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Unformatted text preview: HOMEWORK ASSIGNMENT 3 SOLUTIONS PHYS852 Quantum Mechanics I, Spring 2008 Topics covered: Time-independent perturbation theory up to 2nd order (and higher for 2-level system)and the degenerate case 1. Orthogonality: Start from equation (33) in the lecture notes, and prove that in 2nd order nondegenerate perturbation theory the perturbed states are normalized to second order, by showing explicitly n|n = 1 + O(λ3 ). Similarly, prove that orthogonality is is also guaranteed to second order by showing n|ν = O(λ3 ), where n = ν . Answer: We will start by computing n|ν to second-order, then we can set ν = n at the end for the normalization condition. Eq. (33) gives and |Vmn |2 (0) n |+ 2 Emn λ2 2 µ=ν λ2 n| ≈ 1 − 2 m=n m=n ∗ Vmn −λ + λ2 Emn k =n ∗ ∗ ∗ Vmk Vkn Vmn Vnn (0) m |, − 2 Emn Ekn Emn Vµκ Vκν Vµν Vνν . − 2 Eµν Eκν Eµν ∗ Taking the inner product, keeping only terms to second order, and using Vmn = Vnm and Enm = −Emn gives λ2 |Vmn |2 λ2 |Vµν |2 Vnk Vkν Vnν Vnn Vnν n|ν = δnν 1 − − + λ2 − λ2 + (1 − δnν ) λ 2 2 2 2 Emn 2 Eµν Enν Enk Enν Enν m=n µ=ν k =n Vnm Vmν Vnk Vkν Vnν Vνν Vnν − λ2 + λ2 − λ2 + (1 − δnν ) −λ 2 Enν Enν Eκν Enν Enm Emν κ =ν m=n,ν |ν ≈ |ν (0) 1 − |Vµν + 2 Eµν |2 µ=ν Vµν |µ(0) −λ + λ2 Eµν κ =ν We recognize that the second λ2 term cancels the term in the first sum where m = ν , and likewise with the third and fourth terms. Thus the expression simplifies to n|ν = δnν + (1 − δnν ) Vnm Vmν m=n,ν We proceed by multiplying the last term by δnν + (1 − δnν ) and collecting the δnν and (1 − δnν ) terms seperately. Re-labelling dummy variables, and canceling identical terms then gives Vnm Vmν Vnν Vνν Vnm Vmν Vnn Vnν Vnm Vmν n|ν = δnν +(1−δnν ) λ2 − λ2 + λ2 − λ2 − λ2 . 2 2 Enm Enν Enν Enν Emν Enν Enm Emν m=n m=ν m=n 1 1 1 + − Enm Enν Enν Emν Enm Emν 1 = δnν + (1 − δnν ) = δnν + (1 − δnν ) = δnν + O(λ ), 3 Vnm Vmν m=n,ν Emν + Enm − Enν Enm Emν Enν Em − Eν + En − Em − En + Eν Enm Emν Enν (0) (0) (0) (0) (0) (0) Vnm Vmν m=n,ν (1) which verifies both normalization and orthogonality are correct to second order. 2 2. Two-level system: For the two-level system governed by H = δSz + ΩSx with δ > 0, use perturbation theory to compute E1 and E2 including terms up to fourth-order in Ω. Expand the exact solution in Taylor series around Ω = 0 and compare the two results. Answer: It is straightforward to approach this problem by iterating the general formulas ( Enj ) = n(0) |V |n(j −1) − j −1 k =1 ( Enk) n(0) |n(j −k) n (0) |n (j ) 1 =− 2 j −1 k =1 n(k) |n(j −k) |V |n (j −1) |n (j ) = |n (0) n |n (0 (j ) − m=n |m(0) Emn m (0) − j −1 k =1 ( Enk) m0 |n(j −k) using vector/matrix representation. We start from E1 = − E2 = (0) (0) δ , 2 |1(0) ↔ |2(0) ↔ 2 01 10 0 1 1 0 δ , 2 V= E21 = δ, We then derive the first-order energy shifts E1 (1) E12 = − δ = = 1(0) |V |1(0) 01 2 01 10 0 1 (2) =0 and likewise we can get E2 = 0 Since we know that n(0) |n(1) always vanishes, we can compute the first-order state corrections |1(1) = |1(0) 1(0) |1(1) − 1 δ 1 − 2δ 0 (1) ↔− = and likewise we get 1 0 |2(0) (0) 2 |V |1(0) E21 01 10 2 10 0 1 |2(1) ↔ 0 1 2δ 3 The second order energy-shifts are then E1 (2) = = 1(0) |V |1(1) − E1 01 4δ E2 = (2) (1) 2 01 10 1(0) |1(1) 1 − 2δ 0 =− and likewise 4δ Now we have 1(0) |1(2) 1 (1) (1) 1 |1 2 1 1 ↔− − 2δ 0 2 1 =−2 8δ = − 1 − 2δ 0 so that we can compute |1(2) = |1(0) 1(0) |1(2) − 1 8δ2 0 − 812 δ 2(0) |2(2) = − |2(2) ↔ The third-order shifts are then E1 (3) |2(0) E21 10 2(0) |V |1(1) − E1 2 01 10 (1) 2(0) |1(1) ↔− = and likewise 0 1 − 1 δ 1 − 2δ 0 1 8δ2 812 δ 0 = = 1(0) |V |1(2) − E1 01 2 (1) 01 10 1(0) |1(2) − E1 0 − 812 δ (2) 1(0) |1(1) =0 (3) and likewise Now 1(0) |1(3) 1 2 1 ↔− 2 =0 = − 1(1) |1(2) + 1(2) |1(1) 1 − 2δ E2 = 0 (3) 0 0 − 812 δ + 0 − 812 δ 1 − 2δ 0 4 and |1(3) = |1(0) 1(0) |1(3) − 1 δ 3 16δ3 |2(0) E21 2 2(0) |V |1(2) − E1 01 10 0 (1) 2(0) |1(2) − E1 + δ 4 10 (2) 2(0) |1(1) 0 − 812 δ ↔− = and likewise 1 0 10 − 812 δ 0 |2(3) ↔ Lastly we compute the fourth-order level shift E1 (4) 0 3 − 16δ3 (2) = = = 1(0) |V |1(3) − E1 01 16δ3 2 (1) 01 10 1(0) |1(3) − E1 3 16δ3 1(0) |1(2) − E1 01 (3) 0 + 4δ 0 − 812 δ 1(0) |1(1) and likewise E2 = − Putting the pieces together gives E1 ≈ − E2 ≈ Ω2 Ω4 δ − + 2 4δ 16δ3 (4) 16δ3 Ω2 Ω2 δ + − 2 4δ 16δ3 The exact solutions, expanded to fourth order in Ω are E1 = − E2 = 2 δ 2 + Ω2 ≈= − δ 2 + Ω2 ≈= δ Ω2 Ω4 − + 2 4δ 16δ4 δ Ω2 Ω4 + − 2 2 4δ 16δ3 which agree perfectly with the results from perturbation theory. 5 3. Consider the shifted harmonic oscillator H= P2 M ω2 2 + X + aX. 2M 2 Use perturbation theory to compute the eigenvalues to second order in a and the eigenstates to first order. Compare with the Taylor’s expansions of the exact results. Answer: We can rewrite the Hamiltonian as H= (0) aλ (0) En |n(0) n(0) | + √ A + A† 2 n=0 √ n|(n − 1)(0) ∞ where En = ω n + 1 , and λ = M ω . The operators A and A† satisfy A|n(0) = 2 √ λ and A† |n(0) = n + 1|(n + 1)(0) . With V = √2 A + A† we find λ (1) En = √ n(0) | A + A† |n(0) = 0 2 and |n(1) =− λ m(0) | A + A† |n(0) √ 2 ω (m − n) m=n √ (0) √n|(n − 1)(0) + n + 1|(n + 1)(0) λ (0) m | |m = −√ m−n 2 ω m=n √ √ nδm,n−1 + n + 1δm,n+1 λ |m(0) = −√ m−n 2 ω m=n |m(0) = −√ √ √ λ − n|(n − 1)(0) + n + 1|(n + 1)(0) 2ω (4) For the second order energy shift we find (2) En =− √ √ λ2 (0) n | A + A† − n|(n − 1)(0) + n + 1|(n + 1)(0) 2ω λ2 (0) n| n(n − 1)|(n − 2)(0) − n|n(0) + (n + 1)|n(0) + =− 2ω λ2 λ2 =− (−n + n + 1) = − 2ω 2ω (n + 1)(n + 2)|(n + 2)(0) Putting the pieces together gives En ≈ ω n + and 1 2 − a2 λ2 2ω √ √ aλ n|(n − 1)(0) − n + 1|(n + 1)(0) |n ≈ |n(0) + √ 2ω a2 λ2 2ω The exact solution for the energy is En = ω n + 6 1 2 − so that the second-order result is exact. The exact solution for the wavefunction is |n = UX − aλ2 ω |n(0) , where UX (d) = e−idP/ is the shift operator, defined by UX (d)|x = |x + d . To verify that this is the solution we can compute the wavefunction via ψn (x) = x|n = x|UX − aλ2 ω |n(0) = x + aλ2 (0) aλ2 |n = φn x + ω ω where φn (x) is the unperturbed nth harmonic oscillator wavefunction. This gives the correct result: the wavefunction is simply shifted to the left by the introduction of the linear potential. i To expand the exact solution to first order in a we use P = − √2λ A − A† and the first order expansion UX − aλ2 ω = ei aλ2 P ω ≈1+i aλ2 A − A† −i √ ω 2λ aλ A − A† =1+ √ 2ω thus to first order |n = UX aλ2 |n(0) ω aλ = 1+ √ A − A† |n(0) 2ω √ aλ √ n|(n − 1)(0) − n + 1|(n + 1)(0) = |n(0) + √ 2ω − , which miraculously agrees with the perturbation theory result. 7 4. Find the first non-vanishing corrections to the ground-state energy and wavefunction of the harmonic oscillator when an anharmonic term is added tot he potential. First consider the asymmetric anharmonic oscillator P2 M ω2 2 H= + X + λX 3 , 2M 2 then do the same for symmetric anharmonic oscillator H= M ω2 2 P2 + X + λX 4 . 2M 2 Answer: We can proceed similarly to the last problem, but with V = X 3 . Here there is a problem with notation in that λ is the perturbation parameter, and the same symbol is typically used for the Harmonic oscillator length. For this reason we will use d instead for the oscillator length. For first-order we have (1) En = Vnn = dx x|φn (x)|2 = 0, so the first order terms vanish due to parity. For the first order shift to the state we get |n(1) = − |m(0) m(0) |X 3 |n(0) ω (m − n) m=n The matrix element n(0) |X 3 |m(0) can be computed either as an x-integration or using A and A† . Choosing the latter, we find X3 = = 3 d3 √ A + A† 22 d3 √ AAA + 3(N + 1)A + 3A† (N + 1) + A† A† A† , 22 where N = A† A Thus we have m(0) |X 3 |n(0) = = d3 √ m(0) | AAA + 3(N + 1)A + 3A† (N + 1) + A† A† A† |n(0) 22 √ d3 √ n(n − 1)(n − 2)δm,n−3 + 3 n3 δm,n−1 + 3 (n + 1)3 δm,n+1 22 + (n + 1)(n + 2)(n + 3)δm,n+3 which gives as the first non-vanishing correction |n(1) = d3 √ 6 2ω − √ n(n − 1)(n − 2)|(n − 3)(0) + 9 n3 |(n − 1)(0) − 9 (n + 1)3 |(n + 1)(0) (n + 1)(n + 2)(n + 3)|(n + 3)(3) For the second order energy we now have (2) En = = n(0) |V |n(1) d6 n(0) | AAA + 3(N + 1)A + 3A† (N + 1) + A† A† A† 24 ω 8 n(n − 1)(n − 2)|(n − 3)(0) = d6 n(n − 1)(n − 2) + 27n3 − 27(n + 1)3 − (n + 1)(n + 2)(n + 3) 24 ω d6 =− 30n2 + 30n + 11 9ω √ +9 n3 |(n − 1)(0) − 9 (n + 1)3 |(n + 1)(0) − (n + 1)(n + 2)(n + 3)|(n + 3)(0) For X 4 it is a little easier, since the leading energy-shift is first order. For this case we have X4 = = 4 d4 A + A† 4 d4 AAAA + (AA + A† A† )(2N + 1) + (2N + 1)(AA + A† A† ) + A† A† A† A† + 6N 2 + 6N + 3 , 4 where we used A† A = N and AA† = N + 1, where N is the number operator. Which gives as the first non-zero correction (1) En = = n(0) |X 4 |n(0) d4 6n2 + 6n + 3 . 4 Lasty we need to find the first-order correction to the state |n(1) =− =− |m(0) m(0) |V |n(0) Emn m=n δ4 4 m=n |m(0) m(0) | AAAA + (AA + A† A† )(2N + 1) + (2N + 1)(AA + A† A† ) ω (m − n) +A† A† A† A† + 6N 2 + 6N + 3 |n(0) =− d4 4ω |m(0) (m − n) n! δm,n−4 + (4n − 2) (n − 4)! (n + 4)! δm,n+4 n! n! |(n − 2)(0) (n − 2)! n! δm,n−2 (n − 2)! m=n +(4n + 6) = d4 16 ω (n + 2)! δm,n+2 + n! n! |(n − 4)(0) + (8n − 4) (n − 4)! (n + 2)! |(n + 2)(0) − n! −(8n + 12) (n + 4)! |(n + 4)(0) n! which concludes the problem. 9 5. Consider a 3-dimensional Hilbert space spanned by the states |1 , |2 , and |3 . Let the unperturbed Hamiltonian be 16 4 28 H0 = 4 1 7 . 28 7 49 Let the perturbation operator be 0 3 −5 V = 3 0 3 . −5 3 0 Find the eigenvalues of H = H0 + λV to second order in λ, and find the eigenstates of H to first-order. Answer: To begin with, we find the eigenvalues of H0 . From det |H0 − ω | = 0, we find the characteristic equation is 66ω 2 − ω 3 = 0 Thus the eigenvalues are E1 = 66 and E2 = E3 = 0. The non-degenerate eigenvector in column vector representation is 0.492366 4 1 |1(0) ↔ √ 1 = 0.123091 66 0.86164 7 (0) (0) (0) It can be verified that PD has eigenvalues 1, 1, 0, which is indicative of a projector onto a twodimensional subspace. We can next compute VD = PD V PD by matrix multiplication, yielding 50 −4 −28 0 3 −5 50 −4 −28 1 VD ↔ −4 65 −7 3 0 3 −4 65 −7 662 −28 −7 17 −5 3 0 −28 −7 17 3.09275 1.28834 −1.95133 1.28834 = −1.04913 −0.586318 −1.95133 −0.586318 1.19881 The projector onto the degenerate subspace is therefore PD = 1 − |1(0) 1(0) | sentation gives 100 16 4 28 50 −4 0 1 0 − 1 4 1 7 = 1 −4 65 PD ↔ 66 66 001 28 7 49 −28 −7 which in matrix repre −28 −7 17 The next step is to find the eigenvalues of VD . We know that VD lives in a two-dimensional subspace, so it should have two non-zero eigenvalues, and one zero eigenvalue. This is the case, as the eigenvalues are v1 = −1.4176, v2 = 4.66003, v3 = 0. the normalized eigenvectors corresponding to the non-zero eigenvalues are −0.268324 −0.827996 |v1 ↔ 0.9632 , |v2 ↔ −0.238943 0.0157282 0.507275 We now identify the ‘good’ basis vectors as |2(0) = |v1 and |3(0) = |v2 . 10 The first-order correction to the energy eigenvalues are E1 = 1(0) |V |1(0) = −3.24242 E2 = 2(0) |V |2(0) = −1.4176 and E3 = 3(0) |V |3(0) = 4.66003 The first-order corrections to the energy eigenvectors are |1(1) = + |m(0) m(0) |V |1(0) −0.0354919 = 0.0675929 0.010625 (1) (1) (1) m=1 66 |2(1) = −|1(0) and |3(1) = −|1(0) Lastly, the second-order energy shifts are E1 = − (2) (2) −0.036827 1(0) |V |2(0) = −0.00920675 66 −0.0644473 −0.00917086 1(0) |V |3(0) = −0.00229272 66 −0.016049 | m(0) |V |1(0) |2 = −0.00594137 662 m=1 E2 = − and E3 − Putting the pieces together gives (2) | 1(0) |V |2(0) |2 = −0.00559444 662 | 1(0) |V |3(0) |2 = −0.000346932 662 E1 ≈ 66 − 3.24242λ − 0.00594137λ2 E3 ≈ 4.66003λ − 0.000346932λ2 and 0.492366 − 0.0354919λ |1 ≈ 0.123091 + 0.0675929λ 0.86164 + 0.010625λ −0.268324 − 0.036827λ |2 ≈ 0.9632 − 0.00920675λ 0.0157282 − 0.064447λ −0.827996 − 0.00917086λ |3 ≈ −0.238943 − 0.00229272λ 0.507275 − 0.016049λ E2 ≈ −1.4176λ − 0.00559444λ2 11 ...
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