Graduate Quantum Mechanics I: Prelims and Solutions (Fall 015 Problem 1 (0 points Suppose A and B are two two-level systems represented by the Pauli-matrices σx A,B σ x = ( 0 1 ;σ 1 0 y = ( ( 0 i 1 0 ;σ i 0 z = 0 1,σ A,B y,σz A,B. Recall (1 Suppose you are given the state vector that lives in the A B space, Φ AB = α + A, B + β A,+B ( where ±,A, ±,B represent the eigenvectors of σz A,σz B respectively with eigenvalues ±1 and ± A, ±B is a product state. For simplicity take α,β to be real, and the wave-function to be normalized α + β = 1. a. Write the density matrix ρ AB for the above state. What is Tr [ ρab]? b. Construct the reduced density matrix ρ A = Tr B [ρ AB ]. Write ρ A in the form (1 + P σ A / and give P explicitly in terms of α,β. c. What is the range of Tr [ ρa]?. ] d. Compute the entanglement entropy S A = Tr [ρ A lnρ A. For what values of α,β are the spins maximally entangled? a. In the basis of + A,+B, + A, B, A,+B, A, B 0 0 0 0 0 α ρ AB = Φ AB Φ AB = αβ 0 0 βα β 0 ( 0 0 0 0 Tr [ ( ρab] = α + β = 1 as expected for a pure state. b. In the basis + A, A, ( α 0 ρ A = 0 β = 1 ( 1 + P σ (4 where P = ( α β ẑ. c. Tr [ ρ A] = (1 + P / = 1 α β. The range of Tr [ ρ A] is 1/ Tr [ ρ A ] 1 corresponding to the length of the vector P ranging from 0 P 1. d. The entanglement entropy is S A = [ α lnα + β lnβ ] = [ α lnα + (1 α ln(1 α ] (5 The entanglement is maximum when α = β = 1/, when S A = ln [].
Problem (15 points For the wave-function in Eq., if now an observer in sub-system A uses a Stern-Gerlach set-up to measure the spin A along a direction ˆn, a What is the average outcome of the measurement? b. What is the probability of finding the spin to be aligned along the direction ˆn? What is the probability of finding the spin anti-aligned to the direction ˆn? c. For what values of α,β, will the outcome become independent of ˆn? a. The average outcome of the measurement is [ ( ] nz n Tr [ρ A σ ˆn] = Tr ρ x in y A = n n x + in y n z (α β (6 z b. The probability of finding the spin aligned (P +ˆn or anti-aligned (P ˆn along the direction ˆn is [ ] P ±ˆn = Tr ρ A ± ˆn ±ˆn where ± ˆn are the eigenkets σ ˆn ± ˆn = ± ± ˆn. Thus the projection operator + ˆn +ˆn = 1 ( 1 + nz n x in y ; ˆn ˆn = 1 ( 1 nz (n x in y n x + in y 1 n z (n x + in y 1 + n z (7 (8 Thus the two probabilities are P ±ˆn = 1 ± n z ( α β (9 c. For the maximally entangled state, α = β, the outcome is independent of the orientation ˆn.
Problem (10 points Use the Wigner-Eckart theorem to determine the following ratio l = 5,m = x l = 5,m = l = 5,m = y l = 5,m = 4 (10 where r = (x, y, z is the position operator and l, m is the eigenstate of the orbital angular-momentum operators L,L z. (Useful formulae: J x = (J + + J /,J y = (J + J /(i, J ± j,m = (j m(j ± m + 1 j,m ± 1. According to the Wigner-Eckart theorem, l,m r lm = c l l,m L lm. Thus l = 5,m = x l = 5,m = l = 5,m = y l = 5,m = 4 = l = 5,m = L x l = 5,m = l = 5,m = L y l = 5,m = 4 = 1 l = 5,m = L + l = 5,m = (5 (5 + + 1 i l = 5,m = L l = 5,m = 4 = i = i/ (11 (5 + 4(5 4 + 1
4 Problem 4 (10 points The wave-function of a particle subjected to a spherically symmetric potential V (r is given by ψ(x,y,z = (x + y + z f(r (1 a. Is ψ an eigenfunction of L? If so what is the l value? If not what are the possible values of l that we may obtain when L is measured? b. What are the probabilities for the particle to be found in various m l states? Useful relations Y0 0 (θ,φ = 1 1 ;Y π 1 (θ,φ = 1 π sin θe iφ ;Y 0 1 (θ,φ = 1 1 π cos θ;y1 (θ,φ = 1 π sinθeiφ The wave-function is an eigenfunction of L as [ H,L ] = 0. Recall that the general solution for a particle in a spherically symmetric potential is ψ( r = R nl (ry m l (θ,φ (1 We may write the wave function as ( ψ(x,y,z = sinθ cos φ + sin θ sinφ + cos θ rf(r (14 [ π = (i + 1Y1 1 + (i 1 Y1 1 + ] Y1 0 rf(r (15 The eigenvalue of L is l(l + 1 = 1(1 + 1 =. The probability of measuring m l = 1,1,0 is P(m l = 1 = P(m l = 1 = 1/11,P(m l = = 9/11 (16
Problem 5 (15 points Consider a particle of mass m, in a one-dimensional potential U(x = αδ(x αδ(x a. Solve for all the bound state solutions, and determine the condition for the number of bound states in terms of m,α,a. Note that there are two solutions, one which is a symmetric around x = a/, and the other which is antisymmetric around x = a/. SYMMETRIC SOLUTION: Matching boundary conditions at x = 0, a, the symmetric wave-function is found to be, ψ1 S (x < 0 = B (1 + e ka e kx (17 ψ S (0 < x < a = B (e kx + e ka e kx (18 ψ S (x > a = B (1 + e ka e k(x a (19 where k obeys the condition The above equation always has a solution. k mα = 1 + e ka (0 ANTI-SYMMETRIC SOLUTION: Matching boundary conditions at x = 0, a, the anti-symmetric wave-function is found to be, ψ1 A (x < 0 = B (1 e ka e kx (1 ψ A (0 < x < a = B (e kx e ka e kx ( ( ψ A (x > a = B e ka 1 e k(x a ( where k obeys the condition The above has a solution only if /(mαa < 1. k mα = 1 e ka (4 Thus there is only one bound state (corresponding to ψ S for 0 < mαa/ < 1, while there are two bound states corresponding to ψ S,ψ A, when mαa/ > 1 5