Unformatted text preview: HOMEWORK ASSIGNMENT 3 SOLUTIONS
PHYS852 Quantum Mechanics I, Spring 2008 Topics covered: Timeindependent perturbation theory up to 2nd order (and higher for 2level 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 nn = 1 + O(λ3 ). Similarly, prove that orthogonality is is also guaranteed to second order by showing nν = O(λ3 ), where n = ν . Answer: We will start by computing nν to secondorder, 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 nν = 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 nν = 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. Relabelling dummy variables, and canceling identical terms then gives Vnm Vmν Vnν Vνν Vnm Vmν Vnn Vnν Vnm Vmν nν = 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. Twolevel system: For the twolevel system governed by H = δSz + ΩSx with δ > 0, use perturbation theory to compute E1 and E2 including terms up to fourthorder 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 ﬁrstorder 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 ﬁrstorder 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 energyshifts 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 thirdorder 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 fourthorder 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 An(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) − nn(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 secondorder 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) = xn = xUX − 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 nonvanishing corrections to the groundstate 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 ﬁrstorder we have
(1) En = Vnn = dx xφn (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 xintegration 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 nonvanishing 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 energyshift 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 nonzero correction
(1) En = = n(0) X 4 n(0) d4 6n2 + 6n + 3 . 4 Lasty we need to ﬁnd the ﬁrstorder 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 3dimensional 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 ﬁrstorder. 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 nondegenerate 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 twodimensional subspace, so it should have two nonzero 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 nonzero 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 ﬁrstorder 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 ﬁrstorder 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 secondorder 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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 Spring '10
 MichaelMoore
 mechanics, Work, Schrodinger Equation, å å, A† A†, Enm Enν Enν Enν Emν Enν Enm Emν, Vnm Vmν Vnν Vνν Vnm Vmν Vnn Vnν Vnm Vmν

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