of complex 3D vectors a=(ax.az±bz) also a vector a + (0.az) with ax=Re ax + i Im ax a.λay.0.0) λa=(λax.λaz) is a vector λax = Multiplication of two complex numbers II.ay.ay±by. other mathematical sets (e.0. b are vectors. φ2) => (ψ1±ψ2. =∥∥ 48 .φ1) ± (ψ. if . (ψ.φ) (ψ. (ψ. λφ) = λ(ψ. (φ. Norm („length“): ● ψ.φ) = a = complex number (Mathematics: a is finite) 1.φ) λ = complex number 2.ψ) = (ψ. = 0 ≠ 1 = .φ1±φ2) = (ψ. φ are orthonormal.φ) = (φ. finite (ψ.b) Hilbert space of physicists The Hilbert space is a linear complex space.λψ)* = [λ(φ.ψ) ≥ 0 and real.ψ)]* = λ*(φ.ψ)* = λ*(ψ.ψ) = 0 only if ψ = 0 ● Def.φ) = (ψ1. which has defined additionally a scalar product with the following properties: 0.φ) ● ● (ψ.φ)* = a* => (λψ.φ) ± (ψ2. form a linear space.c) Hilbert space (mathematicians) Additional requirement of completeness Very important for mathematicians but any complex vector space can be completed ● e.g.g. Completing rationale numbers by irrational numbers → limits have to be elements of the space e.. this means that all Cauchy series will converge. is called Hilbert space. set of all real polynomials of order N. scalar product defined as N f N x =∑ a n x n n=0 2 then f N .f M ≝ ∫ f n⋅f m dx 0 sin = limit of the polynomials for N -> ∞ ∞ x3 x 5 sin x=x− ⋯ =∑ a n x n ! 5! n=0Hilbert space: Mathematical correct 3definition of the A linear complex space which is complete with respect to the metric induced by the scalar product. 49 . 2. i=1... ∑ a j j =∑ i ... = i . e. the coefficients ai of the linear combination are given by: i .Basis in Hilbert space There are sets of elements of the Hilbert space. j = ij and complete. The completeness means that any element of the Hilbert space can be represented by a linear combination of the basis. a j j =∑ a j i .g. which are orthonormal i . =∑ a i i i All further properties can be derived from the definition of the scalar product. j =∑ a j ij =a i j j j j 50 . August 1902 Bristol † 20. b=a x b x a y b ya z b z ∥ a∥= ∣a x∣2 ∣a y∣2 ∣a z∣2 real vector The solutions of Schrödingers equation (wavefunctions) are elements of a Hilbert space with a scalar product defined as: . 2 Dirac notation for the scalar product <ψ|φ> bra = <ψ| ket=|φ> 51 . =∫ d r = ∫ d r . * 3 * * * 3 3 * * 3 . 1 ± .t r . 1 ± 2 =∫ 1 ± 2 d r =∫ 1 d r ±∫ 2 d r = . t d 3 r Paul Adrien Maurice Dirac * 8. Oktober 1984 Tallahassee Nobelpreis Physik 1933 . =〈∣ 〉=∫ * r . =∫ d r =∫ d r = . =∫ * d 3 r =∫ * * d 3 r = *∫ * d 3 r * 3 * 3 * 3 .Examples complex vectors ● * * * a . 52 . the dimension of Hilbert space is infinite. The basis is given by a suitable.Comparison N-dimensional vektor space with Hilbert space Vector space a {en} ei·ej=δij a = ∑ ai ei ai = ei·a a·b = b·a Vector Basis Orthogonality linear combination Symmetry scalar product Hilbert space ψ {φ n} <φ i | φj> = δij ψ=∑ a i φi a i = <φi | ψ> <φ | ψ> = < ψ | φ>* The elements of the Hilbert space in QM are normalizable functions. complete and orthonormal set of functions. Mathematical properties 1. . Schwarz inequality ∣ .6. triangle inequality ∥ ∥∥∥∥∥ real vectors = Length a a b ∥ a∥ b a 53 .2. (real vectors ∥ a∥= a a a 2 x 2 y = length of vector) 2 z ∣a b∣=∣a⋅b⋅cos ∣a⋅b ⋅ ∣cos ∣≤1 b a 2.∣∥∥∥∥= . triangle inequality ∥ ∥∥∥∥∥ real vectors = Length a a b ∥ a∥ b a 54 .∣∥∥∥∥= . Mathematical properties 1.6.2. . (real vectors ∥ a∥= a a a 2 x 2 y = length of vector) 2 z ∣a b∣=∣a⋅b⋅cos ∣a⋅b ⋅ ∣cos ∣≤1 b a 2. Schwarz inequality ∣ . z 3 3 ∑ aij b j=ci Einstein: a ij b j =ci Summation over double index (Einstein sum convention) j=1 ● ● ∂ = ∂x Translation operator T x . y . ● A = Matrix in the space of 3D vectors a xx a yx a zx a xy a yy a zy a xz b x cx a yz b y = c y a zz b z cz x 1. z = xa . y 2.3. yb . knowing its result is sufficient! Inversion operator ● P r = −r 55 . Linear operators in Hilbert space Def: An operator maps an element of the Hilbert space on another element.6. zc Operator must not known explicit. p x ]=0 56 . B = x AB≠ B A Def: The combination of two operators A and B in the following way: AB – BA =[A.Def: Operator A is linear if: A 1 1 2 2 =1 A 1 2 A 2 Example for nonlinear operator: Multiplication with ∥∥2 A = . p y ]=0 [ x .B] = C (another operator) is called Commutator. p x ]=i ℏ [ x . x ]=0 [ p x . [ x .g. A = . In general e. A = px. AB= AB . * . px-1 -> integration) AA-1 = 1 e. B = A-1 (only notation not 1/A . −1 = integration ∂ ∂ ∂x ∂x Def: Adjoint operator B is called adjoint operator to A if * . A B = A . B = B A .g.Def: Inverse operator („undo operator“) B is inverse to A if AB = BA = 1. B= A* 1 A1 2 A2 * = *1 A*1 *2 A*2 AB* =B * A* * * * .= A . A = B . AB* =B * A* A** = A Home work 57 . Def: self-adjoint (hermitian) Operator Operator A is called self-adjoint if A* = A.φ) (U*Uψ. All physical meaningful operators are hermitian (self-adjoint).φ) Example: Translation operator Tψ(x) = ψ(x+a) location x shifted by constant a ∞ . = ∫ dx x x −∞ ∞ * * T .Uφ) = (ψ. Def: Unitary Operator Def: Operator U is called unitary if U* = U-1 UU* = U*U = 1 The scalar product is invariant under any unitary transformation.φ) = (ψ. T = ∫ dx xa xa −∞ ∞ = ∫ dy * y y −∞ y= x a dy = dx 58 . (Uψ. x =∫ d 3 r * r x r =∫ d 3 r x r * r position operator x is hermitian Operator (y.ψ) Norm does not change Examples for hermitian operators: 1) . -> r is hermitian partial integration 2) 〈∣ p x 〉= . z ∣xx =∞ =−∞ ℏ ∂ * −∫ d 3 r r r i ∂x = f x g x∣ −∫ dx d f x g x dx 0. =〈 p x ∣ 〉 i ∂x i ∂x 3 59 . y . in case ψ* and/or φ vanish at ∞ * ℏ ∂ ℏ ∂ =∫ d r r r = .Fψ) = (φ. z similar) = x operator . ∫ dx f x dxd g x= ℏ ∂ ℏ =∫ d 3 r * r ∂ r = i ∂x i ∂x =∫ dy dz * x . z x . y .Fourier-transformation F[ψ(x)] = ψ(k) (Fφ. e. ≠0 for . g.px = ℏ ∂ i ∂x p = Is a hermitian operator (py. r =e i k1⋅ r r =e i k2⋅ r r∞ ℏ ∂ 3 −i k ⋅r ℏ ∂ i k ⋅r =∫ d r e e i ∂x i ∂x 1 1 2 =∫ d 3 r e−i k ⋅r x =ℏ k 2 ∫ d 3 r ℏ x i k ⋅r ik e i 2 e 2 −i k1− k2 ⋅r =ℏ k 2x 2 3 3 k1 − k2 60 .pz similar) ℏ grad i is hermitian operator Remark ψ. φ are plane waves. ℏ ∂ . therefore if all components of the momentum operator are hermitian then -> p is hermitian 61 . =∫ d 3 r i ∂x ℏ ∂ i k ⋅r e i ∂x 1 =∫ d 3 r ℏ k 1x e =ℏ k 1 ∫ d r e x 3 i k1⋅ r −i k1⋅r * e * e e i k2⋅ r i k2⋅ r i k2⋅r =ℏ k 1 ∫ d r e x 3 −i k1 − k2 ⋅r =ℏ k 2x 2 3 3 k1 − k2 ' k − k = 3 0 for ≠0 for . k1≠ k2 k1= k2 px is hermitian Similar proof for py and pz. px = p x . Januar 1901 Paris 62 http://turnbull.html .6. The solution of the eigenvalue equation for operator A gives eigenvalues a n and eigenvectors φn. Charles Hermite * 24. A n=a n n In case of an hermitian (self-adjoint) operator A the relation (ψ. 〈 n∣ A n 〉=〈 n∣a n n 〉=a n 〈 n∣ n 〉 〈 * n〈 A n∣ n 〉=〈 a n n∣ n 〉=a n∣ n 〉 a n =a *n The eigenvalues of hermitian operators are always real numbers.mcs.uk/history/PictDisplay/Hermite.ψ) is true. Dezember 1822 Dieuze (Lothringen) † 14.st-and.4 Eigenvalues and expectation values of hermitian operators Measurement values are represented by hermitian operators.Aψ)=(Aψ.ac. If one eigenvalue has several eigenfunctions (degeneracy). By means of normalization of the eigenfunctions we obtain an orthonormal basis in Hilbert space. in case of different eigenvalues an≠am. 63 . 〈 m∣ A n 〉=〈 m∣a n n 〉=a n 〈 m∣ n 〉 〈 A m∣ n 〉=〈a m m∣ n 〉=a *m 〈 m∣ n 〉=a m 〈 m∣ n 〉 The difference of both equations in case of hermitian operators gives: 0=a n −a m 〈 m∣ n 〉 Therefore.The eigenvectors of hermitian operators form a basis in Hilbert space. so that any state can be expressed as a linear combination in that basis. the corresponding eigenfunctions have to be orthogonal (scalar product is zero). it is always possible to build linear combinations of these eigenfunctions which are orthogonal (Gram-Schmidtsch orthogonalisation). A =〈 ∣ A 〉=〈 ∣ A ∑ ci i 〉 =∑ c i 〈 ∣ A i 〉=∑ c i 〈 ∣a i i 〉=∑ c i a i 〈 ∣ i 〉= i i i * i * * =∑ a i c i 〈 ∑ c j j∣ i 〉=∑ ∑ c j a i c i 〈 j∣ i 〉= ∑ ∑ c j a i c i ij =∑ a i c i i j i j i j i c i =∑ a i ∣c i∣2 i That means we will measure the value ai with probability |ci|2. 64 . =∑ ci i =∑ ci ∣ i 〉 i i The expectation value to measure the physical property A in such state will be: 〈 A 〉= .5 Expectation values of hermitian operators Assume we know for an hermitian operator A the eigenvalues and eigenfunctions.6. A n =a n n Then we can represent any state of a physical system by a linear combination using the basis given by the eigenfunctions of A. The eigenvalues of such operators are always real numbers and correspond to the possible measurement values of the physical property. The eigenfunctions of any hermitian operator can be used as basis in the Hilbert space.6 Hermitian operators and measurement All physical properties can be represented by an hermitian (self-adjoint) operator. The measurement of a physical property in a state ψ delivers always an eigenvalue with probability |ci|2. then we will always measure the same eigenvalue corresponding to the eigenstate without uncertainty. The probability is given by the contribution of the corresponding basis function to the state ψ. so that any state can be expressed as linear combination of these basis states.6. 65 . If the system under investigation is already in an eigenstate. B*=B are 2 hermitian linear operators (= physical properties) 1) 1 1 ∣ A . A ∣ 2 2 1 = ∣ . so that we replace these by new ones . B 2 ≥ 1 ∣ .[ A . B − B . A2 . B − A .7. B A . B ] ∣ Because for any complex number z holds: 2 ∣ A . ψ. A B . . φ are arbitrary vectors. ≥ ∣ . AB − . ∣2 Using the hermitian properties and 1) we obtain A . B ∣2 . B ≥ ∣ A . B A ∣ 2 1 = ∣ . B ∣ ≥ ∣z∣= Im z ≥ 2) Schwarz Re z inequality 2 2 1 1 ∣z−z *∣ = ∣2 i Im z∣ = ∣Im z∣ 2 2 Square of the equation ∥∥ ∥∥ ≥ ∣ .∣ ∥∥ = .[ A . B ] ∣2 4 66 . B *∣ = ∣ A . Heisenberg uncertainty principle A*=A. B−number ] = [ A . B ] A B ≥ 4 . number ] ∣ 1 .A.[ A . B are also arbitrary operators. B−〈 B〉 ≥ 2 . if the state ψ is not normalized. B ] − [ A . A2 ⋅ . 2 . ψ)=1. number ] − [number .[ A−〈 A〉 . B B−〈 B 〉 2 1 2 ∣ . B−〈 B〉] ∣ 4 Uncertainty of the measurement A in state ψ. 1 ≥ ∣〈[ A . B ]〉∣ 2 2 ∣ Heisenberg uncertainty relation 67 . which we replace by new ones A A−〈 A〉 . B ] + [number . If ψ normalized then (ψ. 2 .[ A . B ] ∣2 4 [ A−number . A−〈 A〉 A = . A−〈 A〉 . B2 ≥ 1 ∣ . B ]〉∣2 4 2 2 A B ≥ 1 ∣〈[ A . Only if the two operators commute [A. The lower limit is determined by the commutator of the corresponding operators. 68 . B ]〉∣ 2 The uncertainty relation describes the quantum mechanical uncertainty of the measurement process itself. it will be possible to measure the physical properties at the simultaneously without uncertainty. The Heisenberg uncertainty relation gives a lower limit for the product of the uncertainty of physical properties. A B ≥ 1 ∣〈[ A .Heisenberg uncertainty principle The uncertainty is a measure for the deviation of the measurement values from their average.B]=0. p x ] = i ℏ A B ≥ ∣〈[ A . 2 will give reproducible values ℏ z⋅ p z ≥ 2 bigger (or if equal). B. Slit y Δy Δy finite -> Δpy finite Δpy = 0 Δy = ∞ Are exact determined locations with Δx = 0 possible? -> Δpx = ∞ not possible → no ● 69 . which commute ℏ y⋅ p y ≥ [A.g.B. B] = 0 .plane waves: have sharp momentum: Δpx = 0 -> Δx = ∞ Location is completely undetermined. [ x .1 z. depends on the specific state (ψ) ≥ how much e. B ]〉∣ 2 ℏ x⋅ p x ≥ 2 All operators A. Δt ≥ ћ/2 Δt = τ life time of a state I Uncertainty of the spectral line seen as broadening. Therefore.py.E/c) (x.y. because t is not an operator.pz. Origin x = px = 0 corresponds to rest position.ct) -> ΔE. + E 70 .z.px Harmonic Oscillator: x x U E = = k 2 x 2 2 px U 2m Classical mechanic allows any elliptical curve. it must exist a minimal energy. U = 0 and px = 0 is not possible in QM. Theory of relativity -> (px.Δt ≥ ћ/2 can not be derived. because then would be Δx = Δpx = 0. corresponding to zero point motion. px x E min = 1 ℏ 2 = k m Remark: ΔE. Linear harmonic oscillator (1D) classical: E ω = x √ k = 2π f m (frequency) k = 2 m 2 m 2 k 2 -> the ω smaller (deuterated U = x = truexfor all oscillationmaslarger Approximately long as amplitude is small molecules) enough. H can move in x-. z-direction All 12 atoms: 3*12 degrees of freedom 3 translation + 3 rotation -> 36 – 6 = 30 possible vibrations (30 coupled linear harmonic oscillators) 71 .B.8. y-. 2 2 U H z. a n ] 2 a j2 2 j−n = a j1 j2 j ψn ( x) ∼ e −m ω 2 x 2ℏ Hermite Polynomial Hn(y) Hn (√ ) H n x⋅ mω ℏ n d −y = −1 e ne dy n y 2 1 Energy eigenvalues E n = (n+ ) ℏ ω 2 2 72 ...Quantum mechanic: We search stationary states using the stationary Schrödinger equation. H n x = E n n ℏ2 d 2 m 2 2 H n= − x n= E n n 2m dx 2 n 2 substitution = 0 = a 0 e m /ℏ x n x = n −2 2 − 2 2 1 = a1 e ⋅ ⋅ ⋅ −2 2 n n = e [a 0 a 1 a 2 . p x ]= i ℏ=1 2ℏ 2ℏ ℏ ℏ [ A . which requires only the solution of algebraic equations We define two new operator A and A+. x. A+ ] = 73 .Another way to find a solution: Instead of the solution of the differential equation we replace the problem by another one. p x ] [ p x . x]= [ x . px] = 0 we can calculate A* = A = [ ] [ ] [ ][ ] ip −i p x ip m m m −i p x m x. [x. in order to rewrite the Hamilton operator. because A ≠ A+ (not hermitian) from [x. x] = [px. x 2ℏ 2ℏ 2 ℏ 2 ℏ 2 ℏ m 2 ℏ m 2 ℏ m 2 ℏ m −i i −i −i = [ x . A = m x 2ℏ Annihilation operator ip x Creation operator 2 ℏ m ip x m x − 2ℏ 2ℏ m A is not measurable. x x . x . px] = i ħ . Now we rewrite the Hamilton operator with help of A and A+ 2 px m 2 2 H = x 2m 2 2ℏ 1 + x= A A m 2 −i p x = 2 ℏ m A− A + 2 ℏ 2 2 + + +2 x = A AA A A A 2m 1 p 2x =−ℏ m A 2 − AA +− A +A A +2 2 A = A = Inserting in Hamilton operator and using the previous result m x 2ℏ m x − 2ℏ ip x 2 ℏ m ip x 2ℏ m [ A . A +]= A A+− A+ A=1 ℏ ℏ ℏ 2 + + +2 2 + + +2 + + A − AA − A A A A AA A A A = 2 AA A A 4 4 4 ℏ 1 = 2 A +A1=ℏ A+ A 2 2 H = − 74 . This transforms the problem to the solution of the eigenvalue problem of a new operator. N = A +A N is hermitian occupation operator N+ = (A+A)+ = A+ (A+)+ = A+A = N -> real eigenvalues N ∣n 〉 = n∣n 〉 ∣n 〉 Should be a normalized eigenstate = n AA + − A+ A = 1 + = A ∣n 〉 N = A + A A +∣n 〉= A + 1 A + A ∣n 〉= A + 1 N ∣n 〉=n1 A+∣n 〉= n1 therefore A+|n> = ζ is eigenstate of N with eigenvalue (n+1) _ = A ∣n 〉 + + + N _= A A A∣n 〉= AA −1 A∣n 〉= AA A− A ∣n 〉= AN − A ∣n 〉= = A N −1∣n 〉=n−1 A ∣n 〉= n−1 _ 75 . A n ≥ 0 No neigenvalue of N are allowed to be negative! -> There exist a state .In the same way one can show that A∣n 〉 is eigenstate of N with eigenvalue n-1. The Hamilton operator H describes a system of such Vibrational quanta of equal energy (ћω). The operator N just counts the number of vibrational quanta in the state |n> (occupation). where only the number of vibrons n can change. 1 1 H =ℏ A + A =ℏ N 2 2 The vibration of a harmonic oscillator in state |n> contains n vibrational quanta (vibrons). Φ0must different from ∣0 zero 〉 =(||φ 0|| = 1). N n = n . which all have the energy ћω. using A climb down by 1. 0 In state φ0 the operator N has the smallest eigenvalue n = 0. Using A+ one can climb up from n by 1 to the eigenvalue n+1. which will vanish operating on Aφ0 = 0. A A n = A n . because + 〈n∣N∣n〉 = 〈n∣A A∣n〉 = 〈 A n∣ A n〉 + . 76 . Application of A+ (A) on state |n> creates (annihilates) a vibron. Going up is no problem in contrast to going down. Ground state wavefunction φ0(x) = wavefunction of state |0> A 0 x = A ∣0 〉 = 0 m ℏ d x 0 x = 0 2ℏ 2m dx 0 x = c0 e −m 2 x 2ℏ −m d 0 x = c0 x dx ℏ ℏ m = 2m ℏ e −m 2 x 2ℏ = −m x ℏ 1 = ∫ dx *0 x 0 x = c 0 2 −∞ q = 0 m 2ℏ Normalization ∞ ∞ ∫ dx e −m 2 x ℏ −∞ m x ℏ 2 = ∣c0∣ ℏ m dq = ∞ ∫ dq −∞ e m dx ℏ −q 2 c0 = 4 m ℏ 77 . 78 . . . 2. 1.remaining eigenfunctions created by repeatedly operating with A + 1 = 1 4 m 1 d − 0 = d ℏ 2 1 d − d 2 e − 2 2 ⋅ ⋅ ⋅ n = 1 n! 2 n d − d n 0 Hermite Polynomial Hn = e n = x2 2 d x− dx 1 n ! 2n n e m ℏ − 1 4 x2 2 e n = −1 −2 2 H n d n −x e e dx n x2 2 n = 0. . non degenerate with equal energy distance ћω n = 0. we know eigenvalues.Now. … positive integer 1 = ℏ n 2 1 ℏ 2 Zero point energy E0 = 1 E0 = ℏ 2 Not allowed in QM! x px = ℏ 2 79 . eigenvectors of N = A+A 1 = A+ 0 ⋅ ⋅ ⋅ 1 + n n = A 0 n ! En A +∣n 〉 = n1 ∣n1 〉 A∣n 〉 = n ∣n−1 〉 discrete. 1. 2. IR-absorption between states of different parity (dipole selection rules) ~ 〈n∣x∣ m 〉 = ∫ *n x m dx Integral with symmetric limits of an odd function vanishes.The eigenfunction φn has exactly n zeros. n=0 -> n=1 strong transition n=0 -> n=2 forbidden (equal parity) 80 . Importance of the harmonic oscillators for physics Any system which can vibrate (molecules. 81 . 13 C is in higher abundance in biological systems compared to inorganic processes. solids) is approximately a system of coupled harmonic oscillators A transformation to so called „normal coordinates“ changes it into a system of uncoupled 1D harmonic oscillators. and therefore also to the total energy and enthalpy. Confirmation of Planck explanation of black body radiation. Zero point energies are real and contribute to stability (lowering of binding energies). which means is based on the physics of the harmonic oscillator. Quantum electrodynamics is formulated usually in terms of creation and annihilation operators. Two level systems 9. Dipole moment d ∣2 〉 ∣1 〉 82 . but above or below at the same distance and same energy. Ammonia has obviously many different states. The energetic favorable location of N is not in the plane of the H-atoms.9.1 Ammonia (NH3) H-atoms form an equal lateral triangle. however we reduce it to a model with only two states corresponding to the position of the N atom. the states are different and should not overlap. The energy in both of these states will be equal. therefore: 〈1∣H ∣2〉=〈2∣H ∣1〉=V Any state has to be a linear combination of the two basis states ∣ 〉= c1 ∣ 1 〉c 2 ∣ 2 〉 What will be the energy for that state? → Solve stationary Schrödinger equation! H ∣ 〉= E ∣ 〉 We just calculate the scalar product with our two basis functions. so that <1|2>=0.Lets |1> the normalized state with N above the plane of H-atoms and |2> the mirrored one. Assume that there is a finite probability for tunneling of N from |1> to |2> . 〈 1∣ H ∣1 〉=〈 2∣ H ∣2 〉= E 0 However. We can use these two states as basis states in a 2 dimensional Hilbert space. First the left side: 〈 1∣ H ∣ 〉=〈 1∣ H c 1∣1 〉 〈1∣ H c 2∣2 〉= c1 〈 1∣ H ∣1 〉 c2 〈 1∣ H ∣2 〉 =c 1 E 0 c 2 V 〈 2∣ H ∣ 〉=〈 2∣ H c 1∣1 〉〈 2∣ H c 2∣2 〉= c1 〈 2∣ H ∣1 〉 c 2 〈 2∣ H ∣2 〉= c 1 V c 2 E 0 83 . The right side of the equations gives us: 〈 1∣ E∣ 〉=〈 1∣ E c1∣1 〉〈 1∣ E c 2∣2 〉= E c 1 〈1∣1 〉 E c 2 〈1∣2 〉= E c1 〈 2∣ E∣ 〉=〈 2∣ E c 1∣1 〉〈 2∣ E c 2∣2 〉= E c 1 〈 2∣1 〉 E c 2 〈 2∣2 〉= E c 2 This is nothing else than linear set of equations for c 1 and c2. Non-trivial solutions are only possible if the determinant vanishes. c 1 E 0 c 2 V = E c1 c 1 V c 2 E 0 = E c 2 Another. mathematical equal from to express this is based on matrices. E0 V V c1 c =E 1 E 0 c2 c2 E 0− E V V c1 =0 E 0 −E c 2 The solution of the eigenvalue problem gives energy E and the c 1 and c2. det E 0 −E V V = E 0 −E 2−V 2 =0 E 0− E 84 . The energy difference between both states is equal to 2V. than the state with „-“ will be the ground state.87012 GHz (λ=1. because its energy is lower. For NH3 the frequency will be f=23.As a result we find two solutions for the energy E: E +=E 0 V E . 85 .=E 0 −V The corresponding eigenfunctions are: ∣1 〉∣2 〉 ∣ +〉= 2 ∣1 〉−∣2 〉 ∣ . NH3 is absorbing microwaves with that frequency very strongly.〉= 2 If V is positive. 2V=ℏ =h f Electromagnetic waves with frequency f can stimulate transitions between these states.2559 cm) and defines the Frequency standard for microwave technology. det E 0 − F d −E V V = E 0 − F d− E E 0 F d − E −V 2= E 0 −E 2 − F d 2 −V 2 =0 E 0 F d − E E + F = E 0 V 2 F 2 d 2 E .9.2 Stark effect: Ammonia (NH3) in electrostatic field State |1> : Dipole moment d in direction of electric field F. State |2>: Dipole moment against direction of electric field. F = E 0 − V 2 F 2 d 2 86 . This changes the energy E0 in the states: |1> by -F*d and |2> by +F*d E 0 −F d V V c1 c =E 1 E 0 F d c 2 c2 The energy of the stationary state is again given by the vanishing determinant. Compared to electric field strength in atoms these are weak fields. F =E 0 − V 2 F 2 d 2 For small fields the energy changes quadratic with field. The change of energy in electric fields is known as linear and quadratic Stark-effect. Even in case of rather larger electric fields like 10 kV/cm the change corresponds to the quadratic Stark-effect. 87 . for large fields linear.E + F = E 0 V 2 F 2 d 2 E . then there will be some probability for tunneling of the electron between the protons. This lowering of energies is the reason why chemical bonds form.3 Qualitative understanding of chemical bonding E . so that the total energy will be lower. 88 . The basis corresponds to electron at proton one |1> or electron at proton two |2>. ∣1 〉 ∣1 〉−∣2 〉 ∣ . E 0. ∣ +〉= 2 Lets take two protons and one electron (H2+).=E 0 V . ∣2 〉 bonding ∣1 〉∣2 〉 E +=E 0 −V .9.〉= 2 anti-bonding E 0. so that the symmetric state will be the ground state. the sign of V changes to -V (Coulomb potential). In contrast to the NH3 example. If the distance between the protons is small enough. The tunneling of the electrons lowers the kinetic energy of the electron. the square of the momentum goes into kinetic energy. which is always positive. E 0 r 89 . Unfortunately. this would localize the electron in a small volume. Having two protons available then tunneling between these two protons does not localize the electron so much as before and it can access a lower potential energy without paying with increased positive kinetic energy. The electron would prefer to be close to the nucleus.Why has this state a lower energy? The total energy of the electron is the sum of kinetic and potential energy. Now. because the Coulombpotential provides a negative potential energy. Due to Heisenberg uncertainty relation the spatial localization has to be compensated with an increased momentum. The total energy Of the system will be lowered which is the reason why chemical bonds form. Angular momentum operator classical Mechanic: L = r ×p L x = yp z − zp y L y = zp x − xp z L z = xp y − yp x Quantum mechanic: L = r× ℏ y ∂ − z ∂ i ∂z ∂y ℏ Ly = z ∂ − x ∂ i ∂x ∂z ℏ Lz = x ∂ − y ∂ i ∂y ∂x Lx = ℏ ∇ i p ℏ grad i 90 .10. z by there corresponding angular derivatives. --> use spherical coordinates (r. y.Rotations will not change the distance from origin. φ. ∂x ∂ ∂y ∂ ∂z ∂ ∂ = ∂ ∂ ∂ x ∂ ∂ y ∂ ∂ z = −r sin sin ∂ r sin cos ∂ 0 ∂x ∂y = − y∂ x∂ = x∂ − y∂ ∂x ∂y ∂y ∂x Lz = ℏ ∂ i ∂ 91 . ) z r x = r sin cos y = r sin sin z = r cos φ y x We need to replace derivative with respect to x. Now. Lz m = ℏ m m ћm: eigenvalues of Lz Ψm = Ψ0(r. . = 0 r . = Ansatz: ℏ∂ = i ∂ r . ±2... e = i ∂ i ℏ Rotation by 2π should not change the measurement φ and φ + 2π should give the same result e i ℏ = e i 2 ℏ = e i ℏ ⋅e i 2 ℏ 1 i 2 = i 2 m = m ℏ ℏ m = 0. . ) eimφ : eigenfunctions of Lz 92 . we calculate the eigenvalues of Lz : Lz r . . ⋅e i ℏ ℏ∂ ℏ i iℏ = 0 r . ±1. z ] px x [ z . Lx ] = i ℏ Lz = i ℏ Lx = i ℏ Ly The components of the angular momentum operator can not be measured at the same time. p z ] p x x [ z . pz] py = − y [ z . Lz ] [ Lz . pz ] p y iℏ iℏ = i ℏ x p y − y p x = i ℏ Lz [ Lx . one can measure both components sharp. L y ] [ Ly .Can we measure Lx and Ly sharp if we are in an eigenstate Ψm ? Only if the commutatot [Lx. z p x − x pz ] = y pz − z p y z px − x p z − z px − x p z y pz − z p y = y p z z px − y p z x p z − z p y z p x z p y x p z − z p x y p z − z px z p y − x pz y p z x pz z p y = y p z z px − y z p z p x x z p z p y − x p z z p y = y pz z − z p z p x x z p z − p z z p y = y [ pz . L y ] = [ y p z − z p y . Ly] is zero.) 93 . [ Lx . (Vector L can not have a fixed direction. B] = 0 (also true the other way around) 94 . Lz ] 0 = L L z − Lz L L L z − L z L 2y 2 x 2 x 2 y L x Lz L x − L x L z L x L y L z L y − L y L z L y = L x [ L x . How to describe then angular momentum in QM? L2 = L 2x L 2y L2z [ L 2 . L z] Lx L y [ L y . L z ] [ L x . L z ] [ L y .Eigenfunctions of Lx and Ly are not eigenfunctions of Lz. L z ] L y = − i ℏ L x L y − i ℏ L y Lx i ℏ L y Lx i ℏ L x L y = 0 L2 and Lz can be measured at the same time (common set of eigenfunctions) If AΨ =aΨ and B Ψ = b Ψ then: A B Ψ = A (b Ψ) = b A Ψ = b a Ψ B A Ψ = B (a Ψ) = a B Ψ = a b Ψ (A B – B A) Ψ = (b a – a b) Ψ = (a b – b a) Ψ = 0 numbers [A . L z ] = [ L 2x . L z ] [ L 2y . L z ] [ L 2z . = ℏ m Y lm . . . . L = Lx . = ℏ 2 l l 1 Y lm . with l = 0 . 1 . −l 1 . L = . L L y . . . l ∣m∣ ≤ l 95 . Ansatz: r . = f r Y lm . 2 . 0 . Ly 1 ∂ sin ∂ 1 ∂2 sin ∂ ∂ sin 2 ∂ 2 2 z Lz . . . . m = −l . L 2 L2 = −ℏ 2 [ ≥ 0 . . Solving the differential equation (see any text book on QM) gives: L2 Y lm . . . Lz ] L2 contains derivates with respect to both angles -> λ and m are not independent. Lx 2 x 2 y .Eigenvalues of L2 are needed: 2 L = . 3 . . L z Y lm . = spherical harmonics) 2 l1 l −m ! m P l cos ei m 4 l m ! Lz Legendre-Polynome (adjoint to) l+m Y lm .10. = ∓ 5 3cos 2 −1 4 Y 2±1 = ∓ 3 sin e±i 8 15 sin cos e±i 8 15 sin 2 e±2i 2 All spherical harmonics for m≠0 are complex! 96 . = −1m P x = 2l l ! m l m 2 2 1−x d lm d x l m l = 0 1 Y 00 . = l = 2 Y 20 = 1 2 Y 2±2 = 1 4 x 2−1 l see Messiah. Nolting s-state 3 cos 4 Y 1±1 . = 4 l = 1 Y 10 .1 Spherical harmonics Y lm . -1. 0.historical names l l l l l = = = = = 0 1 2 3 4 l=0 s-state p-state d-state f-state g-state ψ(r. 3 . 1. -2. 0. radius ℏ √ 2 L z = −ℏ Lz fixed value. spherical symmetric 2 Classical picture for L und Lz L2 ℏ 2 l l 1 Lz L moves on sphere with radius ∣L∣ ℏ l l1 ℏ l l1 Lz = ℏ Lz = 0 l = 1. 1 m = -2. 0. but smaller than the projection if the radius on L z Lx.φ) = ψ(r) no angular momentum. 2. sharp principal diffuse fundamental m=0 m = -1. 2 m = -3. Ly have no fixed values 97 . -1. 1. Hydrogen atom 2 p H = U r 2 e2 U r = − 4 0 r μ = mass (in pace of m. 2 −ℏ 2 −ℏ 2 ∂ 2 2 ∂ 1 2 p = = − L 2 2 2 ∂ r2 r ∂r r 2 ℏ2 As usual. L z ]=[ L L z ]=0 98 . because m is needed for qunatum number m later on. [H . we search for stationary solutions H ψ = E ψ The Hamilton operator commutes with Lz and L2.Do we need L2 for anything? 2 2 2 2 2 1 2 ∂ ∂ ∂ ∂ = = ∂ − 2 2 L 2 2 2 2 r ∂r ∂x ∂y ∂z ∂r r ℏ 11. which means that the hydrogen wavefunction must contain the already known spherical harmonics. 2 2. L ]=[ H . fnl(r -> ∞) = 0 fnl(r) = Ansatz as polynomial times exponent = e−a r [b nl cnl r d nl r 2 . . but not on the angles itself. 99 . . E nlm 2 ℏ2 ∂ f nl r 2 ∂ ℏ 2 l l1 f nl r = − f r U r f nl r f nl r 2 r ∂ r nl ∂ r2 2 r 2 There is no dependence on m -> Enl We are interested in bound electrons. This gives a condition for allowed energies (see: Mathematica H-Atom. E nlm f nl r Y lm . ] Inserting in differential equation gives constants -> a. [ ] 2 −ℏ 2 ∂ f nl r ℏ2 1 2 ∂ = f nl r U r f nl r Y lm ℏ 2 l l 1 f nl r Y lm 2 2 2 2 r ∂r 2 r ℏ ∂r Differenzial equation for fnl(r) depends on l. = f nl r Y lm . b. .nb).Ansatz: nlm r . -> done nl The wavefunction needs to be normalized so that the series has to be finite.. .. 2 e U r = − 4 0 r In particular for E nl = E n = − 1 E0 n2 no l dependence 2 e 1 E0 = = 13,6 eV = 1 Ry 4 0 2 a0 ℏ2 4 0 −10 a0 = = 0,529177⋅10 m e2 Bohr radius For a general potential the energy will depend on n and l, in case of the Coulomb potential there is only a dependence on n. Main quantum number n = 1, 2, 3, . . . Orbital quantum number l = 0, 1, 2, . . . , n-1 Magnetic quantum number m = -l, . . . , 0, . . . l All ions with 1 electron like He+, Li++, . . . can be described in the same way. Only the potential U(r) changes e2 -> Z e2 Z2 En = − 2 E0 n 100 n=1 l=0 s - state m=0 n=2 l=0 s - state m=0 l=1 p - state m = -1, 0, 1 l=0 s - state m=0 l=1 p - state m = -1, 0, 1 l=2 d - state m = -2, -1, 0, 1, 2 n=3 1 state 4 states 9 states Each state can be occupied by 2 electrons! (Reason is spin → see later chapter) n=1 K – Shell 2 electrons n=2 L – Shell 8 electrons n=3 M – Shell 18 electrons 101 11.1 Radial wave part of hydrogen-like atoms n=1 l=0 n=2 l=0 l=1 fnl(r) r2fnl(r)2 n=3 l=0 l=1 l=2 fnl(r) fnl(r) r2fnl(r)2 r2fnl(r)2 x-axis is for all pictures the radial distance r in units of Bohr radius a 0. The radial functions have n-l-1 zeros and may also become negative. 102 11.2 Construction of real valued angular functions s-wavefunction is spherical symmetric (no angular dependence) l = 0 Y 00 , = 1 4 All spherical harmonics for m≠0 are complex! Building proper linear recombination one can construct completely real valued orbitals. p-wavefunction l = 1 Y 10 , = 3 cos 4 Y 1±1 , = ∓ 3 sin e±i 8 p x= 1 1 3 3 3 x Y 1−1 −Y 11 = sin e−i ei = sin cos = 4 4 r 2 8 2 py= i i 3 3 3 y Y 11Y 1−1 = sin −ei e−i = sin sin = 4 4 r 2 8 2 p z=Y 10 = 3 cos 8 103 s-orbital l=0 p-orbital l=1 Alonso/Finn: Quantenphysik 104 the d-orbitals are constructed the same way: l = 2 Y 20 = 1 2 Y 2±2 = 1 4 5 3cos 2 −1 4 Y 2±1 = ∓ 15 sin cos e±i 8 15 sin 2 e±2i 2 d-wavefunction m =0 d z −r = m =1 d xz = m =1 d yz = 2 2 m =2 dx −y = m =2 d xy 2 2 = 5 2 3 cos −1 16 15 sin cos cos 4 15 sin cos sin 4 15 sin 2 cos 2 4 15 sin 2 sin 2 4 = = = = = 5 3z 2− r 2 16 r2 15 xz 4 r2 15 yz 4 r2 2 15 x − y 2 4 r 15 xy 4 r2 2 105 . d-orbitals l=2 Alonso/Finn: Quantenphysik 106 . ac.f-orbitals l=3 m=±1 m=±2 m=±3 m=0 Many more plots and animations of orbitals and densities: http://winter.group.shef.uk/orbitron/ 107 . nlm r = Z3 2 a30 n2 n−l−1! nl ! l The wavefunctions are orthonormal: 2Z r n a0 L 2l1 n−l−1 2Z r 2Z r exp − Y . n a0 n a 0 lm ∫ d 3 r n*' l ' m ' r nlm r = n ' n l ' l m' m Each l is degenrate with 2l+1 states.605 eV n Solutions with positive energy correspond to scattering of electrons. l and m. With spin one has to multiply the number by108 2! . The complete hydrogen wavefunction depends on 3 quantum numbers n. first excited state n=2 is 4 times degenerated (2s and 3 2p states with equal energy).Summary ● ● ● No planet like orbits There exist bound states with negative energy Z2 En = − 2 E0 E 0 =13. The degeneracy of quantum number n is: n−1 M n=∑ 2 l 1=n 2 l=0 The ground state n=1 is non-degenerate. Energy of the photon must exactly equal to the energy difference of participating levels so that the transition can occur. 109 RH=Rydberg constant . n1=2 Balmer). E =ℏ =2 ℏ =2 ℏ c / 1 E Z 2 e2 1 1 1 1 = = − =R − H 2 2 2 ℏ c 4 a 0 ℏ c n21 n22 n1 n2 For hydrogen gas have been several series observed (n1=1 Lyman. ±1 En/E0 n=3 n=2 n=1 s p d 0 1 2 l An electron transition is always connected with emission or absorption of a photon.Spektral lines ● optical transition are governed by the dipole selection rules l = ±1 m = 0. ● Fine structure: relativistic corrections (Spin-orbit-coupling. d and f-states. 1 1 1 1 = ≈ M Kern me me This allows to observe an isotope effect in optical spectra. Deuterium has been found in 1932 measuring the Rydberg constant very accurately R H/RD=0. which are degenerate in Dirac equation. ● Fine structure constant = 4 0 ℏ2 me 2 ≈ 1 / 137 110 Fließbach:Quantenmechanik .99973.3 further approximation Instead of the electron mass one should use the reduced mass.11. Dirac-equation) Splitting observed in case of p. Hyperfine structure: Interaction of electron spin with nuclear spin ● Lamb-shift: Interaction with vacuum (quantum electrodynamics) gives additional splitting (5μeV) of the 2s1/2 and 2p1/2 states. Particles in magnetic fields Electrodynamics: Charge Q moving in a circular orbit is the same as a current in a loop -> magnetic Dipole Magnetic moment M= current x area M r v Qv Qmv Q r 2= r= rp 2 r 2m 2m Q Q M= r ×p = L 2m 2m The interaction energy of a magnetic dipole M with an external magnetic field B is given by: E =− M⋅ B 111 .12. 27 10-24 J/T The magnetic interaction can be described by an additional term in the Hamilton operator which includes the interaction of the magnetic moment of an electron with a magnetic field: H =− M ⋅ B= B L⋅ B ℏ 112 .QM treatment Replace classical properties by corresponding operators: Electron with charge Q=-e and mass me −e −e ℏ L L M= L= =− B 2 me 2 me ℏ ℏ Bohr magneton µB=eħ/2me=9. -l+1.12..Bz)] Eigenfunctions of Lz are also eigenfunctions ofn H without magnetic field.0.0.. .l Energy levels will then be: Enlm=Enl + m µB Bz 113 ..Lz]=0 (see H-atom) Lzψ=m ħψ mit m=-l.Lz]=[H0.L2]=[L2.. ... because [H0.1 Normal Zeeman-effect Assume hydrogen like atom in a magnetic field with U(r)=U(r) B B Bz H = H 0 L⋅B = H 0 Lz ℏ ℏ [without loss of generality B=(0. atom with d-electrons (d -> l=2) Enlm=Enl + m µB Bz µB Bz E3d m = +2 m = +1 m=0 (2l+1) degnerate states wuth m = -l. 114 .g. This is the reason for calling m magnetic quantum number.…+l B=0 m = -1 m = -2 (2l+1) states with different energy B≠ 0 Splitting in 2l+1 levels by an external magnetic field.normal Zeeman-effect e. 9 transitons allowed by dipole selection rules 3 mesurable spectral lines. 1 15 possible combinations. 0.Numebr of spectral lines m = +2 l=2 m = +1 E3d m=0 m = -1 m = -2 B=0 E2p ΔE B≠ 0 m = +1 l=1 m=0 ΔE+μBB ΔE m = -1 ΔE-μBB Selection rules: Δl=1. 115 . because the levels have equal energy distances. Δm=-1. Fine structure of alkali elements : Yellow D-line of Na is a doublet instead of a single line. Stern-Gerlach experiment Pictures and information on Stern-Gerlach experiment are from Physics Today December 2003: Friedrich and Herschbach. "Stern and Gerlach: How a Bad Cigar Helped Reorient Atomic Physics" http://www.13. Spin Experimental facts: 1. 3.org/vol-56/iss-12/p53.html 116 .physicstoday. Normal Zeeman-effect is the exception: Most often one observes the anormal Zeeman-effect (additional splittings) 2. 13. because F = .grad E =∇ M⋅ F B=∇ − B d Bz L z B z =−m B ℏ dz Therefore. 117 .1 Stern-Gerlach experiment (1921/1922) An inhomogeneous magnetic field produces a force acting on a magnetic moment. particles with different magnetic quantum number m should experience different forces. so that the beam should split. 1969 Berkeley †10.1979 München Physik-Nobelpreis 1943 Post card 8 February 1922 to Niels Bohr Splitting in 2 beams in case of Ag.2.1889 Biebrich am Rhein 17.8. but no splitting in case of Ag+.8. Gerlach Otto Stern 1.Stern-Gerlachexperiment W.1888 Sohrau †17.8. 118 . The calculated magnetic moment for Ag equals 1 µB. 29.g. Phys. we would expect always: • Splitting of the beam in 2l+1 beams • No splitting in case of atoms in an s-like ground state. Cu. Rev.l=0. m=0) 119 .H Phipps and Taylor.Stern-Gerlach experiment for hydrogen Splitting observed in case of alkali elements. Ag. 309-320 (1927) The Magnetic Moment of the Hydrogen Atom Problem: Splitting in two beams l=0 -> 1 beam l=1 -> 3 beams l=2 -> 5 beams So far. hydrogen (n=1. e. m =m s ℏ s . m = ℏ s .=m ℏ Y lm .m 2 3 2 2 2 S s . 1 s= . S y ]=i ℏ S z [ L x . L2 Y lm . m =± ℏ s .2 Explanation of the Stern-Gerlach experiment 1925 by Uhlenbeck and Goudsmit after an idea of Pauli: Elektrons have a permanent magnetic moment = spin In complete analogy to angular momentum we define spin operators with spin quantum numbers s and ms. S y .=l l 1 ℏ2 Y lm . 2 −s≤m s ≤s m s± 1 2 S =S x . L y ]=i ℏ L z 120 Goudsmit Uhlenbeck Pauli . m =s s1 ℏ s . S z Lz Y lm . 1 S z s . m 4 s s s s s s Also identical commutator relations: [ S x .13. t = 1 r .t 1 0 2 r . t + 2 r .t 0 1 121 .13.σz): z= 1 0 0 −1 x= 0 1 1 0 y= 0 −i i 0 ℏ S= 2 These matrices fulfill all commutator relations (home work :-). = 1 r . The wavefunction becomes a vector with two components (spinor).2 Pauli matrices Possible representation of S using Pauli matrices σ = (σx.σy . H= H0 0 0 eB L z 2m 0 H0 0 Lz e ℏ B 1 0 m2 0 −1 Quantitative analysis of experiments The total angular momentum is given by vector addition of orbital momentum and spin J=L+S. Quantum electrodynamics gives g=2.. The total magnetic moment is however not proportional to J. 122 .Using the matrices the Hamilton operator becomes itself matrix form...00231930437. but is given by: e e =− e M L− S = − L g S 2m e me 2m e The Landé-factor g is for an electron equal to 2 within relativistic Dirac theory. ≈ 2 (1+α/2π + ..). Anormal Zeeman-effect: Spin-orbit + Zeeman H SO= S⋅ L l = 1 s = 1/2 Spin-orbit Zeeman j = 3/2 mj mj mj mj = = = = 3/2 1/2 –1/2 –3/2 mj = 1/2 mj = –1/2 j = 1/2 l=0 mj = 1/2 s = 1/2 j = 1/2 mj = –1/2 Quantum numbers mj (j-j coupling) important for elements with large proton number. 123 . 13. orbital quantum number l. magnetic quantum number m and spin quantum number ms. n l m m r = R nl r Y lm . which has no classical counter part. we need the main quantum number n.4 Complete set of quantum numbers for H-atom Hamilton operator for H-atom im magnetic field: B H = H 0 B⋅ L g S ℏ To describe a state of the H-atom uniquely. 124 . The spin represents a fundamental property of the electron. m s 1/ 2 = 1 0 −1/ 2 = s 0 1 The spin functions χ are independent on position and time. positrons etc. Particles with half-integer spin are called Fermions. graviton (s=2) All interactions are exchanged by bosons! Spin can be a result of vector addition of spin of several particles. Ω.5 Fermions and Bosons Not only electrons have a spin but also neutrons. e+ (s=1/2). p.13. Fermions: e. with integer spin bosons. protons. n.g. (e.(s=3/2) All matter is build from fermions! Bosons: photon (s=1). nuclear spins s=3/2 state of 57 Fe) 125 . Particles with half-integer spin are called Fermions.g.(s=3/2) All matter is build from fermions! Bosons: photon (s=1). n. (e. p. Fermions: e. graviton (s=2) All interactions are exchanged by bosons! Spin can be a result of vector addition of spin of several particles. protons.13.5 Fermions and Bosons Not only electrons have a spin but also neutrons. nuclear spins s=3/2 state of 57 Fe) 126 . Ω. positrons etc. with integer spin bosons. e+ (s=1/2). One major result of Pauli principle is an explanation of the order in the periodic table of elements.13. 127 . Fermionic particles are described by anti-symmetric wave functions and can not occupy a single particle state twice. Pauli-Principle: Fermionic particles can not be equal in all quantum numbers.6 Spin and statistic: Pauli principle Bosonic particles can occupy a single particle staremany times and are described by symmetric wavefunctions. Pauli-Principle: Fermionic particles can not be equal in all quantum numbers. One major result of Pauli principle is an explanation of the order in the periodic table of elements.13.6 Spin and statistic: Pauli principle Bosonic particles can occupy a single particle state many times and are described by symmetric wavefunctions. Fermionic particles are described by anti-symmetric wave functions and can not occupy a single particle state twice. 128 . Potential U(r) is not the Coulomb potential anymore! 129 .14 Periodic table 14.1 Atoms with many electrons En/E0 1 E n =− 2 Ry n ground H He Li oder Be n=3 n=2 n=1 0 1 2 l state (1s)1 (1s)2 (1s)2 (1s)2 (1s)2 (1s)1 (1s)2 (2s)1 (2p)1 (2s)2 (2s)1 (2p)2 (2p)1 Something has to be different to the H-atom! Improved approximation: Each electron interacts with the positive nucleus with Z protons and interacts with the other Z-1 electrons of the atom. Result: Enl instead En . En/E0 n=3 Li (1s)2 (2s)1 unique configuration n=2 n=1 0 1 2 l Until Ar (Z = 18) the periodic table behaves normal. then however one finds E3d ≈ E4s 3d – states can be above or below with respect to the 4s – states Similar for 4d above/below 5s. 130 . 5d above/below 6s anomal occupation of side group elements 4f (lanthanide) and 5f (actinide) show anormal behavior. Enl increases mostly with l. Many-body effects and relativistic correction will deliver the correct ordering. Degeneracy for l lifted. then there are no analytic solutions possible.However. One has to use approximations and numerical methods. Degeneracy with respect to m remains. Orginal work of Mendelejew 1869 on the periodic table.Clemens Winkler. In chemistry at Bergakademie Freiberg finds in 1886 the element Germanium in the mineral Argyrodit mined at Grube Himmelsfürst close to Freiberg. 131 . Prof. iap.de/P2K/applets/a3.html 132 .l=0 (s) l=1 (p) l=2 (d) l=3 (f) siehe auch http://www.uni-bonn. electron and neutron) • We need two detectors. rB)|d3 rB 133 .g. Probability to detect particle: dW = |ψ(r)|2 dV 15. Many-body systems (bosons and Fermions) So far: 1 particle in state ψ In order to measure the position we place a detector with volume dV at r. B (e. located at rA and rB Probability. rB)|2 dVA dVB Probability to find particle A at rA dWA = dVA ∫|ψ(rA.1 distinguishable particles • Two distinguishable particles A. that both detectors react at the same time dW = |ψ(rA.15. Similar: Probability. dWB = dVB ∫|ψ(rA. rB .sB) dW A = dV A ∑ ∑ ∫∣ rA . This is fully in accord with our classical expectations. rB. rB) -> ψ(rA. s A . rB)| d3 rA Spin will not really change the discussion.Sum over all possible positions rB for particle B. to detect particle B.sA. ψ(rA. without being interested in particle A. but will just add an additional degree of freedom: 2T : . because B can anywhere in space. . s B ∣ d 3 rB sA sB Up to now there was not really anything new. 134 . . Which electron created the signal at detector 1? Was it electron 1 or electron 2? This question can not be answered.15. that the two-particle wavefunctions ψ(r1. B we will label the electrons with 1 and 2 • at r1 we have positioned detector 1 with dV1 • at r2 we have positioned detector 2 with dV2 If both detectors give a signal at the same time. they are absolutely identical. The main difference in quantum mechanics is.2 Two identical particles (electrons) Instead using A. r1) may be different. ψ(r1. r2): ψ(r2. r2) and ψ(r2. because electrons are really indistinguishable particles. then we know that there has been an electron at r1 and at r2. r1): first electron at r1 and second electron at r2 other way around 135 . . r2) = ψ(r2.Based on the experiment it has to hold that: 2 2 ∣ r1 . . r2 ∣ = ∣ r2 . we have two possible solutions for the wavefunctions which differ in sign. He.). .) This solution corresponds in nature to particles with integer spin (0. r1) (Symmetric functions remain symmetric by additions and multiplication with scalars. A) Symmetric wavefunctions ψ(r1.. = Boson) In particular the many-body wavefunction can be a product of single-particle states. . . (e. 2. r1 ∣ Even if the squares are equal. 1. photonen. r1 r2 = r1 r2 Bosons can occupy single-particle states several times. pion. g. (Bose-Einstein-condensate: Nobel prize 2001) 136 . s2 r2 s2 . s1 = r1 . . s2 1 = -1 Φ=0 ! Fermions can not occupy the same single-particle state several times. . s2 ) = -ψ(r2. s 2 r1 . = Fermion Spin state is important here! ψ(r1. electron. . . . neutrino. s1. myon.B) anti-symmetric wavefunction Particles with half-integer spin (s = ½. proton. r1 s1 = r2 . |ψ| 2 =0 Periodic table of elements 137 . s1 r2 . 3/2. s1 r2 . . r2 s2 = r1 . r2. r1. s1) The anti-symmetric behavior of the many-body wavefunction under particle exchange is nothing else than the well known Pauli principle. It is not possible to have a fermionic state given by a product of one-particle states r1 s1 . s2. ) e.g. m. l. s He: 1s + n. m are equal but s is different allowed forbidden + allowed + + a + r− because 1 a 2 allowed und 2 r a depend on different arguments 2 138 . l.Hydrogen atom n. .15. . Generalization to many particles Notation: r1. . . N) = ψ(2. 2. . .. . s1 = 1 r2. . Definition: Transposition operator Π (exchange of two particles) Π12ψ(1. 2. . . . .N) Eigenvalues needed: Πψ = λψ Π²ψ = Πλψ = λ²ψ = ψ λ² = 1 λ = ±1 Π = Π* Operator Π is self-adjoint and unitary (Π2 = 1 ←→ Π = Π-1 ←→ Π* = Π-1 ) Eigenvalues of Π select between symmetric and anti-symmetric wavefunctions 139 .3. 2.. 1. . . .N) = ψ(1. . 2. . N) (Π83 exchanges 8 with 3) Π12Π12 ψ(1. s2 = 2 etc. . .N) = Π2 ψ(1. 2 .Boson: Every particle can be exchanged with any other particle B 1.2 3.3 .1 ] The bosonic wavefunction ψB has to be normalized.2 .3 1.1 . B= ∑ 140 .2 2. In general one can build a bosonic many-body wavefunction by permutating all coordinates.1 2.1 .2 .3 3.3 .3= [ 1. if Ψ is expressed as product of single-particle functions.3 .3 2.3 − 1.1 3.1 ] χ(Π) = character of permutation +1 = F = -1 for even number of exchanges for odd number of exchanges ∑ This will give determinants.Fermion F = [ 1.3 .2 .2 − 3.1 . 141 .2 − 2.1 .2 . H = ∑ Hi i = 1 1⋅ 2 2 H i i = E i i N=2 1 1 1 2 2 − 2 1 1 2 2 1 Just from normalization if the single-particle 1 1 1 1 2 = 2 wavefunctions are normalized. 2 2 1 2 2 F = ∣ ∣ N=3 F N F ∣ ∣ ∣ ∣ 1 1 1 2 1 3 1 = 1 2 2 2 3 3 ! 2 3 1 3 2 3 3 1 1 1 N 1 = N ! 1 ⋮ N N N are called Slater-Determinants 142 . Hψ = Eψ is known = r1.1 = H 1.2 1 <-> 2 exchange H 2.1 2.1 = E 2.2 1.1 2.2 We want to proof that it holds: H = E H 1. it is true that H 1.Helium atom: 2 electrons 2 2 p p H = 1 2 V r1 V r2 V ∣r1− r2∣ 2 2 Coulomb-interaction Assume the solution.2 2.2 = H 2.1 H 2.2 ] Further. s 2 = 1.2 1. r2 .2 H Π ψ=E Π ψ=Π E ψ (Π−1 H Π) ψ=E ψ=H ψ −1 Π H Π=H H Π=Π H [ H .1 = H 1.2 = E 1.1 = E [ 1. s1 . Π ]=0 143 . There is no superposition of fermionic and bosonic wavefanctions allowed. There will be no mixed character (part boson and part fermion) possible. A particle or system of particles will be either a fermion or a boson. Eigenfunctions have either to be symmetric (boson) or anti-symmetric (Fermion). which means they have a common set of eigenfunctions. 144 .[ H . ]=0 The Hamilton operator H and the operator exchanging particles Π commute. For a small number of . Approximate methods He-atom has 2 electrons 2 H =∑ 2 2 p 2i Ze 1 e 1 − ⋅ 2 4 0 ri 4 0 ∣r1 − r2∣ i=1 Hi H = ∑ H i i Coulomb-between electrons 1 ∑V 2 i≠ j ij What are the stationary solutions ? H =E The solution for independent particles (no e-e interaction) is known form the solution of the Hydrogen atom. electrons one can use numerical methods to solve the The numerical complexity and demand increases exponentially with number of electrons.16. problem exactly. 145 . the e-e interaction depends on the position of both electrons. so that one needs approximate methods to find a solution for realistic problem sizes. However. This does not allow for separation of the problem. Density Functional Theory = DFT The total energy is a functional of the electron density. r1 . r4 . Walter Kohn John Pople 146 .Why do we need approximate methods? Example oxygen: O has 8 electrons The many-body wavefunction depends on position of all 8 electrons. r8 24 coordinates To save the wavefunction we need ψ as function of 24 coordinates. r5 . r6 . r3 . ● 10 values per coordinate 1024 values ● 8 bytes per entry 1025 bytes ● 8. ρ(r) -> 103*8 byte ~ 8 kbyte Nobel prize Chemistry 1998: Walter Kohn and John Pople. r7 . r2 .5*109 bytes per DVD 1015 DVD's ● 10g per DVD 1010 tons If there would be a method which would need only the charge density instead of the full many-body function the problem would be orders of magnitude easier to handle.