Phys 500, Quantum Mechanics Reference solution for Homework 1 Posted: Friday, September 26, 2014 Problem 1 (5 points): In class we introduced the Hermitian adjoint Xy of an operator X through the dual correspondence Xjαi ! hαjXy; 8 jαi 2 H: (1) Denote by [X] the matrix corresponding to the operator X in some ONB B, i.e. [X]ij = haijXjaj i for jaii; jaj i 2 B. (a) Show that for the matrix representation [X] of a linear operator X, taking the Hermitian adjoint according y ∗ to Eq. (1) amounts to the familiar Hermitian transpose, i.e., [X ]kl = ([X]lk) . (b) Show explicitly that taking the Hermitian adjoint of an operator X in some matrix representation [X] is basis- independent. That is, consider the transformation of [X] from one ONB to another, and demonstrate that (i) first taking the Hermitian transpose and then transforming to the other ONB and (ii) first transforming to the other ONB and then taking the Hermitian transpose yield the same result. (c) Likewise, show that taking products of operators in some matrix representation corresponding to an ONB is basis-independent. Solution: (a) We use the property of the scalar product hγjβi = hβjγi and the dual correspondence Eq. (1). Choosing jγi = Xjαi, we find hαjXyjβi = hαjXy jβi = [hβ (jXjαi)]∗ = hβjXjαi∗: Now choose jαi; jβi in some ONB. Then, y ∗ [X ]αβ = ([X]βα) : (b) The transformation from one ONB to another is by conjugation under some unitary matrix U, [X] −! [X]0 = U[X]U y: Then, ([X0])y = U[X]U yy = U[X]yU y. That is, first transforming and then taking the Hermitian transpose is the same as first taking the Hermitian transpose and then transforming. (c) [X]0[Y ]0 := U[X]U yU[Y ]U y = U[X][Y ]U y =: [XY ]0. The second equality follows from U yU = I (U is unitary). Problem 2 (5 points): The Hamiltonian operator for a 2-state system is given by H = a (j1ih1j − j2ih2j + j1ih2j + j2ih1j) ; (2) where a is a constant with the dimension of energy, and the kets j1i and j2i form an orthonormal basis. Find the energy eigenvalues and the corresponding eigenkets, as linear combinations of j1i and j2i. 1 0 Solution: Upon the identification j1i =∼ and j2i =∼ , the Hamiltonian H can be 0 1 represented by a 2 × 2 matrix 1 1 H =∼ a : (3) 1 −1 1 1 −1 p p The eigenvectors of this matrix are p and p , with eigenvalues 2a and − 2a, 2 − 1 2 + 1 respectively. Back in the original orthonormal basis B = fj1i; j2ig we thus have p p j1i+( 2−1)j2i E1 = 2a; jE1i = p p ; 4p−2 2 (4) p j1i−( 2+1)j2i E2 = − 2a; jE2i = p p : 4+2 2 Problem 3. (5 points) The Pauli matrices σx, σy and σz are 2 × 2-matrices defined as follows 0 1 0 −i 1 0 σ = ; σ = ; σ = : (5) x 1 0 y i 0 z 0 −1 T Consider a vector n = (nx; ny; nz) of unit length, and n · ~σ := nxσx + nyσy + nzσz. Are there r 2 R such that I + rn · ~σ O = 2 is a projector? Solution: For O to be a projector we require Oy = O and O2 = O. The first property holds for all r 2 R. Regarding the second property, we find just by expanding (n · ~σ)2 = I; 8n with jnj = 1: Therefore, (1 + r2)I + 2r n · ~σ O2 = ; 4 and O2 = O iff r = ±1: (6) Problem 4. (5 points) Consider a spin-1 particle which can assume the states j+i := js = 1; sz = 1i; j0i := js = 1; sz = 0i; |−i := js = 1; sz = −1i; (7) and any linear combination thereof. The Sz-operator for this system is Sz = ~(j+ih+j − |−ih−|. Consider the 2 sequential measurement of the operator (Sz) (first) and Sz (second) upon the initial state j+i + j0i + |−i j i = p : 3 2 (a) What are the possible measurement outcomes of (Sz) and the probabilities of obtaining them? (b) What are the possible measurement outcomes for Sz in the second measurement, and the conditional prob- abilities for obtaining them? (The outcome probabilities are conditioned upon the outcome of the first measurement.) 2 Solution: (a) Sz = j+ih+j+|−ih−|, which has eigenvalues 0, 1. Thus, the possible outcomes of an 2 S -measurement of a spin-1 particle are 0 and 1. The projectors P 2 and P 2 on the sub-spaces z Sz ;0 Sz ;1 corresponding to the eigenvalues 0 and 1, respectively, are P 2 = j0ih0j;P 2 = j+ih+j + |−ih−|: (8) Sz ;0 Sz ;1 Then, the probabilities p 2 and p 2 for obtaining outcome 0 or 1, respectively, are given by the Sz ;0 Sz ;1 Born rule 1 2 p 2 = h jP 2 j i = ; p 2 = h jP 2 j i = (9) Sz ;0 Sz ;0 3 Sz ;1 Sz ;1 3 (b) The possible measurement outcomes of the Sz-measurement are ±1, 0.The corresponding projectors are PSz;1 = j+ih+j;PSz;0 = j0ih0j;PSz;−1 = |−ih−|: (10) 2 2 Case I: The outcome of the Sz -measurement was 1. Then, the post-measurement state is P 2 j i j+i + |−i ~ Sz ;1 j (1)i = q = p : h jP 2 j i 2 Sz ;1 By the Born rule, the conditional probabilities pSz;1, pSz;0, pSz;−1 for obtaining the outcomes 1,0,-1, respectively, are ~ ~ 1 pSz;1 = h (1)jPSz;1j (1)i = 2 ; ~ ~ pSz;0 = h (1)jPSz;0j (1)i = 0; ~ ~ 1 pSz;−1 = h (1)jPSz;−1j (1)i = 2 : 2 Case II: The outcome of the Sz -measurement was 0. Then, the post-measurement state is P 2 j i ~ Sz ;0 j (0)i = q = j0i: h jP 2 j i Sz ;0 By the Born rule, the conditional probabilities pSz;1, pSz;0, pSz;−1 for obtaining the outcomes 1,0,-1, respectively, are pSz;1 = h0jPSz;1j0i = 0; pSz;0 = h0jPSz;0j0i = 1; pSz;−1 = h0jPSz;−1j0i = 0: Problem 5 (5 points): A spin 1/2 system is known to be in an eigenstate of S · n with eigenvalue ~=2. The unit vector n lies in the xz-plane at an angle φ with the positive z-axis, i.e. S · n = nx(φ)Sx + nz(φ)Sz. Furthermore, the angle φ is oriented such that for φ = π=2, S · n = Sx. (a) Suppose Sx is measured. What is the probability for finding +~=2? 2 (b) Evaluate the dispersion in Sx, that is, h(Sx − hSxi) i. Check your answers for the special cases of φ = 0, φ = π=2 and φ = π. Solution: (a) With the specifications given, n · S^ = cos φS^z + sin φS^x. The eigenvector with eigenvalue +~=2 is φ φ jφ, +i = cos j+i + sin |−i; 2 2 where j+i, |−i are the eigenstates of S^x. (To check the above, you may find the relations sin 2α = 2 2 2 sin α cos α, cos 2α = cos α − sin α useful.) The probability of obtaining the outcomep +~=2 in a 2 measurement of S^x on jφ, +i is p(+) = jhx; +jφ, +ij , where jx; +i = (j+i + |−i)= 2. Thus, 2 p(+) = 1 cos φ + sin φ = 1+sin φ : (11) 2 2 2 2 ^ (b) Then, the probability p(−) for obtaining the outcome −~=2 in the Sx-measurement is p(−) = 1 − p(+) = (1 − sin φ)=2. The expectation value of S^x therefore is hS^ i = ~ sin φ. x 2 2 ^ ^ ^ ^ 2 ^2 ^ 2 ^2 ~ The dispersion (variance) of Sx is Var(Sx) = h(Sx − hSxi) i = hSxi − hSxi . Also, Sx = 4 I, and thus 2 ^ ~ 2 Var(Sx) = 4 cos φ. (12) 2 2 For angles φ = 0; π=2; π the variance is ~ =4; 0; ~ =4, respectively. This agrees with the expectation, since for φ = 0; π the measured state is an eigenstate of S^z, and for φ = π=2 an eigenstate of S^x. Problem 6 (5 points). A beam of spin 1/2 atoms goes through a series of Stern-Gerlach-type measurements as follows: (a) The first measurement accepts sz = ~=2 atoms and blocks sz = −~=2 atoms. 3 (b) The second measurement accepts sn = ~=2 atoms and blocks sn = −~=2 atoms, where n is a unit vector lying in the xz-plane, at an angle β with the positive z-axis. (c) The third measurement accepts sz = −~=2 atoms and blocks sz = ~=2 atoms. What is the intensity of the final sz = −~=2 beam when the sz = ~=2 beam surviving the first measurement is normalized to unity? How must the second apparatus be oriented (i) to maximize, and (ii) to minimize the intensity of the final sz = −~=2 beam? Solution: The state of the atom after passing the fist SG apparatus is j+i, the state after passing the second apparatus is β β jβ; +i = cos j+i + sin |−i; 2 2 (see Problem 4), and the state after passing the third apparatus is |−i. The probability for the atom to pass the second SG apparatus (conditioned upon having passed the first one) is 2 2 β ppass;2 = jh+jβ; +ij = cos 2 . Likewise, the probability for the atom to pass the third SG apparatus 2 2 β (conditioned upon having passed the second one) is ppass;3 = jhβ; +|−i| = sin 2 . The two events are independent, and therefore the probability ppass;23 for passing both SG apparata is 2 β 2 β 1 2 ppass;23 = cos 2 sin 2 = 4 sin β: π The final intensity is the initial intensity × ppass;23.