Ph 219a/CS 219a. Exercises Due: Wednesday 12 November 2008

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1 1 Ph 19a/CS 19a Exercses Due: Wednesday 1 November Whch state dd Alce make? Consder a game n whch Alce prepares one of two possble states: ether ρ 1 wth a pror probablty p 1, or ρ wth a pror probablty p = 1 p 1. Bob s to perform a measurement and on the bass of the outcome to guess whch state Alce prepared. If Bob s guess s rght, he wns; f he guesses wrong, Alce wns. In ths exercse you wll fnd Bob s best strategy, and determne hs optmal probablty of error. Let s suppose (for now) that Bob performs a POVM wth two possble outcomes, correspondng to the two nonnegatve Hermtan operators E 1 and E = I E 1. If Bob s outcome s E 1, he guesses that Alce s state was ρ 1, and f t s E, he guesses ρ. Then the probablty that Bob guesses wrong s p error = p 1 tr (ρ 1 E ) + p tr (ρ E 1 ). (1) a) Show that p error = p 1 + λ E 1, () where { } denotes the orthonormal bass of egenstates of the Hermtan operator p ρ p 1 ρ 1, and the λ s are the correspondng egenvalues. b) Bob s best strategy s to perform the two-outcome POVM that mnmzes ths error probablty. Fnd the nonnegatve operator E 1 that mnmzes p error, and show that error probablty when Bob performs ths optmal two-outcome POVM s (p error ) optmal = p 1 + neg λ. (3) where neg denotes the sum over all of the negatve egenvalues.

2 c) It s convenent to express ths optmal error probablty n terms of the L 1 norm of the operator p ρ p 1 ρ 1, p ρ p 1 ρ 1 1 = tr p ρ p 1 ρ 1 = λ pos neg λ, (4) the dfference between the sum of postve egenvalues and the sum of negatve egenvalues. Use the property tr (p ρ p 1 ρ 1 ) = p p 1 to show that (p error ) optmal = 1 1 p ρ p 1 ρ 1 1. (5) Check whether the answer makes sense n the case where ρ 1 = ρ and n the case where ρ 1 and ρ have support on orthogonal subspaces. d) Now suppose that Alce decdes at random (wth p 1 = p = 1/) to prepare one of two pure states ψ 1, ψ of a sngle qubt, wth ψ 1 ψ = sn(α), 0 α π/4. (6) Wth a sutable choce of bass, the two states can be expressed as ( ) ( ) cos α snα ψ 1 =, ψ snα =. (7) cos α Fnd Bob s optmal two-outcome measurement, and compute the optmal error probablty. e) Bob wonders whether he can fnd a better strategy f hs POVM {E } has more than two possble outcomes. Let p(a ) denote the probablty that state a was prepared, gven that the measurement outcome was ; t can be computed usng the relatons p p(1 ) = p 1 p( 1) = p 1 tr ρ 1 E, p p( ) = p p( ) = p tr ρ E ; (8) here p( a) denotes the probablty that Bob fnds measurement outcome f Alce prepared the state ρ a, and p denotes the probablty that Bob fnds measurement outcome, averaged over Alce s choce of state. For each outcome, Bob wll make hs decson accordng to whch of the two quanttes p(1 ), p( ) (9)

3 3 s the larger; the probablty that he makes a mstake s the smaller of these two quanttes. Ths probablty of error, gven that Bob obtans outcome, can be wrtten as p error () = mn(p(1 ), p( )) = 1 1 p( ) p(1 ). (10) Show that the probablty of error, averaged over the measurement outcomes, s p error = p p error () = 1 1 tr(p ρ p 1 ρ 1 )E. (11) f) By expandng n terms of the bass of egenstates of p ρ p 1 ρ 1, show that p error 1 1 p ρ p 1 ρ 1 1. (1). Fdelty (Hnt: Use the completeness property E = I.) Snce we have already shown that ths bound can be saturated wth a twooutcome POVM, the POVM wth many outcomes s no better. We saw n Exercse 1.4 that the L 1 norm ρ ρ 1 provdes a useful measure of the dstngushablty of the states ρ and ρ. Another useful measure of dstngushablty s the fdelty F(ρ, ρ), whch s defned as ( F(ρ, ρ) ρ 1 1 ρ 1 = tr ρ 1 ρρ 1 ). (13) (Some authors use the name fdelty for the square root of ths quantty.) The fdelty s nonnegatve, vanshes f ρ and ρ have support on mutually orthogonal subspaces, and attans ts maxmum value 1 f and only f the two states are dentcal. a) The fdelty F(ρ, ρ) s actually symmetrc n ts two arguments, although the symmetry s not manfest n eq. (13). To demonstrate the symmetry, show that for any Hermtan A and B, the L 1 norm obeys AB 1 = BA 1. (14) Hnt: Show that BAAB and ABBA have the same egenvalues.

4 4 The overlap of two probablty dstrbutons {p } and { p } s defned as Overlap({p }, { p }) p p. (15) Suppose that we try to dstngush the two states ρ and ρ by performng the POVM {E }. Then the two correspondng probablty dstrbutons have the overlap Overlap(ρ, ρ; {E }) tr ρe tr ρe. (16) It turns out that the mnmal overlap that can be acheved by any POVM s mn {E } [Overlap(ρ, ρ; {E })] = ρ 1 ρ 1 1 = F(ρ, ρ). (17) In ths exercse, you wll show that the square root of the fdelty bounds the overlap, but not that the bound can be saturated. b) The space of lnear operators actng on a Hlbert space s tself a Hlbert space, where the nner product (A, B) of two operators A and B s ( ) (A, B) tr A B. (18) For ths nner product, the Schwarz nequalty becomes ( ) 1/ ( 1/ tr A B tr A A tr B B), (19) Choosng A = ρe 1 1 and B = U ρ E 1 1 (for an arbtrary untary U), use ths form of the Schwarz nequalty to show that c) Now use the polar decomposton (where V s untary) to wrte Overlap(ρ, ρ; {E }) tr ρ 1 U ρ 1. (0) A = V A A (1) ρ 1 ρ 1 = V ρ 1 ρρ 1, () and by choosng the untary U n eq. (0) to be U = V 1, show that Overlap(ρ, ρ; {E }) F(ρ, ρ). (3)

5 5 d) We can obtan an explct formula for the fdelty n the case of two states of a sngle qubt. Usng the Bloch parametrzaton ρ( P) = 1 ( I + σ P ), (4) show that the fdelty of two sngle-qubt states wth polarzaton vectors P and Q s F( P, Q) = 1 ( ) 1 + P Q + (1 P )(1 Q ). (5) Hnt: Frst note that the egenvalues of a matrx can be expressed n terms of the trace and determnant ( of ) the matrx. Then evaluate the determnant and trace of ρ 1 ρρ 1, and calculate the fdelty usng the correspondng expresson for the egenvalues..3 Eavesdroppng and dsturbance Alce wants to send a message to Bob. Alce s equpped to prepare ether one of the two states u or v. These two states, n a sutable bass, can be expressed as u = ( cosα sn α ), v = ( ) snα cos α, (6) where 0 < α < π/4. Suppose that Alce decdes at random to send ether u or v to Bob, and Bob s to make a measurement to determne what she sent. Snce the two states are not orthogonal, Bob cannot dstngush the states perfectly. a) Bob realzes that he can t expect to be able to dentfy Alce s qubt every tme, so he settles for a procedure that s successful only some of the tme. He performs a POVM wth three possble outcomes: u, v, or DON T KNOW. If he obtans the result u, he s certan that v was sent, and f he obtans v, he s certan that u was sent. If the result s DON T KNOW, then hs measurement s nconclusve. Ths POVM s defned by the operators E u = A(I u u ), E v = A(I v v ), E DK = (1 A)I + A ( u u + v v ), (7) where A s a postve real number. How should Bob choose A to mnmze the probablty of the outcome DK, and what

6 6 s ths mnmal DK probablty (assumng that Alce chooses from { u, v } equprobably)? Hnt: If A s too large, E DK wll have negatve egenvalues, and Eq.(7) wll not be a POVM. b) Eve also wants to know what Alce s sendng to Bob. Hopng that Alce and Bob won t notce, she ntercepts each qubt that Alce sends, by performng an orthogonal measurement that projects onto the bass {( ( 1 0), 0 1)}. If she obtans the outcome ( 1 0), she sends the state u on to Bob, and f she obtans the outcome ( 0 1), she sends v on to Bob. Therefore each tme Bob s POVM has a conclusve outcome, Eve knows wth certanty what that outcome s. But Eve s tamperng causes detectable errors; sometmes Bob obtans a conclusve outcome that actually dffers from what Alce sent. What s the probablty of such an error, when Bob s outcome s conclusve?.4 The prce of quantum state encrypton Alce and Bob are workng on a top secret project. I can t tell you exactly what the project s, but I wll reveal that Alce and Bob are connected by a perfect quantum channel, and that Alce uses the channel to send quantum states to Bob. Alce and Bob are worred that an eavesdropper (Eve) mght ntercept some of Alce s transmssons. By measurng the ntercepted quantum state, Eve could learn somethng about what Alce s sendng, and perhaps make an nference about the nature of the project. To protect aganst eavesdroppng, Alce and Bob decde to encrypt the quantum states that Alce sends. They share a secret key, a strng of random bts about whch the eavesdropper knows nothng. By consumng n bts of secret key, Alce can encrypt, and Bob can decrypt, an arbtrary n-qubt state ρ. For every possble state ρ, the encrypted state looks exactly the same to Eve, so she cannot fnd out anythng about ρ. Here s how the encrypton procedure works: We may express the n bt strng x as x = x 1 x x n, where x {0, 1,,3}, and denote a tensor product of n Paul operators as σ(x) = σ x1 σ x σ xn 1 σ xn (8)

7 7 (where σ 0 = I). Note that σ(x) = I n, the dentty operator actng on n qubts. To encrypt, Alce consults her random strng to determne x (whch s chosen unformly at random), and apples σ(x) to the state, obtanng σ(x)ρσ(x). To decrypt, Bob, consults the same strng and apples σ(x) to recover ρ. a) Snce Eve does not know the secret key, to her the encrypted state s ndstngushable from Show that, for any n-qubt state ρ E(ρ) = 1 n σ(x)ρσ(x). (9) x E(ρ) = 1 ni n. (30) Snce E(ρ) s ndependent of ρ, no nformaton about ρ s accessble to Eve. b) Alce wonders f t s possble to encrypt the state usng a shorter key. Alce and Bob could use ther shared randomness to sample an arbtrary probablty dstrbuton. That s, they could agree on a set of N untary matrces {U a, a = 1,, 3,..., N}, and Alce could encrypt by applyng U a wth probablty p a. Then Bob could decrypt by applyng Ua 1. To Eve, the encrypted state would then appear to be E (ρ) = a p a U a ρu 1 a. (31) Show that, f E (ρ) = I n, then p a n for each a. Hnt: Note that E has an operator sum representaton wth Kraus operators {σ(x)/ n } and that E has an operator sum representaton wth Kraus operators { p a U a }. Furthermore E = E. Therefore, there exsts an M M untary matrx V ax (where M = Max(N, n such that p a U a = x V axσ(x)/ n. Now express p a tr U a U a n terms of V ( ). Remark: The result shows that encrypton requres N n, and that at least n bts of key are requred to specfy U a. Thus the

8 8 encrypton scheme n whch σ(x) s appled s the most effcent possble scheme. (For encrypton to be effectve, t s enough for E(ρ) to be ndependent of ρ; t s not necessary that E(ρ) = I n / n. But the same result apples under the weaker assumpton that E(ρ) s ndependent of ρ.).5 Untal maps and majorzaton Recall that the acton of a trace-preservng completely postve (TPCP) map E can be expressed as E(ρ) = µ M µ ρm µ, (3) where M µm µ = I. (33) µ A TPCP map s sad to be untal f E(I) = I, or equvalently f Mµ M µ = I. (34) If A s a nonnegatve Hermtan operator wth unt trace (tr A = 1), let λ(a) denote the vector of egenvalues of A, whch can be regarded as a probablty vector. If A and B are nonnegatve Hermtan operators wth unt trace, we say that A B ( A s majorzed by B ) f λ(a) λ(b). (Recall that for two probablty vectors p and q, we say that p q f there s a doubly stochastc matrx D such that p = Dq.) Show that f ρ s a densty operator and E s a untal map, then E(ρ) ρ. (35) Hnt: Express ρ = U U where s dagonal and U s untary, and express ρ E(ρ) = V V, where s dagonal and V s untary. Then try to show that the dagonal entres of can be expressed as a doubly stochastc matrx actng on the dagonal entres of. Remark: A untal map s the natural quantum generalzaton of a doubly stochastc map (a doubly stochastc map can be regarded as the specal case of a untal map that preserves the bass n whch ρ s dagonal). The result of the exercse shows that a untal map takes an nput densty operator to an output densty operator that s no less random than the nput.