PHYS 455 Short Answers to Homework I

PHYS 455Short Answers to Homework I1. (a) Eigenvalues of A are λ 1 = λ 2 = 1 with degeneracy 2 and λ 3 = −1 with degeneracy1.(b) Two eigenvalues and therefore two projectors to these subspaces. For the casewhere A has only eigenvalues ±1, these projectors can be obtained simply byP ± = (I ± A)/2. (Can you show this?) As a result,⎡P + = 1 ⎣21 0 10 2 01 0 1⎤⎡⎦ P − = 1 ⎣21 0 −10 0 0−1 0 1⎤⎦ .(Note: tr P + = 2=degeneracy of eigenvalue +1 and tr P − = 1=degeneracy of−1.)(c) Check.(d) Various methods. One such possible set is⎡ ⎤ ⎡ ⎤u 1 = √ 1 ⎣ ⎦ u 2 = ⎣2101010⎡⎦ u 3 = √ 1 ⎣210−1⎤⎦ .This particular onb is also the one that is asked in question 4. However, at leastfor A, it is not the only possible one. Apart from the phase arbitrariness of thesevectors, the eigenvectors for λ = 1 can be chosen arbitrarily. For example,u ′ 1 = 1 √2(u 1 + iu 2 ) = 1 2⎡⎣1i √ 21is also a possible answer to this question.2. (a) f(A) = 2I if f(t) = 1 + t 2 + t 137 sin πt.(b) g(A) = A −1 = A (for g(t) = 1/t).(c) h(A) = √ 2P + for h(t) = √ 1 + t.⎤⎦ u ′ 2 = 1 √2(iu 1 + u 2 ) = 1 2⎡⎣i√2i⎤⎦ u ′ 3 = u 33. Suppose that A represents an observable for a three-level system. The system is in thenormalized state⎡ ⎤ψ = √ 1 2⎣ 1 ⎦ , 6iand the observable A is measured in that state.1

(a) Values that can be obtained when A is measured are +1 and −1, namely theeigenvalues of A. In this particular case, both of these values are possible.(b) Probabilities of each possible outcome:P rob(+) = 〈P + 〉 = 〈ψ|P + |ψ〉 = ‖P + ψ‖ 2 = 712P rob(+) = 〈P − 〉 = 〈ψ|P − |ψ〉 = ‖P − ψ‖ 2 = 1 − P rob(+) = 5 12(1). (2)(c) If result +1 is obtained, then collapse is to the state P + ψ (which should berenormalized). In this case, the collapse occurs to⎡ ⎤1ψ + = √ P + ψ = √ 1 2 + i⎣ 2 ⎦ .7/12 142 − iSimilarly, if we obtain −1 as the result of the measurement, then collapse occursto⎡ ⎤1ψ − = √ P − ψ = 2 √ − i 1⎣ 0 ⎦ = e iφ u 3 .5/12 10−1(The value of phase φ in ψ − is unimportant.)Another way to state this. Let us use the onb in 1(c). We can expand ψ in thisbasis asψ = c 1 u 1 + c 2 u 2 + c 3 u 3 = 2 √ + i u 1 + √ 1 u 2 + 2 √ − i u 312 6 12The first two terms is the component of ψ along the eigensubspace correspondingto λ = +1. The last term is the component along the eigensubspace correspondingto λ = −1. The norm squares of these components are probabilities andthe collapse occurs to these components. As a result, probability of obtainingmeasured value +1 is P rob(+) = |c 1 | 2 + |c 2 | 2 = 7/12 and the collapse occurs to√ (112 2 + iψ + = √ (c 1 u 1 + c 2 u 2 ) = √ u 1 + 1 )√ u 2 .P rob(+) 7 12 6The measurement will not destroy the coherence between u 1 and u 2 . The collapsewill occur to the coherent superposition expressed above (not to u 1 and not tou 2 ).Of course, what we described above for the collapse depends on the experimentaldetails of how A is measured. There are many different ways of measuring Awhere the collapsed states are different. (For example, measuring both A andB together as in problem 4. The collapse in that case is to u 1 , u 2 and u 3 withappropriate probabilities.) The mathematical scheme described here is calledprojective measurement and there are experimental methods that realizes this.This is important for example for local measurements done on entangled states.In that case, your local measurements and the collapse caused by them cannotdestroy the coherence in the rest of the wavefunction.2

(d) 〈A〉 = 1/6 and 〈A 2 〉 = 1. You can compute these by using already computedvalues of P rob(±).4. (a) Verify.(b) {u 1 , u 2 , u 3 } given in 1(c) is such an onb. Only the overall phases of these vectorsare arbitrary now. Any superposition of these vectors is either not an eigenvectorof A or not an eigenvector of B. Projection operators can be obtained accordingly.(c) We use the expansionψ = c 1 u 1 + c 2 u 2 + c 3 u 3 = 2 + i √12u 1 + 1 √6u 2 + 2 − i √12u 3(or the projection operators). The possible results obtained for both A and Band their corresponding probabilities are listed below.value of A value of B Probability Collapse to+1 +1 5/12 u 1+1 0 2/12 u 2−1 +1 5/12 u 33