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Quantum Information Theory Solutions 8 HS 2015 Quantum Information Theory Prof. R. Renner Solutions 8. Dr. J.M. Renes Exercise 1. Trace distance and distinguishability Suppose you know the density operators of two quantum states ρ, σ 2 S(H). Then you are given one of the states at random { it may either be ρ or σ with equal probability. The challenge is to perform a single projective measurement of an observable O on your state and then guess which state that is. (a) What is your best strategy? In which basis do you think you should perform the measurement? Can you express that measurement using a single projector P ? (b) Show that the probability of guessing correctly can be written as 1 P (ρ vs: σ) = (1 + tr [P (ρ − σ)]); (1) guess 2 where P is the appropriate projector from (a). Just like in the classical case, that can be shown to be equivalent to 1 P (ρ vs: σ) = [1 + δ(ρ, σ)]; (2) guess 2 where δ(ρ, σ) := 1 kρ − σk is the trace distance between the two quantum states and kSk := trjSj ≡ p 2 1 1 tr[ SyS] the 1-norm for matrices. (You do not have to show this here. Of course you can if you want.) (c) Given a trace-preserving quantum operation E (i.e. a CPTP map) and two states ρ and σ, show that δ (E(σ); E(ρ)) ≤ δ(σ; ρ): (3) (d) What does (3) imply about the task of distinguishing quantum states? Solution. (a) We are looking for a measurement O that maximises our probability of guessing correctly. For each state (say e.g. ρ) the probabilities of obtaining any of the possible outcomes P fygy of the observable O = y yPy that represents the measurement define a classical probability distribution PrO,ρ(y) = tr(Pyρ), Py being the mutually orthogonal projectors onto the eigenspaces of O. Let G = fy : PO,ρ(y) ≥ PO;σ(y)g be the set of outcomes that are more likely to occur when we measure O on ρ than on σ. Naturally, if we obtain y after measuring our unknown state and obtain we should say it was ρ if y 2 G and vice-versa. The probability of guessing 1 correctly is then Pguess = P (ρ) · P (say \ρ"jρ) + P (σ) · P (say \σ"jσ) 1 X 1 X = · P (y) + · P (y) 2 O;ρ 2 O;σ y2G y2G¯ 1 X 1 X = tr(Pyρ) + tr(Pyσ) 2 2 (S.1) y2G y2G¯ 1 h X i 1 h X i = tr P ρ + tr P σ 2 y 2 y y2G y2G¯ 1 = tr (P ρ + P ¯ σ) ; 2 G G P P 1 where PG := y2G Py and PG¯ := y2G¯ Py are projectors too, with PG + PG¯ = . If we explore a little more, we obtain 2Pguess = tr(PG ρ + PG¯ σ) = tr(PG ρ + [1 − PG] σ) (S.2) = tr(PG [ρ − σ]) + tr(1σ) (∗) = tr(PG [ρ − σ]) + 1; where (∗) comes from the fact that σ is a density matrix and therefore tr(σ) = 1. Notice that we have only defined G depending on O so far. Hence, to maximise the guessing probability we need to find the optimal fPygy that maximise tr(PG [ρ − σ]). First we express G in another way using linearity of the trace, G = fy : PO;ρ(y) ≥ PO,σ(y)g = fy : tr(Pyρ) ≥ tr(Pyσ)g (S.3) = fy : tr(Py(ρ − σ)) ≥ 0g : P Now we will try a clever choice of G. Let fygy be the eigenbasis of ρ − σ = y λyjyihyj. Notice that ρ − σ is not a density matrix { in particular it has trace zero. If we choose fPygy to be the projectors on that basis, Py = jyihyj, we obtain G = fy : tr(Py(ρ − σ)) ≥ 0g n X 0 0 o = y : tr jyihyj λy0 jy ihy j ≥ 0 y0 (S.4) = fy : tr (jyihyjλy) ≥ 0g = fy : λy ≥ 0g ; i.e. G is the set of outcomes of O corresponding to projectors on states jyihyj that corre- spond to non negative eigenvalues of ρ − σ. In this case, tr(PG [ρ − σ]) is the sum of all positive eigenvalues of ρ − σ. This result is promising, but now we have to prove that is is indeed optimal, i.e. that no other choice of projector P could give better results. We can write ρ − σ as R − S, where P P R = y2G λyjyihyj and S = y2G¯ −λyjyihyj. Both operators R and S are positive and diagonal. Furthermore they are mutually orthogonal because fjyig is an orthogonal basis. 2 We have that X tr(PG [ρ − σ]) = λy = tr(R): (S.5) y2G For any other projector P 0, however, tr(P 0 [ρ − σ]) = tr(P 0 [R − S]) = tr(P 0 R) − tr(P 0 S) (S.6) ≤ tr(R) − tr(P 0 S) (∗∗) ≤ tr(R); (∗∗∗) where (∗∗) stands because projectors can only decrease the trace and (∗∗∗) because P 0S is positive by assumption. P We have proved that a measurement represented by O = y yjyihyj, where fjyigy is the eigenbasis of ρ − σ optimises the probability of guessing correctly which state we were given. This solution corresponds to the following strategy. We measure our state (ρ or σ) in the eigenbasis of ρ − σ. If we obtain a state that corresponds to a positive eigenvalue of ρ − σ (i.e. y 2 G) then it is more likely that we have measured ρ. If we get a negative eigenvalue of ρ − σ (i.e. y 2 G¯) we should say the state was σ. In the particular case where the two density operators share the same eigenbasis, this cor- responds to following the classical strategy for distinguishing two probability distributions after measuring the state in their common eigenbasis. (b) We already proved that in the previous exercise, (S.2) and the following. (c) In (a) we have shown constructively how to write the difference between two quantum states, ρ−σ, as R −S, where R and S are two positive operators with orthogonal support. We now use this fact to write jρ − σj = R + S and obtain 1 δ(σ; ρ) = tr(jρ − σj) 2 1 = (tr(R) + tr(S)) 2 = tr(R) (∗) = tr [E(R)] (S.7) ≥ max ftr [P E(R)] − tr [P E(S)]g (∗∗) P = max tr [P (E(R − S))] (∗∗∗) P = δ (E(σ); E(ρ)) ; where (∗) stands because tr(R) − tr(S) = tr(R − S) = tr(ρ − σ) = tr(ρ) − tr(σ) = 1 − 1 = 0; and the inequality (∗∗) follows from tr(P E(R)) ≤ tr(E(R)) and tr(P E(S)) ≥ 0 for any pro- jector P , since projectors are positive operators and can only decrease the trace. Finally, linearity of TPMs allows us to perform step (∗∗∗). We also used the characterization of the trace distance in terms of a maximization over projectors, see (2), in the very last step. 3 (d) This result implies that there is no experimental setup that allows us to distinguish non- orthogonal states with certainty (because whatever this setup is, its action on the quantum states can be described by some CPTPM). If there was such a setup, we could copy (clone) the states perfectly, hence the contradiction. In fact, the trace distance (as we have seen in the lecture) gives us an upper limit on our ability to distinguish them. If there were quantum operations that increase the distance between two states, we could design measurement devices such that this upper limit no longer holds. Exercise 2. Fidelity and Uhlmann's Theorem Given two states ρA and σA on HA with fixed basis fjiiAgi and a reference Hilbert space HB with fixed basis fjiiBgi, which is a copy of HA, Uhlmann's theorem claims that the fidelity can be written as F (ρA; σA) = max jhΨjΦij ; (4) jΨiAB ;jΦiAB where the maximum is over all purifications jΨiAB of ρA and jΦiAB of σA on HA ⊗HB. Let us introduce the state j iAB as p X j i = ( ρ ⊗ UB) jΩi; jΩi = jiiA ⊗ jiiB; (5) i where UB is any unitary on HB. We have seen in Exercise Sheet 6 that j iAB is a purification of ρA and that any purification of ρA can be written in this form. (a) Use the construction presented in the proof of Uhlmann's theorem to calculate the fidelity between 0 1 0 σA = 2=2 and ρA = pj0ih0jA + (1 − p)j1ih1jA in the 2-dimensional Hilbert space. Hint: Convince yourself that the vector jΩi has the property that 1 ⊗ S jΩi = ST ⊗ 1jΩi for all linear operators S on HA. (b) Give an expression for the fidelity between any pure state and the completely mixed state 1n=n in the n-dimensional Hilbert space. Hint: You may want to use a different characterization of the fidelity than the one by Uhlmann for this exercise. Solution. (a) Because of the special form in which we can write purifications, it is apparent that it is sufficient to maximise over one set of purifications only. We set p 0 jΨi = ( ρ ⊗ VB)jΩi ; 1 1 (S.8) jΦi = p (1A ⊗ 1B)jΩi = p (j00i + j11i) 2 2 for some unitary V on B (which can also be seen as a unitary on A because A and B are isomorphic). 4 It follows that 1 p 0 jhΨjΦij = p hΩj ρ ⊗ VBjΩi 2 1 hp 0 i = p tr ρ ⊗ VBjΩihΩj 2 1 hp 0 T i (∗) = p tr ρ · V ⊗ 1BjΩihΩj 2 A (S.9) 1 h i = p tr pρ0 · V T 2 A 1 h i ≤ p tr pρ0 2 1 p = p p + p1 − p : 2 For (∗) we used the fact that the state jΩi is such that applying V to its B part is equivalent to applying V T to its A part.
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