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Problem Set 68.04 Spring 2013 Solutions April 2, 2013 Problem 2. (10 points) Finding Meaning in the Phase of the Wavefunction (a) (3 points) We calculate the expectation value of xˆ in the usual way: ∗ (ˆ x)ψnew = dxψnew (x)ˆ xψnew (x) = dxx(eiqx/n ψ(x))∗ eiqx/n ψ(x) = dxxe−iqx/n ψ ∗ (x)eiqx/n ψ(x) = dxxψ ∗ (x)ψ(x) = xo . So we conclude that (xˆ)ψnew = (xˆ)ψ . (b) (4 points) The expectation value of pˆ is given by ∂ ∗ (ˆ p)ψnew = dxψnew (x)ˆ pψnew (x) = −in dx(eiqx/n ψ(x))∗ (eiqx/n ψ(x)) ∂x iq iqx/n ∂ ∗ −iqx/n e ψ(x) + eiqx/n ψ(x) = −in dxψ (x)e n ∂x iq ∂ = −in dxψ ∗ (x) ψ(x) − in dxψ ∗ (x) ψ(x) = q + po , n ∂x which means p)ψnew = (pˆ)ψ + q. (ˆ (c) (3 points) We see that adding an overall factor eiqx/n is equivalent to shift the momen tum expectation value by q, while the position expectation value remains unchanged. This is not to be confused with the multiplication of the wavefunction by a constant complex phase eiα . The latter is x-independent, and does not change anything in the state. 1 Problem Set 6 8.04 Spring 2013 Solutions April 2, 2013 Problem 3. (15 points) Relation between Wavefunction Phase and Probability Current (a) (4 points) The probability current is given by in ∂ψ ∗ ∗ ∂ψ . J (x, t) = ψ −ψ 2m ∂x ∂x (1) Our wavefunction is ψ(x) = A(x)eiθ(x) , so ∂ψ ∂θ ∂A iθ(x) = A(x)i eiθ(x) + e ∂x ∂x ∂x ∂θ ∂A −iθ(x) ∂ψ ∗ = −A(x)i e−iθ(x) + e . ∂x ∂x ∂x and Putting everything together gives in ∂A ∂A n ∂θ 2 ∂θ 2 ∂θ +A − |A| i −A = |A(x)|2 J (x, t) = −|A| i , 2m ∂x ∂x ∂x ∂x m ∂x (2) (3) which is the result we wanted. (b) (4 points) If the wavefunction is real, then ψ = ψ ∗ . The two terms in the probability ∗ current J (i.e. ψ ∂ψ and ψ ∗ ∂ψ ) thus become identical, and when we subtract them ∂x ∂x from each other we get zero. Note that in general, the two terms are complex conjugates of each other, which means the probability current can be written in a slightly more compact form: n ∂ψ ∗ J (x, t) = − Im ψ . (4) m ∂x ∗ has no This conﬁrms our earlier conclusion, because if ψ is purely real, then ψ ∂ψ ∂x imaginary part to it, and J = 0. (c) (3 points) For a plane wave of wavevector k, we have ψ(x) = Aeikx ⇒ dψ = Aikeikx , dx (5) which, upon substitution into our previous equation, gives J (x, t) = nk 2 |A| . m (6) Note that this result holds even if we re-insert the time dependence into our wavefunction, so that ψ(x, t) = Aei(kx−ωt) , because the time-dependent exponential factors cancel. This is simply a statement of the fact that a plane wave represents a constant rate of probability ﬂow. 2 because the combination B ∗ C − C ∗ B is in fact already imaginary. our condition is just that B ∗ C − C ∗ B = 0. t) = − Im (B ∗ C − C ∗ B) = 0. the “Im” label is actually unnecessary. (7) and ∂ψ ∗ ψ = |B|2 e2αx − |C|2 e−2αx + α(B ∗ C − C ∗ B). then ψ ∗ = B ∗ eαx + C ∗ e−αx . because our wavefunction may be complex. If ψ = Beαx + Ce−αx . we get (8) n nα Im |B|2 e2αx − |C|2 e−2αx + α(B ∗ C − C ∗ B) = − Im (B ∗ C − C ∗ B) .(d) (4 points) It is not correct to say that since e±αx is real that the current inside the barrier is zero. m m (9) For the current to vanish. (10) Note that in this particular case. ∂x Plugging this into our expression for the probability current. Thus. since B and C are complex numbers. To see this. we get a minus sign: (B ∗ C − C ∗ B)∗ = BC ∗ − CB ∗ = −(B ∗ C − C ∗ B). (12) 3 . we thus require J (x. (11) which is a smoking gun evidence for a purely imaginary number ((ai)∗ = −ai). note that if we complex conjugate the entire expression. we must have A = 0. We can now construct a dimensionless parameter. ΔxΔp ≥ n2 ). po 2mVo (13) where we’ve kept the factor of 2 for later convenience. (15 points) Odd-parity Energy Eigenstates in the ﬁnite square well (a) (3 points) The parameters which characterize this system are the length L. This means Beκx for x < −L ψ(x) = (18) Ae−κx for x > L.g. We can build a characteristic momentum by taking p2o = 2mVo . the energy V0 and the mass m. where V = 0. 2m dx2 dx2 2m dx2 √ where κ ≡ −2mE/n.Problem Set 6 8. (15) We now proceed to solve the Schr¨odinger equation in the diﬀerent regions. 4 . Note that this obtains when the well is suﬃciently deep and wide. for x > L we must have B = 0 to prevent a divergence as x → ∞. the the evenescent tails are negligibly small and quantum eﬀects can be generally ignored. Recalling that n has units of momentum times length (e. (17) For x < −L. We’ll come back to this in part (c) below. First we deal with |x| > L. Similarly. so E < 0 and κ > 0. the length Ro sets a lower limit on the size for the evanescent tails of bound states in the well: a state with energy E = −Vo would decay as e−x/Ro outside the well. we can trade this momentum for a second characteristic length Ro as.04 Spring 2013 Solutions April 2. all bound states must decay more slowly. This gives n2 d2 d2 ψ n2 d2 (16) − + V ψ(x) = − ψ(x) = Eψ(x) ⇒ = κ2 ψ. 2013 Problem 4. Meanwhile. Ro is thus (up to factor of 2π) the quantum wavelength of a particle of mass m with energy Vo . The general solution to this equation is ψ(x) = Ae−κx + Beκx . We choose to write κ this way because we are searching for bound states. (b) (6 points) The ﬁnite square well has potential V (x) = −V0 0 for − L ≤ x ≤ L for |x| > L. the dimensionless constant go is a measure of how classical the well is – if go >> 1. because otherwise ψ would diverge as x → −∞ and be unnormalizable. Ro = n n =√ . go = L . Ro (14) Physically.. V0 . The sine function is odd while the cosine function is even. with go set to 10. and constant factors like n. This allows to avoid having to deal with the x < 0 region separately. for example. one can show that our transcendental equation thus becomes a pair of equations to be solved simultaneously: y = −z cot z go2 = y 2 + z 2 . which means −Aκe−κL = Dl cos(lL). (23) Taking the ratio of these two equations gives κ = −l cot(lL). Our next step is to match boundary conditions. y ≡ −2mE. A solution exists whenever the curves intersect. The diﬀerential equation is the same as that for a free particle. Putting everything together. In the plot below. what we have here is a transcendental equation for the permissible energies. we see that for the odd-parity states we are considering in this problem. (20) In this problem we are searching for solutions with odd parity. We can therefore say that D = 0. we show y = −z cot z in blue and go2 = y 2 + z 2 in red. At this point we non dimensionalize by letting L L √ L z≡ 2m(E + V0 ). (26a) (26b) Given a value for go . where the Schr¨odinger equation takes the form d2 ψ n2 d2 ψ(x) = Eψ(x) ⇒ = −l2 ψ. one can go back to the deﬁnition of z shown above to calculate the bound state energies E. the A’s) are the same.g. we have ⎧ −κx ⎪ for x > L ⎨Ae ψ(x) = D sin(lx) for − L < x < L (21) ⎪ ⎩ for x < −L. First we require that the wavefunction be continuous at x = L: Ae−κL = D sin(lL). 5 .Now consider the region −L ≤ x ≤ L. Once the values of z at the intersections have been found. so the general solution is ψ(x) = C sin(lx) + D cos(lx). there exist three bound states. In the example shown below. these equations can be solved graphically. (24) Since l and κ contain only E. (22) We also require the derivative to be continuous there. −Aeκx where we have taken advantage of the fact that we are dealing with odd-parity states to insist that the normalization constants on either side of the origin (e. and go = 2mV0 (25) n n n With a little algebra. (19) − + V 0 dx2 2m dx2 where l ≡ 2m(E + V0 )/n. the number of bound states decreases. This makes sense — if go is large. we get (n + 1)2 π 2 n2 En − (−V0 ) ≈ . at multiples of π. where we have set go = 100. because we have only considered solutions with odd parity. we show a variety of go ’s. then we have a very deep well. y 200 100 10 20 30 40 z !100 Substituting this back into our deﬁnition of z. y 10 5 2 4 6 !5 6 8 10 z . The left hand side is the energy of the state above the bottom of the well (which is at energy −V0 ). that is. This can be seen in the plot below. Note that we only get the odd energies here. then the intersections occur very close to the locations of the discontinuities in the cotangent function. In the plot below. 2m(2L)2 (27) where n is odd. Referring back to Problem set 4.y 20 10 2 4 6 8 10 12 z !10 !20 (c) (4 points) If go is very large. we see that the right hand side precisely the energy of the nth energy eigenstate for an inﬁnite potential well of width 2L. As go is decreased. one which is an excellent approximation to an inﬁnite well. i. the evanescent tails become narrower as the well becomes deeper. and conversely. in order to have an odd bound state L has to be large. 7 . holding Ro ﬁxed. Remember that. Ro has to be small. holding L ﬁxed. the square well has to be deeper than a certain value. Since the blue curve crosses the axis at z = π/2. as we saw in class. there is always at least one even bound state. 2 (28) The latter inequality tells us that. This is in agreement with (??). This occurs when go is so small that the red curves hit the horizontal axis while the blue curve is still negative.e. the red curves “sweep left”. the condition for a bound state to exist is go ≥ π 2 ⇒ L≥ π Ro . and eventually we no longer have any bound states.As go decreases. (d) (2 points) As seen from the pictures below. 8 . and we have excluded the decaying exponential solution in the leftmost region and the growing exponential solution in the rightmost region in order to ensure that our wavefunction is normalizable. −Ce−κx (31) (Recall that the hyperbolic cosine and hyperbolic sine functions are the even and odd combinations respectively of real exponentials). we can make our life simpler by remembering that the energy eigenfunctions must be either even or odd: ⎧ κx ⎪ for − ∞ < x < −L ⎨Ae ψeven (x) = B cosh κx for − L < x < L (30) ⎪ ⎩ −κx for L < x < ∞. this is an implicit equation for the bound state energies. where E < form: ⎧ κx ⎪ ⎨Ae ψ(x) = Ceκx + De−κx ⎪ ⎩ −κx Fe 0. (32) dψ dψ 2mW0 − = − 2 ψ(a).Problem Set 6 8. Ae and ⎧ κx ⎪ for − ∞ < x < −L ⎨Ce ψodd (x) = D sinh κx for − L < x < L ⎪ ⎩ for L < x < ∞. Continuity of the wavefunction requires B cosh κL = Ae−κL . 2013 Problem 5. Because our potential is even. dx + dx − n (33) The jump condition is a a so we have −κAe−κL − Bκ sinh κL = − 2mW0 −κL Ae .04 Spring 2013 Solutions April 2. (35 points) Quantum Glue (a) (10 points) For bound states. We can n 0 non-dimensionalize the equation by letting ξ ≡ κL and g0 ≡ mLW : n2 ξ + ξ tanh ξ = 2g0 9 (36) . the wavefunction takes the following for − ∞ < x < −L for − L < x < L for L < x < ∞. First let us deal with the even solutions. tanh κL = n2 κ (34) (35) √ Since κ ≡ −2mE . (29) √ where κ ≡ −2nmE . n2 Eliminating A and B from these equations gives 2mW0 −1 . and the lines corresponding to 2g0 equal to 2 · 0. i. We can then expand ξ + ξ tanh ξ ≈ ξ + ξ 2 ≈ ξ. the greater the value of ξ. the lower (the more negative) the bound state energy. ξ + ξ coth ξ in blue.As for the odd solutions. 2 · 0. only one bound state (the even one) exists.5. grey and light grey. while intersections between the blue and horizontal curves give values of ξ corresponding to odd bound solutions. Note that a larger value of g0 corresponds to a larger value of L. (38) (39) (b) (6 points) These transcendental equations can be solved graphically. respectively. a wider separation of the two delta’s. 2 · 4 in black. (37) while the jump in the slope tells us that Cκe−κL − Dκ cosh κL = Eliminating C and D gives −1 2mW0 tanh κL = −1 n2 κ 2mW0 −κL Ce n2 ⇒ ξ + ξ coth ξ = 2g0 . g0 → 0. Shown below is a plot of ξ + ξ tanh ξ in red. and the more tightly bound the particle is. • As L → 0.e. continuity requires D sinh κL = −Ce−κL . 10 .2. 8 6 4 2 1 2 3 4 There are several noteworthy features here: • For two wells that are close together. Intersections between the red and horizontal curves give values of ξ corresponding to even bound solutions. and the intersection between the red curve and the horizontal lines approaches ξ = 0. The energies can then be found using the deﬁnition of ξ: n2 ξ 2 E=− . (40) 2mL2 Thus. From Equation 40. Indeed. when we get the ﬁrst odd bound state) can be computed by the following technique. coth ξ ≈ 1. the energies of the two states are almost degenerate. The wavefunction may thus be Taylor expanded in κ: ⎧ ⎪ for − ∞ < x < −L ⎨C(1 + κx) ψodd (x) = Dκx (43) for − L < x < L ⎪ ⎩ −C(1 − κx) for L < x < ∞. • As the separation between the wells is increased (i. The continuity and jump conditions are therefore DκL = −C(1 − κL) and Cκ − Dκ = − 11 2mW0 DκL. it’s easy to see from the plot that large g0 gives large values of ξ at the intersections. • For widely separated wells.e. and both the trascendental equations become 2ξ = 2g0 . we see that m(2W0 )2 =− . so we can approximate tanh ξ.so that the corresponding trascendental equation simpliﬁes to ξ = 2g0 ⇒ ξ = 2. n2 (44) . this is because the wells are so close together that they eﬀectively look like one well. so it must be even.e. g0 is large and both the even and odd solutions ap proach ξ = g0 . we eventually get to a point where there are two bound states. as L. this is because two widely-separated wells eﬀectively operate as independent single wells. we eventually get to a point where κ = n2 is approximately zero (the odd “bound-state-in-waiting” is barely bound). Physically. • When there are two bound states. This makes sense because we know the ground state cannot have any nodes by the node theorem. we thus see the even state is always more tightly bound than the odd state. Correspondingly. • The separation L at which the second bound state appears (i. with energy close to mW02 Ef ar = − . 2n2 Eclose (41) which is precisely the ground state energy for a single delta function well of strength 2W0 ! Physically. and therefore g0 is increased). As we √ 2mE increase g0 (and therefore L). the even bound state always has a higher value of ξ than the odd bound state. (42) 2n2 which is the ground state energy of a single delta function potential. g0 Plugging this value into our formula for the energy. for g0 = 10. 2 2mW0 (46) (c) (5 points) From part (b) we know that g0 = 0. (45) We wish to take the limit κ → 0.5 0.8 1.0 Finally. Ñ2 Ñ2 Ψ´ Ψ´ m W0 m W0 0. because we want to ﬁnd the g0 (and therefore L) when the second bound state ﬁrst appears. but if we study g0 = 0.4 -4 0.6 0.0 0.2 -10 -5 -2 2 4 xL -0.8 0. barely-normalizable state is called a threshold bound state. it becomes normalizable.Solving these gives (1 − 2g0 )(κL − 1) = κL. a second odd solution appears with zero energy (and thus zero curvature). both even and odd solutions exist and are in fact all but degenerate: 12 .6 0. Technically this is not normalizable.5 + f for any inﬁnitessimal f. Such a barely-bound. Doing so tells us that g0crit 1 n2 = ⇒ Lcrit = .1 gives a single even bound state: Ñ2 Ψ´ m W0 1.5 xL 5 10 -1.0 0.5.4 -20 -10 10 20 xL For g0 = 0. 4 0.3 -1 1 2 xL -0.8 1.6 0. Doing so yields the blue line in the plot below.6 0.4 0.6 xL (d) (3 points) From the plots.2 0. we can see that the ground state becomes more tightly bounded.Ñ2 Ñ2 Ψ´ Ψ´ m W0 m W0 0.2 0. Using Equation 40. 0 E m W0 2 2 Ñ2 1 0. we can see that as the wells as brought closer together. the value of ξ for the ground state decreases. The induced force between wells is: F =− d d 1 d E(x) = − E(L) = − E(L).2 0.4 0.4 -2 0.0 L Ñ2 m W0 -1 -2 -3 -4 (e) (5 points) Going to lower L minimizes the energy.5 0. We can (numerically) solve for ξ as a function of L.2 -0. so the wells “want” to be close together. where the energy is shown in mW 2 n2 units of 2n2 0 and L is in units of mW .1 -2 -1 1 2 -0.6 0.7 0. dx d(2L) 2 dL (47) In terms of dimensionless variables (indicated in parentheses) the force induced by the ground state looks like: 13 . and then use Equation 40 to plot the ground state energy E as a function of L. whereas when they are far apart the energy levels are approximately the same as those of a single well of width L. say. 14 . one sees that there exists an equilibrium L. the energy levels move from the eigenvalues of the well of width 2L to the eigenvalues of the well of width L. increasing the distance between the two wells.8 1. and the net result would be the red curve.F m2 W0 3 2 Ñ4 0. In that case. The two pictures below show the situation.6 0.8 1. Thus.4 0. E m W0 2 2 Ñ2 1 0.6 0. of width L.2 0. are very close to each other. We then expect that increasing the separation between the two wells. the repulsive eﬀect between the nuclei would dominate at really small L. some of the levels will converge to the same eigenvalue at large enough distance.2 0.4 0. we can consider them as one only well of width 2L. Note the decrease in the number of energy levels as the distance of the wells is increased.0 L Ñ2 m W0 -20 -40 -60 -80 -100 -120 -140 (f) (3 points) If delta functions were real protons.0 L Ñ2 m W0 -1 -2 -3 -4 (g) (3 points) When the two wells. From Problem 4 we know that a square well of width 2L has more energy eigenstates than a well of width L of the same depth. The reason this does not violate the Node Theorem is that the antisymmetric combination has an extra zero half way between the two wells. etc. being the symmetric superposition of the second excited states of the individual wells.For two separated wells. the second two states. which are the 2nd and 3rd excited states. 15 . the ﬁrst two states are symmetric and antisymmetric super positions of the ground states of the individual wells. are symmetric and antisymmetric superpositions of the ﬁrst excited states of the individual wells. but it diﬀers from the number of nodes of the 4th excited state. so we have the same number of nodes inside each well for the 2nd and the 3rd excited state.. 16 . ˆ )† = B ˆ † Aˆ† . (Bf (55) Putting everything together. (15 points) Further Facts about Hermitian Operators and Commutators (a) (2 points) Note the following useful property: Z ∗ Z ∗ ∗ (f |g) ≡ f gdx = g ∗ f dx = (g|f ).Problem Set 6 8. treating it as an operator that acts on the state Aˆ† g: Now we do the same for B ˆ |Aˆ† g) = (f |B ˆ † Aˆ† g). If we let Aˆ = in we have Z Z ∗ (51) f ingdx = (−inf )∗ gdx. 48 gives us the answer: ˆ |g) = (g|Af ˆ )∗ . (Aˆ(Bf (54) ˆ .04 Spring 2013 Solutions April 2. (AˆB 17 (56) . it is useful to go back to writing out the integrals that deﬁne the adjoint. where we have simply used the deﬁnition of complex conjugation. So we see that the adjoint of Aˆ = in must be (52) Aˆ† = −in. where we thought of AˆB ˆ and to use the The trick now is to treat Aˆ as an operator acting on the state Bf deﬁnition of the adjoint to “bring Aˆ back”: ˆ )|g) = (Bf ˆ |Aˆ† g). (f |(AB (53) ˆ as a single operator and moved it to the left as one “package”. (c) (2 points) First we use the deﬁnition of the adjoint: ˆ ˆ )† g) = (AˆBf ˆ |g). (48) But we already know that: ˆ |g). (f |Aˆ† g) = (Af (49) which combined with Eq. we see that in general. 2013 Problem 6. (f |Aˆ† g) = (Af (50) (b) (2 points) To ﬁnd the adjoint of in. while.ˆ ]. The ﬁnal expression ˆ J] ˆ = −J. eigenvalues of iCˆ are real. using hermiticity of A. our assumption that Jˆ is Hermitian). (1 point) The operator J easily proven by contradiction. Let us rewrite Jˆ as − 1s [K. Z Z Z ∗ ˆ ∗ dx φa A φa = dx φa a φa = a dx |φa |2 (60) ˆ we also have that. Since we have a contradiction. ie ˆ a = aφa . (57) We now use hermiticity of A and B to conclude that this equals ˆ Aˆ − AˆB ˆ = −[A. ˆ B ˆ ]† = (AˆB ˆ −B ˆ Aˆ)† = (AˆB ˆ )† − (B ˆ Aˆ)† = B ˆ † Aˆ† − Aˆ† B ˆ† Cˆ † = [A. (59) Aφ To see that a must be real. (63) i.e. our algebra implies that Jˆ† = −J. ˆ† J =− 18 . † 1 ˆ ˆ 1 ˆ ˆ ˆ ˆ [K. which is something that is mostly i. J] = − K J − JK s s † 1 ˆ † Jˆ† ) = 1 (K ˆ ˆ† − K ˆ Jˆ − JˆK). Let us consider Cˆ † : ˆ B (d) (1 point) The operator Cˆ is deﬁned as Cˆ ≡ [A. Z Z Z Z ∗ ˆ ∗ ∗ ∗ ˆ dx |φa |2 . contradicts our assumption of Jˆ† = Jˆ (i. Suppose we were to assume that Jˆ were Hermitian. on the one hand. (g) ˆ is not Hermitian. ˆ Thus.e a is real. ˆ which is equal to + 1s [K. ˆ Jˆ] and consider Jˆ† : so that Jˆ† = Jˆ. so eigenvalues of Cˆ must be purely imaginary. As a result. we are forced to conclude that Jˆ is not Hermitian. dx φa A φa = dx (A φa ) φa = dx (a φa ) φa = a (61) This can only be true if a = a∗ (62) ˆ (iCˆ )† = −iCˆ † = iC. ˆ B] ˆ = −Cˆ =B (58) (e) (2 points) Let φa be an eigenfunction of the hermitian operator Aˆ with eigenvalue a. (f) (1 point) Consider the operator (iCˆ ): which means that iCˆ is Hermitian. note that. = − (Jˆ† K s s (64) ˆ is Hermitian and our (soon to where in the last equality we used the fact that K be proved incorrect!) assumption that Jˆ is also Hermitian. are two results (which really imply each other): ˆ Jˆ] = −sJˆ and [K. [K. ˆ Jˆ − JˆK) ˆ † = − [K.e. then. ˆ J] ˆ + JK ˆ ˆ )ϕk = −sJϕ ˆ k + k Jϕ ˆ k = (k − s)(Jϕ ˆ k ). ˆ J] ˆ = −sJˆ. (66) ˆ k = kϕk (i. (65a) (65b) where ˆ to replace K ˆ with K ˆ †. ˆ Jˆ† ] = +sJˆ† . as we now show: ˆ (Jϕ ˆ k ) = K ˆ Jϕ ˆ k = ([K. (1 point) Similarly. we see that Jˆ is a lowering operator that acts on an ˆ and gives a result that is directly proportional to an eigenstate of eigenstate of K ˆ with an eigenvalue that is s lower than the eigenvalue of the original eigenstate. we have: ˆ (Jˆ† ϕk ) = K ˆ Jˆ† ϕk = ([K. ˆ Jˆ† ] = [K ˆ † . Jˆ† ] = K ˆ † Jˆ† − Jˆ† K ˆ J) ˆ † = (JˆK) ˆ † − (K ˆ† [K. we have ii. ˆ Jˆ† ]+ Jˆ† K)ϕ ˆ k = +sJˆ† ϕk +kJˆ† ϕk = (k+s)(Jˆ† ϕk ). (1 point) Starting with [K. • In the ﬁrst equality we used the Hermiticity of K • In the 3rd equality of the ﬁrst line we used Equation 56 from part (a). (1 point) If Kϕ ˆ ˆ the state (Jϕk ) is also an eigenfunction of K with eigenvalue (k − s). then iii. (1 point) From part iii.ˆ Jˆ† ]. ˆ J] ˆ = −(K † = −(−sJˆ)† = sJˆ† . [K. The algebraic structure explored in this problem is precisely analogous to the one in the quantum harmonic oscillator. K (67) iv. What we have. One simply makes the following identiﬁcations: Harmonic Oscillator ˆ K Jˆ (Lowering) ˆ ≡a Eˆ or N ˆ† a ˆ a ˆ 19 Jˆ† (Raising) a ˆ† . • In the penultimate equality we used our deﬁnition of Jˆ. (68) K v.e. K The operator Jˆ† “raises” in an analogous way. i. ϕk is an eigenstate of K ˆ with eigenvalue k). visit: http://ocw.MIT OpenCourseWare http://ocw. .edu/terms.mit.mit.edu 8.04 Quantum Physics I Spring 2013 For information about citing these materials or our Terms of Use.