Quantum Mechanics 1 - Homework 2 with Solutions | PHYS 5260, Assignments of Quantum Mechanics

Download Quantum Mechanics 1 - Homework 2 with Solutions | PHYS 5260 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics II: HW 2 Solutions Leo Radzihovsky February 13, 2008 1 Anharmonic Oscillator We shall examine the anharmonic oscillator with perturbation theory. The Hamiltonian is: H = p2 2m + 1 2 mω2x2 + λx4. (1) (a) Energy correction The exact energy can be written: En = E 0 n + E 1 n + · · · , (2) with E0n = ~ω ( n + 12 ) , and the first order correction given by: E1n = 〈n|λx4 |n〉 . (3) We will use the second quantization formalism to evaluate E1n. Now, x = 1√ 2 x0 ( a† + a ) , (4) with x0 ≡ √ ~ mω . Thus, x4 = x40 4 [ (a†)4 + (a†)3a + (a†)2aa† + a†a(a†)2 + a(a†)3 + (a†)2a2 + a2(a†)2 + a(a†)2a + a†a2a† + · · · ] . (5) The other terms can easily be determined. This is quite a large collection of terms. It is easy to see that the only ones that contribute are those with equal numbers of creation and annihilation operators, since we are taking the expectation value of the above expression. So we get can write our correction term as: 1 E1n = λx40 4 〈n| a† aa†−1 ︷︸︸︷ a†a a + aaa†a† + a†a+1 ︷︸︸︷ aa† a†a + a†a a†a+1 ︷︸︸︷ aa† +a†aa†a + (a†a+1)(a†a+1) ︷ ︸︸ ︷ aa†aa† |n〉 (6) = λx40 4   n2 − n + (n + 2)(n + 1) ︸ ︷︷ ︸ n2+3n+2 + 2(n + 1)n ︸ ︷︷ ︸ 2n2+2n + (n + 1)2 ︸ ︷︷ ︸ n2+2n+1 +n2    (7) = 3~3λ 4m2ω2 [ 1 + 2n + 2n2 ] (8) ⇒ En = ~ω ( n + 1 2 ) + 3~3λ 4m2ω2 [ 1 + 2n + 2n2 ] , (9) to first order in λ. (b) Limit of validity For large enough n, perturbation theory clearly breaks down, since: E1n E0n = 3~λ 4m2ω3 1 + 2n + 2n2 n + 1/2 n→∞≈ 3~λ 2m2ω3 nλ  1, (10) Thus, if n is large, E1n E0n  1, no matter how small λ is. Physically, for large n, E is large, and thus 〈 x2 〉1/2 is large: Classically For large n, we can treat the oscillator classically: En = ~ω(n + 1/2) ≈ ~ωn ≈ 1 2 mω2x2n (11) ⇒ xn ≈ x0 √ n (12) Quantum mechanically Since Psi goes like xn exp (−x2), our variance 〈 x2 〉1/2 ≡ x̄ is set by: n ln x̄ ≈ x̄2 ⇒ n x̄ ≈ x̄ ⇒ x̄ ≈ √ n (13) Thus, in either case, we can see: E1n = λ 〈 x4 〉 ≈ λx40n2. (14) Since E0n goes like ~ωn, we can see that E 1 n  E0n as n increases. Thus, at large energies the harmonic approximation breaks down. 2 where we get the last line from the Wigner-Eckart theorem. So the first-order correction term vanishes. We move on to the second-order term: E2Sz = ∑ S′z ∣ ∣ ∣ 〈 S0z ∣ ∣− γ~S · ~B⊥ ∣ ∣ ∣S′ 0 z 〉∣ ∣ ∣ 2 E0Sz − E0S′z (36) E2+z = γ2B2⊥ 4 ∣ ∣ 〈 +0z ∣ ∣S+ + S− ∣ ∣−0z 〉∣ ∣ 2 E0+z − E0−z (37) = γ2B2⊥ −4 (~γB0) ~ 2 ( 1 2 3 2 + 1 2 1 2 ) ︸ ︷︷ ︸ 1 = −~ 4 γB2⊥ B0 (38) Recalling the form of the energy E±z = E0+z + E 1 +z + E 2 +z + · · · , we have: E±z ≈ ∓ 1 2 ~γB0 ∓ 1 4 ~γB2⊥ B0 + · · · (39) ≈ ∓1 2 ~γB0 ( 1 + 1 2 ( B⊥ B0 )2 + · · · ) , (40) which matches our expansion in part (a). We finally find the eigenstates to first order: |Sz〉 = ∣ ∣S0z 〉 + ∑ S′z 〈 S0 ′ z ∣ ∣ ∣− γ~S · ~B⊥ ∣ ∣S0z 〉 E0Sz − E0S′z ∣ ∣ ∣S0 ′ z 〉 , (41) which leads to expressions in the ± basis: ∣ ∣+~B 〉 = ∣ ∣+0z 〉 + γB⊥ ~γB0 〈 −0z ∣ ∣ Sx ︸︷︷︸ 1/2(S++S−) ∣ ∣+0z 〉 ∣ ∣−0z 〉 (42) = ∣ ∣+0z 〉 + B⊥ 2B0 ( 1 2 3 2 − +1 2 −1 2 )1/2 ︸ ︷︷ ︸ 1 ∣ ∣−0z 〉 , (43) where we have used the fact that S+ ∣ ∣+0z 〉 = 0, and S− ∣ ∣+0z 〉 = ~ (S (S + 1) − Sz (Sz − 1))1/2 ∣ ∣−0z 〉 . So our final expression for the eigenstates to first order are: ∣ ∣±B̂ 〉 ≈ ∣ ∣±0z 〉 ± B⊥ 2B0 ∣ ∣∓0z 〉 , (44) which agrees with what we found in part (a). 5 3 The Thomas-Reiche-Kuhn sum rule We first prove the Thomas-Reiche-Kuhn sum rule: ~ 2 2m = ∑ n′ (En′ − En) |〈n′| x̂ |n〉|2 ︸ ︷︷ ︸ 〈n|x̂|n′〉〈n′|x̂|n〉 (45) = 1 2 ∑ n′   〈n| x̂Ĥ − Ĥx̂ ︸ ︷︷ ︸ −[Ĥ,x̂] |n′〉 〈n′| x̂ |n〉 + 〈n| x̂ |n′〉 〈n′| Ĥx̂ − x̂Ĥ ︸ ︷︷ ︸ [Ĥ,x̂] |n〉    (46) = 1 2 〈n| [ x̂, [ Ĥ, x̂ ]] |n〉 (47) = 1 2 〈n| [ x̂, [ p̂2 2m + V (x̂) , x̂ ]] |n〉 = 1 2 〈n|   x̂, p̂ m [p̂, x̂] ︸ ︷︷ ︸ −i~    |n〉 (48) = −i ~ 2m 〈n| [x̂, p̂] ︸ ︷︷ ︸ i~ |n〉 = ~ 2 2m . (49) Now we test it with the harmonic oscillator, V (x̂) = 12mω 2x̂2. We first note that x̂ = 1√ 2 x0 ( a† + a ) and En = ~ω ( n + 1 2 ) . (50) 〈n′| x̂ |n〉 = 1√ 2 x0 (√ n + 1δn′,n+1 + √ nδn′,n−1 ) , (51) since a† |n〉 = √ n + 1 |n + 1〉 (52) a |n〉 = √ n |n − 1〉 . (53) (54) So, |〈n′| x̂ |n〉|2 = x 2 0 2 ((n + 1) δn′,n+1 + nδn′,n−1) . (55) We can thus write the Thomas-Reiche-Kuhn sum rule as: ∑ n′ (En′ − En) |〈n′| x̂ |n〉|2 = x20 2 [~ω (n + 1) − ~ωn] (56) = ~ωx20 2 = ~ω~ 2mω (57) = ~ 2 2m , (58) So it works! 6 4 Band structure We examine the following Hamiltonian: H = p2 2m + Vbox(x) ︸ ︷︷ ︸ H0 +Vions(x), (59) with Vions(x) a generic periodic potential. (a) Perturbation theory on unspecified Vions(x) We first need to find the zeroth order eigenenergies and eigenstates. The eigenvalue equation H0Ψ 0 = E0Ψ0 gives us the following energies and states: E0k = ~ 2k2 2m , Ψ0k(x) = 1√ L eikx (60) k = 2π L n, L → ∞. (61) The first order corrections are given by: E1k = 〈 k0 ∣ ∣Vions(x) ∣ ∣k0 〉 (62) = 1 L ∫ L 0 dx Vions(x) = Ṽ Q=0 ions (63) E1k = ṼQ=0, (64) where from now on the “ions” subscript is understood. It is easy to see from the above that the first-order shift is independent of k. We now calculate the second-order shift: E2k = ∑ k′ ∣ ∣ 〈 k′0 ∣ ∣V ∣ ∣k0 〉∣ ∣ 2 E0k − E0k′ . (65) The matrix element is given by: 〈k′|V |k〉 = 1 L ∫ L 0 dx ei(k−k ′)xV (x) (66) = ˜V (k′ − k) = ∑ Qn ṼQnδQn,k′−k, (67) where V (x) = ∑ Qn ṼQne iQnx. (68) So our second order energy is given by: E2k = ∑ k′ ∑ Qn ∣ ∣ ∣ṼQn ∣ ∣ ∣ 2 δQn,k′−k ~2k2 2m − ~ 2k′2 2m (69) = ∑ Qn ∣ ∣ ∣ṼQn ∣ ∣ ∣ 2 ~2k2 2m − ~2(k+Qn) 2 2m . (70) 7 Ek k −Q 2 Q 2 Figure 4: Splitting the degeneracy at k = k±. Each point is a separated from the parabola by V1/2. (e) Degenerate perturbation theory at k± Recalling equations 68 and 75, we can write our degeneracy matrix as: 〈 k′0 ∣ ∣Vions ∣ ∣k0 〉 ︸ ︷︷ ︸ 2×2 matrix = V1 2      0 δQ1,k+ − k− ︸ ︷︷ ︸ Q1 + δ−Q1,k+ − k− ︸ ︷︷ ︸ Q1 δQ1,k− − k+ ︸ ︷︷ ︸ −Q1 + δ−Q1,k− − k+ ︸ ︷︷ ︸ −Q1 0      = V1 2 ( 0 1 1 0 ) = V kk ′ ions. (79) Diagonalization leads to the following eigenvectors and eigenvalues: |±〉 = 1√ 2 ( 1 ±1 ) , λ± = ± V1 2 . (80) Thus, the eigenstates and eigenenergies at k± are given by: ΨQ1+ (x) = √ 2 L cos (Q1x) (81) ΨQ1− (x) = i √ 2 L sin (Q1x) (82) EQ1± = ~ 2Q21 2m ± 1 2 V1. (83) 10 Ek k −Q 2 Q 2 Q Figure 5: Two points near k±, separated by Q1. These points are typical the study in part(f). (f) Behavior near k± We generalize to nearly degenerate perturbation theory for pairs of k’s connected by Q1, i.e. k ≈ k+ and k − Q1. The matrix we concern ourselves with is: Hkk′ =          ~ 2k2 2m ︸ ︷︷ ︸ E0+ 0 0 ~ 2 (k − Q1)2 2m ︸ ︷︷ ︸ E0−          + V kk ′ ions = ( E0+ 1 2V1 1 2V1 E 0 − ) . (84) The characteristic equation is: ( E0+ − E±k ) ( E0− − E±k ) − 1 4 V 21 = 0 (85) E0+E 0 − − E±k ( E0+ − E0− ) + ( E±k )2 − 1 4 V 21 = 0. (86) Solving for E±k , we get: E±k = 1 2 ( E0+ + E 0 − ) ± 1 2 √ √ √ √ √ ( E0+ + E 0 − )2 − 4E0+E0− ︸ ︷︷ ︸ (E0+−E0−) 2 +V 21 (87) = 1 2 ( E0+ + E 0 − ) ± 1 2 √ ( E0+ − E0− )2 + V 21 . (88) 11 Ek k −Q 2 Q 2 Figure 6: Taylor expansion gives the shape of the dispersion near k±. In this case, the deviation is parabolic. We first Taylor expand about k+ = Q1 2 : E±k ≈ Ek+ + 1 2 ( E0+ − Ek+ ) + 1 2 ( E0− − Ek+ ) ± 1 2 V1 ( 1 + ( E0+ − E0− )2 2V 21 ) (89) ≈ ~ 2Q21 8m + ~ 2 (k − Q1/2)2 2m ± V1 2 ± ( ~ 2m )2 1 4V1 (2Q1 (k − Q1/2))2 ︸ ︷︷ ︸ 1 V1 ~2Q2 1 2m ~2(k−Q1/2) 2 2m (90) E±k ≈ ~ 2Q21 8m ± V1 2 ± ( ~ 2Q21 2mV1 ± 1 ) ︸ ︷︷ ︸ ≈ ~ 2Q2 1 2mV1 ~ 2 (k − Q1/2)2 2m , (91) with the factor to the right of the parenthesis indicating a parabolic deviation from the k± position; see figure 6. Now we Taylor expand far from k±: E±k ≈ 1 2 ( E0+ + E 0 − ) ± ∣ ∣E0+ − E0− ∣ ∣ 2 ( 1 + V 21 2 ( E0+ − E0− )2 ) (92) ≈ 1 2 ( E0+ + E 0 − ) ± ∣ ∣E0+ − E0− ∣ ∣ 2 ± (V1/2) 2 E0+ − E0− (93) ≈ E0± ± V 21 /4 E0+ − E0− , (94) 12 h h h 2 (A−B) (A+B) 0 E m=0 m=+1 A A=0, B=0 m=+1 m=−1 m=0 m=+1 m=0 A=0, B=0 A=0, B=0 Figure 8: Degeneracy splitting as parameters A and B are “turned on.” since S± |m〉 = ~ √ S(S + 1) − m(m ± 1) |m ± 1〉, and S = 1. upon diagonalization, H (1)mm′ has the following eigenstates and eigenvalues: |±〉 = 1√ 2   1 0 ±1   , E (1) ± = ±B~2 (114) ⇒ E± = (A ± B) ~2, E0 = 0. (115) 6 Anisotropic 2D harmonic oscillator (a) Isotropic oscillator We first look at the unperturbed Hamiltonian: H0 = p2x 2m + p2y 2m + 1 2 mω2 ( x2 + y2 ) . (116) i. Eigenenergies and eigenstates The eigenvalue equation is H0Ψnxny (x, y) = EnxnyΨnxny (x, y). (117) Since H0 = H0x + H0y, i.e., it is separable into two simple harmonic oscillators, we can immediately write down the eigenenergies and eigenstates: Enxny = ~ω (nx + ny + 1) (118) Ψnxny (x, y) = NnxNnyHnx (x/l0) Hny (y/l0) exp ( −x 2 + y2 2l20 ) =< xy |nxny〉 = 〈x| 〈y| |nx〉 |ny〉 ,(119) 15 hω |00> |10>, |01> |20>, |11>, |02> Figure 9: The isotropic energy spectrum. The degeneracy grows with energy. where l0 = √ ~ mω , and Hi are the Hermite polynomials. ii. Spectrum sketch see figure 9. (b) Symmetry-breaking perturbation We now add a perturbation H1 = λxy to H0. i. Shift in spectrum of |00〉, |10〉, and |01〉 in perturbation theory We first look at the ground state, |00〉. Since E000 is non-degenerate, we use non-degenerate perturbation theory: E00 = E 0 00 + E 1 00 + E 2 00 (120) = ~ω + 〈00|λxy |00〉 + ∑ nx, ny 6=0 |〈00|λxy |nxny〉|2 −~ω (nx + ny) . (121) We use the second quantization formalism, with x = l0√ 2 ( a†x + ax ) , y = l0√ 2 ( a†y + ay ) (122) where [ ai, a † j ] = δij , |nxny〉 = NnxNny ( a†x )nx ( a†x )ny |00〉 . (123) So we can immediately find the first order correction: E100 = 〈00| ( a†x + ax ) ( a†y + ay ) |00〉 = 0, (124) since a† |n〉 = √ n + 1 |n + 1〉, a |n〉 = √n |n − 1〉, and the states are orthogonal. The second order correction is more involved: E200 = − ∑ nx, ny 6=0 ∣ ∣〈00| ( a†x + ax ) ( a†y + ay ) |nxny〉 ∣ ∣ 2 ~ω (nx + ny) l40λ 2 4 (125) = − ∑ nx, ny 6=0 ∣ ∣〈00|a†xa†y + axa†y + a†xay + axay |nxny〉 ∣ ∣ 2 nx + ny l40λ 2 4~ω (126) = − l 4 0λ 2 8~ω = − ~ 2λ2 8~ωm2ω2 = − ~λ 2 8m2ω3 , (127) 16 |10>, |01> |00> 0 0 0 λ=0λ=0 |00> |−> |+> Figure 10: Splitting the degeneracy with λ. where we have used 〈0| a† = 0 to eliminate the first three terms in equation 126. So the ground state energy, to second order in λ, is: E00 = ~ω − ~λ2 8m2ω3 = ~ω ( 1 − λ 2 8m2ω4 ) . (128) A note on validity: since we require that the correction be much smaller than the zeroth order term, we can cast that as a constraint on λ: 1 8 ( λl20 ~ω )2  1. (129) The correction to the eigenstate is now straightforward: |00〉 = ∣ ∣000 〉 − 〈 110 ∣ ∣λxy ∣ ∣000 〉 2~ω ∣ ∣110 〉 (130) = ∣ ∣000 〉 − λl 2 0 4~ω ∣ ∣110 〉 . (131) Now for the lowest excited states |10〉 and |01〉. Since the two states have the same zeroth order energy, we must use degenerate perturbation theory. Thus, we need to compute the matrix 〈nxny|H1 ∣ ∣n′xn ′ y 〉 ≡ H1~n~n′ , with |nxny〉 the degenerate subset: H1~n~n′ = ( 0 λl20/2 λl20/2 0 ) . (132) The upper-right element, corresponding to 〈10|H1 |01〉, comes from the a†xay term of xy, since all others vanish in this 2×2 subspace. Similarly, the lower-left element arises from the axa†y term. We can diagonalize this matrix with the eigenvectors: |±〉 = 1√ 2 (|10〉 ± |01〉) , (133) leading to the resulting energies: E± = 2~ω ± λl20 2 = 2~ω ± λ~ 2mω = 2~ω ( 1 ± λ 2mω2 ) . (134) ii. Spectrum sketch See figure 10. 17