1 Ph 219a/CS 219a Exercses Due: Wednesday 23 October 2013 1.1 How far apart are two quantum states? Consder two quantum states descrbed by densty operators ρ and ρ n an N-dmensonal Hlbert space, and consder the complete orthogonal measurement {E a, a = 1, 2, 3,...N}, where the E a s are onedmensonal projectors satsfyng N E a = I. (1) a=1 When the measurement s performed, outcome a occurs wth probablty p a = tr ρe a f the state s ρ and wth probablty p a = tr ρe a f the state s ρ. The L 1 dstance between the two probablty dstrbutons s defned as d(p, p) p p 1 1 N p a p a ; (2) 2 ths dstance s zero f the two dstrbutons are dentcal, and attans ts maxmum value one f the two dstrbutons have support on dsjont sets. a) Show that d(p, p) 1 2 a=1 N λ (3) where the λ s are the egenvalues of the Hermtan operator ρ ρ. Hnt: Workng n the bass n whch ρ ρ s dagonal, fnd an expresson for p a p a, and then fnd an upper bound on p a p a. Fnally, use the completeness property eq. (1) to bound d(p, p). b) Fnd a choce for the orthogonal projector {E a } that saturates the upper bound eq. (3). =1
2 Defne a dstance d(ρ, ρ) between densty operators as the maxmal L 1 dstance between the correspondng probablty dstrbutons that can be acheved by any orthogonal measurement. From the results of (a) and (b), we have found that d(ρ, ρ) = 1 2 N λ. (4) =1 c) The L 1 norm A 1 of an operator A s defned as [ A 1 tr (AA ) 1/2]. (5) How can the dstance d(ρ, ρ) be expressed as the L 1 norm of an operator? Now suppose that the states ρ and ρ are pure states ρ = ψ ψ and ρ = ψ ψ. If we adopt a sutable bass n the space spanned by the two vectors, and approprate phase conventons, then these vectors can be expressed as ψ = where ψ ψ = sn θ. ( ) cosθ/2 sn θ/2, ψ = ( ) sn θ/2 cos θ/2 d) Express the dstance d(ρ, ρ) n terms of the angle θ., (6) e) Express ψ ψ 2 (where denotes the Hlbert space norm) n terms of θ, and by comparng wth the result of (d), derve the bound d( ψ ψ, ψ ψ ) ψ ψ. (7) f) Bob thnks that the norm ψ ψ should be a good measure of the dstngushablty of the pure quantum states ρ and ρ. Explan why Bob s wrong. Hnt: Remember that quantum states are rays. 1.2 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 2 wth a pror probablty
3 p 2 = 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 2 = I E 1. If Bob s outcome s E 1, he guesses that Alce s state was ρ 1, and f t s E 2, he guesses ρ 2. Then the probablty that Bob guesses wrong s p error = p 1 tr (ρ 1 E 2 ) + p 2 tr (ρ 2 E 1 ). (8) a) Show that p error = p 1 + λ E 1, (9) where { } denotes the orthonormal bass of egenstates of the Hermtan operator p 2 ρ 2 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. (10) where neg denotes the sum over all of the negatve egenvalues of p 2 ρ 2 p 1 ρ 1. c) It s convenent to express ths optmal error probablty n terms of the L 1 norm of the operator p 2 ρ 2 p 1 ρ 1, p 2 ρ 2 p 1 ρ 1 1 = tr p 2 ρ 2 p 1 ρ 1 = pos λ λ, (11) neg the dfference between the sum of postve egenvalues and the sum of negatve egenvalues. Use the property tr (p 2 ρ 2 p 1 ρ 1 ) = p 2 p 1 to show that (p error ) optmal = 1 2 1 2 p 2ρ 2 p 1 ρ 1 1. (12)
4 Check whether the answer makes sense n the case where ρ 1 = ρ 2 and n the case where ρ 1 and ρ 2 have support on orthogonal subspaces. d) Now suppose that Alce decdes at random (wth p 1 = p 2 = 1/2) to prepare one of two pure states ψ 1, ψ 2 of a sngle qubt, wth ψ 1 ψ 2 = sn(2α), 0 α π/4. (13) Wth a sutable choce of bass, the two states can be expressed as ( ) ( ) cos α snα ψ 1 =, ψ snα 2 =. (14) 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(2 ) = p 2 p( 2) = p 2 tr ρ 2 E ; (15) 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(2 ) (16) 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(2 )) = 1 2 1 p(2 ) p(1 ). (17) 2 Show that the probablty of error, averaged over the measurement outcomes, s p error = p p error () = 1 2 1 tr(p 2 ρ 2 p 1 ρ 1 )E. (18) 2
5 f) By expandng n terms of the bass of egenstates of p 2 ρ 2 p 1 ρ 1, show that p error 1 2 1 2 p 2ρ 2 p 1 ρ 1 1. (19) (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. 1.3 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, (20) 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 2A)I + A ( u u + v v ), (21) where A s a postve real number. How should Bob choose A to mnmze the probablty of the outcome DK, and what 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.(21) 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
6 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? 1.4 What probablty dstrbutons are consstent wth a mxed state? A densty operator ρ, expressed n the orthonormal bass { α } that dagonalzes t, s ρ = p α α. (22) We would lke to realze ths densty operator as an ensemble of pure states { ϕ µ }, where ϕ µ s prepared wth a specfed probablty q µ. Ths preparaton s possble f the ϕ µ s can be chosen so that ρ = µ q µ ϕ µ ϕ µ. (23) We say that a probablty vector q (a vector whose components are nonnegatve real numbers that sum to 1) s majorzed by a probablty vector p (denoted q p), f there exsts a doubly stochastc matrx D such that q µ = D µ p. (24) A matrx s doubly stochastc f ts entres are nonnegatve real numbers such that µ D µ = D µ = 1. That the columns sum to one assures that D maps probablty vectors to probablty vectors (.e., s stochastc). That the rows sum to one assures that D maps the unform dstrbuton to tself. Appled repeatedly, D takes any nput dstrbuton closer and closer to the unform dstrbuton (unless D s a permutaton, wth one nonzero entry n each row and column). Thus we can vew majorzaton as a partal order on probablty vectors such that q p means that q s more nearly unform than p (or equally close to unform, n the case where D s a permutaton).
7 Show that normalzed pure states { ϕ µ } exst such that eq. (23) s satsfed f and only f q p, where p s the vector of egenvalues of ρ. Hnt: Recall that, accordng to the Hughston-Jozsa-Wootters Theorem, f eq. (22) and eq. (23) are both satsfed then there s a untary matrx V µ such that qµ ϕ µ = p V µ α. (25) You may also use (but need not prove) Horn s Lemma: f q p, then there exsts a untary (n fact, orthogonal) matrx U µ such that q = Dp and D µ = U µ 2.