Phys 500, Quantum Mechanics Reference Solution for Homework 1

Reference solution for Homework 1

Posted: Friday, September 26, 2014

Problem 1 (5 points): In class we introduced the Hermitian adjoint X† of an operator X through the dual correspondence X|αi ←→ hα|X†, ∀ |αi ∈ H. (1)

Denote by [X] the matrix corresponding to the operator X in some ONB B, i.e. [X]ij = hai|X|aj i for |aii, |aj i ∈ B. (a) Show that for the matrix representation [X] of a linear operator X, taking the Hermitian adjoint according † ∗ 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 hγ|βi = hβ|γi and the dual correspondence Eq. (1). Choosing |γi = X|αi, we ﬁnd hα|X†|βi = hα|X† |βi = [hβ (|X|αi)]∗ = hβ|X|αi∗.

Now choose |αi, |βi in some ONB. Then,

† ∗ [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 †.

Then, ([X0])† = U[X]U †† = U[X]†U †. 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 †U[Y ]U † = U[X][Y ]U † =: [XY ]0. The second equality follows from U †U = I (U is unitary). Problem 2 (5 points): The Hamiltonian operator for a 2-state system is given by H = a (|1ih1| − |2ih2| + |1ih2| + |2ih1|) , (2) where a is a constant with the dimension of energy, and the kets |1i and |2i form an orthonormal basis. Find the energy eigenvalues and the corresponding eigenkets, as linear combinations of |1i and |2i. 1 0 Solution: Upon the identification |1i =∼ and |2i =∼ , the Hamiltonian H can be 0 1 represented by a 2 × 2 matrix 1 1 H =∼ a . (3) 1 −1

1 1 −1 √ √ The eigenvectors of this matrix are √ and √ , with eigenvalues 2a and − 2a, 2 − 1 2 + 1 respectively. Back in the original orthonormal basis B = {|1i, |2i} we thus have

√ √ |1i+( 2−1)|2i E1 = 2a, |E1i = √ √ , 4√−2 2 (4) √ |1i−( 2+1)|2i E2 = − 2a, |E2i = √ √ . 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 ∈ R such that I + rn · ~σ O = 2 is a projector? Solution: For O to be a projector we require O† = O and O2 = O. The ﬁrst property holds for all r ∈ R. Regarding the second property, we ﬁnd just by expanding

(n · ~σ)2 = I, ∀n with |n| = 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

|+i := |s = 1, sz = 1i, |0i := |s = 1, sz = 0i, |−i := |s = 1, sz = −1i, (7) and any linear combination thereof. The Sz-operator for this system is Sz = ~(|+ih+| − |−ih−|. Consider the 2 sequential measurement of the operator (Sz) (ﬁrst) and Sz (second) upon the initial state |+i + |0i + |−i |ψi = √ . 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 probabilities for obtaining them? (The outcome probabilities are conditioned upon the outcome of the ﬁrst measurement.)

2 Solution: (a) Sz = |+ih+|+|−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 = |0ih0|,P 2 = |+ih+| + |−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 = hψ|P 2 |ψi = , p 2 = hψ|P 2 |ψi = (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 = |+ih+|,PSz,0 = |0ih0|,PSz,−1 = |−ih−|. (10)

2 2 Case I: The outcome of the Sz -measurement was 1. Then, the post-measurement state is

P 2 |ψi |+i + |−i ˜ Sz ,1 |ψ(1)i = q = √ . hψ|P 2 |ψi 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 = hψ(1)|PSz,1|ψ(1)i = 2 , ˜ ˜ pSz,0 = hψ(1)|PSz,0|ψ(1)i = 0, ˜ ˜ 1 pSz,−1 = hψ(1)|PSz,−1|ψ(1)i = 2 . 2 Case II: The outcome of the Sz -measurement was 0. Then, the post-measurement state is

P 2 |ψi ˜ Sz ,0 |ψ(0)i = q = |0i. hψ|P 2 |ψi 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 = h0|PSz,1|0i = 0, pSz,0 = h0|PSz,0|0i = 1, pSz,−1 = h0|PSz,−1|0i = 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.

2 p(+) = 1 cos φ + sin φ = 1+sin φ . (11) 2 2 2 2

ˆ (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 Sˆx therefore is

hSˆ i = ~ sin φ. x 2

2 ˆ ˆ ˆ ˆ 2 ˆ2 ˆ 2 ˆ2 ~ The dispersion (variance) of Sx is Var(Sx) = h(Sx − hSxi) i = hSxi − hSxi . Also, Sx = 4 I, and thus 2 ˆ ~ 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 Sˆz, and for φ = π/2 an eigenstate of Sˆx. 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.

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 |+i, the state after passing the second apparatus is β β |β, +i = cos |+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 = |h+|β, +i| = 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 = |hβ, +|−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. The output intensity is maximized for β = ± 2 and minimized for β = 0, π.

Problem 7 (5 points). (Continuation from class) We noted the Born rule for measurement for the general (potentially degenerate) case. If Pa is the projector on the subspace corresponding to the measurement outcome a for the observable A, then the probability p(a) for obtaining the outcome a given the quantum state |Ψi is

p(a) = hΨ|Pa|Ψi.

P Show that p(a) ∈ R and p(a) ≥ 0 for all possible measurement outcomes a, and that a p(a) = 1. † ∗ ∗ Solution. p(a) = hΨ|Pa|Ψi = hΨ|Pa |Ψi = hΨ|Pa|Ψi = p (a). Hence, p(a) ∈ R. Furthermore, 2 † † p(a) := hΨ|Pa|Ψi = hΨ|Pa |Ψi = hΨ|Pa P |Ψi = hΨ|Pa (P |Ψi). Hence, p(a) ≥ 0. Finally, denote by Pa the projector on the subspace of eigenstates of A with eigenvalue a, Pa = P i |a, iiha, i| where A|a, ii = a|a, ii. The key point is that BA := {|a, ii, ∀a, i} is an ONB of H . You may use this fact without proof, since its pure linear algebra, but a proof is given below. P P P P Then, a,i |aiiha, i| = I, and hence a Pa = I. Therefore, a p(a) = ahΨ|Pa|Ψi = 1.

Why BA is a basis of H, for any observable A: First, H is by deﬁnition complete, hence has a basis. Then, it also has an ONB, B0, which can be constructed e.g. by the Gram-Schmidt procedure. We may compute the matrix [A] corresponding to A in the basis B0. Now, [A] may be brought into Jordan normal form [J] by a unitary transformation U,[J] = U[A]U †. Since [A] is a matrix over the complex numbers, the Jordan normal form always exists. Furthermore, since A is Hermitian, [A] is a Hermitian matrix, and so is [J]. Therefore, [J] is actually a diagonal matrix. Its eigenvectors ej span the full space. Therefore, the † † eigenvectors of [A], {U ej} also span the full space. {U ej} corresponds to BA in the matrix representation induced by B0. Total: 35 points.