PHYS85 Quantum Mechanics II, Spring 00 HOMEWORK ASSIGNMENT 6: Solutions Topics covered: Time-independent perturbation theory.. [0 Two-Level System: Consider the system described by H = δs z + ΩS x, with δ > 0, where S x S z are components of the spin vector of an s = / particle. Treat the S z term as the bare Hamiltonian. (a) [5 Use perturbation theory to compute the eigenvalues eigenvectors of H. Compute all terms up to fourth-order in Ω. The bare eigenstates are with bare eigenvalues we therefore need to iterate the following equations up to j = 4: E (j) n j k= (0) = () (0) = () = δ/ () = δ/ (4) = n (0) V n (j ) E n (k) n(0) n (j k) (5) [ n (j) = n (0) j n (k) n (j k) [ m (0) m (0) V n (j ) j E n (k) m (0) n (j k) E mn E mn k= k= (6) for j = we have for the states we have n () = E () n = Ω n(0) S x n (0) = 0 (7) m (0) Ω m (0) S x n (0) = Ω m (0) (8) E mn E mn () = Ω (9) δ () = Ω (0) δ for j = we have E () n = Ω n(0) S x n () E () n = Ω n(0) S x n () () E () = Ω δ S x = Ω 4δ ()
E () = Ω δ S x = Ω () 4δ for the states [ n () = n (0) n() n () [ m (0) Ω m (0) S x n () E() n m (0) n () E mn E mn = Ω 8δ n(0) For j = we have for the energies (4) E () n = Ω n(0) S x n () E () n n(0) n () E () n n(0) n () = 0 (5) for the states we have n () = [ n(0) n () n () + n () n () [ m (0) Ω m (0) S x n () E() n m (0) n () E() n m (0) n () E mn E mn E mn = [ m (0) Ω 6δ + E() n m (0) n () E mn E mn (6) [ Ω () = 6δ + Ω 8δ = Ω (7) 6δ And finally, for j = 4, we have () = [ Ω 6δ Ω 8δ = Ω (8) 6δ E (4) n = Ω n (0) S x n () E () n n (0) n () E () n n (0) n () E () n n (0) n () for the states, we have = Ω n (0) S x n () + Ω E n () 8δ (9) E (4) = Ω4 δ Ω4 δ = Ω4 6δ (0) E (4) = Ω4 δ + Ω4 δ = Ω4 6δ () n (4) = [ n(0) n () n () + n () n () + n () n () [ m (0) Ω m (0) S x n () E() n m (0) n () E() n m (0) n () E() n m (0) n () E mn E mn E mn E mn = n (0) [ n () n () Ω4 8δ 4 ()
Putting the pieces together gives [ Ω (4) 4 = δ 4 Ω4 8δ 4 = Ω4 () 8δ4 [ Ω (4) 4 = δ 4 Ω4 8δ 4 = Ω4 (4) 8δ4 [ Ω E δ Ω 4δ + Ω4 6δ (5) E δ + Ω 4δ Ω4 6δ (6) 8δ + Ω4 8δ 4 [ Ω δ Ω 6δ + + Ω [ + Ωδ 6δ [ Ω 8δ + Ω4 8δ 4 (b) [5 Exp the exact eigenvalues eigenvectors around Ω = 0 compare to the perturbation theory results. The exact results are E = δ + Ω δ Ω 4δ + Ω4 6δ (9) E = δ + Ω δ + Ω 4δ Ω4 6δ (0) = (δ + (δ + Ω ) Ω (δ + δ + Ω ) + Ω [ Ω 8δ + Ω4 8δ 4 + [ Ωδ + Ω 6δ = ω + (δ + δ + Ω [ ) Ω Ω + (δ + δ + Ω ) δ Ω 6δ + [ Ω 8δ + Ω4 8δ 4 (7) (8) () () (c) [0 Verify that the states computed in (a) are normalized to unity orthogonal up to fourthorder. = = [ Ω 8δ + Ω4 Ω 8δ 4 + [ + Ωδ 6δ + O(Ω 5 ) = Ω 4δ + Ω4 64δ 4 + Ω4 64δ 4 + Ω 4δ Ω4 6δ 4 + O(Ω5 ) = + O(Ω 5 ) () [ Ω δ Ω 6δ + [ Ω 8δ + Ω4 8δ 4 + O(Ω 5 ) = Ω 4δ Ω4 6δ 4 + Ω 4δ + Ω4 64δ 4 + Ω4 64δ 4 + O(Ω5 ) = + O(Ω 5 ) (4) It is clear by inspection of Eqs. (7) (8) that = 0.
. [0 Resonance-frequency shifts: Consider a system with a -dimensional Hilbert space spanned by states a, b, c. In the basis { a, b, c}, let the bare Hamiltonian of the system be H 0 =. (5) For the case where the system is perturbed by the operator 0 V = χ, (6) 0 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 H 0 are the solutions to det H 0 ω = 0, which yields the characteristic equation which simplifies to ( ω)[( ω)( ω) [ + ω + [ ( ω) = 0 (7) ( ω)ω(ω 4) = 0 (8) so that the eigenvalues of H 0 are = 0, =, = 4. The three bare resonance frequencies are therefore The eigenvectors satisfy taking a n (0) = gives for n =, ω (0) = E(0) ω (0) = E(0) ω (0) = E(0) E(0) n E(0) n E(0) n =, (9) =, (40) = 4. (4) which has the solution b (0) = c (0) = 0, so that a n (0) b n (0) c n (0) = 0 (4) b (0) + c (0) = (4) b (0) + c (0) = (44) (0) = [ a + b (45) for n = the equations are b (0) + c (0) = 0 (46) 4
c (0) = (47) so we see that for n =, we have (0) = [ a b c (48) b (0) + c (0) = (49) b (0) c (0) = (50) (0) = 6 [ a b + c (5) Switching to the bare-eigenstate basis, we have we have from which we can obtain V = H 0 = 0 0 0 0 0 0 0 4 (0) V (0) (0) V (0) (0) V (0) (0) V (0) (0) V (0) (0) V (0) (0) V (0) (0) V (0) (0) V (0) V (0) = χ V (0) = χ V (0) = χ 6 V = χ 0 0 6 (5) (5) (54) (55) 0 0 0 (56) The first-order energy shifts are zero, while the second-order terms are given by (57) E () = V V 4 = χ 6χ 8 = 9χ 4 (58) E () = V V = χ 9χ 6 = 0 (59) E () = V 4 + V = 6χ 8 + 9χ 6 = 9χ 4 (60) 5
Thus the resonance frequency shifts are therefore ω () = E() E () = 9χ 4, (6) ω () = E() E () ω = E() E () = 9χ 4, (6) = 9χ. (6) 6
. [5 Consider a pair of quantum harmonic oscillators, described by the bare Hamiltonian H 0 = ω(a A + /) + Ω(B B + /). Assume that ω < Ω < ω, determine the three lowest bare energy eigenvalues eigenvectors. Consider the perturbation ( ) V = g A A B + B AA. Show that two of the three lowest levels are exact eigenstates of H = H 0 + V. For the remaining bare level, compute the first non-vanishing corrections to the eigenvalue eigenvector. The three lowest energy levels are (0) = 00 with energy = ω+ω, (0) = 0, with energy = ω+ω, (0) = 0 with energy = ω+ω. We have V (0) = 0, V (0) = 0 so that (0) (0) are eigenstates of V with eigenvalue 0. For level we have V (0) = g 0. This means that E () = 0 (64) so that are the first non-vanishing corrections. E () = g (ω Ω) g () = 0 (ω Ω) (65) (66) 7
4. [5 Consider a particle of mass M confined to a -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 ( x V tilt (x) = β L ) to the box potential, V box (x) = { 0; 0 < x < L ; else Calculate the three lowest perturbed eigenstates to first-order their corresponding eigenvalues to second-order. We have = π ML (67) As well as V mn = β L = 4 π ML (68) = 9 π ML (69) x (0) = sin(πx/l) (70) L x (0) = sin(πx/l) (7) L x (0) = 0 sin(πx/l) (7) L The first-order energy shifts are all zero because V tile (x) is odd with respect to the center of the box. We will need the matrix elements L ( ) x dx sin(mπx/l) sin(nπx/l) The first-order states are () = () = n (0) = n = β 0 ( dy sin(mπy) y L ) sin(nπy) = 4βmn (( )m+n ) (m n ) π (7) n (0) V n n= E n (0) n (0) n (0) V n n V n E n (0) E e (0) = βml π 4 = βml π 4 8 = 96βML π 4 m (0) m ((m) ) (74) m= m (0) m ((m ) 4) (75) m= m (0) m (m) 9) (76) m=
the corresponding first-order energies are The results are then E () = (0) V () = 8β ML π 6 E () = (0) V () = 5β ML π 6 m= m= E () = (0) V () = 5β ML π 6 (m) ((m) ) 5 = (5 π )β ML 4π 4 (77) (m ) ((m ) 4) 5 = (4π 5)β ML 84π 4 (78) m= (m) ((m) 9) 5 = (9π 5)β ML 944π 4 (79) E π ML (5 π )β ML 4π 4 (80) E 4π ML + (4π 5)β ML 84π 4 (8) E = 9π ML + (9π 5)β ML 944π 4 (8) (0) + βml π 4 m (0) m ((m) ) (8) (0) + βml π 4 (0) + 96βML π 4 m= m (0) m ((m ) 4) (84) m= m (0) m (m) 9) (85) m= 9