852_2010hw6_Solutions - PHYS852 Quantum Mechanics II,...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 6: Solutions Topics covered: Time-independent perturbation theory. 1. [30] Two-Level System: Consider the system described by H = δSz + ΩSx , with δ > 0, where Sx and Sz are components of the spin vector of an s = 1/2 particle. Treat the Sz term as the bare Hamiltonian. (a) [15] Use perturbation theory to compute the eigenvalues and eigenvectors of H . Compute all terms up to fourth-order in Ω. The bare eigenstates are |1(0) = |↓ (1) = |↑ (2) = − δ/2 (3) = (4) (0) |2 with bare eigenvalues (0) E1 (0) E2 δ/2 we therefore need to iterate the following equations up to j = 4: j −1 ( Enj ) = |n (j ) n(0) |V |n(j −1) − = |n 1 − 2 (0) ( Enk) n(0) |n(j −k) k =1 (5) j −1 n k =1 (k ) |n (j −k ) − m=n |m (0) m(0) |V |n(j −1) − Emn j −1 k =1 (k ) En m(0) |n(j −k) Emn (6) for j = 1 we have (1) En = Ω n(0) |Sx |n(0) = 0 (7) and for the states we have |n(1) = − m=n |m(0) Ω m(0) |Sx |n(0) = − Emn which gives m=n Ω |m(0) 2Emn Ω |↑ 2δ Ω = |↓ 2δ |1(1) = − |2(1) (8) (9) (10) for j = 2 we have (2) (1) En = Ω n(0) |Sx |n(1) − En = Ω n(0) |Sx |n(1) which gives (2) E1 = − Ω2 Ω2 ↓ |Sx | ↑ = − 2δ 4δ 1 (11) (12) (2) E2 = Ω2 Ω2 ↑ |Sx | ↓ = 2δ 4δ (13) and for the states |n(2) = |n(0) =− Ω2 8δ2 − (1) 1 (1) (1) n |n 2 − m=n |m(0) En Ω m(0) |Sx |n(1) − m(0) |n(1) Emn Emn |n(0) (14) For j = 3 we have for the energies (3) (1) (2) En = Ω n(0) |Sx |n(2) − En n(0) |n(2) − En n(0) |n(1) = 0 (15) and for the states we have |n(3) 1 = − |n(0) 2 n(1) |n(2) + n(2) |n(1) (1) − m=n (2) |m(0) En En Ω m(0) |Sx |n(2) − m(0) |n(2) − m(0) |n(1) Emn Emn Emn |m(0) En Ω3 + m(0) |n(1) 2E 16δ mn Emn (2) = m=n which gives Ω3 Ω3 3Ω3 + 3= |↑ 16δ3 8δ 16δ3 |1(3) = | ↑ |2(3) = | ↓ (16) − (17) Ω3 Ω3 3Ω3 − 3 =− |↓ 16δ3 8δ 16δ3 (18) And finally, for j = 4, we have (4) (1) (2) (3) En = Ω n(0) |Sx |n(3) − En n(0) |n(3) − En n(0) |n(2) − En n(0) |n(1) (2) = Ω n(0) |Sx |n(3) + Ω2 En 8δ2 which gives (4) E1 = (4) E2 = − (19) 3 Ω4 Ω4 Ω4 − = 32δ3 32δ3 16δ3 (20) Ω4 Ω4 3 Ω4 + =− 32δ3 32δ3 16δ3 (21) and for the states, we have |n(4) 1 = − |n(0) 2 n(1) |n(3) + n(2) |n(2) + n(3) |n(1) (1) − m=n = |n(0) |m(0) (2) (3) En En En Ω m(0) |Sx |n(3) − m(0) |n(3) − m(0) |n(2) − m(0) |n(1) Emn Emn Emn Emn − n(1) |n(3) − Ω4 128δ4 (22) 2 which gives |1(4) = | ↓ 3Ω4 Ω4 11Ω4 − |↓ = 32δ4 128δ4 128δ4 (23) |2(4) = | ↑ 3Ω4 Ω4 11Ω4 − |↑ = 32δ4 128δ4 128δ4 (24) Putting the pieces together gives E1 ≈ − E2 ≈ and |1 ≈ | ↓ |2 ≈ | ↓ 1− δ Ω2 Ω4 − + 2 4δ 16δ3 (25) δ Ω2 Ω4 + − 2 4δ 16δ3 Ω2 11Ω4 + +|↑ 8δ2 128δ4 3Ω3 Ω − +|↑ 2δ 16δ3 1− − (26) 3Ω3 Ω + 2δ 16δ3 11Ω4 Ω2 + 8δ2 128δ4 (27) (28) (b) [5] Expand the exact eigenvalues and eigenvectors around Ω = 0 and compare to the perturbation theory results. The exact results are Ω2 Ω4 δ + E1 = − δ 2 + Ω2 ≈ − − (29) 2 2 4δ 16δ3 Ω2 Ω4 δ + − E2 = δ 2 + Ω2 ≈ (30) 2 2 4δ 16δ3 (δ + (δ2 + Ω2 )| ↓ − Ω| ↑ 11Ω4 3Ω3 Ω Ω2 +|↑ − + ≈|↓ 1− 2 + |1 = (31) √ 8δ 128δ4 2δ 16δ3 (δ + δ2 + Ω2 )2 + Ω2 √ ω | ↓ + (δ + δ2 + Ω2 )| ↑ Ω 3Ω3 11Ω4 Ω2 |2 = (32) ≈|↓ − +|↑ 1− 2 + √ 2δ 16δ3 8δ 128δ4 Ω2 + (δ + δ2 + Ω2 )2 (c) [10] Verify that the states computed in (a) are normalized to unity and orthogonal up to fourthorder. 2 1|1 2 2|2 2 Ω2 Ω 11Ω4 3Ω3 = 1− 2 + +− + + O(Ω5 ) 8δ 128δ4 2δ 16δ3 11Ω4 Ω4 Ω2 3Ω4 Ω2 + + 2− + O(Ω5 ) = 1− 2 + 4δ 64δ4 64δ4 4δ 16δ4 = 1 + O(Ω5 ) (33) 2 Ω Ω2 3Ω3 11Ω4 + 1− 2 + + O(Ω5 ) − 2δ 16δ3 8δ 128δ4 3Ω4 Ω2 11Ω4 Ω4 Ω2 − +1− 2 + + + O(Ω5 ) = 4δ2 16δ4 4δ 64δ4 64δ4 = 1 + O(Ω5 ) = It is clear by inspection of Eqs. (27) and (28) that 1|2 = 0. 3 (34) 2. [20] Resonance-frequency shifts: Consider a system with a 3-dimensional Hilbert space spanned by states |a , |b , and |c . In the basis {|a , |b , |c }, let the bare Hamiltonian of the system be 1 −1 1 H 0 = ∆ −1 1 −1 . (35) 1 −1 3 For the case where the system is perturbed by the operator 1 10 V = χ 1 −3 1 , 0 12 (36) also given in {|a , |b , |c } basis. Calculate the shifts in the resonance frequencies of the full system relative to those of the unperturbed system, to second-order in χ. The eigenvalues of H0 are the solutions to det |H0 − ω | = 0, which yields the characteristic equation (1 − ω ) [(1 − ω )(3 − ω ) − 1] − [−1 + 3 − ω ] + [1 − (1 − ω )] = 0 (37) (1 − ω )ω (ω − 4) = 0 (38) which simplifies to (0) (0) (0) so that the eigenvalues of H0 are E1 = 0, E2 = ∆, and E3 = 4 ∆. The three bare resonance frequencies are therefore (0) (0) = E2 − E1 ω2 = (0) E3 − E2 (0) E3 − E1 (0) ω1 (0) and (39) = 3∆, (40) = 4∆. (41) (42) (0) (0) ω3 = = ∆, (0) The eigenvectors satisfy 1− (0) En ∆ −1 1 −1 1− 1 (0) En ∆ −1 −1 3− taking a|n(0) = 1 gives for n = 1, (0) En ∆ a|n(0) b|n(0) = 0 c|n(0) − b|1(0) + c|1(0) = −1 (43) − b|1(0) + 3 c|1(0) = −1 (44) 1 |1(0) = √ [|a + |b ] 2 (45) − b|2(0) + c|2(0) = 0 (46) which has the solution b|1(0) = 1 and c|1(0) = 0, so that for n = 2 the equations are 4 − c|2(0) = 1 1 |2(0) = √ [|a − |b − |c ] 3 (48) − b|3(0) + c|3(0) = 3 so we see that (47) (49) and for n = 3, we have −3 b|3(0) − c|3(0) = 1 1 |3(0) = √ [|a − |b + 2|c ] 6 which gives (50) (51) Switching to the bare-eigenstate basis, we have and we have 000 H0 = ∆ 0 1 0 004 1(0) |V |1(0) V = 2(0) |V |1(0) 3(0) |V |1(0) 1(0) |V |2(0) 2(0) |V |2(0) 3(0) |V |2(0) 2 χ = √ −2 2 1 0 χ =√ 3 3 −3 V |1(0) V |2(0) and V= 0 χ =√ 6 6 3 (53) (54) (55) (56) √ 2 0 1 √ 3 χ√ 1 0 − 3 √ 2 2−3 0 (57) V |3(0) from which we can obtain 1(0) |V |3(0) 2(0) |V |3(0) 3(0) |V |3(0) (52) The first-order energy shifts are zero, while the second-order terms are given by (2) E1 = − (2) 3 χ2 6 χ2 9 χ2 |V21 |2 |V31 |2 − =− − =− ∆ 4∆ 2∆ 8∆ 4∆ 3 χ2 9 χ2 |V12 |2 |V32 |2 − = − =0 ∆ 3∆ 2∆ 6∆ (59) 6 χ2 9 χ2 9 χ2 |V13 |2 |V23 |2 + = + = 4∆ 3∆ 8∆ 6∆ 4∆ (60) E2 = (2) E3 = (58) 5 Thus the resonance frequency shifts are therefore (2) (2) = E2 − E1 ω2 = E3 − E2 (3) ω1 (2) and (2) (2) ω3 = 9χ2 , 4∆ (61) = 9χ2 , 4∆ (62) (2) (2) E3 − E1 6 = = 9χ2 . 2∆ (63) 3. [15] Consider a pair of quantum harmonic oscillators, described by the bare Hamiltonian H0 = ω (A† A + 1/2) + Ω(B † B + 1/2). Assume that ω < Ω < 2ω , and determine the three lowest bare energy eigenvalues and eigenvectors. Consider the perturbation V = g A† A† B + B † AA . Show that two of the three lowest levels are exact eigenstates of H = H0 + V . For the remaining bare level, compute the first non-vanishing corrections to the eigenvalue and eigenvector. (0) The three lowest energy levels are |1(0) = |00 with energy E1 = ω+Ω , |2(0) = |10 , with energy 2 (0) E2 = 3ω +Ω 2, (0) and |3(0) = |01 with energy E3 = ω +3Ω 2. We have V |1(0) = 0, and V |2(0) √ 0 so that |1(0) and |2(0) are eigenstates of V with eigenvalue 0. = (0) = 2g|20 . This means that For level |3 we have V |3 (1) E3 = 0 so that (2) E3 = − and (1) |3 are the first non-vanishing corrections. 2 g2 (2ω − Ω) √ 2g = −|20 (2ω − Ω) 7 (64) (65) (66) 4. [15] Consider a particle of mass M confined to a 1-dimensional box of length L. Use perturbation theory to caculate the effects of adding a tilt to the box, represented by adding the linear potenital Vtilt (x) = β x1 − L2 to the box potential, 0; 0 < x < L ∞; else Vbox (x) = Calculate the three lowest perturbed eigenstates to first-order and their corresponding eigenvalues to second-order. We have 2 π2 (0) E1 = (67) 2M L2 4 2 π2 2M L2 (68) 9 2 π2 2M L2 (69) x|1(0) = 2 sin(πx/L) L (70) x|2(0) = 2 sin(2πx/L) L (71) x|3(0) = 2 sin(3πx/L) L (72) (0) E2 = and (0) E3 = As well as and The first-order energy shifts are all zero because Vtile (x) is odd with respect to the center of the box. We will need the matrix elements L 2β L x1 − sin(nπx/L) L2 0 1 1 dy sin(mπy ) y − sin(nπy ) = 2β 2 0 4 β mn ((−1)m+n − 1) = (m2 − n2 )2 π 2 Vmn = dx sin(mπx/L) (73) The first-order states are ∞ |1(1) = − |2(1) = − (0) |3 n=2 n =2 =− |n(0) |n(0) n =3 |n Vn1 (0) En − (0) E1 Vn2 (0) En (0) − (0) E2 = = Vn 3 (0) En − (0) Ee 8 32βM L2 π4 32βM L2 π4 ∞ m=1 ∞ m=1 96βM L2 = π4 m ((2m)2 − 1)3 (74) 2m − 1 ((2m − 1)2 − 4)3 (75) m (2m)2 − 9)3 (76) |2m(0) |m(0) ∞ m=1 |m(0) and the corresponding first-order energies are (2) E1 = 1(0) |V |1(1) = − (2) E2 = 2(0) |V |2(1) = − 512β 2 M L2 π6 (2) E3 = 3(0) |V |3(1) = − ∞ 128β 2 M L2 π6 (77) (4π 2 − 15)β 2 M L2 (2m − 1)2 = ((2m − 1)2 − 4)5 384π 4 (78) m=1 ∞ m=1 ∞ 1152β 2 M L2 π6 The results are then (15 − π 2 )β 2 M L2 (2m)2 =− ((2m)2 − 1)5 24π 4 m=1 (2m)2 (9π 2 − 15)β 2 M L2 = ((2m)2 − 9)5 1944π 4 (79) E1 ≈ π2 2 (15 − π 2 )β 2 M L2 − 2M L2 24π 4 (80) E2 ≈ (4π 2 − 15)β 2 M L2 4π 2 2 + 2M L2 384π 4 (81) E3 ≈= (9π 2 − 15)β 2 M L2 9π 2 2 + 2M L2 1944π 4 32βM L2 + π4 (0) |1 ≈ |1 32βM L2 + π4 (0) |2 ≈ |2 (0) |3 ≈ |3 ∞ m=1 m=1 96βM L2 + π4 9 m ((2m)2 − 1)3 (83) 2m − 1 ((2m − 1)2 − 4)3 (84) m (2m)2 − 9)3 (85) |2m(0) ∞ |m(0) ∞ m=1 (82) |m(0) ...
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This note was uploaded on 12/21/2011 for the course PHYS 852 taught by Professor Moore during the Spring '11 term at Michigan State University.

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