Quantum Information (Instructor's Solution Manual) Stephen M. Barnett

SOLUTION MANUAL FROM TEXT BOOK

iv

Quantum Information: Solutions to Exercises

Stephen M. Barnett

Preface The following pages present outline, or sketch solutions, to the exercises in my book Quantum Information, published by Oxford University Press in June 2009. The solutions are arranged into eight ‘chapters’ to match the chapters in the book. Equation numbers refer to the correspondingly numbered equations in the book.

Glasgow

March 2019

Errata Shannon taught us that all communications channels contain errors and, inevitably, Quantum Information is no exception. Following is a list of corrections that I have come across since the initial publication. p. 74 There is an error in Fig. 3.7. The final Jones matrix in the table should be

cos2 θ − sin2 θ 2 cos θ sin θ .

2 cos θ sin θ sin2 θ − cos2 θ p. 112 p. 98

In exercise (4.5) “density” should be “density operator” There is an h̄ missing in eqn 4.36, which should be

Z pm x (x − xm )2 + i . |xm , pm i = (2πσ 2 )−1/4 dx exp − 4σ 2 h̄

Also on this page, there a normalization factor missing in eqn 4.39, which should be

Z (x − xm )2 2 −1/2 P(xm ) = (2πσ ) dx hx|ρ̂|xi exp − .

2σ 2 p. 113

In exercise (4.24) the operator Υ̂ should be defined as Υ̂ =

M X

π̂j ŵj .

j=1

p. 135 In eqn 5.63, there is an error in the normalization of the wavefunction. The prefactor should be (πσ 2 )−1/2 p. 162 In exercise (6.23) there is an error in the transformations given They should be: 1/2 Ûswap |00i = |00i

vi Preface

e−iπ/4 √ (|01i + i|10i) 2 −iπ/4 e 1/2 Ûswap |10i = √ (|10i + i|01i) 2 1/2 Ûswap |11i = |11i

1/2 Ûswap |01i =

p.155 Figure 6.26 has the incorrect powers on the final X gates. The correct from of the figure is given here in Fig. ??. p. 194 In exercise (7.20) the condition for the two strings to have the same parity should be b · c + b · c = 0 mod 2. p. 228 In exercise (8.21) the inequality is quoted the wrong way around. It should be S(C) + S(B) ≤ S(AB) + S(AC) . p. 230

In exercise (8.44) the two signal states should be ρ̂1 = q|0ih0| + (1 − q)|1ih1| ρ̂2 = (1 − q)|0ih0| + q|1ih1| .

Probability and information (1.1) by

We know that the joint and single-event probabilities are related P (ai ) =

X

P (ai , bj )

j

and that the joint probabilities are necessarily greater than or equal to zero. It follows immediately that P (ai ) ≥ P (ai , bj ) with equality occuring only if P (ai , bk ) = 0

k 6= 0

⇒ P (ai , bk ) = P (ai )δjk .

(1.2) No it does not. The conditional probabilities are related by Bayes’ theorem and are not in general equal.

(1.3)

We can read off from the probability tree the probabilities

1 1 P (a2 ) = 2 3 and the conditional probabilities P (a1 ) =

1 4 2 P (b1 |a2 ) = 3 1 P (b1 |a3 ) = 3 P (b1 |a1 ) =

1 4 1 P (b2 |a2 ) = 3 1 P (b2 |a3 ) = 3 P (b2 |a1 ) =

P (a3 ) =

1 6

P (b3 |a1 ) =

1 2

P (b3 |a2 ) = 0 P (b3 |a3 ) =

1 .

3

It is now straightforward to construct the joint probabilities P (ai |bj ) and then extract the required probabilities: 29 7 P (b2 ) = 72 24 and conditional probabilities: P (b1 ) =

2 29 3 P (a1 |b2 ) = 7 9 P (a1 |b3 ) = 11 P (a1 |b1 ) =

16 29 8 P (a2 |b2 ) = 21

P (a2 |b1 ) =

P (a2 |b3 ) = 0

P (b3 ) =

11 36

4 29 4 P (a3 |b2 ) = 21 2 P (a3 |b3 ) = .

11 P (a3 |b1 ) =

1

2 Probability and information

From this we can readily read off the entries to construct the required probability tree.

(1.4) The problem gives the probabilities for arriving on time given that the long and shirt routes are taken: P (bO |as ) = 1

P (bO |al ) =

3 4

and probabilities for taking the long and short routes: P (al ) =

1 4

P (as ) =

3 .

4

From the conditional probabilities we have P (bL |as ) = 1 − P (bO |as ) = 0 1 P (bL |al ) = 1 − P (bO |al ) = .

4 The required conditional probabilities are, therefore, P (bL |al )P (al ) =1 P (bL |al )P (al ) + P (bL |as )P (as ) P (as |bL ) = 1 − P (al |bL ) = 0 .

P (al |bL ) =

(1.5) Each particle is detected with probability η and missed with probability 1 − η. This means that if we detect n given that N were ! present, then we also miss N − n particles. There are n!(NN−n)! was for N! n N −n . We solve this to happen and hence P (n|N ) = n!(N −n)! η (1 − η) for P (N |n) using Bayes’ theorem: P (N |n) =

P (n|N )P (N ) .

P (n)

In each case we need to find P (n) using P (n) =

∞ X

P (n|N )P (N ) .

n=0

(i)

For the Poisson distribution we find P (n) =

∞ X N =n

N! N̄ N η n (1 − η)N −n e−N̄ n!(N − n)!

N!

= e−ηN̄

(η N̄ )n . n!

Note that this is a Poisson distribution with mean η N̄ . Bayes’ theorem gives the answer P (N |n) = e−N̄ (1−η)

[N̄ (1 − η)]N −n .

(N − n)!

Probability and information 3

(ii) The case in which all P (N ) are equal needs careful handling as each of the probabilities is zero! We can overcome this problem by use fo a suitable limit. Let the probabilities, P (N ), for N = 0, 1, · · · , s each be 1/(s + 1) and the probabilities for N > s be zero. We shall take the limit s → ∞ at the end of the calculation. We start by calculating P (n): P (n) =

s X

1 N! η n (1 − η)N −n . n!(N − n)! s+1

N =n

Hence N! n N −n 1 n!(N −n)! η (1 − η) s+1 .

0 N ! n (1 − η)N 0 −n 1 η 0 0 N =n n!(N −n)! s+1

P (N |n) = P∞

Taking the limit as s → ∞ then gives P (N |n) =

N! η n+1 (1 − η)N −n . n!(N − n)!

(iii) The problem here is that we need the probability distribution P (N ) but we know only that the mean number is N̄ : ∞ X

N P (N ) = N̄ .

N =0

In the absence of any further information we use Jaynes’s Max Ent method to maximize X He = − P (N ) ln P (N ) N

subject to the constraints that the probabilities sum to unity and give the correct value for N̄ . It is straightforward to use Lagrange’s method of undertermined multipliers to get P (N ) =

N̄ N .

(N̄ + 1)N +1

We can now calculate P (n) and hence the required conditional probability: P (n) =

∞ X N 0 =n

0

N̄ N N 0! n N 0 −n η (1 − η) n!(N 0 − n)!

(N̄ + 1)N 0 +1

(η N̄ )n (1 + η N̄ )n+1 N! (1 + η N̄ )n+1 ⇒ P (N |n) = [(1 − η)N̄ ]N −n . n!(N − n)!

(N̄ + 1)N +1 =

It is interesting to compare this result with that from part (ii). In the limit N̄ → ∞, the conditional probabilities become equal, as they should.

4 Probability and information

(1.6) P (ai , bj , ck ) = P (ai |bj , ck )P (bj , ck ) = P (ck |ai , bj )P (ai , bj ) P (ck |ai , bj )P (ai , bj ) ⇒ P (ai |bj , ck ) = .

P (bj , ck ) (1.7)

We start from eqn 1.21: P (ai |bj ) = `(ai |bj )P (ai ) P (ai , bj ) P (ai |bj ) = , ⇒ `(ai |bj ) = P (ai ) P (ai )P (bj )

which is manifestly symmetric in ai and bj .

(1.8)

The prior probabilities are P (BB) =

1 3

P (Bb) =

2 , 3

and the likelihoods are `(BB|xi = black) = 1

`(Bb|xi = black) = 0 .

It then follows that 1 P (BB|xi = black) ∝ 1n n 3 1 2 P (Bb|xi = black) ∝ .

2 3 Normalizing then gives P (BB|xi = black) =

1n 13 1n 13 +

=1− (1.9) are

1 n 2 2 3

1 2n−1 + 1

.

The probabilities that each of the players wins in the first round P (Amy) =

1 2

P (Barbara) =

1 4

P (Claire) =

1 .

8

The probability that each wins on the second round are 1/8 multiplied by these same probbabilities. In each round the probabilities are in the ration 4:2:1. It follows that the total probabilities that each player will win are 2 1 4 P (Barbara) = P (Claire) = .

P (Amy) = 7 7 7 (1.10) This is the famous Monty Hall problem. Keeping the originally chosen box does not take into account the additional information

Probability and information 5

provided by displaying an empty box. You should change to the remaining unopen box and will then win with probability 2/3. If this seems strange, then consider that another way of stating the problem is for the host to offer you to keep your one box or take both the remaining boxes.

There is much scope here for classroom discussion!

(1.11) (a) The possibilities left are (Boy, Boy), (Boy, Girl) and (Girl, Boy) and all are equally probable. Hence the probability that they are both boys is 1/3. (b) The possibilities are (Reuben, Boy not Reuben). (Boy not Reuben, Reuben), (Reuben, Girl), (Girl, Reuben) and (Reuben, Reuben).

The probability that they are both boys is P (R, B(6 R)) + P (B(6 R), R) + P (R, R) P (R, B(6 R)) + P (B(6 R), R) + P (R, R) + P (R, G) + P (G, R) 2P (R, B) − P (R, R) = 2P (R, B) − P (R, R) + 2P (R, G) P (R) − P (R, R) = 2P (R) − P (R, R) 1 − P (R) = 2 − P (R) 1 ≈ , 2

P (BB) =

where the last line follows because Reuben is an unusual name. This problem (and other teasers) is treated in more depth in the book by Mlodinow.

(1.12) The strategy is for each player to look at the other players cards and to make a guess or to decline to guess on the basis of what he/she sees. If a player sees that the colours of the other two players cards are different then he/she declines to guess. If he/she sees that they are both red (black) then he/she bets that he/she has a black (red) card. This strategy wins if there are two cards of the same colour and one that is different. It loses if all three cards are the same colour. Hence the winning probability is 34 .

It is true that any individual player will guess the colour of his/her card correctly with probability 12 , but this strategy ensures that when a player guesses correctly they are the only player making a guess, but when they guess incorrectly all the players are guessing. It works by correlating the guesses.

(1.13)

It follows from the definition of the logarithm that x = aloga x = blogb x .

Taking the logarithm in base b gives logb x = logb a loga x ,

6 Probability and information

which is the required identity. The conversion factor between bits and nats then follows from the fact that ln 2 ≈ 0.693.

(1.14) H = −p log p − (1 − p) log(1 − p) 1 = [−p ln p − (1 − p) ln(1 − p)] . ln 2 Differentiating with respect to p gives

1 p dH =− ln dp ln 2 1−p which is zero only at p = 1/2. This is the only stationary point. Differentiating a second time gives

d2 H 1 1 1 + =− , dp2 ln 2 p 1 − p which is clearly negative at p = 1/2 confirming that this is a maximum.

The fact that H is a continuous function means that the minimum occurs at the edge of the allowed range, and the fact that it is symmetric in p and 1 − p means that the minima occur at p = 0 and p = 1.

(1.15)

Let us start by assuming that the inequality holds for k terms: ! k k X X f pki xi ≤ pki f (xi ) , i=1

i=1

where the subscript serves to remind us that k X

pki = 1 .

i=1

Let k X

pki xi = Xk .

i=1

Eqn 1.37 then means that f [λXk + (1 − λ)xk+1 ] ≤ λf (Xk ) + (1 − λ)f (xk+1 ) k X ≤ λpki f (xi ) + (1 − λ)f (xk+1 ) . i=1

If we introduce the probabilities pk+1 = λpki i

pk+1 k+1 = 1 − λ ,

as we clearly can (normalization is correct). Then we have ! k+1 k+1 X X f pk+1 xi ≤ pk+1 f (xi ) . i i i=1

i=1

Probability and information 7

Hence if the inequality is true for k terms then it also holds for k + 1 terms. It clearly holds for 2 terms and therefore for 3,4, and so on.

(1.16) Let us choose three points x, x0 , y such that x < x0 < y and introduce the parameter t such that x0 = tx + (1 − t)y (0 < t < 1). It then follows that from Taylor’s theorem that f 00 (x̃) (x − x0 )2 2 f 00 (ỹ) f (y) = f (x0 ) + f 0 (x0 )(y − x0 ) + (y − x0 )2 2 f 00 (x̃) (x − x0 )2 ⇒ tf (x) + (1 − t)f (y) = f (x0 ) + t 2 f 00 (ỹ) (y − x0 )2 . +(1 − t) 2 f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +

If the second derivative is greater than or equal to zero then it follows that tf (x) + (1 − t)f (y) ≤ f (x0 ) = f [tx + (1 − t)y] , and the function is convex.

(1.17) Recall that l’Hôpital’s rule works for the ratio of two functions that both tend to zero or infinity (from the same side).

1 lim x ln x ln 2 x→0+ 1 ln x = lim+ 1 ln 2 x→0 x

lim+ x log x =

x→0

=

d ln x 1 lim+ dx−1 dx ln 2 x→0 dx

x−1 1 lim = + ln 2 x→0 −x−2 = 0.

(1.18) It is simplest to work with He (A) and to use Lagrange’s method of undetermined multipliers, with the constraint that the probabilities sum to unity. To this end we introduce the quantity !

X X H̃e (A) = − P (ai ) ln P (ai ) + λ P (ai ) − 1 , i

i

in which we vary the probabilities to give X dH̃e (A) = dP (ai ) (− ln P (ai ) − 1 + λ) i

⇒ P (ai ) = λ − 1 . this means that the probabilities all take the same constant value, which normalization determines to be n1 . Clearly, this also maximizes H(A).

8 Probability and information

(1.19) We use the concavity of H(A) or, more precisely, of the function −x log x. First write X H(A0 ) = − P (a0i ) log P (a0i ) i

=−

X

  X X  λij P (aj ) log λik P (ak ) .

i

j

k

Concavity then tells us that X X XX − λij P (aj ) log λik P (ak ) ≥ − λij P (aj ) log P (aj ) j

i

k

=−

X

j

P (aj ) log P (aj )

j

⇒ H(A0 ) ≥ H(A) .

The equality occurs if the mixing occurs only between probabilities that are already equal.

(1.20) (a) For the linear entropy, pi (1 − pi ) is greater than or equal to zero and takes P the value zero only if pi = 0, 1. For the Tsallis and Rényi q entropies, i pi ≤ 1. It takes the value unity only if just one of the probbabilities is unity. (b) P Let us use Lagrange’s method to determine the minimum value of i pqi . We seek to vary this subject to the constraint that the probabilities sum to unity. Hence we form the quantity !

X q X Q= pi + λ pi − 1 . i

i

Independent variation of the probabilities then gives

X dQ = dpi qpiq−1 + λ = 0 i

⇒ pi = −

λ q

1−q ,

so that the probabilities are equal. It follows that T and R are maximized when the probabilities areP equal. That L is also maximized follows from writing it in the form 1 − i p2i and using the value q = 2. (c) He − L = −

X

pi (ln pi + (1 − pi )) .

i

We can use the Taylor series for ln(1 − x) to write ln pi + (1 − pi ) = ln[1 − (1 − pi )] + (1 − pi ) (1 − pi )3 (1 − pi )2 − · · · + (1 − pi ) = −(1 − pi ) − 2 3 ≤ 0.

Probability and information 9

It follows that pi (ln pi + (1 − pi )) = 0 only if pi = 0, 1 for which He = 0 = L. (d)

We can solve this problem using l’Hôpital’s rule.

P d k pk exp[(q − 1) ln pk ]) dq (1 − lim T = lim d q→1 q→1 dq (q − 1) X = lim − pk ln pk exp[(q − 1) ln pk ] q→1

k

= He .

Similarly d dq ln (

P

lim R = lim

q→1

q→1

k pk exp[(q − 1) ln pk ]) d dq (1 − q)

P

pk ln pk exp[(q − 1) ln pk ] = lim − kP q→1 k pk exp[(q − 1) ln pk ] P − k pk ln pk P = k pk = He .

(1.21)

Eqn 1.42 states that H(A) + H(B) ≥ H(A, B) .

By considering the B and C events as a single event we can write H(A) + H(B, C) ≥ H(A, B, C) ⇒ H(A) + H(B) + H(C) ≥ H(A, B, C) .

The first inequality is satisfied (as shown in the text) if P (ai , bj , ck ) = P (ai )P (bj , ck ) and the second is then satisfied if P (ai , bj , ck ) = P (ai )P (bj )P (ck ).

(1.22) Let us start by showing that Q is an acceptable joint probability distribution. We note that it is strictly greater than or equal to zero and summation over P i, j, k clearly gives unity. The simplest way to see that it is acceptable probability distribution is simply to write is as Q(ai , bj , ck ) = P (ai |bj )P (ck |bj )P (bj ) .

We can prove the desired inequality by using the fact that the relative entropy is greater than or equal to zero: X H(P kQ) = P (ai , bj , ck ) (log P (ai , bj , ck ) − log Q(ai , bj , ck )) ijk

=

X

P (ai , bj , ck ) (log P (ai , bj , ck ) − log P (ai , bj )

ijk

− log P (bj , ck ) + log P (bj )) ≥ 0 ⇒

H(A, B) + H(B, C) − H(B) − H(A, B, C) ≥ 0 .

10 Probability and information

(1.23) H(A : B) = H(A) + H(B) − H(A, B) X =− P (ai , bj ) (log P (ai ) + log P (bj ) − log P (ai , bj )) ij

=

X

=

X

P (ai , bj ) log

ij

P (ai , bj ) P (ai )P (bj )

P (ai , bj ) log `(ai |bj ) .

ij

(1.24) If the balls are identical then the situation of the balls is completely determined by the number of balls in box A. Let this number be n. The numbers of balls in the two boxes are completely correlated in that P (A = n) = P (B = N − n) = P (A = n, B = N − n) .

It follows that H(A, B) = H(A) = H(B) ⇒ H(A : B) = H(A) .

The maximum value clearly occurs if all the probabilities P (A = n) are equal so that H(A : B) = log N .

If the balls are not identical then we can treat each of them separately.

Consider the first ball, which will be in box A with probability p and in box B with probability 1 − p. Clearly for this single ball H(A, B) = H(A) = H(B) and H(A : B) takes its maximum value of 1 bit. For N balls the mutual information is then H(A : B) = N .

(1.25) We can follow the analysis in the text for the normal die, but take account of the novel numbering on the faces of the die by varying !

5 X X H̃ = − pn ln pn + λ 1 − pn n

n=2

+µ (3.52 − 2p2 − 3 × 2p3 − 4 × 2p4 − 5p5 ) .

Variation of this e−2µ e−2µ + 2e−3µ + 2e−4µ + e−5µ 2e−3µ p3 = −2µ −3µ e + 2e + 2e−4µ + e−5µ 2e−4µ p4 = −2µ −3µ e + 2e + 2e−4µ + e−5µ −5µ e p5 = −2µ . −3µ e + 2e + 2e−4µ + e−5µ p2 =

Solving (numerically) to find the correct mean gives µ = −0.20.

Probability and information 11

(1.26) We need to maximize P P H(C, D) subject to the constraints that P (c , d ) = P (d ) and i j j i j P (ci , dj ) = P (ci ). The relevant quantity for varying is   X X H̃ = − P (ci , dj ) ln P (ci , dj ) + λ 1 − P (ci , dj ) ij

ij

#

" +

X

µj P (dj ) −

X

j

P (ci , dj )

i



 +

X

νi P (ci ) −

X

P (ci , dj )

j

i

⇒ dH̃ =

X

dP (ci , dj ) [− ln P (ci , dj ) − 1 − λ − µj − νi ] .

ij

It follows that P (ci , dj ) = e−(1+λ) e−νi e−µj , which clearly factorizes into the product of a function depending on i and another that depends only on j. Imposing the constraints clearly shows that this product is P (ci , dj ) = P (ci )P (dj ).

(1.27) (a) E0 e−βE0 + E1 e−βE1 E0 + E1 e−β(E1 −E0 ) = −βE −βE 0 + e 1 e 1 + e−β(E1 −E0 )

1 E1 − Ē . ⇒β= ln E1 − E0 Ē − E0 Ē =

(b) P∞ −βEn h̄ω h̄ω n=0 En e P = + βh̄ω ∞ −βE n 2 e −1 e n=0

1 Ē + 2 h̄ω 1 ⇒β= ln . h̄ω Ē − 12 h̄ω Ē =

(1.28)

Eqn 1.61 is pn =

e−βEn Z(β)

Z(β) =

X

e−βEn .

n

It follows that the thermodynamic entropy is X

S = −kB pn ln e−βEn − ln Z = kB β Ē + kB ln Z . n

12 Probability and information

Differentiating with respect to T gives

dS dβ dĒ kB dZ = kB Ē + kB β + dT dT dβ Z dβ

dβ kB dĒ = kB Ē + kB β − ĒZ dT dβ Z dβ dĒ = kB dT dβ dĒ = kB β . dT Comparing this with the the formula given in the hint, we see that β = (kB T )−1 .

(1.29) In order to be specific, let V1 ≤ V0 /2. If the intelligent being finds that the molecule is in volume V1 , then this allows him/her to extract an amount of work given by Z V0 Z V0 V0 WV 1 = , P dV = kB T V −1 dV = kB T ln V 1 V1 V1 which is greater that kB T ln 2. If, however, he/she finds it in the larger volume then he/she extracts less work: Z V0 Z V0 V0 .

WV0 −V1 = P dV = kB T V −1 dV = kB T ln V 0 − V1 V0 −V1 V0 −V1 The probabilities for finding the molecule in either of the sides are proportional to their respective volumes. It follows that the average work extracted is

V1 V0 V0 − V1 V0 W̄ = kB T ln + ln .

V0 V1 V0 V0 − V1 It is straightforward to see that this is the product of the temperature with the thermodynamic entropy W̄ = T S = −T kB (p1 ln p1 + (1 − p1 ) ln(1 − p1 )) , where p1 = V1 /V0 is the probability for finding the moleclue in volume V1 . Clearly this is maximized if we take equal volumes (V1 = V0 ).

(1.30)

The probability that there are n zeros in the string is P (n) =

N! pn (1 − p)N −n n!(N − n)!

1

2 n N N −p

≈p exp − 2p(1 − p) 2πN p(1 − p)

! ,

where we have used the Gaussian approximation to the binomial distribution (see M. L. Boas, Mathematical methods in the physical sciences

Probability and information 13

2nd ed. (Wiley, New York, 1983) p. 730). This is a Gaussian with a variance inversely proportional to N . As N → ∞ this width tends to zero and hence the probability that p is arbitrarily close to n/N tends to unity.

(1.31) Because then the code would not be decodable. If for example we set A=0, B=1, C=00 and D=11, then the short message 00 could be AA or C. (1.32)

The entropy associated with these probabilities is

H = −(0.49 log 0.49 + 0.42 log 0.21 + 0.09 log 0.09) ≈ 1.763 bits .

The average length of a character is with the proposed coding is 1 × 0.49 + 2 × 0.21 + 3 × 0.21 + 3 × 0.09 = 1.81 bits .

It is perhaps surprising that this value is so close to the Shannon limit.

(1.33) This famous problem is actually treated in Shannon’s original paper and is, therefore, in the book by Shannon and Weaver.

H = −p log p − (1 − p) log(1 − p) = −p log p − (1 − p) log e ln(1 − p)

p2 − ··· = −p log p − (1 − p) log e −p − 2

e . ≈ p log p Let A occur with probability p. It follows that we will have a long sequence of Bs between each A. A good coding scheme would simply list the number of Bs in the a sequence then a special sequence, say 0000, would denote the letter A followed by the next number of Bs. Naturally, we would need to avoid the use of the sequence 0000 in the number of Bs so, for example, a sequence of 16 Bs would be represented by 10001 and other similar modifications apply to larger numbers.

(1.34) (a) The entropy for the letters occuring with the same probabilities as in English is H ≈ 4.17 bits. (b) For Morse code we find, on weighting the length of the letters by their probabilities for occuring, the average length 2.54 bits. (c) The conundrum, of course, is that the length of the letters in Morse is less than the entropy, which should be the Shannon coding limit. The resolution comes from realizing that each symbol is necessarily followed by a space so that the average length of a letter should be 3.54 characters (each of which is a dot, dash or a space). The fact that there are three symbols in Morse code, dot, dash and space, means that 3.54 symbols is really 3.54 × log2 3 = 5.61 bits > H. (1.35) We envisage a long sequence of observers each correcting the errors on the channel of his/her predecessor. In this way we find that

14 Probability and information

the total number of bits required on all the channels is

N = N0 1 + H(q) + H 2 (q) + H 3 (q) + · · · =

N0 .

1 − H(q)

(1.36) It is zero for q = 21 because the output of the channel is then random and carries no information about the input. For q > 12 a bit flip is more likely to have occured than not occured. We can reduce the single-bit error rate to 1 − q < 12 by flipping each bit as we receive it.

(1.37)

The minimum redundancy is N H(q) . −1= N0 1 − H(q)

This is a smooth function of q so there is no hard cut off. An error rate of 10% corresponds to a redundancy of about 0.3, so that each message is increased in length by about 30%.

(1.38)

The conditional probabilities are

(1 − q)P (a1 ) P (b1 |a1 )P (a1 ) = = 1 − P (a2 |b1 ) P (b1 ) (1 − q)P (a1 ) + qP (a2 ) (1 − q)P (a2 ) P (a2 |b2 ) = = 1 − P (a1 |b2 ) .

(1 − q)P (a2 ) + qP (a1 )

P (a1 |b1 ) =

Here we have used the probabilities P (b1 ) = (1 − q)P (a1 ) + qP (a2 ) P (b2 ) = (1 − q)P (a2 ) + qP (a1 ) .

We can use H(A : B) = H(A) − H(A|B) X X X = P (ai ) log P (ai ) + P (bj ) P (ai |bj ) log P (ai |bj ) . i

j

i

Substituting the calculated probabilites gives the required value. (It is, of course, easier to get to this result using the alternative form H(A : B) = H(B) − H(B|A).)

(1.39)

It suffices, because of the symmetry in the problem, to set P (a1 ) = P ⇒

(a)

P (a2 ) = Q = P (a3 )

P + 2Q = 1 .

The mutual information is H(A : B) = −P log P − 2Q log Q − 2QH(q) .

Probability and information 15

We can maximize this quantity, and so determine the channel capacity using Lagrange’s method of undetermined multipliers: H̃ = H + λ(P + 2Q − 1) dH̃ = −1 − log P + λ ⇒ dP dH̃ = −2 − 2 log Q − 2H(q) + 2λ . dQ Eliminating λ between these equations gives log P = log Q + H(q) ⇒ P = 2H(q) Q 1 2H(q) Q = H(q) . ⇒ P = H(q) 2 +2 2 +2 Hence the channel capacity is

1 2H(q) 1 H(q) H(q) C = − H(q) 2 log H(q) − 2 log H(q) −2 H(q) 2 +2 2 +2 2 +2

= log 1 + 21−H(q) . (b) For q = 0, 1 the error-entropy H(q) = 0 and C = log 3. If q = 0 then we can achieve this error-free value by choosing the three inputs with equal probability and interpreting b1 , b2 , b3 as a1 , a2 , a3 respectively.

If q = 1 then we can achieve this error-free value by again choosing the three inputs with equal probability and interpreting b1 , b2 , b3 as a1 , a3 , a2 respectively.

For q = 21 the error-entropy H(q) = 1 and C = 1. We can achieve this value by choosing a1 with probability 12 and the remaining two in any way such that the sum of the probabilities is 12 (P (a2 ) + P (a3 ) = 12 ). In this way the channel becomes equivalent to an ideal one bit channel.

(1.40) By the symmetry of the problem, it is clear that we need to take P (a1 ) = 21 = P (a2 ). This then gives the probabilities 1 6 1 P (a1 , b3 ) = P (a1 , b4 ) = P (a2 , b1 ) = P (a2 , b1 ) = 12 1 P (b1 ) = P (b2 ) = P (b3 ) = P (b4 ) = .

4

P (a1 , b1 ) = P (a1 , b2 ) = P (a2 , b3 ) = P (a2 , b4 ) =

Hence the channel capacity is C = H(A) + H(B) − H(A, B) =

5 − log 3 ≈ 0.082 bits .

3

(1.41) (a) The symmetry of the problem means that it sufficies, for calculating the channel capacity, to consider all 27 inputs to be equally likely.

16 Probability and information

It follows that all the outputs are equally likely and that there are 8 × 27 non-zero joint probabilities, each of which is equal. It follows that C = H(A) + H(B) − H(A, B) = log 27 + log 27 − log(8 × 27 ) = 4 bits. (b) ity.

The Hamming [7,4] code achieves this maximum channel capac-

(1.42) The first three are trivial. The first holds as the Hamming distance is the number of places at which the strings are different, which is clearly greater than or equal to zero. The second is also clear; if there are no differences then the strings are the same. The third is also clearly true as if a bit in x is different to the corresponding bit in y then the converse is clearly true.

To prove the final property we need consider only single bit in x and the corresponding bit in y and z. If the bits are different in the strings x and y then there is a unit contribution to dH (x, y). The bit in z must be different to only one of x and y so there is a unit contribution to dH (x, z) + dH (y, z). If the bits are the same in the strings x and y then there is no contribution to dH (x, y). The right hand side will be zero if the bit is the same in z but it will be two if it is different. Clearly the single bit either satisfies the inequality and so, therefore, must the whole string.

(1.43) (a) 1110100 → 1110 (b) 1001101 → 1001 (c) 0100111 → 0100 (d) 0110010 → 0110 (1.44) If the codewords have a minimum Hamming distance d then it takes at least d errors in order to transform one valid codeword into another one. If we have d − 1 errors or fewer then the received string will not be a valid codeword and will indicate the presence of errors.

If (d − 1)/2 or fewer errors have occured then we can correct there by replacing the received string by its nearest valid codeword.

Elements of quantum theory (2.1) It is straightforward, if a bit labourious to explicitly check the properties. It is quicker, in fact to show them in general: ∗ ∗ (B̂ † )†ij = (B̂ † )∗ji = (Bij ) = Bij

(B̂ + Ĉ)†ij = (B̂ † + Ĉ † )ij X X † † (B̂ Ĉ)†ij = (B̂jk Ĉki )∗ = Ĉik B̂kj = (Ĉ † B̂ † )ij k † † ∗ (λB̂)ij = λ∗ B̂ji = λ∗ B̂ij .

(2.2)

k

For a general (pure) state |ψi we find

2 ∆A = hψ| Â − hÂi |ψi h

i h

i = hψ|  − hÂi  − hÂi |ψi ,

which is zero if and only if

 − hÂi |ψi = 0 , so that |ψi is an eigenvalue of  with eigenvalue hÂi.

(2.3) Let us start by considering the case in which the eigenvalues of  are non-degenerate, that is they are all different.

Â|an i = an |an i ⇒ B̂ Â|an i = B̂an |an i .

The fact that the operators commute then means that ÂB̂|an i = B̂an |an i , so that B̂|an i is also an eigenstate of  with eigenvalue an . As the eigenvalues are non-degenerate, it then follows that B̂|an i ∝ |an i , so that |an i is an eigenstate of B̂.

Let us turn now to the degenerate case. It suffices to consider a single eigenvalue am and let this have M -fold degerenacy so that there are M

2

18 Elements of quantum theory

orthogonal eigenvectors |aim i where i = 1, · · · , M . The argument follows that given above, except that ÂB̂|aim i = B̂am |aim i , does not mean that B̂|aim i is proportional to |aim i but rather that it lies within the degenerate subspace so that B̂|aim i = bij |ajm i .

We can diagonalize the, necessarily Hermitian, matrix bij and so construct a complete set of eigenstates for B̂ for the degenerate subspace.

(2.4) Any skew-Hermitian operator can be written as the product of i and a Hermitian operator. It follows that the skew-Hermitian operators have imaginary eigenvalues.

(2.5) (a)

 Ĉ[B̂, D̂] + [B̂, Ĉ]D̂ + [Â, Ĉ]D̂ + Ĉ[Â, D̂] B̂ = Â[B̂, Ĉ D̂] + [Â, Ĉ D̂]B̂ = [ÂB̂, Ĉ D̂] . (b) ÂB̂ Ĉ + B̂ ÂĈ + B̂ Ĉ  − B̂ ÂĈ = {Â, B̂ Ĉ} . (c)

Expanding give B̂ ÂĈ D̂ − 2B̂ Ĉ D̂Â + ÂB̂ Ĉ D̂ = {Â, B̂} − 2B̂ Ĉ D̂Â .

(d) {Â, {B̂, Ĉ}} .

(2.6) The uncertainty relation was derived using the Cauchy-Schwarz inequality and the first requirement for an intelligent state, therefore, is for equality in the Cauchy-Schwarz inequality. This requires that

 − hÂi |ψi ∝ B̂ − hB̂i |ψi

⇒ Â + iλB̂ |ψi = hÂi + iλhB̂i |ψi .

We also require for equality that n

o

2 hψ| Â − hÂi , B̂ − hB̂i |ψi = (iλ − iλ∗ )hψ| Â − hÂi |ψi = 0.

Hence λ must be real. The real part of the eigenvalue of  + iλB̂ is simply hÂi and the imaginary part is iλhB̂i.

Elements of quantum theory 19

(2.7) The intelligent states (minimum uncertainty states) must be eigenstates of the operator x̂ + iλp̂. We can find these most easily by using the position representation and so need to solve the differential equation

d ψ(x) = (x̄ + iλp̄) ψ(x) . x + λh̄ dx The solution is straightforward:

(x − x̄)2 ψ(x) = (πh̄λ)−1/4 exp − eip̄x/h̄ , 2h̄λ where the prefactor has been chosen so as to normalize the wavefunction.

(2.8) It is clear that, in order to maximize this probability we require the width, or uncertainty, to derive entirely from the probabilites at x̄ ± µ∆x. It is clear that the probability in question will be maximized by a probability distribution of the form P (x) = pδ(x + µ∆x) + (1 − 2p)δ(x) + pδ(x − µ∆x) , for which the variance is 2pµ2 (∆x)2 . For this to equal (∆x)2 we clearly require 2p = µ−2 , which is the upper bound on the probability in question. This useful result is Chebychev’s theorem.

(2.9) A positive operator  is defined by the conditions that the expectation value is always real and greater than or equal to zero. To establish the proof we write the operator in terms of its Hermitian and skew-Hermitian parts and take the expectation value hÂi =

1 1 h( + † )i + h( − † )i .

2 2

The first term is real, but the second is imaginary. Hence, the operator can only be positive if for all states |ψi we have 1 hψ|( − † )|ψi = 0 2 ⇒  = † .

(2.10) f (Â) =

X

f (λn )|λn ihλn |

n

" ∞ # X X f (m) (0) m = λn |λn ihλn | m! n m=0 =

∞ X f (m) (0) m=0

m!

Âm .

A difficulty would arise if the Taylor series were not to converge for some or all of the eigenvalues.

20 Elements of quantum theory

(2.11) The minimum value occurs when all of the eigenvalues of the density operator are equal. If the system has a d-dimensional state space then the minimum value of the purity is 1/d. This happens when the density operator is Î/d.

(2.12)

p(1 − p) p2 Î ⊗ Î + + (1 − p)2 |Ψ− ihΨ− | 16 2 2 p(1 − p) p ⇒ Tr(ρ̂2 ) = + + (1 − p)2 4 2 3 = 1 − p(2 − p) .

4 This clearly takes its maximum value of unity at p = 0 and its minimum value of 1/4 at p = 1 as it should.

P (2.13) The ij component of ρ̂2 is (ρ̂2 )ij = k ρ̂ik ρ̂kj . It follows that XX XX Tr(ρ̂2 ) = ρ̂ik ρ̂ki ≤ ρ̂ii ρ̂kk = 1 . ρ̂2W =

i

i

k

k

The equality holds if and only if ρ̂ik ρ̂ki = ρ̂ii ρ̂kk for all i and k. The required result then follows from the fact that the purity is unity if and only if the state is pure.

(2.14) The Cauchy-Schwarz inequality is |hφ1 |φ2 i|2 ≤ hφ1 |φ1 ihφ2 |φ2 i.

We can always write |φ2 i = a|φ1 i + |χi, where hχ|φ1 i = 0. With this ansatz, the two sides of the inequality are |hφ1 |φ2 i|2 = |a|2 hφ1 |φ1 i2 hφ1 |φ1 ihφ2 |φ2 i = hφ1 |φ1 i(|a|2 hφ1 |φ1 i + hχ|χi) .

These are clearly equal if and only if |χi = 0 so that |φ2 i ∝ |φ1 i.

(2.15) It follows from the unitarity of Û (t) and the cyclic property of the trace that Tr(ρ̂n (t)) = Tr(ρ̂n (0)).

(2.16) We can always diagonalize the density operator. Let us suppose that the two eigenvalues are ρ0 and ρ1 = 1 − ρ0 . It then follows that Tr(ρ̂) = 1 and Tr(ρ̂2 ) = ρ20 + ρ21 = 1 − 2ρ0 (1 − ρ0 ). (a) Tr(ρ̂3 ) = ρ30 + ρ31 = ρ30 + (1 − ρ0 )3 = 1 − 3ρ0 (1 − ρ0 )

1 = 3Tr(ρ̂2 ) − 1 .

2 (b) det(ρ̂) = ρ0 ρ1 = ρ0 (1 − ρ0 )

1 = 1 − Tr(ρ̂2 ) .

2

Elements of quantum theory 21

For the extra challenge, we need only note that the properties Tr(ρ̂) = Tr(ρ̂2 ) =

d X n=1 d X

ρn ρ2n

n=1

.. .

Tr(ρ̂d ) =

.. . d X

ρdn

n=1

provide sufficient information to determine the values of all the eigenvalues of ρ̂ (although not, of course, their order). It follows that all higher order moments of ρ̂ must be expressible in terms of these first d.

(2.17) The easiest way to establish this is to use the idea of purification. To this end let us introduce two different complete orthonormal bases {|χn i} and {|νl i} for a second quantum system. It then follows that the state Xp Xp Pn |ψn i ⊗ |χn i = Ql |φl i ⊗ |νl i |Ψi = n

l

can be constructed so that the expectation value of any opertor  acting in the first state space alone is the same as it would be for the density operator. Acting on this expression with hχn | then gives X p p hχn |νl i Ql |φl i . hχn ||Ψi = Pn |ψn i = l

It only remains to show that the matrix unl = hχn |νl i is unitary: X X u†mn unl = hνm |χn ihχn |νl i = hνm |νl i = δml . n

n †

The proof that uu = 1 follows in the same way.

(2.18) It is possible to construct an inductive proof for this, but there is a quicker way. Let us start by showing that Û † (t)Û (t) = Î.

1 1 d † Û (t)Û (t) = − Û † (t)Ĥ(t)Û (t) + Û † (t)Ĥ(t)Û (t) = 0 dt ih̄ ih̄ ⇒ Û † (t)Û (t) = Û † (0)Û (0) = Î .

To prove that Û (t)Û † (t) = Î we can construct its Taylor series t2 d2 † d Û (t)Û † (t) = Û (0)Û † (0) + t Û † (t)Û (t) + Û (0)Û (0) + ··· dt 2! dt2 t=0 t=0 h i t Ĥ(0), Û (0)Û † (0) + · · · = Î − ih̄ = Î

22 Elements of quantum theory

(2.19) Orthonomality:

hλm |Û † Û |λn i = hλm |λn i = δmn .

Completeness: X

Û |λn ihλn |Û † = Û ÎÛ † = Î .

n

(2.20)

Consider the operator expression

F̂ (λ) = exp −iλĈ Â exp iλĈ .

The Taylor series of this around λ = 0 is λ2 d2 d + + ··· F̂ (λ) F̂ (λ) dλ 2! dλ2 λ=0 λ=0 i λ2 h h ii h = Â − iλ Ĉ, Â − Ĉ, Ĉ, Â + · · · , 2!

F̂ (λ) = F̂ (0) + λ

which, for λ = 1, is the desired expression.

(2.21) (a)

We make use of the identity

n x lim 1 + + O(n−2 ) = ex . n→∞ n

It then follows that " ! !#n !n Ât Ât B̂t B̂t −2 lim exp i exp i = lim Î + i +i + O(n ) n→∞ n→∞ n n n n i h = exp i  + B̂ t . (b) For finite n we need also to consider the effect of the O(n−2 ) terms. For the expression in part (a) we find ! ! !2 !2 Ât B̂t Ât B̂t 1 Ât 1 B̂t exp i exp i = Î + i +i − − n n n n 2! n 2! n −

Ât B̂t + ··· n n

which, last term, is not the correct expansion of h because of the i exp i  + B̂ (t/n) . For that in part (b), however, we find B̂t exp i 2n

!

Ât exp i n

!

B̂t exp i 2n

!

!2 Ât B̂t 1 Ât = Î + i +i − n n 2! n !2 o 1 B̂t t2 n − − Â, B̂ + · · · , 2! n 2n

Elements of quantum theory 23

h

i which is the correct expansion of exp i  + B̂ (t/n) to O(n−2 ).

(2.22)

Ûn · · · Û2 Û1

†

= Û1† Û2† · · · Û1† .

Unitarity then follows from Ûi† Ûi = Î = Ûi Ûi† .

(2.23) They are manifestly Hermitian, so that σ̂i† = σ̂i . The fact that the square of each of the operators is the identity operator is, therefore, a demonstration of unitarity.

(2.24) ˆ~b · ~σ ˆ= ~a · ~σ

3 X 3 X

aj bk σ̂j σ̂k

j=1 k=1

=

3 3 X X

aj bk δjk Î + iεjkl σ̂l

j=1 k=1

ˆ. = ~a · ~b + i(~a × ~b) · ~σ (2.25) ∆σx ∆σy =

1 |h[σ̂x , σ̂y ]i| = |hσ̂z i| .

2

The intelligent states are the eigenstates of the operators

0 1+λ σ̂x + iλσ̂y = .

1−λ 0 It is straightforward to find the eigenstates:

0 1+λ a a =α .

1−λ 0 b b The eigenstates are trivial intelligent states, for which the uncertainty product is zero. The only non-trivial ones have eigenvalue α = 0, corresponding to λ = ±1, which leads us to the eigenstates of σ̂z .

(2.26) We require to minimize ∆σy for a given value of ∆σx . For the general pure qubit state, eqn 2.89, we have ∆σx2 = 1 − sin2 θ cos2 ϕ

∆σy2 = 1 − sin2 θ sin2 ϕ .

We can minimize the value of ∆σy for fixed ∆σx by choosing the angles such that sin θ = 1, which corresponds to a Bloch vector in the equatorial plane. This gives an the uncertainty product ∆σx ∆σy = |sin ϕ cos ϕ| .

24 Elements of quantum theory

(2.27) We begin by noting that the probability density P (θ, φ) must, of course, be normalized so that integration over the surface of the Bloch sphere gives unity. (a) Z 2π

Z π

θ iϕ θ sin θ dθ ρ̂ = cos |0i + e |1i 2 2 0 0

θ θ h0| cos + h1|e−iϕ sin 2 2

Z π 1 θ θ = sin θ cos2 |0ih0| + sin2 |1ih1| 2 2 2 0 1 = Î .

2 1 dϕ 4π

(b) Let us start by find the correct form for the probability distribtion. Let the constant non-zero value be P. Z µ Z 2π P sin θ dθ dϕ = P 2π(1 − cos µ) 0

0

1 ⇒P = .

2π(1 − cos µ) It then follows that Z µ Z 2π ρ̂ = sin θ dθ dϕ

1 θ θ cos |0i + eiϕ |1i 2π(1 − cos µ) 2 2 0 0

θ θ h0| cos + h1|e−iϕ sin 2 2

Z π 1 θ θ = sin θ cos2 |0ih0| + sin2 |1ih1| 1 − cos µ 0 2 2

i 1 h 2 µ 2 µ = 1 + cos |0ih0| + sin |1ih1| .

2 2 2

(2.28) (a) It is straightforward to show that a general 2 × 2 unitary matrix has the form

cos θeip sin θeiq Û = . sin θeir − cos θei(q+r−p) Let us compare this to the unitary form given in (2.91).

ˆ Û = eiα cos β Î + sin β~a · ~σ iα

e (cos β + iaz sin β) eiα sin β(iax + ay ) = eiα sin β(iax − ay ) eiα (cos β − iaz sin β) We can identify, with each other, the parameters used in these two expressions (with care). For example: q sin θ = sin β a2x + a2y q cos β = ± cos2 β + sin2 β a2z .

Elements of quantum theory 25

(b)

We require to show that

ν µ

λ ˆ . exp i σ̂z exp i σ̂y exp i σ̂z = cos β Î + sin β~a · ~σ 2 2 2

To this end, let us multiply out the exponentials to give

ν µ µ λ λ ν cos Î + i sin σ̂z = cos Î + i sin σ̂z cos Î + i sin σ̂y 2 2 2 2 2 2 µ ν+λ µ ν−λ cos cos Î + sin sin iσ̂x + 2 2 2 2 µ ν−λ µ ν+λ sin cos iσ̂y + cos sin iσ̂z .

2 2 2 2

ˆ is clear from That this can always be made equal to cos β Î + sin β~a · ~σ the fact that the sum of the squared moduli of the coefficients of the four Pauli operators is unity for both expressions. This is, of course, the Euler deomposition of a rotation.

(2.29) The eignevalues of the density operator are from eqn 2.94 (1 ± r)/2, where r is the length of the Bloch vector. We also have (1 + r)2 (1 − r)2 1 + r2 + = 4 2 p 4 ⇒ r = 2Tr(ρ̂2 ) − 1 .

Tr(ρ̂2 ) =

(2.30) For a product state of the form of eqn 2.100, the must exist a state of the first qubit that is orthogonal to the state and so has a zero overlap. For the entangled state in eqn 2.101, however, we find

θ 1 −iϕ cos |0ib + e sin θ|1ib , a hλ||ψi = √ 2 2 which is non-zero for all θ and ϕ. Hence the state cannot be written as a product state.

(2.31) (a) The minimum and maximum values are simply d and 1. The minimum occurs only if precisely one of the amplitudes an has a modulus of unity, with the others being zero. Inspection of the state in eqn 2.105 clearly shows that the state is then not entangled. (b) The Bell states are entangled states of a pair of 2-state systems, so the maximim value of K in this case is 2. The two non-zero amplitudes for each of the Bell states are ±2−1/2 and it then follows that K = 2.

(2.32) (a)

No. ∆(A − B)2 =

X

|an |2 (λn − φn − hÂi + hB̂i)2 ,

n

which will be zero only if the eigenvalues are such that λn = φn ∀ n.

26 Elements of quantum theory

(b)

We require the probabilities P (λn , φm ) = |an |2 δnm

P (λn ) = |an |2 = P (φn ) .

From these we find X X P (λn , φm ) H(A : B) = P (λn , φm ) log − P (λn ) log P (λn ) P (λn )P (φm ) nm n = H(A) = H(B) .

This is the maximum allowed value, for given H(A) = H(B), and this reflects the perfect correlation.

(2.33) We note that the three Bell states |Φ+ i, |Φ− i and |Ψ+ i are all symmetric under exchange of the systems. This means that interchanging the states leaves them unchanged. The final state, |Ψ− i, is antisymmetric, which means that it changes sign under interchange. It follows that we can write the unitary swap operator in the form Ûswap = |Φ+ ihΦ+ | + |Φ− ihΦ− | + |Ψ+ ihΨ+ | − |Ψ− ihΨ− | = Î ⊗ Î − 2|Ψ− ihΨ− | , as may be readilly verified by direct calculation.

(2.34) [σ̂x ⊗ σ̂x , σ̂y ⊗ σ̂y ] = (iσ̂z ) ⊗ (iσ̂z ) − (−iσ̂z ) ⊗ (−iσ̂z ) = 0 .

The others are similary zero. Checking that the Bell states are the common eigenstates is straightforward and we present here only one example 1 σ̂x ⊗ σ̂x |Ψ− i = √ (|1i ⊗ |0i − |0i ⊗ |1i) = −|Ψ− i .

2 The product of the eigenvalues is always −1 because (σ̂x ⊗ σ̂x ) (σ̂y ⊗ σ̂y ) (σ̂z ⊗ σ̂z ) = −Î ⊗ Î , which clearly has eigenvalue −1.

(2.35) ρ̂a = (|α|2 + |β|2 )|0ih0| + |γ|2 |1i h1| ρ̂b = (|α|2 + |γ|2 )|0ih0| + |β|2 |1i h1| ρ̂c = (|β|2 + |γ|2 )|0ih0| + |α|2 |1i h1| ⇒ Tr(ρ̂2a ) = 1 − 2|γ|2 (1 − |γ|2 ) Tr(ρ̂2b ) = 1 − 2|β|2 (1 − |β|2 ) Tr(ρ̂2c ) = 1 − 2|α|2 (1 − |α|2 ) .

(2.36)

Elements of quantum theory 27

(a) Qubit a is entangled with qubit b and qubit c is entangled with qubit d. (b) The symmetry of the state means that we need consider only one of the qubits. Let us suppose that a measurement reveals qubit a to be in the state θ θ |ψi = cos |0i + eiϕ sin |1i .

2 2 It follows that the measurement will change the state in the following manner

θ θ + + −iϕ |Φ iab |Φ icd → cos |0i + e sin |1i |Φ+ icd , 2 2 so that the combined state of the three remaining qubits has changed. (c) Each of the four qubits is in a maximally entangled state and it follows then that Tr(ρ̂2i ) = 12 .

(2.37) (a)

Both are. In fact, the state is a maximally entangled one.

(b) Note that this state is deliberately not normalized. (This is a trap and a warning about simply calculating Tr(ρ̂2i ).)

1 (5|0ih0| + |1ih1|) 6 1 ρ̂b = (|0ih0| + 5|1ih1|) 6 1 ρ̂c = (|0ih0| + 2|1ih1|) .

3

ρ̂a =

It follows, from the fact that the combined state is pure and that there are only three qubits, that all three are mutually entangled. (c)

Inspection of the state reveals that we can write it in the form 1 (|0ia |0ic + i|1ia |1ic ) (|0ib + i|1ib ) , 2

so that qubits a and c and (maximally) entangled and qubit b is entangled with neither. (d) This state is simply |Φ+ i ⊗ |Φ+ i so that the first two qubits are mutually entangled and the last two qubits are also mutually entangled.

(2.38) Let the two sets of component density operators have the diagonal forms X X j ρ̂ia = ρia (m)|ρia (m)ihρia (m)| ρ̂jb = ρb (n)|ρjb (n)ihρjb (n)| . m

n

It then follows that X ρ̂ = P (ai , bj )ρia (m)ρjb (n)|ρia (m)ihρia (m)| ⊗ |ρjb (n)ihρjb (n)| . ijmn

28 Elements of quantum theory

If we combine the i and m summations into a sum over a single index, k, and the j and n summations into a single index, l, then we recover the desired form.

(2.39) (a) The density matrix is Hermitian and therefore the transpose of the density matrix is also its complex conjugate. It follows, therefore, that X X ρ̂ = ρm |ρm ihρm | ⇒ ρ̂T = ρm |ρ∗m ihρ∗m | . m

m

The complex conjugate states are also valid states and hence ρ̂T , being a mixture of allowed pure states, is an allowed density operator. (b) The transpose of the density matrix is also its complex conjugate and therefore its eigenvalues will be the complex conjugates of the original density matrix. The eigenvalues of the density operator are real, however, and it follows that the eigenvalues of ρ̂T and the same as those of ρ̂. (c) As may be seen in part (a), the eigenvectors of ρ̂T are the complex conjugates of those of ρ̂.

(2.40) 

p 4

 −i p ρ̂ =  p 4 4 −i p4

i p4 1 − 3p 4 i p4 p 4

p 4

−i p4 p 4 −i p4

i p4  p 4  . i p4 p 4

We can construct the partial transpose by making the transpose operation on the second qubit alone. This corresponds to interchanging the elements of this 4 × 4 matrix as follows: ρ(12) ↔ ρ(21), ρ(14) ↔ ρ(23), ρ(41) ↔ ρ(32) and ρ(34) ↔ ρ(43). For the density matrix in question, this results simply in the complex conjugate of the original density matrix and this, as we have seen, is an acceptable state.

(2.41) Let us consider an equally weighted mixture of anticorrelated eigenstates of the three Pauli operators σ̂i . The resulting, explicitly unentangled, density operator is 1 (|0ih0| ⊗ |1ih1| + |1ih1| ⊗ |0ih0| 6 1 + (|0i + |1i)(h0| + h1|) ⊗ (|0i − |1i)(h0| − h1|) 4 1 + (|0i − |1i)(h0| − h1|) ⊗ (|0i + |1i)(h0| + h1|) 4 1 + (|0i + i|1i)(h0| − ih1|) ⊗ (|0i − i|1i)(h0| + ih1|) 4

1 (|0i − i|1i)(h0| + ih1|) ⊗ (|0i + i|1i)(h0| − ih1|) 4 1 1 = Î ⊗ Î + |Ψ− ihΨ− | .

6 3

ρ̂unent =

Elements of quantum theory 29

Clearly this corresponds to the unentangled Werner state with p = 23 .

It then follows that we can write Werner states for larger values of p as ρ̂W = 3(1 − p)ρ̂unent +

3p − 2 Î ⊗ Î , 4

which is clearly an acceptable density operator if p ≥ 23 .

Quantum cryptography NOTE. The task of deciphering ciphertexts is complicated, of course, by the fact that they inevitably contain errors. In the solutions to the first few problems I shall denote errors (at least the ones I have found) using parentheses. I also restore the punctuation, so as to make the messages easier to read.

(3.1) IS THIS A DAGGER I SEE BEFORE ME. (From Shakespeare’s Macbeth.)

(3.2) BE PATIENT FO(R) THE WORLD IS BROAD AND WIDE .

(From the opening to Abbot’s Flatland.)

(3.3)

Because the 26th is the plaintext.

(3.4) IF, WITHOUT IN ANY WAY DISTURBING A SYSTEM, WE CAN PREDICT WITH CERTAINTY, I.E. WITH A PROBABILITY EQUAL TO UNTIY, THE VALUE OF A PHYSICAL QUANTITY, THEN THERE EXISTS AN ELEMENT OF PHYSICAL REALITY CORRESPOND(I)NG TO THIS PHYSICAL QUANTITY. (The definition of an element of physical reality from the famous 1935 paper by Einstein, Podolsky and Rosen.)

(3.5) IT IS ARGUED THAT COMPUTING MACHINES INEVITABLY INVOLVE DEVICES WHICH PERFORM LOGICAL FUNCTIONS THAT DO NOT HAVE A SINGLE-VALUED INVERSE. THIS LOGICAL IRREVERSIBILITY IS ASSOCIATED WITH PHYSICAL IRREVERSIBILITY AND REQUIRES A MINIMAL TYPICALLY A MINIMAL HEAT GENERATION PER MACHINE CYCLE, TYPICALLY OF THE ORDER OF KT, FOR EACH IRREVERSIBLE FUNCTION. (This is the abstract of Landauer’s famous paper on the thermodynamic limits to computation.)

(3.6)

This follows as a simple consequence of Bayes’ theorem: P (Pi , Cj ) = P (Pi |Cj )P (Cj ) = P (Pi )P (Cj ) P (Pi , Cj ) = P (Cj |Pi )P (Pi ) = P (Pi )P (Cj .

(3.7) Because if we have two messages enciphered using the same key, then the modulo addition of the two removes the key. We are then left with the modulo sum of two plaintexts and can use the properties of the English language (and any other knowledge) to assist in the task of obtaining the two messages. Once we have these, moreover, we can also find the key.

3

32 Quantum cryptography

(3.8) 21 mod 13 = 2,

22 mod 13 = 4,

23 mod 13 = 8,

24 mod 13 = 3,

25 mod 13 = 6,

26 mod 13 = 12,

7

2 mod 13 = 11,

28 mod 13 = 9,

29 mod 13 = 5,

210 mod 13 = 10,

211 mod 13 = 7,

212 mod 13 = 1.

41 mod 13 = 4,

42 mod 13 = 3,

43 mod 13 = 12,

44 mod 13 = 9,

45 mod 13 = 10,

46 mod 13 = 1,

47 mod 13 = 4,

48 mod 13 = 3,

9

4 mod 13 = 12,

410 mod 13 = 9,

411 mod 13 = 10,

412 mod 13 = 1.

(3.9) Let us suppose that a = 2 and b = 5 (other examples are acceptable, of course). Our Diffie-Hellman key is then K = 310 mod 17 = 8. If either a = 0 or b = 0, then the key is simply 0. If a = 1 and b = 1 then either Alice or Bob transmits the key in open transmission!

(3.10) Recall that ϕ(N ) in the number of positive integers less than or equal to N which are prime to N . (a) The integers relatively prime to a prime p are (obviously) 1, 2, · · · , p − 1. Hence ϕ(p) = p − 1. (b) For the product of two primes pq, the numbers that are not prime to pq are q, 2q, · · · , (p − 1)q, pq and p, 2p, · · · , (q − 1)p, of which there are p + q − 1. It follows that ϕ(pq) = pq − p − q + 1 = (p − 1)(q − 1). (c) For the square of a prime p2 , the numbers that are not prime to p are p, 2p, · · · , p2 , of which there are p. It follows that ϕ(p2 ) = p2 −p = p(p − 1).

2

(3.11) (a)

Let us take P = 101 (in base 10). The ciphertext is then C = 1015 mod 247 = 43 .

(b) A search among the primes reveals that p = 13 and q = 19. It follows that ϕ(N ) = (p − 1)(q − 1) = 12 × 18 = 216 .

From this we can find, very simply, that 4 × 216 + 1 = 865 = 5 × 173

⇒ d = 173 .

Quantum cryptography 33

We can now readily perform the deciphering operation 43173 mod 247 = 101 .

(3.12) We require a two-outcome projective measurement, with projectors P̂1 and P̂2 = Î − P̂1 , such that P̂1 ρ̂2 = 0 = ρ̂2 P̂1 and P̂2 ρˆ1 = 0 = ρˆ1 P̂2 . A simpler and equivalent requirement is

ρ̂1 ρ̂2 = ρ̂1 P̂1 + P̂2 ρ̂2 = 0 .

Hence discrimination with certainty is possible only if the two density operators are orthogonal.

(3.13) (a) B̂ = (1 − |α|2 )|ψ1 ihψ1 | − |β|2 |ψ1⊥ ihψ1⊥ | − αβ ∗ |ψ1 ihψ1⊥ | − α∗ β|ψ1⊥ ihψ1 | = |β|2 |ψ1 ihψ1 | − |β|2 |ψ1⊥ ihψ1⊥ | − αβ ∗ |ψ1 ihψ1⊥ | − α∗ β|ψ1⊥ ihψ1 | .

We can write this as a 2 × 2 matrix and thereby find the eigenvalues, which are ±|β|. (b) hψ1 |B̂|ψ1 i = |β|2 = |β| {P [(+|β|)|ψ1 ] − P [(−|β|)|ψ1 ]} 1 ± |β| . ⇒ P [(±|β|)|ψ1 ] = 2 2 hψ2 |B̂|ψ2 i = −|β| = |β| {P [(+|β|)|ψ2 ] − P [(−|β|)|ψ2 ]} 1 ∓ |β| ⇒ P [(±|β|)|ψ2 ] = .

2 (3.14) This instructive problem is, perhaps, more subtle than it might at first appear to be.

Let p be the probability that a single photon survives propagation through the fibre. This probability is, of course, the ratio of the output to the input power. Hence p = 10−0.1L/10 = 10−L/100 , where L is the length of the fibre in km. It then follows that photons leave the fibre at its end at the rate Rout = pRin = p MHz .

The photon deletions are, of course, random and this means that we need to exploit error-correction. If a photon is missing, then we can only guess the intended bit value. It follows that the single-photon error probability is (1 − p)/2, that is one half of the absorption probability. If we use an optimal coding scheme (operating at the Shannon limit) then

34 Quantum cryptography

the number of bits per photon will be H[(1 − p)/2]. The maximum bit rate is then

1−p Rbit = Rout 1 − H 2 p = [(1 − p) log(1 − p) + (1 + p) log(1 + p)] MHz .

2 If p 1 then we can further simplify this by expanding the logarithms to find Rbit =

p3 ≈ p3 MHz , 2 ln 2

which is smaller that the rate of receiving photons by a factor of p2 . We are now in a position to calculate each of these quantities for the three given fibre links. (a) p = 10−0.1 = 0.794 , so about 80% of the photons make it through the fibre. It follows that the rate at which photons make it through the fibre is Rout = 794 kHz but the maximum bit rate is only 0.794 (0.206 log 0.206 + 1.794 log 1.794) MHz 2 = 414 kHz .

Rbit =

(b) p = 10−6.72 = 1.92 × 10−7 , so that only one in every five million photons arrives. It follows that the rate at which photons make it through the fibre is Rout = 0.192 Hz , corresponding to about one photon every five seconds. The maximum bit rate is much lower: Rbit ≈ p3 = 7 × 10−15 Hz , corresponding to one bit every 1.4×1014 seconds, or about 500,000 years! (c) p = 10−55.85 = 1.4 × 10−56 .

Essentially no photons make it to New York. To emphasize the point, the arrival rate is Rout = 1.4 × 10−50 Hz .

Quantum cryptography 35

This corresponds to one photon every 2 × 1042 years, which is many orders of magnitude longer than the age of the universe. The bit rate is unimaginably smaller even than this!

(3.15) Any single-qubit operator can be written as a linear combination of the identity and the three Pauli-spin operators. It follows that any operator acting on a qubit and an ancilla has the form of eqn 3.28.

The operator will be unitary if  !  X X Î ⊗ Â0 + σ̂i ⊗ Âi × Î ⊗ †0 + σ̂j ⊗ †j  = Î ⊗ Î . i

j

Multiplying this out gives   !

X X X Âi †i + Î ⊗ Â0 †0 + σ̂k ⊗ Âk †0 + Â0 †k + i εkij Âi †j  . i

ij

k

It follows that the requirements are Â0 †0 +

X

Âi †i = Î

i

Âk †0 + Â0 †k + i

X

εkij Âi †j = 0 .

ij

(3.16) If two measurements of σ̂z (respectively σ̂x ) give different answers then we identify the state as the measured eigenstate of σ̂x (respectively σ̂z ). It follows that we can only incorrectly identify the state if all the measurements of both components of spin are the same. A moments thought shows this probability to be 2N −1 . If this happens, we need to guess one of the two possible eigenstates to be the true state.

Hence the probability of making an error is 21 × 2N −1 = 2−N . Hence the probability of correctly identifying the state is 1 − 2−N .

(3.17) Let us suppose that Alice and Bob each have a qubit and that these are prepared in the Bell state |Ψ− i and let them be seprated by a space-like interval. Let Alice choose to measure either σ̂z or σ̂x . If Bob can make a single extra perfect copy then he would have two qubits each of which is in the same eigenstate of one of these operators. If he measures σ̂z on both his qubits and finds different answers then he knows for certain that Alice chose to measure σ̂x . In this way he would have gained some information superluminally.

(3.18) We should start by pointing out that the Gaussian form for the electric field is not an exact solution to Maxwell’s equations, but it is a very useful form within the paraxial approximation, in which beams are not too tightly focussed. (You can find more about paraxial optics and Gaussian beams in the book Lasers by Siegman.) (a)

We can find Ez by using the first Maxwell equation: ~ ·E ~ =0 ∇

36 Quantum cryptography

⇒

2 ∂Ez x + y2 ei(kz z−ωt) = w−2 (αx + βy)E0 exp − ∂z 2w2 2

x + y2 (αx + βy) ~ exp − ei(kz z−ωt) . ⇒E= ikz w2 2w2

We can see that this is much smaller than the other components by the following line of reasoning. The Gaussian form of the beam means that we need only consider values of |x| and |y| up to of order w, hence the z-component of the electric field is smaller than the component in the x, y plane by a factor of order kz w. This is necessarily large because, by p Fourier theory, w ≈ 2π/ k 2 − kz2 or, in words, the beam waist greatly exceeds the wavelength. Hence −1/2 2 k − 1 1. kz w ≈ 2π kz2 (b) tion

We can find the magnetic field by using the third Maxwell equa∂ ~ ~ =∇ ~ ×E ~ H = iωµ0 H ∂t 2

kz x + y2 ⇒ Hx = − βE0 exp − ei(kz z−ωt) ωµ0 2w2 2

x + y2 kz αE0 exp − Hy = ei(kz z−ωt) ωµ0 2w2 2

x + y2 −xβ + yα exp − Hz = ei(kz z−ωt) , ωµ0 w2 2w2

−µ0

where we have omitted the very small contributions the x and y derivatives of Ez . (This is actually a requirement of the paraxial approximation.)

(3.19)

X ∂ w= εEi Ėi + µ0 Hi Ḣi ∂t i i Xh ~ × H) ~ i − Hi (∇ ~ × E) ~ i = Ei (∇ i

~ ·S ~. = −∇ Integrating over a volume V and then using Gauss’s theorem gives the required result.

(3.20) Half of the unpolarized light will pass through the filter whatever its orientation. For maximum transmission all of the polarized light will pass and for minimum transmission none of it will. It then follows that 1 1 Imin = Iunpol Imax = Ipol + Iunpol 2 2 Imax − Imin Imax − Imin ⇒P = = .

Imax − Imin + 2Imin Imax + Imin

Quantum cryptography 37

(3.21)

From the third Maxwell equation we have ~ = − k (ı̂Ey − ̂Ex ) H ωµ0 k = ∓i E0 (ı̂ ± i̂) exp [i(kz − ωt)] . ωµ0

(3.22) By extracting an unimportant phase factor, we can always write the electric field amplitude in the form ~ 0 = Ex~ı + (E 0 + iE 00 )~ .

E y y where Ex is real and where Ey0 and Ey00 are the real and imaginary parts of Ey . We can rewrite this as a linear combination of a circular polarization and a linear polarization ~0 = E

√

1 2Ey00 √ (ı̂ + i̂) + (Ex − Ey00 )ı̂ + Ey0 ̂ , 2

which is, of course, an eliptical polarization.

(3.23) (a)

1 0 Î + σ̂z = 0 2

1 0 0 = Î − σ̂z 0 1 2

2 1 cos θ cos θ sin θ Î + cos(2θ)σ̂ + sin(2θ)σ̂ = z x cos θ sin θ sin2 θ 2

1 1 i 1 = Î − σ̂y −i 1 2 2

1 1 −i 1 Î + σ̂y = 2 i 1 2

i 1h 1 0 = (1 − i)Î + (1 + i)σ̂z 0 −i 2

i 1h 1 0 = (1 + i)Î + (1 − i)σ̂z 0 i 2

1h cos2 θ + i sin2 θ cos θ sin θ(1 − i) = (1 + i)Î+ 2 2 cos θ sin θ(1 − i) sin θ + i cos θ 2 (1 − i) (cos(2θ)σ̂z + sin(2θ)σ̂x )]

1 0 = σ̂z .

0 −1 1 0

There is an error in the final Jones matrix, which should be

cos2 θ − sin2 θ 2 cos θ sin θ = cos(2θ)σ̂z + sin(2θ)σ̂x .

2 cos θ sin θ sin2 θ − cos2 θ

38 Quantum cryptography

(b) Only the final five are unitary. This reflects the fact that they pass all of the photons. The remaining, non-unitary, matrices are associated with devices that can absorb some of the light.

(3.24)

Energy conservation requires that 2

2

2

2

|E1 in | + |E2 in | = |E1 out | + |E2 out |

2

= (|t1 |2 + |r2 |2 ) |E1 in | + 2

2

2

(|t2 | + |r1 | ) |E2 in | + (t∗1 r1 + t2 r2∗ )E1∗ in E2 in + (t1 r1∗ + t∗2 r2 )E1 in E2∗ in .

This must be true for all input fields and it then follows that |t1 |2 + |r2 |2 = 1 = |t2 |2 + |r1 |2

t∗1 r1 + t2 r2∗ = 0 .

To see that |t1 | = |t2 | we need only note that the final condition can only hold if |t1 |2 |r1 |2 = |t2 |2 |r2 |2 ⇒ |t1 |2 (1 − |t2 |2 ) = |t2 |2 (1 − |t1 |2 ) ⇒ |t1 |2 = |t2 |2 .

It is now straightforward to invert the beam splitter relations to find E1 in = r1∗ E1 out + t∗2 E2 out E2 in = r2∗ E2 out + t∗1 E1 out .

(3.25)

We can use the formlae (G.14) to write 1 H† V † V† V† âH† â â + iâH† 1 in â1 in |vaci = 2 in â1 in 2 1 out 1 out

V† H† V † +iâH† 1 in â2 in − â2 in â2 in |vaci 1 H† V † V† V† â â + iâH† âH† â |vaci = 2 in â2 in 1 in 2 in 2 1 out 2 out

V† H† V † +iâH† 1 in â1 in − â2 in â1 in |vaci

i H† V † V† V † H† H† V † √ |vaci = â â + â â â â + â 2−1/2 âH† 1 in 2 in 1 in 1 in 2 in 2 in |vaci 1 in 2 in 2

1 H† V † V† V † H† V† 2−1/2 âH† â1 in â2 in − âH† 1 in â2 in − â1 in â2 in |vaci = √ 2 in â1 in |vaci .

2

In the first two of these, the photons are clearly distinguishable by their polarizations and each is reflected with probability 21 and reflected with the same probability.

The remaining two exhibit a more interesting behaviour. We can understand this as a consequence of Bose-symmetry. Because the photons are bosons, their state vector must be unchanged if be interchange them.

We can do this in two steps. First interchange the polarization labels and then interchange the input directions (1 and 2). In both cases this

Quantum cryptography 39

combined procedure leaves the state unchanged, as it should. In the first case, each interchange leaves the state vector itself unchanged, but in the second each introduces a sign change. These properties must be retained by the action of the beam splitter and it is these that are responsible for the very different behaviours (both photons leaving in the same direction or one leaving in each of the two possible directions).

(3.26) As the state chosen is simply a guess, each guessed photon will pass as a genuine one with probability 12 . It follows that the note will pass as genuine with probability 2−20 ≈ 10−6 .

(3.27) If the photon polarization state is |ψi then it will give the result + (−) with probability |h+, θ|ψi|2 (|h−, θ|ψi|2 ). The couterfeiter’s prepared photon will then pass a test as genuine with same probability.

Each of the four states is prepared with equal probability and it follows that a single photon will pass the test as genuine with probability

1 X |hi, θ|0i|4 + |hi, θ|1i|4 + |hi, θ|00 i|4 + |hi, θ|10 i|4 4 i=+,− "

4 θ 1 1 θ θ θ cos + sin = 2 cos4 + 2 sin4 + 2 √ 4 2 2 2 2 2

4# θ θ 1 cos − sin +2 √ 2 2 2

P=

=

3 .

4

(3.28) It makes a difference if Bob can store the photons until after Alice has announced the basis she used and only then make his measurement. You might like to think what difference this might make to Eve. (3.29) (a) hni = e

−n̄

=e

−n̄

∞ X n̄n n n! n=0 ∞

d X n̄n n̄ dn̄ n=0 n!

= e−n̄ n̄ = n̄ .

d n̄ e dn̄

To calculate the uncertainty, we first find the mean square photon number: hn2 i = e−n̄

∞ X n=0

n2

n̄n n!

40 Quantum cryptography

2 X ∞

n̄n n! n=0

d2 d = e−n̄ n̄2 2 + en̄ dn̄ dn̄ = n̄(n̄ + 1) .

=e

−n̄

d n̄ dn̄

It follows that ∆n =

p

hn2 i − hni2 =

√

n̄ .

(b) The probability that a pulse contains more than one photon is 1 − P (0) − P (1). If there are no photons present, however, then this will not contribute to the key. It follows that the probabilities that Bob’s came from a single-photon pulse p1 or a mulitphoton pulse p>1 are p1 =

P (1) 1 − P (0)

p>1 =

1 − P (0) − P (1) .

1 − P (0)

It follows that the fraction of key bits that could be determined by a sufficiently technologically advanced Eve is p>1 =

1 − e−n̄ (1 + n̄) .

1 − e−n̄

For n̄ 1, this takes the simple form p>1 ≈

n̄ P (2) = .

2 P (1)

(c) As no errors are detected, Alice and Bob may safely assume that Eve has only managed to determine the key bits that were encoded on multiphoton pulses. From the above formula, this corresponds to 0.1 2 , or one bit in twenty.

(3.30) (a) The probability that a single observation produces no detected error is 43 . It follows that no errors will be found by Alice and Bob with M probability 34 . (b) Her bits would be correct when she had measured in the correct basis and, with probability 12 , those bits she had measured in the wrong basis. This corresponds to 34 of the bits. (c) Yes. Before the discussion, Eve knows that each bit is correct with probability 34 . After the discussion she knows half of her bits for certain (and the remaining half with probability 12 ).

(3.31) (a)

The probability that the measured bit value is correct is P=

1 |h0B |0i|2 + |h0B |00 i|2 + |h1B |1i|2 + |h1B |10 i|2 4

Quantum cryptography 41

π 2 1 π cos + sin 2 √ 8 8 2+ 2 = ≈ 0.85 .

4

=

(b) is 34 .

This is the same calculation as in exercise (3.27). The answer

(3.32) Because Eve could use this redundancy to determine some of the key bits.

(3.33) We can use Shannon’s noisy-channel coding theorem to place a bound. The corrected key will not be longer than N [1 − H(q)] bits in length.

(3.34) (a)

The mutual information per bit between Eve and the key is

X P (ki , ej ) H(E : K) = P (ki , ej ) log P (ki )P (ej ) ij 1 1 = (1 + ε) log(1 + ε) + (1 + ε) log(1 + ε) .

2 2

For small, this becomes H(E : K) ≈

ε2 .

2 ln 2

(b) We showed in the book that privacy amplification simply replaces ε by εm . It follows that following privacy amplification, the mutual information is H(E : K) ≈

ε2m 2 ln 2

which is exponential in m.

(3.35) Correcting for general eavesdropping attacks in the presence of noise is a challenging problem. For the simplifying assumptions, however, we can obtain an answer. Let us suppose that we have N bits in the final key and that Eve has measured n of these. Her measurements will produce an error in Bob’s key for any bit she has measured with probability 14 . It then follows that the number or errors will be 14 n, so that n n q= ⇒ = 4q .

4N N For a measurement in the Breidbardt basis, √ Eve will discover the correct bit value with a probability of about (2 + 2)/4. It follows that Alice and Bob can expect Eve, in this simple case, to have measured a fraction √ √ 2+ 2 n = (2 + 2)q ≈ 3.4q , 4 N

42 Quantum cryptography

of the bits. Finally, because the measurement introduces the very errors that have been corrected, one quarter of these bits will have been removed during the error correction. It follows that Alice and Bob determine that Eve knows a fraction √ 3(2 + 2) q. f= 4 of the bits in the corrected key.

(3.36) Eve has to guess the value of any bit she has not measured. It follows that the probability she knows the value of any given bit is √ 1 1 3(2 + 2) P = f + (1 − f ) = (1 + f ) → ε = f = q.

2 2 4 Hence we require privacy amplification to reduce the length of key by a factor m such that εm > µ log µ ⇒m> log ε =

log µ √ . log[3(2 + 2)] − 2

(3.37) (a) The probability that a photon will survive to make it into the raw key is (from Fig. 3.14) 1 θ 1 1 × |h000 |0i|2 + |h100 |1i|2 = sin2 , 2 2 2 2 which gives a mamimum value (consistent with the protocol) of 41 , corresponding to θ = π/2. (b) If Eve carries out the same measurement strategy as Bob, then she will get the bit with probability 41 . If she can replace Alice and Bob’s lossy channel with a lossless one, then she can mimic its behaviour perfectly and have all the key bits, if the losses on the original channel exceed 75%. Actually, Eve can do a little better than this by using unambiguous state discrimination, described in Chapter 4. For smaller angles, higher losses are required to mask Eve’s activities.

(3.38) Let Eve perform a measurement with outcomes 0 and 1 corresponding to the two orthogonal state vectors |e0 i and |e1 i. Each of these occurs with probability 12 and Alice’s qubit is left in the orthogonal state. Hence the density operator for Alice’s qubit and the one prepared

by Eve following her measurement is ρ̂ =

1 (|e1 ihe1 | ⊗ |e0 ihe0 | + |e0 ihe0 | ⊗ |e1 ihe1 |) , 2

which clearly has the form of a correlated but not entangled state.

Generalized measurements (4.1) I would hope that it is (almost) self-evident that they are. If it is not, then one line should suffice for each property.

(4.2)

We require that X

P (λn ) = 1

n

! ⇒ Tr ρ̂

X

|λn ihλn |

= 1.

n

P If the projectors are complete then ρ̂ n |λn ihλn | = ρ̂ and the condition is satisfied. If they are not complete then they span only a subspace of the total Hilbert space. For a state outside this subspace the trace of the product of the density operators and the sum of the projectors will be zero, and not unity as required.

(4.3)

Yes. It corresponds to not performing a measurement.

(4.4) (a) P̂0 : P̂0 |0i = |0i . |0i → q h0|P̂0 |0i The other outcome is impossible. (b) P̂0 : 3−1/2 (|0i + |1i + |2i) → q

P̂0 (|0i + |1i + |2i) (h0| + h1| + h2|) P̂0 (|0i + |1i + |2i)

= |0i .

P̂1,2 : −1/2

3

(|0i + |1i + |2i) → 2−1/2 (|1i + |2i) .

(c) P̂0 : ρ̂ →

P̂0 ρ̂P̂0 Tr(ρ̂P̂0 )

= |0ih0| .

4

44 Generalized measurements

P̂1,2 : ρ̂ →

P̂1,2 ρ̂P̂1,2 Tr(ρ̂P̂1,2 )

=

1 1 |1ih1| + (|1i − |2i)(h1| − h2|) , 3 3

that is, a mixture of the states |1i and 2−1/2 (|1i + |2i) with probabilities 1 2 3 and 3 respectively.

(4.5) It will be unchanged if the density operator commutes with the projectors P̂n , that is, if it is diagonal in the basis formed by thse projectors.

(4.6) X 2 X

Tr ρ̂02 = P (λn )Tr ρ̂02 P (λn )Tr ρ̂02 n ≤ n . n

n

The physical meaning is that the state is more mixed if we do not know the measurement outcome than the average degree of mixing if we do know the outcome.

(4.7) To be projectors, we require them to satisfy the conditions π̂l π̂m = π̂l δlm .

X π̂l π̂m = P (r = λl |i = λn )P (r = λm |i = λn )P̂n . n

Hence they will be projectors only if the conditional probabilities are all zero or unity and if each operator is a sum of distinct projectors.

(4.8) If there are n photons present then we can detect m by also not n! ways that this can happen. detecting n − m of them. There are m!(n−m)!

It follows that the probability for detecting m photons is P (m) =

∞ X

n! η m (1 − η)n−m Tr ρ̂P̂n m!(n − m)! n=m

= Tr (ρ̂π̂m ) , where π̂m =

∞ X

n! η m (1 − η)n−m P̂n . m!(n − m)! n=m

It is straightforward to check that these satisfy the required properties.

They are clearly Hermitian and positive because the projectors are. For the third property we have ∞ X

π̂m =

m=0

=

∞ X n X

n! η m (1 − η)n−m P̂n m!(n − m)! n=0 m=0 ∞ X

(η + 1 − η)n P̂n

n=0

= Î .

Generalized measurements 45

They do not obey the final property of the projectors because π̂l π̂m =

∞ X sup (m,l)

n! n! η m+l (1 − η)2n−m−l P̂n l!(n − l)! m!(n − m)!

6= π̂l δlm .

(4.9) If the density oeprator transforms in the manner suggested, then after the first measurement gives the result λm the density operator then becomes P P (r = λm |i = λn )P̂n ρ̂P̂n . ρ̂0m = n P (r = λm ) A second measurement will also give the result λm with probability P (λm |λm ) = Tr (π̂m ρ̂0m ) !

P 0 0 0 n0 P (r = λm |i = λn )P̂n ρ̂m = Tr P (r = λm ) P 2 P (r = λm |i = λn )P (i = λi ) . = n P (r = λm ) Hence the probability that the two measurement results are the same is X X P (λm |λm )P (r = λm ) = P 2 (r = λm |i = λn )P (i = λi ) . m

n

(4.10) Let us denote by ρ̂S the state of the qubit to be measured, and let the ancilla be prepared in a state with density operator

1 ˆ . ρ̂A = Î + ~r · ~σ 2 It then follows that the results for the proposed simultaneous measurement will be si = Tr (ρ̂S ⊗ ρ̂A σ̂i ⊗ σ̂i ) = hσ̂i iS ri

i = x, y, z ,

where ri are the components of the Bloch vector.

PAs the length of the Bloch vector cannot exceed unity, it follows that i ri2 ≤ 1, as required.

(4.11) Û =

X

|Φi ihi|

i

=

X

∗ ψji |jihi| .

ij

To demonstrate untarity, we first construct the Hermitian conjugate of this operator X Û † = |iihj|ψji . ij

46 Generalized measurements

It then follows that Û † Û =

X

=

X

∗ ψki ψji |jihk|

ijk

|jihj|

j

= Î , where we have used eqn 4.27. The fact that Û Û † = Î may be established in the same way.

(4.12)

There is a typographical error in eqn 4.36, which should be

Z pm x (x − xm )2 + i |xm , pm i = (2πσ 2 )−1/4 dx exp − .

4σ 2 h̄

The following answer and those of the following questions are based on this correct form.

∂ hx| (x̂ + iλp̂) |xm , pm i = x + λh̄ hx|xm , pm i ∂x

pm x − xm + i hx|xm , pm i . = x + λh̄ − 2σ 2 h̄ This is clearly an eigenstate if we set λh̄ = 2σ 2 . It then follows that the state is an intelligent state (and hence a minimum-uncertainty-product state) with mean position and momentum given by xm and pm . We can calculate the uncertainty in x by integration: Z ∞ 2 ∆x = dx (x − xm )2 |hn|xm , pm i|2 −∞

Z ∞ (x − xm )2 2 −1/2 2 = (2πσ ) dx (x − xm ) exp − 2σ 2 −∞ 2 =σ ⇒ ∆x = σ .

(4.13) (a) Z ∞ hxm , pm |x0m , p0m i = (2πσ 2 )−1/2 dx exp [i(p0m − pm )x] −∞

(x − xm )2 + (x − x0m )2 × exp − 4σ 2

0 2 (xm − xm ) σ 2 (p0m − pm )2 = exp − − 8σ 2 2

(xm + x0m )(p0m − pm ) +i .

2

Generalized measurements 47

(b) Z Z 1 dxm dpm |xm , pm ihxm , pm | = 2πh̄

Z Z Z (x − xm )2 + (x − xm )2 2 −1/2 0 (2πσ ) dx dx dxm exp − 4σ 2 0 ×δ(x − x )|x0 ihx| Z = dx|xihx| = Î .

(4.14)

There is a typographical error in eqn 4.39, which should be

Z (x − xm )2 P(xm ) = (2πσ 2 )−1/2 dx hx|ρ̂|xi exp − .

2σ 2

We use this corrected form in this solution. The average value is Z Z Z dxm P(xm )xm = (2πσ 2 )−1/2 dxm dx hx|ρ̂|xi

(x − xm )2 (xm − x + x) . × exp − 2σ 2 If we perform, first, the xm -integration then we find Z Z dxm P(xm )xm = dx hx|ρ̂|xix = hxi .

It follows that the measured variance is Z Z 2 −1/2 Var(xm ) = (2πσ ) dxm dx hx|ρ̂|xi

(x − xm )2 (xm − hxi)2 × exp − 2σ 2

Z (x − xm )2 = (2πσ 2 )−1/2 dxm exp − (xm − x)2 2σ 2 Z + dx hx|ρ̂|xi(x − hxi)2 = σ 2 + ∆x2 , which is the sum of the variances for the state measured and for the state |xm , pm i. The parallel analysis for the momentum gives h̄2 + ∆p2 .

4σ 2 It is clear that a necessary condition for the product of the measured variances to take its smallest possible value is for the measured state itself to be a minimum uncertainty state so that we can write ∆p2 = h̄2 /(4∆x2 ). With this value, the product of variances becomes

h̄2 h̄2 + Var(xm )Var(pm ) = σ 2 + ∆x2 4σ 2 4∆x2

h̄2 σ2 ∆x2 = 2+ + 2 .

4 ∆x2 σ Var(pm ) =

48 Generalized measurements

It is now straightforward to show that the minimum value occurs if σ 2 = ∆x2 , which gives the smallest possible value as h̄2 . This value occurs only if the state being measured is a minimum uncertainty state and if its uncertainties in the position and momentum match, exactly, those for the state |xm , pm i.

(4.15)

The condition in eqn 4.47 is π̂j (pj ρ̂j − pk ρ̂k )π̂k = 0

∀j, k .

If we sum this over j and k then we find X X π̂j pj ρ̂j − pk ρ̂k π̂k = Γ̂† − Γ̂ = 0 , j

k

so that Γ̂ is necessarily Hermitian.

(4.16) They are manifestly Hermitian and also, by their construction, also positive. It remains, only, to show that they sum to the identity.

We can prove this by using the fact that the operators π̂i form a POM:

X π̂i0 = Î − εP̂λ Î Î − εP̂λ + ε(2 − ε)P̂λ i

= Î − 2εP̂λ + ε2 P̂λ + ε(2 − ε)P̂λ = Î , as required.

(4.17) In BB84, the four possible states are selected with equal probabilities of 14 . It follows that the Γ̂ operator is 1 (µ|0ih0| + µ|1ih1| + (1 − µ)|00 ih00 | + (1 − µ)|10 ih10 |) 4 Î = , 4 which is clearly the same for all values of µ. It follows immediately that the rquired inequality for a minimum-error measurement, given in eqn 4.48, is satisfied for all values of µ. It only remains to check the required conditions, in eqn 4.47, for the six distinct pairs of states: Γ̂ =

1 µ|0ih0| (|0ih0| − |1ih1|) µ|1ih1| = 0 4 1 µ(1 − µ) µ|0ih0| (|0ih0| − |00 ih00 |) (1 − µ)|00 ih00 | = (h0|00 i − h0|00 i) |0ih00 | 4 4 =0 1 0 0 0 0 µ|0ih0| (|0ih0| − |1 ih1 |) µ|1 ih1 | = 0 4 1 µ|1ih1| (|1ih1| − |00 ih00 |) µ|00 ih00 | = 0 4 1 µ|1ih1| (|1ih1| − |10 ih10 |) µ|10 ih10 | = 0 4 1 µ|00 ih00 | (|00 ih00 | − |10 ih10 |) µ|10 ih10 | = 0 .

4

Generalized measurements 49

It follows that all values of µ give the same (minimum) error probability.

(4.18) The two required probability operators are pure-state projectors onto the two eigenstates of p1 |ψ1 ihψ1 | − p2 |ψ2 ihψ2 | =

p1 − p2 (p1 − p2 ) cos(2θ) sin(2θ) Î + σ̂z + σ̂x .

2 2 2

Let the two states eigenstates be denoted

φ φ |π1 i = cos |0i + sin |1i 2 2

φ φ |0i + cos |1i . |π1 i = − sin 2 2 Explicit calculation of the eigenvectors reveals that these will be the required states if we impose the condition tan φ =

tan(2θ) . p1 − p2

(4.19) It is clear that they are Hermitian. They are also clearly positive because the density operators ρ̂i are themselves positive operators:

hψ|π̂i |ψi = pi hψ|ρ̂−1/2 ρ̂i ρ̂−1/2 |ψi .

The operators sum to give the identity (in the space spanned by ρ̂) because X π̂i = ρ̂−1/2 ρ̂ρ̂−1/2 = Î , i

as required.

(4.20) To answer this question we could determine the probability operators for the square-root measurement and to compare these with those found in question (4.20). This is somewhat labour intensive, however, so here we sketch a quicker way to show that, in general, the optimum measurement is not a square-root measurement. Recall that any minimum-error POM will satisfy the condition given in eqn 4.47.

If the square-root measurement were optimal then it would then follow that hψ1 |ρ̂−1/2 (p1 |ψ1 ihψ1 | − p2 |ψ2 ihψ2 |) ρ̂−1/2 |ψ2 i = 0 ⇒ hψ1 |ρ̂−1/2 (ρ̂ − 2p2 |ψ2 ihψ2 |) ρ̂−1/2 |ψ2 i = 0 ⇒ 2p2 hψ1 |ρ̂−1/2 |ψ2 ihψ2 |ρ̂−1/2 |ψ2 i = hψ1 |ψ2 i .

For this to be true for all states and probabilities, it must at least be true that the left-hand side of this equation is independent of p2 . The eigenvalues of the density operator are

q 1 2 1 ± 1 − sin (2θ)4p2 (1 − p2 ) , ρ± = 2

50 Generalized measurements

and it necessarily follows that the left-hand side will not be, in general, independent of p2 .

(4.21) Let is start by calculating the density operator for this ensemble of states: N

ρ̂ =

1 X j−1 V̂ |ψ1 ihψ1 |V̂ †(j−1) .

N j=1

It immediately follows that V̂ m ρ̂V̂ †m = ρ̂ ⇒ hψk |ρ̂|ψk i = hψj |ρ̂|ψj i

∀k, j .

The first condition for minimum error, given in equation 4.47, then follows: hψk |ρ̂−1/2 (|ψk ihψk | − |ψj ihψj |) ρ̂−1/2 |ψj i =

hψk |ρ̂−1/2 |ψj i hψk |ρ̂−1/2 |ψk i − hψj |ρ̂−1/2 |ψj i = 0 .

We also require that the inequality 4.48 be satisfied. To show this we first calculate Γ̂ = =

N 1 X j−1 V̂ |ψ1 ihψ1 |V̂ †(j−1) ρ̂−1/2 V̂ j−1 |ψ1 ihψ1 |V̂ †(j−1) ρ̂−1/2 N 2 j=1

hψ1 |ρ̂−1/2 |ψ1 i 1/2 ρ̂ .

N

There is nothing special, of course, about |ψ1 i, so we cann also write Γ̂ =

hψi |ρ̂−1/2 |ψi i 1/2 ρ̂ , N

for any of the states |ψi i. The inequality 4.48 requires that the operator Γ̂ − pi |ψi ihψi | must be positive. We can prove that this is so with the assistance of the Cauchy-Schwartz inequality:

1 hφ| Γ̂ − pi ρ̂i |φi = hψi |ρ̂−1/2 |ψi ihφ|ρ̂1/2 |φi − |hψi |φi|2 N

1 = hψi |ρ̂−1/4 ρ̂−1/4 |ψi ihφ|ρ̂1/4 ρ̂1/4 |φi − |hψi |φi|2 N

1 ≥ |hψi |ρ̂−1/4 ρ̂1/4 |φi|2 − |ψi |φi|2 N = 0, as required.

(4.22) It is clear that such a strategy, in which there is only one nonzero probability operator, will always satisfy the condition given in eqn 4.47. All that is necessary, therefore, for the no-measurement POM to be optimal is that pi ρ̂i − pj ρ̂j ≥ 0

∀j .

Generalized measurements 51

Futher details and elaborations may be found in K. Hunter, Phys. Rev. A 68, 012306 (2003).

(4.23) form

It is natural to try, as suggested in the hint, an ansatz of the π̂i = |πi ihπi |

where |π1 i = 2−1/2 (a|0i + |1i) |π2 i = 2−1/2 (a|0i − |1i) |ψ3 i = (1 − a2 )|0i .

If p is large enough, then it is reasonable to expect that the optimal strategy for discriminating between just two equiprobable states, |ψ1 i and |ψ2 i, will be optimal. In this case we set a = 1. The remaining twoelement POM satisfies the required conditions for minimum probability of error so long as 1 p≥ , 2 + cos θ(cos θ + sin θ) where we have set both cos θ and sin θ to be positive. For smaller values of p we need all three probability operators with the parameter a given by p cos θ sin θ .

1 − p(2 + cos2 θ)

a=

Note that had we started with the three element POM then we would have been led to the same result, as this value of a has to exceed unity if p satisfies the inequality given above. Further details may be found in E. Andersson et al Phys. Rev. A 65, 052308 (2002).

(4.24) There is a typographical error in the question. The operator Υ̂ should be defined as Υ̂ =

M X

π̂j ŵj .

j=1

The proof follows, closely, that presented in the text for minimizing the error probability and we only sketch it here. Let us suppose that there is a primed strategy with probability operators {π̂i0 }. The difference between the Bayes cost and the minimum will then be !

X C̄ 0 − C̄ = Tr ŵi π̂i0 − C̄  i    X X ŵi − ŵj π̂j  π̂i0  = Tr  " = Tr

i

j

X

i

#

ŵi − Υ̂ π̂i0 .

52 Generalized measurements

For this to be be greater than or equal to zero for all possible primed strategies, then we require the operators ŵi − Υ̂ to be positive. For this to be possible, the operator Υ̂ must be Hermitian, and this is a consequence of the equality condition in the question.

(4.25) We can set this up as a two-state minimum-error problem by noting that Eve’s task is to learn the bit value. Hence she needs to determine whether the state carries the bit value ‘0’ or ‘1’. Hence we can think of her task as being that of discriminating between two mixed states 1 (|0ih0| + |00 ih00 |) 2

1 3 1 = 4 1 1 1 ρ̂1 = (|1ih1| + |10 ih10 |) 2

1 1 −1 = .

4 −1 3 ρ̂0 =

which have been prepared with equal probability. The minimum-error measurement will be realized by a projective measurement onto the eigenstates of the operator 1 1 (ρ̂0 − ρ̂1 ) = (σ̂z + σ̂x ) , 2 4 which is proportional to the component of spin that lies halfway between the z- and x-components. The eigenstates, therefore, are clearly those of the Breidbart basis.

(4.26) We require that each of the probability operators in eqn 4.64 is positive. The first two clearly are, but the final one will not be if we make A too large. This final probability operator, which corresponds to the inconclusive result, is

π̂? = Î − A |ψ1⊥ ihψ1⊥ | + |ψ2⊥ ihψ2⊥ |

1 − 2A sin2 θ 0 = .

0 1 − 2A cos2 θ We have chosen θ, in the text, such that cos2 θ ≥ sin2 θ. Hence there will be a negative eigenvalue for this operator if 2A cos2 θ > 1. It follows that the three operators will form a valid POM only if 2A cos2 θ ≤ 1 ⇒A≤

1 1 = .

2 cos2 θ 1 + |hψ1 |ψ2 i|

Clearly, the maximum allowed value of A will minimize the probability of an inconclusive outcome.

(4.27) A rather direct way to approach this problem is to construct the Naimark extension of the probability operators in an extended state

Generalized measurements 53

space. To this end we introduce a third basis state |2i and replace the probability operators corresponding to unabiguously identifying the two states by projectors onto the two orthogonal states √ √ |φ1 i = A|ψ2⊥ i + 1 − A|2i √ √ |φ2 i = B|ψ1⊥ i + 1 − B|2i , where we have defined the states so that hψ2⊥ |ψ1⊥ i is real and negative.

The orthogonality of these states imposes the constraint that (1 − A)(1 − B) = AB|hψ1⊥ |ψ2⊥ i|2 = AB|hψ1 |ψ2 i|2 .

The probability for getting the inconclusive result is

P? = 1 − (Ap1 + Bp2 ) 1 − |hψ1 |ψ2 i|2 .

Our task is to minimize this subject to the above constraint (which also embodies, of course, the requiment that 0 ≤ A, B ≤ 1). We can do this using Lagrange’s method of undetermined multipliers not forgetting, of course, to look at the boundaries A = 1, B = 0 and A = 0, B = 1. The result is that the minimum value for P? is the smallest of √ 2 p1 p2 |hψ1 |ψ2 i| ,

p2 + p1 |hψ1 |ψ2 i|2 ,

p1 + p2 |hψ1 |ψ2 i|2 .

A related analysis and further elaborations can be found in G. Jaeger and A. Shimony Phys. Lett. A 197, 83 (1995).

(4.28) By making it less likely that a measurement by Bob, or by Eve, will give an inconclusive result. Hence both can get a greater number of bits.

(4.29) The probability operators for unambiguous state elimination are proportional to projectors that are orthogonal to the states themselves. Hence for unambiguous state elimination of the mirror symmetric states, we have the probability operators π̂1̄ = a2 (sin θ|0i − cos θ|1i) (sin θh0| − cos θh1|) π̂2̄ = b2 (sin θ|0i + cos θ|1i) (sin θh0| + cos θh1|) π̂3̄ = c2 |1ih1| π̂? = Î − π̂1̄ − π̂2̄ − π̂3̄ .

In order to do this without a probability operator corresponding to the inconclusive outcome, we need to be able to choose a, b, c such that the first three operators sum to the identity. Clearly, for the sum of these operators not to contain a term proportional to |0ih1| we require that b2 = a2 . The three operators will then sum to the identity if 2a2 sin2 θ|0ih0| + 2a2 cos2 θ|1ih1| + c2 |1ih1| = Î .

For this to be possible, we require sin2 θ ≥ cos2 θ.

54 Generalized measurements

(4.30) They are manifestly Hermitian and they are also positive as hπ̂ij |ψi is the square of the magnitude of the vector B̂j Âi |ψi. It remains, only to show that they sum to the identity: X † † X † Âi B̂j B̂j Âi = Âi Î Âi ij

i

= Î .

(4.31) (a) (i)

P (i, j, k) = Tr Ĉk B̂j Âi ρ̂†i B̂j† Ĉk† = Tr †i B̂j† Ĉk† Ĉk B̂j Âi ρ̂ . (ii) P (j) =

X

X

Tr †i B̂j† B̂j Âi ρ̂ .

Tr †i B̂j† Ĉk† Ĉk B̂j Âi ρ̂ = i

ik

(iii)

† † † Tr  B̂ Ĉ Ĉ B̂  ρ̂ k j i i j k P (i, j, k)

.

P (k|i, j) = = P (i, j) Tr †i B̂j† B̂j Âi ρ̂ (iv)

† † † Â B̂ Ĉ Ĉ B̂ Â ρ̂ k j i i j k P (j, k) .

=P P (j|k) = † † † P (k) Ĉ B̂ Â ρ̂ Ĉ B̂ Â Tr k j i i j k ij P

i Tr

(b) (i) Ĉk B̂j Âi ρ̂†i B̂j† Ĉk† .

Tr †i B̂j† Ĉk† Ĉk B̂j Âi ρ̂ (ii) X

Ĉk B̂j Âi ρ̂†i B̂j† Ĉk† .

ijk

(iii) † † † k Ĉk B̂j Âi ρ̂Âi B̂j Ĉk

P

Tr †i B̂j† B̂j Âi ρ̂

.

(iv) † † † j Ĉk B̂j Âi ρ̂Âi B̂j Ĉk

P P

j 0 Tr

†i B̂j†0 Ĉk† Ĉk B̂j 0 Âi ρ̂

.

Generalized measurements 55

(4.32) A necessary and sufficient condition for the two-state density operator to be positive is for ρ00 ρ11 − ρ01 ρ10 ≥ 0. We can use the equations of motion to write an equation of motion for this quantity d (ρ00 ρ11 − ρ01 ρ10 ) = 2Γρ211 − 2Γρ00 ρ11 + 2γρ01 ρ01 . dt This will become negative if this derivative is negative for any state for which ρ00 ρ11 − ρ01 ρ10 = 0, that is, for any special pure state. Let us consider a pure state with ρ11 ρ00 . It is clear that for such a state, with ρ11 arbritrarily small but not zero, the derivative will be negative unless γ ≥ Γ.

(4.33) If the flips occur (or do not occur) in the order x, y and then z, then the density operator evolves according to ρ̂ → (1 − qx )ρ̂ + qx σ̂x ρ̂σ̂x → (1 − qy )(1 − qx )ρ̂ + (1 − qy )qx σ̂x ρ̂σ̂x + qy (1 − qx )σ̂y ρ̂σ̂y +qx qy σ̂z ρ̂σ̂z → (1 − qz )(1 − qy )(1 − qx )ρ̂ + (1 − qz )(1 − qy )qx σ̂x ρ̂σ̂x +(1 − qz )qy (1 − qx )σ̂y ρ̂σ̂y +

(1 − qz )qx qy σ̂z ρ̂σ̂z

qz (1 − qy )(1 − qx )σ̂z ρ̂σ̂z + qz (1 − qy )qx σ̂y ρ̂σ̂y +qz qy (1 − qx )σ̂x ρ̂σ̂x + qx qy qz ρ̂ = [qx qy qz + (1 − qx )(1 − qy )(1 − qz )]ρ̂ [qz qy (1 − qx ) + (1 − qz )(1 − qy )qx ]σ̂x ρ̂σ̂x [qz qx (1 − qy ) + (1 − qz )(1 − qx )qy ]σ̂y ρ̂σ̂y [qx qy (1 − qz ) + (1 − qx )(1 − qy )qz ]σ̂z ρ̂σ̂z .

Note that the order in which the spin flips have been applied has turned out not to be important and this transformation has, in fact, precisely the same for as that given in eqn 4.91.

(4.34) Let us begin by noting the action of the transforming terms on the Pauli operators: σ̂j Î σ̂j = Î σ̂j σ̂j σ̂j = σ̂j σ̂j σ̂k σ̂j = −σ̂k

j 6= k .

Let is use these to calculate the effect of the transformation on a density operator written in terms of its Bloch-vector components.

1 Î + uσ̂x + vσ̂y + wσ̂z ρ̂ = 2 1 0 ⇒ ρ̂ = [Î + (1 − 2py − 2pz )uσ̂x + (1 − 2px − 2pz )vσ̂y 2 +(1 − 2px − 2py )wσ̂z ] .

We found in exercise (2.29) that Tr(ρ̂2 ) = (1 + r2 )/2, where r is the length of the Bloch vector. It then follows that 1 Tr ρ̂02 = 1 + (1 − 2py − 2pz )2 u2 + (1 − 2px − 2pz )v 2 2

56 Generalized measurements

+ (1 − 2py − 2pz )2 w

It follows that

Tr ρ̂02 − Tr ρ̂2 = −2(py + pz )(1 − py − pz )u2 −2(px + pz )(1 − px − pz )v 2 −2(px + py )(1 − px − py )w2 , which is clearly less than or equal to zero. The transformed density operator, above, will be the maximally mixed state for all initial states only if 1 − 2px − 2py = 0 1 − 2px − 2pz = 0 1 − 2py − 2pz = 0 ⇒ px = py = pz =

1 .

4

(4.35) A very simple and explicit construction serves to illustrate the point. Consider the effect comprising the two operators Â1 = |0ih0| and Â2 = |0ih1|. It is clear that this opertion changes all qubit states into the same state, |0i.

(4.36) For optmium state separation, the operator ÂS changes a normalized superposition of the states |ψ1 i and |ψ2 i into a different normalized superposition. This operator and the failure effect operator must satisfy the condition †S ÂS + †F ÂF = Î .

It necessarily follows that the failure operator, ÂF , annihilates this same superposition. The only way that this can happen is if ÂF |ψ1 i ∝ ÂF |ψ2 i , so the failure operator must produce the same state from |ψ1 i and |ψ2 i.

Further details of this problem and the next one can be found in A. Chefles and S. M. Barnett, J. Phys. A: Math. Gen. 31, 10097 (1998).

(4.37) Making N copies is equivalent to state separation in which |hψ10 |ψ20 i| = |hψ1 |ψ2 i|N .

The probability for successfully making these copies is then bounded by PS ≤

1 − |hψ1 |ψ2 i| .

1 − |hψ1 |ψ2 i|N

Taking the N → ∞ limit gives PS ≤ 1 − |hψ1 |ψ2 i| ,

Generalized measurements 57

which is the maximum probability for unambiguously discriminating between the original two states. This is reasonable, for if we can distinguish between the states unambiguously, then we can make as many copies as we please.

(4.38) (a) A necessary and sufficient condition for the transformed density operator to be positive is that ρ000 ρ011 − ρ001 ρ010 ≥ 0. It follows that for all possible initial density operators we require that 1 − κ2 ρ01 ρ10 ≥ 0 .

4 The largest allowed value of ρ01 ρ10 is 1/4 and it follows that positivity requires |κ| ≤ 1. (b)

In the basis used in eqn J.28, the transformation matrix is  1 0 0 12 2 0 κ 0 0 L= .

0 0 κ 0 1 1 0 0 2 2

It follows that the matrix for the associated transformation is  1 0 0 κ 2 1 0 2 0 0 Lass =  .

0 0 12 0 1 κ 0 0 2 The determinant of this matrix is 41 ( 14 − κ2 ). For this to be positive, and so correspond to a physically allowed transformation, we need |κ| ≤ 21 .

(4.39)

It is simple enough to find a suitable set of three effect operators √ Â0 = κÎ 1 Â1 = √ |0ih1| 2 1 Â2 = √ |1ih0| 2 r 1 Â3 = − κ2 |0ih0| r2 1 Â4 = − κ2 |1ih1| .

2 P It is straightforward to confirm that i †i Âi = Î as is required, and that these operators perform the desired transformation. The set of effect operators is, of course, not unique.

Entanglement (5.1) There exist a number of ways of showing this. Perhaps the simplest is simply to note that they are all pure states for which the reduced density operator for each individual qubit is that for a maximally mixed state Î/2.

(5.2) The square of the total angular momentum for the two spin-half particles is the operator 2 2 2 i h̄2 h A σ̂x + σ̂xB + σ̂yA + σ̂yB + σ̂zA + σ̂zB Jˆ2 = 4

h̄2 3Î ⊗ Î + σ̂x ⊗ σ̂x + σ̂y ⊗ σ̂y + σ̂z ⊗ σ̂z . = 2 The Bell state |Ψ− i is an eigenstate of the three operators σ̂i ⊗ σ̂i with the common eigenvalue −1. It follows, therefore, that Jˆ2 |Ψ− i = 0.

(5.3) The square of the total angular momentum for a single spin-1 particle is Jˆ2 = Jˆx2 + Jˆy2 + Jˆz2 = 2h̄2 Î .

It then follows that the square of the total angular momentum for the two particles is Jˆ2 = 4h̄2 Î ⊗ Î + 2Jˆx ⊗ Jˆx + 2Jˆx ⊗ Jˆx + 2Jˆz ⊗ Jˆz .

Let us take as an ansatz the state a superposition of the states |1i⊗|−1i, | − 1i ⊗ |1i and |0i ⊗ |0i: |Ψi = α|1i ⊗ | − 1i + β|0i ⊗ |0i + γ| − 1i ⊗ |1i .

We know that the required eigenstate must be of this form as all other combinations give a non-zero total z-component of angular momentum.

We can determine the forms of the coefficients by solving the eigenvalue equation: Jˆ2 |Ψi = 0

= 4h̄2 Î ⊗ Î + 2Jˆx ⊗ Jˆx + 2Jˆy ⊗ Jˆy + 2Jˆz ⊗ Jˆz |Ψi = 4h̄2 (α|1i ⊗ | − 1i + β|0i ⊗ |0i + γ| − 1i ⊗ |1i) +h̄2 [(α + γ)|0i ⊗ |0i + β (|1i ⊗ |1i + |1i ⊗ | − 1i +| − 1i ⊗ |1i + | − 1i ⊗ | − 1i)]

5

60 Entanglement

+h̄2 [(α + γ)|0i ⊗ |0i + β (−|1i ⊗ |1i + |1i ⊗ | − 1i +| − 1i ⊗ |1i − | − 1i ⊗ | − 1i)] 2

+2h̄ (−α|1i ⊗ | − 1i − γ| − 1i ⊗ |1i) 2

= h̄ [2(α + β)|1i ⊗ | − 1i + 2(2β + α + γ)|0i ⊗ |0i +2(γ + β)| − 1i ⊗ |1i] ⇒ β = −α = −γ .

The normalized solution is therefore 1 |Ψi = √ (|1i ⊗ | − 1i − |0i ⊗ |0i + | − 1i ⊗ |1i) .

3 That this is a maximally entangled state follows from the fact the state is pure but the reduced density operators for each spin are maximally mixed, Î/3.

(5.4) ˆ ⊗ ~a · ~σ ˆ |Ψ− i = √1 {[az |0i + (ax + iay )|1i] ⊗ ~a · ~σ 2 [−az |1i + (ax − iay )|0i] − [−az |1i + (ax − iay )|0i] ⊗ [az |0i + (ax + iay )|1i]} 2 2 = −(ax + ay + a2z )|Ψ− i = −|Ψ− i , as required.

(5.5) Let Bob perform a measurement with probability operators π̂iB .

The probability that he will get measurement outcome i if Alice chooses operation k is

X AB Ak† P (i|1) = Tr π̂iB ÂAk Âj j ρ̂ j

=

X

AB Tr π̂iB ÂAk† ÂAk j ρ̂ j

j

= Tr π̂iB ρ̂AB .

This probability is clearly independent of Alice’s choice of operation and so is the required result.

(5.6) (a)

The Schmidt decomposition is simply

1 1 |Ψi = cos θ|0iA √ (|0iB + |1iB ) + sin θ|1iA √ (|0iB − |1iB ) .

2 2

(b)

The reduced density operator for the second system is

1 1 cos(2θ) ρ̂B = , 1 2 cos(2θ)

Entanglement 61

and the no-signalling theorem requires that no measurement performed on the first qubit can change this. If Alice can determine, unambiguously, the state of the first qubit then she also determines for certain that the second qubit is in the corresponding state |0iB or |1iB . Hence, if P? is the probability for an the result to give an ambiguous result, the density matrix for Bob’s qubit must be

1 − P? 0 0 1 − P? 1 0 + + P? ρ̂? , ρ̂B = 0 0 0 1 2 2 where ρ̂? is the state that the second qubit is left in if Alice’s measurement gives the ambiguous result. Our two expressions for ρ̂B must be the same and also, ρ̂? must be an allowed density operator. It follows, from the requirements that ρ00 ρ11 ≥ ρ01 ρ10 and Tr(ρ̂? ) = 1 that P? ≥ | cos(2θ)|.

A slightly more complete treatment of this problem can be found in S. M. Barnett and E. Andersson, Phys. Rev. A 65, 044307 (2002).

(5.7) ~

~

~

~

~ ab ~ ab ~ ab ~ ab PAB (+1, +1) + PAB (−1, −1) − PAB (+1, −1) − PAB (−1, +1) Z 1 dλ ρ(λ)[(1 + A)(1 + B) + (1 − A)(1 − B) = 4 −(1 + A)(1 − B) − (1 − A)(1 + B)] Z = dλ ρ(λ)AB .

(5.8)

No, because hσ̂x ⊗ σ̂x i = −1 = hσ̂z ⊗ σ̂z i hσ̂x ⊗ σ̂z i = 0 = hσ̂x ⊗ σ̂z i .

Combining these in a Bell inequality gives a maximum value of only S = 2.

(5.9) The four Bell states are common eigenstates of the three operators σ̂j ⊗ σ̂j with eigenvalues |Ψ− i |Ψ+ i |Φ− i |Φ+ i

σ̂x ⊗ σ̂x −1 +1 −1 +1

σ̂y ⊗ σ̂y −1 +1 +1 −1

σ̂z ⊗ σ̂z −1 −1 +1 +1

The expectation values of all other pairs of operators is zero for all the Bell states, hσ̂j ⊗ σ̂k i = 0 for j 6= k. It then follows that ˆ ⊗ ~b · ~σ ˆ |Ψ− i = −~a · ~b hΨ− |~a · ~σ ˆ ⊗ ~b · ~σ ˆ |Ψ+ i = ~a · ~b − 2az bz hΨ+ |~a · ~σ ˆ ⊗ ~b · ~σ ˆ |Φ− i = ~a · ~b − 2ax bx hΦ− |~a · ~σ ˆ ⊗ ~b · ~σ ˆ |Φ+ i = ~a · ~b − 2ay by . hΦ+ |~a · ~σ

62 Entanglement

(5.10) S = ~a · (~b − ~b0 ) + ~a0 · (~b + ~b0 ) .

To maximize this, we first need to choose ~a to be parallel to ~b − ~b0 and ~a0 to be parallel to ~b + ~b0 , so that we have S = ~b − ~b0 + ~b + ~b0 , √ the maximum value of which is 2 2 which occurs only if ~b is perpendicular to ~b0 .

(5.11) Let us choose our bases such that our non-maximally entangled state has the form |ψi = c1 |0i ⊗ |1i − c2 |1i ⊗ |0i , where the amplitudes, cj , are both real and positive. The non-zero expectation values of the operators hσ̂j ⊗ σ̂k i for this state are hσ̂x σ̂x i = −2c1 c2 = hσ̂y σ̂y i hσ̂z σ̂z i = −1 .

It follows, therefore, that E(~a, ~b) = −2c1 c2~a · ~b − (1 − 2c1 c2 )az bz .

If we choose, as for the singlet state, the vector ~a to be parallel to ~b − ~b0 and the vector ~a0 to be parallel to ~b + ~b0 then we find

S = −2c1 c2 (bx − b0x )ı̂ + (by − b0y )̂ − (bz − b0z )k̂

+ −2c1 c2 (bx + b0x )ı̂ + (by + b0y )̂ − (bz + b0z )k̂ .

It suffices, in finding the maximum value, to restrict ourselves to the x − z plane so that S = −2c1 c2 (bx − b0x )ı̂ − (bz − b0z )k̂ + −2c1 c2 (bx + b0x )ı̂ − (bz + b0z )k̂ .

If we now choose ~b = k̂ and ~b0 = ı̂ we find q S = 2 4c21 c22 + 1 , which clearly exceeds 2 unless either c1 or c2 is zero, in which case the state is not entangled.

Further details can be found in N. Gisin Phys. Lett. A 154, 201 (1991) and in S. M. Barnett and S. J. D. Phoenix, Phys. Lett. A 167, 233 (1992).

(5.12) Let |j, k, li represent the states of the three qubits in the computational basis with j, k, l = 0, 1. The three operators have the effect

Entanglement 63

that σ̂x ⊗ σ̂y ⊗ σ̂y |j, k, li = (−1)1+k+l |j̄, k̄, ¯li σ̂y ⊗ σ̂x ⊗ σ̂y |j, k, li = (−1)1+j+l |j̄, k̄, ¯li σ̂y ⊗ σ̂y ⊗ σ̂x |j, k, li = (−1)1+j+k |j̄, k̄, ¯li σ̂x ⊗ σ̂x ⊗ σ̂x |j, k, li = |j̄, k̄, ¯li .

It follows, therefore, that the eight entangled states

1 |ψ(k, l, ±)i = √ |0, k, li ± |1, k̄, ¯li 2 are the common eigenstates of the four operators. That any of these states allows us to demonstrate a conflict with local realism follows directly from eqn 5.19.

(5.13) (a) Let us start with an entangled state with Schmidt decomposition |ψi = α|0̃i ⊗ |1̃0 i + β|1̃i ⊗ |0̃0 i .

This will be orthogonal to the state |1i ⊗ |1i if we choose the computational basis such that αh1|0̃ih1|1̃0 i + βh1|1̃ih1|0̃0 i = 0 .

We are then left with our state in the required form. If b or c are zero then we have a product state. If a is zero then the state has the form given above and we need to select a different computational basis. (b)

As with the presentation in Appendix K, the form of the state |ψi = a|0iA |0iB + b|1iA |0iB + c|0iA |1iB

leads us immediately to the three probabilities, which we can write (in a hopefully transparent notation) as P (|1iA , |1iB ) = 0 , ! a|0iB + c|1iB p |1iA = 1 , |a|2 + |c|2 ! a|0iA + b|1iA p |1iB = 1 . |a|2 + |b|2

P P

Viewed from a local realistic perspective, the first condition tells us that at least one of the properties σzA and σzB must take the value +1. It is a prediction of local realism, therefore that the state must be orthogonal to the state in which qubit A is orthogonal to a|0iA + b|1iA and qubit B is orthogonal to a|0iB + c|1iB . Hence 1 p

|a|2 + |b|2

1 p

|a|2 + |c|2

(A h0|b − A h1|a) ⊗ (B h0|c − B h1|a) |ψi

64 Entanglement

should, according to local realism, be zero. The quantum calculation, however, gives the value −abc p

(|a|2 + |b|2 )(|a|2 + |c|2 )

.

This will be zero only if at least one of the amplitudes a, b, c is zero. (c) For a maximally entangled state or for a product state, at least one of the the amplitudes a, b, c is zero.

Further details and elaborations may be found in L. Hardy, Phys.

Rev. Lett. 71, 1665 (1993) and S. Goldstein, Phys. Rev. Lett. 72, 4951 (1994).

(5.14) For the Hamiltonian in eqn 5.23 and interaction time T such that gT /h̄ = π/4 we have !

ĤT 1 (|0i + |1i) ⊗ (|0i + i|1i) exp −i h̄ 2

1 = (|0i + |1i) ⊗ eiπ/4 |0i + ie−iπ/4 |1i 2 eiπ/4 = (|0i + |1i) ⊗ (|0i + |1i) 2 !

ĤT 1 exp −i (|0i − |1i) ⊗ (|0i + i|1i) h̄ 2

1 = (|0i − |1i) ⊗ e−iπ/4 |0i + ieiπ/4 |1i 2 e−iπ/4 = (|0i − |1i) ⊗ (|0i − |1i) .

2 Hence a measurement of σ̂x on the ancilla gives the value of this component of spin for the original system qubit.

(5.15)

We start by considering powers of Hamiltonian in eqn 4.24.

Ĥ 2 = g 2 2 Î ⊗ Î − σ̂y ⊗ σ̂y Ĥ 3 = g 3 4 (σ̂z ⊗ σ̂x − σ̂x ⊗ σ̂z ) = 4g 2 Ĥ .

Expanding the unitary operator as a Taylor series, we find that

θ2 Û = Î ⊗ Î − iθ (σ̂z ⊗ σ̂x − σ̂x ⊗ σ̂z ) − 2 Î ⊗ Î − σ̂y ⊗ σ̂y 2! θ3 +i 4 (σ̂z ⊗ σ̂x − σ̂x ⊗ σ̂z ) + · · · 3!

1 1 = [cos(2θ) + 1]Î ⊗ Î − [cos(2θ) − 1]σ̂y ⊗ σ̂y 2 2 sin(2θ) −i (σ̂z ⊗ σ̂x − σ̂x ⊗ σ̂z ) 2 = cos2 θÎ ⊗ Î − i sin θ cos θ (σ̂z ⊗ σ̂x − σ̂x ⊗ σ̂z ) + sin2 θσ̂y ⊗ σ̂y .

Entanglement 65

(a) The action of this unitary transfromation on a state in which the first qubit is an eigenstate of σ̂z or σ̂x and the second is in the state 2−1/2 (|0i + i|1i) is 1 1 Û |0i ⊗ √ (|0i + i|1i) = cos2 θ|0i ⊗ √ (|0i + i|1i) 2 2 1 + sin θ cos θ|0i ⊗ √ (|0i − i|1i) 2 1 +i sin θ cos θ|1i ⊗ √ (|0i − i|1i) 2 1 2 +i sin θ|1i ⊗ √ (|0i + i|1i) 2 1 1 Û |1i ⊗ √ (|0i + i|1i) = cos2 θ|1i ⊗ √ (|0i + i|1i) 2 2 1 − sin θ cos θ|1i ⊗ √ (|0i − i|1i) 2 1 +i sin θ cos θ|0i ⊗ √ (|0i − i|1i) 2 1 −i sin2 θ|0i ⊗ √ (|0i + i|1i) 2 1 1 0 2 0 Û |0 i ⊗ √ (|0i + i|1i) = cos θ|0 i ⊗ √ (|0i + i|1i) 2 2 1 + sin θ cos θ|10 i ⊗ √ (|0i − i|1i) 2 1 +i sin θ cos θ|00 i ⊗ √ (|0i − i|1i) 2 1 2 0 −i sin θ|1 i ⊗ √ (|0i + i|1i) 2 1 1 0 2 0 Û |1 i ⊗ √ (|0i + i|1i) = cos θ|1 i ⊗ √ (|0i + i|1i) 2 2 1 − sin θ cos θ|00 i ⊗ √ (|0i − i|1i) 2 1 0 −i sin θ cos θ|1 i ⊗ √ (|0i − i|1i) 2 1 +i sin2 θ|10 i ⊗ √ (|0i + i|1i) 2 From these we can read off the probabilities that a measurement of the ancilla gives the correct state of the original qubit: P (correct) =

1 2 (cos θ + sin θ) .

2

If we set cos θ = sin θ = 2−1/2 then these two probabilities are unity and we can determine the bit value with certainty, once we know the basis in which to measure the qubit. (b) The answer to the preceeding part means that we should ask what effect this would have if we adopted it as an eavesdropping strategy

66 Entanglement

to attack quantum key distribution. From the above transformed states we see that the probability that this strategy would generate an error in Bob’s qubit is P (error) = sin2 θ .

For the optimal strategy for learning the qubit, this value is 1/2. Hence this corresponds to Eve keeping the qubit sent by Alice and then sending to Bob a qubit prepared in a state selected at random. If we adjust the parameters so that the error probability is 14 , as encountered in the intercept, measure and resend strategy, then we find !2 √ 1 3 1 P (correct) = + ≈ 0.93 , 2 2 2 which is better than found for the Breidbart measurement (for which the value is ≈ 0.85). We can understand this improvement as arising from the additional information provided by knowing in which basis it is best to perform the measurement.

(5.16) (a) An error will occur if a state with σz = ±1 produces a state with ∓ momentum. If the two states were prepared with probabilities p0 and p1 then the error probability will be Z 0 a 1/2

Perror = p0 dp exp −a(p − gT )2 π −∞ Z ∞ a 1/2

exp −a(p + gT )2 +p1 dp π 0 Z ∞ a 1/2

= dp exp −a(p + gT )2 π 0

√ 1 = 1 − erf( agT ) .

2 (b) The minimum error for discriminating between two pure states |ψ1,2 i with minimum error is given by the Helstrom bound:

p 1 Hel Perror = 1 − 1 − 4p1 p2 |hψ1 |ψ2 i|2 | .

2 For our two momentum states we have a 1/2 Z ∞ a

a

hψ1 |ψ2 i = dp exp − (p − gT )2 exp − (p + gT )2 π 2 2 −∞

2 2 = exp −ag T .

Hence the Helstrom bound is

p 1 Hel Perror = 1 − 1 − 4p1 p2 exp (−ag 2 T 2 ) .

2 (5.17)

No it is not because it requires a quantum channel.

Entanglement 67

(5.18) All states work as well as |0iC . The probability that Claire’s state will open the lock is 1/4. If the lock is based on n ebits then the probability that Claire’s n-qubit state will open the lock is (1/4)n .

(5.19) If Claire performs a measurement that shows her qubit to be in the state cos(θ/2)|0i + eiϕ sin(θ/2)|1i then Alice and Bob’s qubits will be left in the state θ θ cos |0iA |0iB + e−iϕ sin |1iA |1iA .

2 2 Clearly, by varying θ, which is determined by Claire’s choice of basis, she can generate a pure state of Alice and Bob’s qubits with any desired degree of entanglement. She can also fix the Schmidt basis by sending instructions to both Alice and to Bob to perform the relevant local unitary transformations.

(5.20) (a) It suffices to show that the three symmetric states can be prepared by applying Pauli operators to either of the qubits in the antisymmetric state. |Ψ+ i = σ̂z ⊗ Î|Ψ− i = −Î ⊗ σ̂z |Ψ− i |Φ− i = −σ̂x ⊗ Î|Ψ− i = Î ⊗ σ̂x |Ψ− i |Φ+ i = −iσ̂y ⊗ Î|Ψ− i = iÎ ⊗ σ̂y |Ψ− i .

It follows, of course, that a product of Pauli operators (which is also a Pauli operator) allows us to perform the other transformations. (b) For a maximally entangled state, any basis for a single qubit can form the Schmidt decomposition. The various maximally entangled states differ only in the Schmidt basis that is partnered by, say, the computational basis for first (second) qubit. All that we need do in order to transform between these states is to apply the required unitary transformation to the second (first) qubit so as to modify the basis correlated with the Schmidt basis.

(5.21) (a) The four signal states are generated by Alice performing one of the four unitary transformations, Î, σ̂x , σ̂y and σ̂z on her qubit , which she then sends to Bob. The four signal states generated in this way are: p ρ̂(I) = Î ⊗ Î + (1 − p)|Ψ− ihΨ− | 4 p ρ̂(X) = Î ⊗ Î + (1 − p)|Φ− ihΦ− | 4 p ρ̂(Y) = Î ⊗ Î + (1 − p)|Φ+ ihΦ+ | 4 p ρ̂(Z) = Î ⊗ Î + (1 − p)|Ψ+ ihΨ+ | .

4 Obviously Bob’s best strategy is to measure these signal states in the Bell basis (its the only basis defined) and on doing so he finds:

1 3p P (I, I) = P (X, X) = P (Y, Y) = P (Z, Z) = 1− 4 4

68 Entanglement

P (I, X) = P (I, Y) = · · · = P (Z, Y) =

p .

16

P Note that ij P (i, j) = 1 as it should. The mutual information associated with these probabilities is 1 1 1 1 log − 4 × log 4 4 4

4

1 1 3p 1 1 3p +4 × − log − 4 4 16 4 4 16 p p +12 × log 16 16

3p 3p 3p p =2+ 1− log 1 − + log .

4 4 4 4

H(A : B) = −4 ×

(b) We saw in Chapter 2 that the Werner state is not entangled if p ≥ 23 . For p = 23 we have 1 1 1 1 log + log 2 2 2 6

1 1 1 1 =2− + log + log 2 2 2 3 1 1 = 1 + log < 1 .

2 3

H(A : B) = 2 +

This value monotonically decreases as p increases. There is no reason to expect this to give the classical value as quantum dense coding has not been optimized to work with unentangled states. It is clear, however, that it should be less than 1.

(5.22)

The state is |ψi = α|0i + β|1i .

To send instructions to Bob, Alice needs to send one complex coefficient, either α or β. (Normalization and the irrelevance of the global phase takes care of the other coefficient.) Let us suppose that Alice sends to Bob a finite digit approximation βapp to Bob who then constructs the state q |ψBob i = 1 − |βapp |2 |0i + βapp |1i .

The fidelity is then F = |hψ|ψBob i|2 =

2 q p 1 − |β|2 1 − |βapp |2 + β ∗ βapp .

Let βapp = β + δ, with δ being small. If we Taylor expand F to second order in δ we find F ≈ 1 − |δ|2 +

1 (δβ ∗ + βδ ∗ )2 1 + (|β ∗ δ − δ ∗ β|2 ) .

4 1 − |β|2 4

Entanglement 69

(Note that this requires |β| to be not too close to 1. However if |β| is very close to unity then a special sequence of numbers, or coding, can be used.) The smallest value of this arises as β → 0 for which F ≈ 1 − |δ|2 .

Hence we need |δ|2 < 10−6 which means that the squares of both the real and imaginary parts of δ need to be fixed to better than 12 × 10−6 .

Hence 1 1 |δ 0 |, |δ 00 | < √ × 10−3 ≈ .

1414 2 In binary we have 1 1024 1 0.00000000001 = .

2048 Hence we need to fix both the real and imaginary parts of βapp to 11 binary places, corresponding to 22 bits of information. We need a further 2 bits to fix the signs of δ 0 and δ 00 , so we need a total of 24 bits.

0.0000000001 =

(5.23) 1 |ψia |Ψ− iAB = √ (α|0ia |0iA |1iB − α|0ia |1iA |0iB 2 +β|1ia |0iA |1iB − β|1ia |1iA |0iB )

1 = α |Φ+ iaA + |Φ+ iaA |1iB − α |Ψ+ iaA + |Ψ+ iaA |0iB 2

+β |Ψ+ iaA − |Ψ+ iaA |1iB − β |Φ+ iaA − |Φ+ iaA |0iB 1 − |Ψ iaA (−α|0iB − β|1iB ) + |Ψ+ iaA (−α|0iB + β|1iB ) = 2

+|Φ− iaA (α|1iB + β|0iB ) + |Φ+ iaA (α|1iB − β|0iB ) Q.E.D. (5.24) It is straightforward to create a protocol for each of these states.

The simplest way to show that it is possible, however, is to note that Alice or Bob can simply change the prepared state to |Ψ− iAB by applying, locally, a Pauli matrix to their qubit before commencing the protocol.

(5.25) This is, in essence a corollary to the no-cloning theorem. To see this let us suppose that this operation works for the two states |0i and |1i: |0i ⊗ |Bi → |B0 i ⊗ |0i

|1i ⊗ |Bi → |B1 i ⊗ |1i .

For a more general α and β we have, from linearty: (α|0i + β i) ⊗ |B 0 i → α|B0 i ⊗ |0i + β|B1 i ⊗ |1i , which gives the desired transformation only if |B0 i = |B1 i. This is the SWAP operation encountered in Chapter 6.

70 Entanglement

(5.26) Z π Z 2π 1 dθ sin θ(cos4 θ + sin4 θ) dφ 4π 0 0 Z 1 π dθ sin θ(cos4 θ + 1 − 2 cos2 θ + cos4 θ) = 2 0 11 = , 15

F̄ =

which is between 12 and 1 as it should be.

(5.27) We can think of the Werner state as one in which the state |Ψ− i is prepared with probability (1−p) with the fully mixed state 41 Î⊗ Î with probability p. If the state |Ψ− i is prepared then the fidelity is 1, but if the state 14 Î ⊗ Î then the outcome is a random state and the fidelity takes its minimum value, 21 . Hence the fidelity in this case will be p 1 F = (1 − p) + p = 1 − .

2 2 (5.28) To evaluate this consider teleporting a general state cos(θ/2)|0i+ sin(θ/2)eiφ |1i:

θ θ |aABi = cos |0ia + sin eiφ |1ia ⊗ (cos ϕ|0iA |1iB − sin ϕ|1iA |0iB ) .

2 2 Alice measures in the Bell basis and the associated states obtained by Bob are found by projection:

1 θ θ iφ − − cos sin ϕ|0iB − sin e cos ϕ|1iB aA hΨ |aABi = √ 2 2 2

1 θ θ iφ + √ hΨ |aABi = − cos sin ϕ|0i + sin e cos ϕ|1i aA B B 2 2 2

θ θ 1 − cos cos ϕ|1iB + sin eiφ sin ϕ|0iB aA hΦ |aABi = √ 2 2 2

θ θ iφ 1 + cos cos ϕ|1iB − sin e sin ϕ|0iB . aA hΦ |aABi = √ 2 2 2 In each case Bob performs the associated transformation corresponding to the result announced by Alice:

1 θ θ Ψ− : P −1/2 (Ψ− ) √ cos sin ϕ|0iB + sin eiφ cos ϕ|1iB 2 2 2

1 θ θ Ψ+ : P −1/2 (Ψ+ ) √ cos sin ϕ|0iB + sin eiφ cos ϕ|1iB 2 2 2

1 θ θ iφ − −1/2 − Φ : P (Φ ) √ cos cos ϕ|0iB + sin e sin ϕ|1iB 2 2 2

1 θ θ + −1/2 + iφ √ Φ : P (Φ ) cos cos ϕ|0iB + sin e sin ϕ|1iB 2 2 2

Entanglement 71

where P (Ψ− ) is the probability that Alice’s measurement gives the result |Ψ− i. Hence the fidelity expected is the average of the fidelities for these states weighted by the probability that Alice gets the associated measurement result. The probabilities P (Ψ− ) etc. drop out and we are left with 2 2

θ θ θ θ F = cos2 sin ϕ + sin2 cos ϕ + cos2 cos ϕ + sin2 sin ϕ 2 2 2 2 1 2 2 = (1 + cos θ) + sin θ sin ϕ cos ϕ .

2 If we average this by integrating over the Bloch sphere we find Z 2π Z π 1 F̄ = dφ sin θF dθ 4π 0 0 2 = (1 + sin ϕ cos ϕ) .

3 Note that for a true singlet state sin ϕ = 2−1/2 = cos ϕ and this fidelity becomes 1 as it should.

(5.29)

A little algebra leads us to rewrite the state as |Ψ− iCa |Ψ− iAB =

1 −|Ψ− iaA |Ψ− iCB + |Ψ+ iaA |Ψ+ iCB 2

+|Φ− iaA |Φ− iCB − |Φ+ iaA |Φ+ iCB .

Hence if Alice measures her two qubits in the Bell basis then she can tell Bob and Claire which Bell state they have and either of them can create the state |Ψ− iCB by applying a local unitary transformation.

(5.30) Yes it is optimal. If Alice and Bob each swap a qubit that is maximally entangled to another one that they have locally, then after the operation they will have a pair of entangled states and hence share two ebits. We know that ebits cannot be created using LOCC operations and hence creating them must have consumed at least two ebits. Two ebits are consumed in the pair of teleportations and it follows that this process is optimal.

(5.31)

It is maximally entangled if ρ̂A is the maximally mixed state: d−1

1 1X |jihj| = Î ρ̂A = TrB |Ψ ihΨ | = d j=0 d 00

00

which is the maximally mixed state Q. E. D. (5.32)

For d = 2 we have 1 |Ψ00 i = √ (|0i|0i + |1i|1i) = |Φ+ i 2 1 |Ψ01 i = √ (|0i|1i + |1i|0i) = |Ψ+ i 2

72 Entanglement

1 |Ψ10 i = √ (|0i|0i − |1i|1i) = |Φ− i 2 1 11 |Ψ i = √ (|0i|1i − |1i|0i) = |Ψ+ i .

2 (5.33) The simplest way is to first note that these states are orthonormal and that there are precisely d2 states and that they resolve the identity. First note that d−1 X

ei2π(j−k)n/d = dδj,k .

n=0

Hence X X1X ei2π(j−k)n/d |ji|(j + m)moddihk|h(k + m)modd| |Ψnm ihΨnm | = d n,m n,m j,k X = |jihj| ⊗ |(j + m)moddih(j + m)modd| m,j

=

X

|jihj| ⊗ |lihl|

j,l

= Î ⊗ Î .

(5.34) The simple answer is log d2 = 2 log d. The careful answer is the smallest integer greater than or equal to 2 log d.

(5.35) (a) † Ûnm =

d−1 X

|(k 0 + m)moddihk 0 | ,

k0 =0

hence † Ûnm Ûnm =

d−1 X

0

ei2π(k−k )n/d |kih(k + m)modd|(k 0 + m)moddihk 0 |

k,k0 =0

=

d−1 X

|kihk| = Î .

k=0

(b) Ûnm Ûpq =

d−1 X

ei2πkn/d |kih(k + m)modd|

k=0

×

d−1 X

ei2πlp/d |lih(l + q)modd|

l=0

=

d−1 X

ei2π[kn+(k+m)p]/d |kih(k + m + q)modd|

k=0

= Û(n+p)(m+q) ei2πmp/d .

Entanglement 73

(5.36) First, with apologies, please note that there is an error in the normalization of the wavefunction in equation 5.63. The prefactor should be (πσ 2 )−1/2 :

(xA + xB )2 k(xA − xB + L)2 − . ψAB (xA , xB ) = (πσ 2 )−1/2 exp − 4σ 2 4kσ 2 To calculate the desired variances it is convenient to change coordinates: 1 x+ = √ (xA + xB ) 2

1 x− = √ (xA − xB + L) , 2

so that

kx2 x2+ ψAB (x+ , x− ) = (πσ 2 )−1/2 exp − − − 2σ 2 2kσ 2 R and dx+ dx− |ψ|2 = 1. From these it is straighforward to evaluate the desired variances: hx− i = 0 Z 2 hx− i = dx+ dx− x2− |ψ|2 σ2 k 2σ 2 2 ⇒ ∆(xA − xB ) = k hp+ i = 0 Z 2 ∂ψ hp2+ i = dx+ dx− −ih̄ ∂x+ 2 h̄ k = 2 4σ 2 h̄ k ⇒ ∆(pA + pB )2 = 2 .

2σ =

(5.37) It is possible to tackle this problem by working through the teleportation protocol with the state provided. An alternative (shorter) method is to note that the entangled state 5.63 provides a distribution of xA − xB and pA + pB as investigated in the preceding question. To evaluate the fidelity we need to compare the desired state with that produced by Bob. Consider first an ideal teleportation but with the values of the momentum and position shifts somewhat inaccurate. The overlay between the desired state ψa (xB ) and the state produced ψapp (xB ) is then 1/2

Z b b hψa (xB )|ψapp (xB )i = dxB exp − [(xB − x0 )2 + (xB − x00 )2 ] π 2

(p00 − p0 )xB +i h̄

b 02 b i(p00 − p0 ) 2 0 2 = exp − (x0 + x0 ) + + (x0 + x0 ) .

2 4 h̄b

74 Entanglement

To obtain the fidelity we take the modulus squared of this overlap and then integrate over a Gaussian distribution of x00 and p00 centred on x0 and p0 respectively and with variances as calculated in the preceding question. We find

Z 2kσ 2 (p0 − p00 )2 k k(x0 − x00 )2 0 0 dp0 dx0 exp − − F= πh̄ 2σ 2 h̄2 2 ×|hψa (xB )|ψapp (xB )i| k =q k .

1 b 2kσ 2 + 2b + 2 2σ 2 Note that in the limit as k → ∞ this tends to 1 as it should as the Gaussian state approaches the ideal delta-function state. In the opposite limit, k → 0, the fidelity tends to 0 as there is no correlation to work with and we might as well guess; the state space is infinitely large and a guess is overwhelmingly likely to produce a state that is orthogonal to the desired one.

iv

Quantum Information: Solutions to Exercises

Stephen M. Barnett

Quantum information processing (6.1) These can best be understood using the logical statements OR, AND and NOT: (a) A OR A = A, i.e. if A OR A are one then the value is 1 and if A OR A are zero, then the value is zero. (b)

A OR NOT A = 1, clearly one of A and NOT A must be 1.

(c)

A AND A = A. This is 1 if A = 1 and 0 if A = 0.

(d)

A AND NOT A = 0, as A and NOT A cannot both be 1.

(6.2) (a) A · B · C + A · B · C̄ = A · B · (C + C̄) =A·B (b) A · (Ā + B) = A · Ā + A · B =A·B (c) (A + B) · (A + D) · (B + C) · (B + D) = [A + C · (A + D)] · [B + C · (B + D)] = [A + C · D] · [B + C · D] = A · B + C · D · (A + B) + C · D =A·B+C·D (6.3) A + (A · B) = (A + A) · (A + B) = A · (A + B) = A A · (A + B) = (A · A) + (A · B) = A + (A · B) = A (6.4)

The simplest way is to use the truth tables. See Fig. 6.1.

6

12 Quantum information processing

Fig. 6.1

Fig. 6.2

(6.5)

From the first De Morgan theorem we have A + B = A + B.

In Fig. 6.5 in the book we see how to realize a NOT gate using a NAND. Hence we can realize this using NAND gates as depicted in Fig. 6.2.

(6.6) It is sensible, first, to try to simplify the expressions. See the solution to all parts in Fig. 6.3.

(6.7) There are a number of ways of achieving this. We depict one for each gate in Fig. 6.4. Note that the final one gives the gate iY but that this differs from Y only by an unimportant global phase.

(6.8) Yes it is possible. One way to see this is that the action of the Pauli-X gate rotates the Bloch vector through π about the x-axis. Hence a rotation through ± π2 represents the desired square-root. The following two unitary operations, when acting twice produce the Pauli-X:

e−iπ/4 eiπ/4 √ √ Î + iσ̂x Î − iσ̂x 2 2 (6.9)

Note that each the operators is Hermitian and †i Âi = Â2i = 13 Î.

Quantum information processing 13

Fig. 6.3

Hence 3 X

†i Âi = Î

i=1

as required for a valid operation.

(6.10)

Acting on just one of the qubits in the Bell state gives

3 X

1 1 1 Âi ⊗ Î|Ψ− ihΨ− |Âi ⊗ Î = |Φ− ihΦ− | + |Φ+ ihΦ+ | + |Ψ+ ihΨ+ | 3 3 3 i=1

1 = Î ⊗ Î − |Ψ− ihΨ− | .

3

If we act also on the second qubit then the mapping is simply repeated to give: 3 X i=1

Î ⊗ Âi

1 1 Î ⊗ Î − |Ψ− ihΨ− | Î ⊗ Âi = 2Î ⊗ Î + |Ψ− ihΨ− | .

3 9

(6.11) For the transformations to be unitary it suffices that they maintain the orthogonality of the states in the computational basis. If they do not then the transformation is not unitary.

14 Quantum information processing

Fig. 6.4

(a) |0i ⊗ |0i → |0i ⊗ |1i |0i ⊗ |1i → |0i ⊗ |1i |1i ⊗ |0i → |0i ⊗ |0i |1i ⊗ |1i → |0i ⊗ |0i .

Hence this is unitary. It is a combination of a Pauli-X and a SWAP. (b)

This is clearly not unitary as |0i ⊗ |0i → |1i ⊗ |0i |0i ⊗ |1i → |1i ⊗ |0i .

(c) |0i ⊗ |0i → |0i ⊗ |0i |0i ⊗ |1i → |1i ⊗ |0i |1i ⊗ |0i → |1i ⊗ |1i |1i ⊗ |1i → |0i ⊗ |1i .

Hence this is unitary. It is a combination of a CNOT and a SWAP. (d)

This is clearly not unitary as |0i ⊗ |0i → 2−1/2 (|0i ⊗ |0i + |1i ⊗ |1i) |1i ⊗ |1i → 2−1/2 (|0i ⊗ |0i + |1i ⊗ |1i) .

(e) |0i ⊗ |0i → |0i ⊗ |0i |0i ⊗ |1i → −|0i ⊗ |1i |1i ⊗ |0i → −|1i ⊗ |0i |1i ⊗ |1i → |1i ⊗ |1i .

Hence this is unitary. It is a combination of two controlled Z gates, with each qubit playing the role of the control once.

Quantum information processing 15

Fig. 6.5

(6.12) As shown in the text, this is not possible as a unitary transformation, but the logical transformation is possible classically. Hence the simplest way to implement this quantum transformation is by measuring the qubits in the computational basis and, depending on the measurement outcome, performing the associated transformation. This corresponds to a pair of effect operators: Â00 = |0ih0| ⊗ |0ih0|

Â01 = |0ih0| ⊗ |0ih1|

Â1 = |1ih1| ⊗ Î .

(6.13)

In general the answer is no; the no-cloning theorem forbids it.

(6.14)

Please see Fig 6.5.

(6.15) The transformation given in eqn 6.20 can be checked by explicit calculation for each of the possible input states but there is a quicker way, based on writing ÛCNOT in terms of the Pauli operators: 1 1 ÛCNOT = (Î + σ̂z ) ⊗ Î + (Î − σ̂z ) ⊗ σ̂x 2 2 1 1 = Î ⊗ (Î + σ̂x ) + σ̂z ⊗ (Î − σ̂x ) .

2 2 Note that interchanging σ̂x and σ̂z corresponds to interchanging the roles of the control and target qubits.

The action of the CNOT gate on a pair of qubits prepared in eigen-

16 Quantum information processing

Fig. 6.6

states of σ̂y generates a maximally entangled state. For example: 1 1 ÛCNOT (|0i + i|1i) ⊗ (|0i + i|1i) = [|0i(|0i + i|1i) − |1i(|0i − i|1i)] .

2 2 (6.16)

This transformation generates the four Bell states: 1 ÛCNOT √ (|0i + |1i) ⊗ |0i = |Φ+ i 2 1 ÛCNOT √ (|0i + |1i) ⊗ |1i = |Ψ+ i 2 1 ÛCNOT √ (|0i − |1i) ⊗ |0i = |Φ− i 2 1 ÛCNOT √ (|0i − |1i) ⊗ |1i = |Ψ− i .

2

(6.17) The control qubit must be an equally weighted superpostion of the two eigenstates of σ̂z and the target qubit must be an equally weighted superposition of the two eigenstates of σ̂x .

(6.18)

Please see Fig. 6.6

iv

Quantum Information: Solutions to Exercises

Stephen M. Barnett

Quantum information processing (6.19)

Please see Figs. 6.1 and 6.2

(6.20) To realize a controlled unitary with these resources we need the circuit depicted in Fig. 6.3. ˆ where ~a is The most general CU realizable in this was has Û = ~a · ~σ any unit vector.

(6.21)

The required transformation is |0i ⊗ |0i → |0i ⊗ |0i |0i ⊗ |1i → |0i ⊗ |1i |1i ⊗ |0i → eiα |1i ⊗ |0i |1i ⊗ |1i → eiα |1i ⊗ |1i .

So no controlled dynamics are required! All that is needed is to apply the one-qubit gate to the first (control) qubit:

1 0 .

0 eiα (6.22) One way is to build the circuit given in Fig 6.16 with the W and W† gates prepared using the circuit given in Fig 6.15. To achieve this we can follow the prescription given on p. 149 of the text with, for example, ~b = ~yˆ and ~c = 2−1/2 (~yˆ + ~zˆ) for W and ~b = ~zˆ and ~c = 2−1/2 (~yˆ + ~zˆ) for W† .

(6.23) Apologies, there is an error in the transformations given in the question. They should be: 1/2 Ûswap |00i = |00i

e−iπ/4 √ (|01i + i|10i) 2 −iπ/4 e 1/2 Ûswap |10i = √ (|10i + i|01i) 2 1/2 Ûswap |11i = |11i 1/2 Ûswap |01i =

There exist a number of ways of realizing the desired circuit. One, perhaps the simplest, is depicted in Fig. 6.4

6

12 Quantum information processing

(6.24) (a) Yes universal - applying CS twice gives a CZ and if we act in the target bit with V̂ = 2−1/2 (Î − iσ̂y ) before applying CS twice and the operator V̂ † after applying the CS twice then be get the CNOT. (b) No, not universal. The simplest way to see this is that thee gates cannot create an entangled state between initially unentangled qubits. (c)

Yes universal - see the solution to part (a).

(6.25)

Please see Fig. 6.5

(6.26)

Please see Fig. 6.6

Quantum information processing 13

Fig. 6.1

Fig. 6.2

14 Quantum information processing

Fig. 6.3

Fig. 6.4

Fig. 6.5

Fig. 6.6

iv

Quantum Information: Solutions to Exercises

Stephen M. Barnett

Quantum information processing (6.27) Let cos θ and sin θ both be positive with cos θ > sin θ. The transformation we require is in two steps: p (cos θ|0i + sin θ|1i) ⊗ |0i → cos θ 1 − tan2 θ|0i ⊗ |1i + sin θ(|0i + |1i) ⊗ |0i p 1 → √ cos θ 1 − tan2 θ(|0i + |1i) ⊗ |1i 2 √ + 2 sin θ|0i ⊗ |0i p (cos θ|0i − sin θ|1i) ⊗ |0i → cos θ 1 − tan2 θ|0i ⊗ |1i + sin θ(|0i + |1i) ⊗ |0i p 1 → √ cos θ 1 − tan2 θ(|0i − |1i) ⊗ |1i 2 √ + 2 sin θ|0i ⊗ |1i .

Hence we require a controlled unitary and a Hadamard gate as depicted in Fig. 6.1.

(6.28) The only element in the circuit that is not already either a CNOT or a single qubit gate is the controlled-controlled-Z gate. We can construct this from the Toffoli gate built from CNOTs and single qubit gates in problem (6.22) as depicted in Fig. 6.2.

(6.29) If we add a single Y gate to Fig. 6.23, acting on the uppermost qubit as the final operation then we produce the following transformation: ρ̂ → (1 − p)(1 − q)σ̂y ρ̂σ̂y + pσ̂z ρ̂σ̂z + q(1 − p)σ̂x ρ̂σ̂x .

If we set p = 13 and q = 12 then we produce the desired transformation.

(6.30)

The circuit produces the transformation |0i ⊗ |ψ in i → |0i ⊗ P̂~a |ψ in i + |1i ⊗ P̂−~a |ψ in i ,

where P̂~a (P̂−~a ) is the projector onto the +1 (-1) eigenstate of Û . Hence the circuit produces, in general, an entangled state in which the Schmidt basis is the computational basis for the first qubit and the eigenbasis of Û for the second qubit.

6

12 Quantum information processing

Fig. 6.1

Fig. 6.2

(6.31) The essential idea is that if we make a measurement and then apply an operation to another qubit then we can simply introduce a controlled gate in its place. For example, if we measure qubit 1 in the computational basis and, if we get the result 1 apply a NOT to qubit 2 and if we get the result 1 then apply a Z gate to qubit 2, then rather than make the measurement we can link qubit 1 to qubit 2 with a CNOT gate and qubit 1 to qubit 3 with a CZ gate (with the operation applied if the control qubit is in the state |0i.

(6.32) The required probability is the sum of the probabilities that no error occurs and the probability that precisely one error occurs. This is P = (1 − p)n + np(1 − p)n−1 n(n − 1) 2 ≈1− p .

2 (6.33) This form of error differs from that discussed on p. 154 by the addition of a phase shift to the bit flip |0i ↔ |1i. Hence we can use the three qubit states |0`3 i and |1`3 i and the error detection and correction scheme given in the text, followed by a single Z gate to undo the phase shift.

(6.34) First apologies; the figure 6.26 has the incorrect powers on the final X gates. The correct from of the figure is given in Fig. 6.3.

We can replace the measurements and correcting X gates with Toffoli gates as depicted in Fig. 6.4 (6.35) (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, +1, +1, +1) ⇒ |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, +1, +1, −1) ⇒ X̂1 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, +1, −1, +1) ⇒ X̂2 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, +1, −1, −1) ⇒ X̂3 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, −1, +1, +1) ⇒ X̂4 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, −1, +1, −1) ⇒ X̂5 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, −1, −1, +1) ⇒ X̂6 |ψ`7 i

Quantum information processing 13

Fig. 6.3

Fig. 6.4

(J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, +1, −1, −1, −1) ⇒ X̂7 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, −1, +1, +1, −1) ⇒ Ŷ1 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, −1, +1, +1, −1, +1) ⇒ Ŷ2 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, −1, −1, +1, −1, −1) ⇒ Ŷ3 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, +1, +1, −1, +1, +1) ⇒ Ŷ4 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, +1, −1, −1, +1, −1) ⇒ Ŷ5 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, −1, +1, −1, −1, +1) ⇒ Ŷ6 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, −1, −1, −1, −1, −1) ⇒ Ŷ7 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, +1, −1, +1, +1, +1) ⇒ Ẑ1 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, −1, +1, +1, +1, +1) ⇒ Ẑ2 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (+1, −1, −1, +1, +1, +1) ⇒ Ẑ3 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, +1, +1, +1, +1, +1) ⇒ Ẑ4 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, +1, −1, +1, +1, +1) ⇒ Ẑ5 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, −1, +1, +1, +1, +1) ⇒ Ẑ6 |ψ`7 i (J1 , J2 , J3 , K1 , K2 , K3 ) = (−1, −1, −1, +1, +1, +1) ⇒ Ẑ7 |ψ`7 i .

(6.36) The easiest way to see this is to note that the density operator produced (with a new qubit added in the sixth place prepared in a random state) can be understood as an equally-weighted mixture of four possibilities: no error, an X error, a Y error and a Z error. The error detection scheme will determine which one of these applies and allow the original state to be recovered by performing the appropriate error correction.

14 Quantum information processing

Fig. 6.5

Fig. 6.6

(6.37) This circuit is sufficiently large that I need to draw it in two sections. The error detection part is given in Fig. 6.5. The error corrections to the seven qubits can be made post-measurement, as shown in Fig. 6.6 or via controlled gates.

(6.38)

The cluster state is 1 |C4 i = (|0000i + |0001i + |0010i − |0011i 4 +|0100i + |0101i − |0110i + |0111i +|1000i − |1001i + |1010i + |1011i −|1100i + |1101i + |1110i + |1111i) .

The state is symmetric under cyclic permutation of the qubits so we can choose any qubit to measure. Consider then the effect of measuring the first qubit. Following are the states produced on finding the first qubit

Quantum information processing 15

in the indicated state. |0i :

|1i : 1 √ (|0i + |1i) : 2 1 √ (|0i − |1i) : 2 1 √ (|0i ± i|1i) : 2

1 √ (|000i + |001i + |010i − |011i 2 2 |100i + |101i − |110i + |111i) 1 √ (|000i − |001i + |010i + |011i 2 2 −|100i + |101i + |110i + |111i) 1 (|000i + |010i + |101i + |111i) 2 1 (|001i + |011i + |100i − |110i) 2 1 h √ e±iπ/4 (|000i + |010i + |101i + |111i) 2 2 i

+e∓iπ/4 (|001i − |011i + |100i − |110i) .

(6.39)

First apply a Hadamard gate to qubit 1 to give:

1 |C3 i → √ (|0i1 |0i2 |00 i3 + |1i1 |1i2 |10 i3 ) .

2 √ √ Next, measure the third qubit using the basis { 12 ( 3|00 i+|10 i), 21 ( 3|10 i− |00 i)}. If you get the first result then the remaining qubits are in the desired state. If you get the second result then apply an X gate to qubit 1 and a Y gate to qubit 2 (or the other way around).

(6.40) The state |GHZi is given in eqn 5.25. There is a clear symmetry between the four qubits in the state |C4> i, which is clear if we apply a Hadamard gate to qubits 1, 3 and 4: 1 |C4> i → √ (|0i1 |0i2 |0i3 |0i4 + |1i1 |1i2 |1i3 |1i4 ) .

2 If we measure σ̂x for any one of these qubits and get the result +1 then the remaining three qubits will be left in the desired GHZ state. If we get the result − then we can recover the desired state by applying a Z gate to any of the remaining three qubits.

iv

Quantum Information: Solutions to Exercises

Stephen M. Barnett

Quantum computation (7.1) (a) After the operation of the Turing machine the two tapes are left in the states:

(b)

.

:

0

:

0

:

1

: ∅

: ∅

:

···

.

:

0

:

1

:

0

:

: ∅

:

···

1

The program performs the function multiply by 2.

(7.2) (S, .) ⇒ (I, .), (I, 0) ⇒ (I, 0), (I, 1) ⇒ (II, 1), (II, 0) ⇒ (II, 0), (II, 1) ⇒ (I, 1), (I, ∅) ⇒ (F, 0), (II, ∅) ⇒ (F, 1) .

(7.3) (a) 0

T1 = b1 τ n = b1 τ (n + 1)

⇒

T2 = b2 τ nk = b2 τ 0 (n + 1)k

⇒

T3 = b3 τ 2kn = b3 τ 0 2k(n+1)

⇒

1 n ≈τ 1− τ =τ n+1 n

k

n k τ0 = τ ≈τ 1− n+1 n 0

τ 0 = τ 2−k .

(7.4) We are required to show that T (n) diverges more slowly than n` as n → ∞ for all positive `. The easiest way to do this is to use l’Hôpital’s rule: lim

n→∞

(7.5)

T (n) (3 log e/n) 3 log e = = = 0. n` `n`−1 n`

We can tackle both at once by writing T (n) = aT (n/2) + Θ(n) .

7

14 Quantum computation

It is immediately clear that T (n) grows at least as fast as Θ(n). If it is growing faster than this then in the limit of large n we approach T (n) = aT (n/2) , the solution of which is T (n) = bnlog a . It follows that T (n) = Θ(nlog a ) , which gives N.8 and N.11 for a = 4 and a = 3 respectively.

(7.6) If in each operation we compare two elements that have not been previously considered then the number of orderings we know about groups exponentially, i.e. as 2` . To explore all possible orderings we will need, also, to compare elements that we have previously examined, so 2` is an upper bound. To complete the problem we need to compare 2` with n!: 2` = n!

⇒

` = log n! ≈ n log n .

As this is an upper bound it follows that ` grows at most like n log n so that T (n) = Ω(n log n).

(7.7) Checking all the numbers up to the square root of an n bit number to factored requires 2n/2 trials. The number of primes less that a number N scales (for large numbers as N/ ln N , hence checking only the prime numbers up to the square root of the n bit number requires 2n/2 / ln 2n/2 ∼ 2n/2 2− log(n/2) trials which is still not polynomial in n.

(7.8) The density operator for a general n qubit state may be represented as a 2n ×2n matrix, with the diagonal elements being positive and summing to 1, and the opposite diagonal elements being the complex conjugates of each other. Hence we need to specify 22n − 1 numbers. For the state of a single qubit we need to specify only 3 numbers and hence a tensor product state of n qubits requires us to specify 3n numbers.

(7.9) Ûf =

X

|aiha| ⊗ |b ⊕ f (a)ihb|

a,b

Ûf† =

X

|aiha| ⊗ |bihb ⊕ f (a)|

a,b

⇒ Ûf Ûf† =

X

|aiha| ⊗ |bihb| = Î ⊗ Î

a,b

Ûf† Ûf =

X

|aiha| ⊗ |b ⊕ f (a)ihb ⊕ f (a)| = Î ⊗ Î ,

a,b

where the final step follows from the fact that the elements of b ⊕ f (a) give all the possible values of b.

(7.10)

Quantum computation 15

Fig. 7.1

(a) No there is no conflict as this“copying” process works only in the computational basis. (b) All that is required is a circuit in which a CNOT couples each of the qubits making up |ai to the corresponding qubit in the initially blank state |0i. This will leave 0 as 0 if the qubit in |ai has the value 0 but change it to 1 if the he qubit in |ai has the value 1.

(7.11) (a) If the f (a) are all different then |ai and |f (a)i form the Schmidt basis so that ρ̂1 is diagonal on the |ai basis: −n

ρ̂1 = 2

n 2X −1

⇒ Tr(ρ̂21 ) = 2−n .

a=0

(b) Let a0 and a00 denote numbers for which f (a) takes the same values. Hence we can write the state in the form   X X |ai ⊗ |f (a0 )i .

2−n/2  (|a0 i + |a00 i) ⊗ |f (a0 )i + a0

a∈{a / 0 ,a00 }

There are m terms in the first sum and 2n − 2m in the second. It follows that Tr(ρ̂21 ) = 2−n + m2−(2n−1) .

The two registers will be unentangled only if there is just a single value of f (a): f (a) = f ∀ a.

(7.12) The required transformation is |Ai ⊗ |0i → |Ai ⊗ |f (A)i. The simple circuits associated with these are depicted in Fig. 7.1.

(7.13) Let the input consist of n bits. There are two possibilities, either the function is constant or it is balanced. If it is constant then we need to check one more than half the possible inputs to be certain of this, i.e. 2n−1 + 1 tests. If the function is balanced then we will know this as soon as we get both possible answers. The probability that this first happens after 2 tests is 21 , after 3 tests 14 and after m tests it is

16 Quantum computation

2−(m−1) . Hence the average number of trials required is n−1

2X (n−1) 1 m̄ = × (2n−1 + 1) + m2−(m−1) + 22 (2n−1 + 1) , 2 m=2

where the final term accounts for the possibility that the function is balanced but the first 2(n−1) trials give the same result. Note that the (n−1) probability of this happening is 22 , which becomes very small even for moderate values of n. (For n = 10, for example, it is approximately 7 × 10−155 !) Hence, to an excellent approximation we have m̄ ≈ =

∞ X 1 × (2n−1 + 1) + m2−(m−1) 2 m=2

1 × (2n−1 + 1) + 3 , 2

which is clearly dominated by the vast number of trials needed for the constant case.

(7.14) The error arises if we adopt a (very reasonable) strategy of allowing only a finite number, M , of trials, concluding if all M give the same result that the function is constant. The probability of error arising in this case is the product of the probability that the function is constant multiplied by the probability that the first M trials give the same result. This error probability is =

1 × 2−(M −1) = 2−M .

2

Hence M is the smallest integer larger than − log .

M X M m2−(m−1) + 2 m=2 M −1 M 1 = + 3 − (M + 2) .

2 2

m̄ =

(7.15) (a)

Let f (a) be a balanced function. Then the overlap is −n

hconst|balancei = 2

n 2X −1

(−1)f (a) = 0

a=0

because function f (a) has an equal number of 0s and 1s. (b) No they are not. For example, if we have two balanced functions that differ at only two places in the bit string then the overlap will be 1 − 2−(n−1) .

(7.16)

We start with with each qubit in input register in the state |ini = (cos θ|0i + sin θ|1i)⊗n ⊗ (sin θ|0i − cos θ|1i) .

Quantum computation 17

We are interested in values of θ very close to π4 and so can neglect the small error in the preparation of the final qubit: 1 |ini ≈ (cos θ|0i + sin θ|1i)⊗n ⊗ √ (|0i − |1i) .

2 This is transformed into |outi =

n 2X −1

1 (−1)f (a) A(a)|ai ⊗ √ (|0i − |1i) , 2 a=0

where A(a) = cosm θ sinn−m θ for a having m 0s and n − m 1s. Hence the required conditional probability is 2 −n/2

P (result balanced|balanced) = 2

X

ha|

X

0

A(a )|a i

a0

a n X

0

n! cos θ sin −mθ =2 m!(n − m)! m=0

n 1 + sin(2θ) = , 2 −n

m

2

n

For θ = π4 + this probability becomes P (result balanced|balanced) ≈ (1 − 2 2 )n ≈ 1 − 4n 2 .

Hence for this to tend to a non-zero constant as n → ∞ we need ∼ n−1/2 .

(7.17) The Bernstein-Vazirani algorithm subject to the same errors in the preceding question performs he transfromation 1 (cos θ|0i + sin θ|1i)⊗n ⊗ √ (|0i − |1i) → 2 1 (cos θ|0i + (−1)Bn ) ⊗ · · · (cos θ|0i + (−1)B1 ) ⊗ √ (|0i − |1i) .

2 If we apply Ĥ ⊗n ⊗ I to this we find

cos θ sin θ 1 √ (|0i + |1i) + (−1)Bn √ (|0i − |1i) ⊗ · · · ⊗ √ (|0i − |1i) .

2 2 2 Hence √ the amplitude for each qubit to be in the correct state is (cos√ θ+ sin θ)/ 2 and the amplitude for it to be incorrect is (cos θ − sin θ)/ 2. (a)

Hence the probability for there to be no errors is

n 2n cos θ + sin θ 1 + sin(2θ) √ P (no errors) = . = 2 2

(b)

The probability for just one error is P (one error) = n2−n [1 + sin(2θ)]n−2 cos2 (2θ) .

18 Quantum computation

(c)

The probability for ` errors is P (` errors) =

n!

2−n [1 + sin(2θ)]n−2` cos2` (2θ) . `!(n − `)!

(7.18) (a) We submit one string, chosen at random and then a second string. The probability that the second string gives the same output as the first is 1/(2n −1). If we then choose a second number at random, the probability that it matches either of the first two is 2/(2n − 2). Carrying on in this way, the probability that the mth trial matches any of the preceding m − 1 trials is (m − 1)/(2n − (m − 1)). Adding these together we find (for large m but m n) that the probability for successfully discovering the period in m or fewer trials is approximately m2 /2n+1 .

Hence the number of trials required to give the (small) probability p of discovering the period is p m ≈ 2p2n/2 , which scales as 2n/2 as required. (b) The problem is O(2n/2 ) as the numbers selected above for testing are chosen at random. A more efficient approach would be to ensure that each number selected tested a new set of possible periods.

(7.19) Let us start with the effect of the Hadamard gate on a single qubit: 1

1 1 X Ĥ|Ai = √ (|0i + (−1)A |1i) = √ (−1)A·C |Ci .

2 2 C=0 This applies to each bit in the string and hence Ĥ

⊗n

−n/2

|ai = 2

n 2X −1

(−1)a·c |ci ,

c=0

where a · c is the sum over all the bits forming a and c: a · c =

P

i Ai · Ci .

(7.20) Apologies this question as stated is nonsense, it doesn’t even work for a one-bit string! If B = 1 and C = 1 then B · C = 1! The question should state the condition as b · c + b · c = 0 mod 2.

For the corrected condition the solution is as follows: For each bit B · C + B · C = 1 if B 6= C and B · C + B · C = 0 if B = C. Hence b · c + b · c equals the number of places where b 6= c. If this is 0(1) mod 2 then the strings have the same (opposite) parity.

(7.21)

The reduced density operator is X 1 ρ̂ = 2−(n−1) (|ai + |a ⊕ bi)(ha| + ha ⊕ b|) .

2 a∈f (a)

This is a mixture of superposition states with each state in ever superposition separated by b. Hence we can follow the procedure outlined in the

Quantum computation 19

text, applying Hadamard gates to each qubit and then measuring in the computational basis to give a single value for c for which b · c = 0 mod 2.

(7.22)

N −1 N −1 X −2πab 1 X 2πa0 c exp i ha| |a0 i exp i N a=0 N N a0 =0

N −1 X 1 2πa(c − b) = exp i N a=0 N

hb̃|c̃i =

= δbc .

(7.23)

The discrete Fourier transform is

N −1 2πab 1 X exp i F (a) .

F̃ (b) = √ N N a=0

(a)

N −1 2πab 1 X 1 1 − γ N +1 √ exp i γa = √ .

N N a=0 N 1 − γei2πb/N (b) N −1

iN −1 (N − 1)!

1 h i2πb/N 1 X i2πab/N √ e pa q N −1−a = √ pe +q . a!(N − 1 − a)!)

N a=0 N (c) T

1 1 1 X i2πab/N √ √ e =p ei2πa0 b/N T +1 N a=a0 N (T + 1) (7.24) (a)

1 − ei2πb(T +1)/N 1 − ei2πb/N

From eqn 7.31 we have

N −1 2πab 1 X exp i |aihb| ÛQFT = N N a,b=0

N −1 1 X 2π(b + c)a 2 ⇒ ÛQFT = exp i |cihb| N N a,b,c=0

=

N −1 X

| − b mod N ihb| .

b=0

(b) (7.25)

This follows directly form part (a).

Please see Figs. 7.2 and 7.3.

.

20 Quantum computation

Fig. 7.2

(7.26)

For N = 8 the QFT circuit is as depicted in Fig. 7.4.

(7.27) (a) We need to perform a Hadamard transform on each qubit, therefore we need n Hadamard gates. Further, for n qubits we require 1 + 2 + · · · n − 1 = 12 n(n − 1) controlled phase gates. (b) To correctly order the qubits we need to invert the list, hence we swap the first with the last, the second with the penultimate etc..

Hence if n is even then we require n/2 SWAP gates and if n is odd then we require (n − 1)/2: the middle qubit is already in the correct place.

(7.28) This is straighforward as the QFT is a unitary transformation.

Hence all that is required is to reverse the order of operations in the QFT circuit and replace each unitary gate by its conjugate.

(7.29) (a) N −1 X

P (b) =

b=0

N −1 X b=0

=

=

1 N2

N −1 1 X iϕ(a−a0 ) −i2πb(a−a0 )/N e e N2 0 a,a =0

N −1 X

0

eiϕ(a−a ) N δa,a0

a,a0 =0 N −1 X

1 N = 1.

N 2 a=0

(b) To consider the values of 2πb/N closest to ϕ let ϕ = 2πb/N + , where | | ≤ 2π/N is small. If we substitute this into eqn 7.46 and use

Quantum computation 21

Fig. 7.3

the fact that is small we find 1 sin2 (N /2) N 2 sin2 ( /2) ≈ sinc2 (N /2) .

P (b) =

The smallest value for the sum of the probabilities for the two nearest values to ϕ occurs when they are equidistant the required value, i.e. for = π/N . For this value we find 2 2 .

P (b) = sinc2 (π/2) = π If we add both the smaller and larger value then we find the required value of 8/π 2 ≈ 0.81.

(7.30) The trivial case is when the two occurences of f (a) are precisely half a period appart and in this case the period is simply half that originally stated. In all other cases we can proceed following eqn 7.49 adding to |a0 + mri a second ket |a1 + mri corresponding to the second appearance of f (a) in the period. Hence the, correctly normalized state in place of eqn 7.49 is N/r−1 r X (|a0 + mri + |a1 + mri) .

2N r=0

If we apply the QFT to this we find r−1

`N 1 X i2πa0 `/r e + ei2πa1 `/r 2r r `=0

.

22 Quantum computation

Fig. 7.4

A measurement will give one of b = `N/r as before but the interference between the phase factors in the sum means that some values are more likely to be found than others.

(7.31)

Approximate values suffice to make the point.

FT FFT QFT

(7.32) (a) are

n = 10 n = 100 1, 000, 000 1.6 × 1060 10, 000 1.3 × 1032 100 1.0 × 104

n = 1000 1.1 × 10602 1.1 × 10304 1.0 × 106

The numbers between 1 and N that are have p or q as factors p, 2p, · · · , (q − 1)p

q, 2q, · · · , (p − 1)q .

There are p + q − 2 of these. Hence the probability that a number chosen at random in the range is relatively prime to N is P (rel. prime) = 1 −

p+q−2 1 1 ≈1− − .

N −2 p q

(b) If the two numbers chosen at random are coprime then they have no common factors. A number chosen at random will be divisible by a much smaller prime p with probability 1/p, (only every pth number is divisible by p). Hence the probability that two numbers chosen at random will be divisible by p is 1/p2 . Hence the probability that the two numbers are coprime is

Y 1 P (coprime) = 1− 2 , pi i

Quantum computation 23

where the product covers all primes up to the square root of the smaller randomly chosen number. For large numbers this function converges rapidly to the value over all primes:

Y 1 1 , P (coprime) → 1− 2 = p ζ(2) i i where ζ(s) =

∞ X 1 n=1

ns

is the Riemann zeta function. For s = 2 the sum is π 2 /6 and hence the probability for large numbers tends to 6/π 2 .

(7.33) (a) 110 mod 133 = 1 111 mod 133 = 11 112 mod 133 = 121 113 mod 133 = 1 so the period is 3, which is not helpful as 3 is odd. (b)

After much calculating we find that the period is 24 hence 212 − 1 = 4095 = 3 × 3 × 5 × 7 × 13 212 − 1 = 4097 = 17 × 241 .

The greatest common divisor of 221 and 4095 is 13 and gcd(221,4097) = 17 and 221 = 13 × 17.

(7.34) First note that Q is a large number and is approximately equal to L/r, which means that 1 sin2 [π(rb mod L)(Q + 1)/L] l(Q + 1) sin2 [π(rb mod L)/L] 2 r sin [π(rb mod L)/r] ≈ 2 .

L sin2 [π(rb mod L)/L]

P (b) =

This takes its maximum value 1/r for rb mod L = 0 and falls off as rbmodL approaches the edges of its allowed values. For rbmodL = ±r/2 we have r 1

2 πr 2 L sin 2L r 1 > 2 2 πr L

P (b) →

2L

4 = 2 π r

24 Quantum computation

by virtue of the fact that | sin x| < x. There are r of these likely values and therefore the probability of a successful measurement exceeds r × 4/π 2 r = 4/π 2 .

(7.35) Recall that Shor’s requires two registers, one of size L and the other at least of size N , the number to be factorized. Moreover, L must be at least as large as N 2 Hence the smaller register must have a number of qubits greater than or equal to log N and the larger register must have at least twice as many. All that we need to do is to determine log N , round this up to the next highest integer and multiply by 3 to give a lower bound on the number of qubits required: RSA − 704 ≈ 7.4037 × 10187 ⇒ log10 RSA − 704 ≈ 187 log1 07.4037 187 log1 07.4037 ⇒ log2 RSA − 704 ≈ log10 2 ≈ 540.1 .

Hence we require a minimum of 3 × 541 = 1623 qubits.

(7.36) (a) D̂† = D̂

⇒ D̂† D̂ = 2|ψihψ| − Î

2|ψihψ| − Î

= 4|ψihψ| − 2|ψihψ| − 2|ψihψ| + Î = Î and hence the diffusion operator is unitary. (b) 

 N −1 N −1 X X 2 ca |ai =  ca |ai D̂ |bihb0 | − I N 0 a=0 a=0 N −1 X

b,b =0

=

=

2 N

N −1 X

b=0 N −1 X

|bi

N −1 X

ca −

a=0

N −1 X

ca |ai

a=0

(2c̄ − ca )|ai .

a=0

(7.37)

It is perhaps easiest to work from eqn. 7.84:

D̂ = Ĥ ⊗n 2|0ih0| − Iˆ⊗n Ĥ ⊗n .

The central operator gives a different sign for the state |0i compared to all the other n-qubit states. Apart from an irrelevant overall sign, this is equivalent to the circuit given in Fig. 7.5.

Quantum computation 25

Fig. 7.5

(7.38)

We require ` to be the integer closest to

π 1 θ≈ `+ 2 2 π ⇒θ≈ .

2` + 1 If we take the cosine of both sides of this equation then we find

π 2 cos θ = 1 − ≈ cos N 2` + 1 π . ⇒ ` = the closest integer to −1 2 cos 1 − N2 For N = 100 we find that this function is approximately 7.3 so we require 7 iterations. The approximate formula gives 7.8 i.e. 8 iterations. For N = 1000 we find that the function is approximately 24.3 so we require 24 iterations. The approximate formula gives 24.8 i.e. 25 iterations.

Note that in both cases the accuracy of the approximate solution is greatly improved simply by adding the 0.5 omitted in the approximation.

(7.39) If there are M elements that are acceptable then we can associate these with the kets |b1 i, |b2 i, · · · , |bM i and write the normalized superposition of these as M

1 X |a0 i = √ |bm i .

M m=1 With this identification eqn 7.87 becomes ! r r 1 M M ⊥ 1 |ψi ⊗ √ (|0i − |i) = |a0 i + 1 − |a i ⊗ √ (|0i − |i) , N N 0 2 2 which has the same p form as eqn 7.87 but √ with N replaced by N/M .

Hence require O( N/M ) rather than O( N ) operations.

(7.40) None at all! The Vernam cipher, in the absence of the one-time pad contains no information, so no information processing, be it classical or quantum, can reveal the message.

Quantum information theory (8.1) 1 1 ρ̂ = |0ih0| + (|0i + |1i)(h0| + h1|) 2 3 1 4 4 4 . = 1 1 4

4

The eigenvalues of this matrix are

1 1 λ± = 1± √ , 2 2 and it follows that the von Neumann entropy is S(ρ̂) = −λ+ log λ+ − −λ− log λ− = 0.60 bits .

(8.2) p ρ̂W = Î ⊗ Î + (1 − p)|Ψ− ihΨ− | 4

3p p |Ψ− ihΨ− | + (|Ψ+ ihΨ+ | + |Φ− ihΦ− | + |Φ+ ihΦ+ |) . = 1− 4 4 Hence the eigenvalues of ρ̂W are 1 − 3p/4, p/4, p/4 and p/4 so the von Neumann entropy is

3p 3p p 3p log 1 − − log .

S(ρ̂W ) = − 1 − 4 4 4 4 (8.3) The general density operator for a single qubit can be written in the form

1 ˆ . ρ̂ = Î + ~r · ~σ 2 ˆ i = r̂, so rz = s, and the eigenvalues of ρ̂ are 1 (1 + |~r|) and Here h~σ 2 1 r|). Hence to maximize (minimize) the von Neumann entropy we 2 (1 − |~ need to make |~r| as small (large) as possible.

8

28 Quantum information theory

(a) The maximum von Neumann entropy corresponds to |~r| = s so the corresponding density operator is

1 Î + sσ̂x . ρ̂ = 2 (b) The minimum von Neumann entropy corresponds to |~r| = 1 so the corresponding density operator is any state of the form

1 ρ̂ = Î + sσ̂x + vσ̂y + wσ̂z , 2 where v 2 + w2 = 1 − s2 . These are all pure states, of course, so the entropy is zero.

(8.4) We saw already in Chapter 1 that the entropy takes its minimum value (of 0) when one of the probabilities is 1 and its maximum value when all of the probabilities are equal. The eigenvalues of the density operator, {ρm }, are all positive or zero and sum 1 and hence are probabilities. The two conditions in eqns 8.5 and 8.6 follow directly.

(8.5)

The three density operators are

1 1 12 ρ̂1 = 1 1 2 2

1 1 − 2i ρ̂2 = i 1 2 2 ρ̂ = p1 ρ̂1 + p2 ρ̂2

1 1 1 2 (p1 − ip2 ) = , 1 1 2 2 (p1 + ip2 )

with corresponding eigenvalues λ± 1 =

1 1 ± = λ± 2 2 4

λ± =

1 ± 2

p p21 + p22 .

4

It follows that S(ρ̂) = −

1 + 2 −

! ! p p p21 + p22 p21 + p22 1 log + 4 2 4 ! ! p p p21 + p22 p21 + p22 1 1 − log − 2 4 2 4

is greater than or equal to 1 1 3 3 p1 S(ρ̂1 ) + p2 S(ρ̂2 ) = − log − log , 4 4 4 4 as the pair of probabilities are closer in value to each other in S(ρ̂). The two sides of the inequality become equal only if p21 + p22 = 1 meaning that either p1 = 0 or p2 = 0.

Quantum information theory 29

(8.6)

We start by writing the density operator in the form X ρ̂ = pi ρ̂i i

= p1 ρ̂1 + (1 − p1 )ρ̂01 , where ρ̂01 =

X i6=1

pi ρ̂i .

1 − p1

Applying the inequality given in eqn 8.12 we then have S(ρ̂) ≥ p1 S(ρ̂1 ) + (1 − p1 )S(ρ̂01 ) .

We can apply the same procedure to ρ̂01 by writing X pi p2 ρ̂2 + ρ̂i ρ̂01 = 1 − p1 1 − p1 i6=1,2

1 − p1 − p2 0 p2 ρ̂2 + ρ̂2 , = 1 − p1 1 − p1 from which we obtain p2 1 − p1 − p2 S(ρ̂2 ) + S(ρ̂02 ) 1 − p1 1 − p1 ⇒ S(ρ̂) ≥ p1 S(ρ̂1 ) + p2 S(ρ̂2 ) + (1 − p1 − p2 )S(ρ̂02 ) .

S(ρ̂01 ) ≥

Continuing in this way with ρ̂3 , ρ̂4 , · · · yields the desired result.

(8.7) (a) The Fano entropy is a monotonically decreasing function of Tr(ρ̂2 ). It follows that F has its maximum (minimum) value when Tr(ρ̂2 ) takes its smallest (largest) value. If the quantum system has a d-dimensional state space then 1/d ≤ Tr(ρ̂2 ) ≤ 1. Hence Fmax = log d and Fmin = 0. (b) A simple sketch shows that − log x is a convex function and it follows that − log[λx1 + (1 − λ)x2 ] ≤ −λ log x1 − (1 − λ) log x2 . " # X X ⇒ − log pi xi ≤ − pi log xi . i

i

The required inequality follows on setting xi = pi , where the pi are the eigenvalues of ρ̂.

(8.8) First note that ρ̂1 is proportional to the identity operator and this makes it straightforward to evaluate the eigenvalues of ρ̂1 . ρ̂2 and ρ̂ = p1 ρ̂1 + p2 ρ̂2 . The eigenvalues of ρ̂1 and both 21 , those of ρ̂2 are √ √ 1 5 and hence those of ρ̂ and 12 1 ± p2 35 . Hence the three 2 1± 3

30 Quantum information theory

quantities we require are √ ! √ !

5 5 log 1 + pi S(ρ̂i ) = p1 − p2 1+ 3 3 i √ ! √ !# 5 5 + 1− log 1 − 3 3 ! √ ! √ !

X 5 5 1 1 S pi ρ̂i = − 1 + p2 log 1 + p2 2 3 2 3 i √ ! √ !

1 5 1 5 − 1 − p2 log 1 − p2 2 3 2 3 " √ ! √ !

X X 5 5 log 1 + pi S(ρ̂i ) − pi log pi = p1 − p2 1+ 3 3 i i √ ! √ !# 5 5 + 1− log 1 − 3 3 "

X

−p1 log p1 − p2 log p2 .

The simplest way to proceed might be to plot these, together with the final quantity on the right-hand side of the inequality as functions of p1 remembering, of course, that p2 = 1 − p1 .

(8.9) Let ρ̂ be a density operator formed by removing the off-diagonal elements of σ̂ in a chosen basis so that X σ̂ = ρnm |λn ihλm | m,n

ρ̂ =

X

ρmm |λm ihλm | .

m

It follows that the relative entropy is S(σ̂||ρ̂) = Tr [σ̂ log σ̂ − σ̂ log ρ̂] X = −S(σ̂) − hλm |σ̂|λm i log ρmm m

=−

X

ρmm log ρmm − S(σ̂) .

m

The required inequality then follows from the positivity of the relative entropy.

(8.10) (a) The relative quantum entropy will diverge if ρ̂ has a zeroeigenvalue eigenvector |ψi (or more than one) for which hψ|σ̂|ψi is not zero. Then the trace of −σ̂ log ρ̂, when evaluated in a basis that includes |ψi will include the divergent term hψ| − σ̂ log ρ̂|ψi = −hψ|σ̂|ψi log 0 .

Quantum information theory 31

(b) No it does not. The fact that ρ̂ has a zero eigenvalue and, moreover, that there is a non-zero overlap between σ̂ and this eigenvalue does not mean that the converse is true.

(8.11) Recall that x log x is a convex function as is, necessarily x ln x.

Therefore Klein’s inequality tells us that h i Tr σ̂ ln σ̂ − ρ̂ ln ρ̂ − (σ̂ − ρ̂)(ln ρ̂ − Î) ≥ 0 ⇒ Tr [σ̂ ln σ̂ − σ̂ ln ρ̂)] ≥ 0 , which is the required result.

(8.12) (a) Consider the variation of the following quantity with respect to σ̂ constrained by the fact that its trace must be 1: S̃e = Tr [σ̂ ln σ̂ − σ̂ ln ρ̂)] + λ(Trσ̂ − 1) nh i o ⇒ dS̃e = Tr ln σ̂ + Î − ln ρ̂ + λ dσ̂ .

It follows that for the extremum (which is clearly the minimum value) ln σ̂ and ln ρ̂ can differ only by a constant. This means that σ̂ ∝ ρ̂. As both density matrices must have unit trace this means that σ̂ = ρ̂. (b) Consider the variation of the following quantity with respect to ρ̂ constrained by the fact that its trace must be 1: S̃e = Tr [σ̂ ln σ̂ − σ̂ ln ρ̂)] + λ(Trρ̂ − 1) −1

σ̂ ρ̂ + λ dρ̂ ,

⇒ dS̃e = Tr

where the trace is limited to the space of non-zero overlap with ρ̂ so that the inverse is properly defined. For this to be zero for all variations we require σ̂ ∝ ρ̂ which again gives σ̂ = ρ̂.

(8.13) (a)

This is done in the text: eqns 8.12 to 8.15.

(b) There is a subtlety involved in evaluating the derivative of the logarithm and this is the ordering of the operators in the general case in which ρ̂ and σ̂ do not commute. We shall see, however, that this is not an issue for the problem as posed. First let us work with Se rather than S to make differentiation easier: Sp = log e {Se [pσ̂ + (1 − p)ρ̂] − pSe (σ̂) − (1 − p)Se (ρ̂)} .

Near to p = 0 we can write ln[pσ̂ + (1 − p)ρ̂)] = ln[ρ̂ + p(σ̂ − ρ̂) ≈ ln ρ̂ + pρ̂−1 (σ̂ − ρ̂) .

Note that we could have used an anticommutator so that the operator is Hermitian and therefore real-valued but, as we shall see, the ordering of the operators is not an issue because the trace that we need to take

32 Quantum information theory

in the calculation is unchanged by cyclic permutation of the operators.

The derivative at p = 0 is

dSp = log e −Tr(σ̂ ln ρ̂) − Se (ρ̂) − Tr[ρ̂(ln ρ̂ + ρ̂−1 (σ̂ − ρ̂)] dp p=0 −Se (σ̂) + Se (ρ̂)} = −Tr(σ̂ log ρ̂) = S(σ̂||ρ̂) + S(ρ̂) , which is greater than or equal to zero as it must be because Sp=0 = 0 and Sp ≥ 0.

For p = 1 we simply need to note that changing p to 1 − p corresponds to interchanging ρ̂ with σ̂. Note, however, that this also changes the sign of the derivative with respect to p. Hence dSp = −S(ρ̂||σ̂) − S(σ̂) , dp p=1 which is less than or equal to zero as it must be because Sp=1 = 0 and Sp ≥ 0.

(8.14) (a) The two qubits, A and B, are in a pure state so S(A, B) = 0. The reduced density operators are ρ̂A = 21 Î = ρ̂B hence S(A) = 1 = S(B). (b) This is a pure state of three qubits hence to evaluate S(A, B) we need to find ρ̂AB : ρ̂AB = TrC (|GHZihGHZ|) 1 = (|00ih00|AB + |11ih11|AB ) 2 1 ⇒ ρ̂A = Î = ρ̂B .

2 Hence S(A, B) = 1 = S(A) = S(B). (c) For this Werner state the reduced density operator for the A and B qubits is ρ̂AB = TrC (|WihW|) = |α|2 (|00ih00|)AB + (β|01i + γ|10i)(β ∗ h01| + γ ∗ h10|) , the eigenvalues of which are clearly |α|2 and 1 − |α|2 so S(A, B) = −|α|2 log |α|2 − (1 − |α|2 ) log(1 − |α|2 ) .

To evaluate S(A) and S(B) we require the corresponding density operators: ρ̂A = TrB ρ̂AB = (|α|2 + |β|2 )|0ih0| + |γ|2 |1ih1|

Quantum information theory 33

⇒ S(A) = −|γ|2 log |γ|2 − (1 − |γ|2 ) log(1 − |γ|2 ) ρ̂B = TrA ρ̂AB = (|α|2 + |γ|2 )|0ih0| + |β|2 |1ih1| ⇒ S(B) = −|β|2 log |β|2 − (1 − |β|2 ) log(1 − |β|2 ) .

(8.15)

3p 3p 3p p S(A, B) = − 1 − log 1 − − log 4 4 4 4 1 ρ̂A = Î = ρ̂B ⇒ S(A) = 1 = S(B) .

2 Hence: (a)

We require S(A, B) = 2

⇒

p = 1,

corresponding to the fully mixed product state 41 Î ⊗ Î. (b)

We require S(A, B) = 0

⇒

p = 0,

corresponding to the maximally entangled pure state |Ψ− i.

(8.16)

Applying the formulae derived in question (8.15) we have

3p 3p 3p p − 1− log 1 − − log ≥ 1 .

4 4 4 4

The condition for the equality can be found numerically. The value I found was p ≈ 0.252, which is much less that 23 . Any value of p in excess of this critical value will satisfy the inequality. For p = 23 we find S(A, B) = 1 + 12 log 3 ≈ 1.79 which is greater than 1: unentangled states must satisfy the inequality, but some entangled states will also satisfy it.

(8.17) (a) S(A : B) = 0 means that the two systems are not correlated, i.e. ρ̂AB = ρ̂A ⊗ ρ̂B . (b) S(A : B) = 2Inf(S(A), S(B))

⇒

S(A, B) = |S(A) − S(B)| .

(8.18) To maximize S(A : B) we seek a pure state for which S(A) and S(B) take the maximum value allowed. If there is a total of n quanta then the maximum values for S(A) and S(B) will be log n for which each number of quanta in each oscillator has the same probability of occurring. A pure state with tis property is n

1 X √ |mi ⊗ |n − mi . n m=0

34 Quantum information theory

and this maximizes S(A : B).

(8.19) (a) We can use an adaptation of the maximum entropy method from Chapter 1 to first find the maximum entropy for the reduced states of each of the oscillators give the constraints Trρ̂ = 1 and Tr(ρ̂n̂) = n̄/2, where n̂ is the number operator. To this end we perform the variation S̃e = −Tr(ρ̂ ln ρ) + λ[Trρ̂ − 1] + α[Tr(ρ̂n̂) − n̄/2] ⇒ dS̃e = Tr[(− ln ρ̂ − 1 − λ − αn̂)dρ̂] = 0

n̂ 1 n̄ ⇒ ρ̂ = 1 + n̄ 2 + n̄

2 n̄ n̄ n̄ n̄ log 1 + − log . ⇒ Se = 1 + 2 2 2 2 If we can find a pure state for which each of the individual modes has this entropy then this will clearly give the maximum value for S(A : B).

Such a state is n/2 ∞

n̄ −1/2 X n̄ |ψi = 1 + einθ |n, ni , 2 2 + n̄ n=0 where the state |n, ni has n quanta in each oscillator and θ is any phase. (These states are referred to, in quantum optics, as the twomode squeezed states.) (b) For this problem we need only note that if the first oscillator is in a maximum entropy state with a mean number of quanta m̄ and the second is also in a maximum entropy state with a mean number of quanta n̄ − m̄, then the total entropy for the two individual oscillators will be S(A) + S(B) = (1 + m̄) log(1 + m̄) − m̄ log m̄ + (1 + n̄ − m̄) log(1 + n̄ − m̄) + (n̄ − m̄) log(n̄ − m̄) .

Differentiating with respect to m̄ to find the maximum we arrive at m̄ = n̄/2. As there is a pure state with these maximum-entropy reduced states, this is indeed the state with maximum S(A : B), subject to the constraint.

(8.20) Let ρ̂ABC be a pure state such that ρ̂AB = TrC ρ̂ABC . This is always possible by making the three-system state a purification of ρ̂AB .

It follows that S(B) ≤ S(AB) + S(BC) ⇒ S(AB) ≥ S(B) − S(BC) .

As ρ̂ABC is a pure state it necessarily follows, via the Schmidt decomposition, that the eigenvalues of ρ̂BC are the same as those of ρ̂A and hence that S(BC) = S(A). It follows that S(AB) ≥ S(B) − S(A) .

Quantum information theory 35

There is nothing special about the choice of system A and system B and hence it must also be true that S(AB) ≥ S(A) − S(B) ⇒ S(AB) ≥ |S(A) − S(B)| .

(8.21) Apologies the inequality in the question is the wrong way round!

For a pure state ρ̂ABCD we have S(ABC) = S(D) and S(BC) = S(AD).

Substituting these into eqn 8.46 gives S(D) + S(B) ≤ S(AB) + S(AD) .

Relabelling D → C gives S(C) + S(B) ≤ S(AB) + S(AC) .

(8.22) (a)

1 ˆ |0i Î + ~r · ~σ 2 1 + rz = .

2

F = h0|

(b) 1 1 1 F = hψ| Î|ψi = hψ|ψi = .

2 2 2 (8.23) (a) F = hΨ− |

p

Î ⊗ Î + (1 − p)|Ψ− ihΨ− | |Ψ− i

4 p 3p . = +1−p=1− 4 4

(b) p

F = h01| Î ⊗ Î + (1 − p)|Ψ− ihΨ− | |01i 4 p = + (1 − p)|hΨ− |01i|2 4 p 1−p 2−p = + = .

4 2 4 (c) p

F = h00| Î ⊗ Î + (1 − p)|Ψ− ihΨ− | |00i 4 p = .

4

36 Quantum information theory

(8.24) (a)

Let ρ̂0 = Û ρ̂Û † and σ̂ 0 = Û σ̂ Û † . Then p 2 F (ρ̂0 , σ̂ 0 ) = Tr ρ̂01/2 σ̂ 0 ρ̂01/2 2 q 1/2 † † 1/2 † = Tr Û ρ̂ Û Û σ̂ Û Û ρ̂ Û q 2 1/2 1/2 † = Tr Û ρ̂ σ̂ ρ̂ Û

Note that if ρ̂01/2 = Û ρ̂1/2 Û † looks a bit strange, try squaring it.

This also means that 2 p Û M̂ Û † = Û M̂ 1/2 Û †

2 p ⇒ F (ρ̂0 , σ̂ 0 ) = TrÛ ρ̂1/2 σ̂ ρ̂1/2 Û † p 2 = Tr ρ̂1/2 σ̂ ρ̂1/2 = F (ρ̂, σ̂) , by virtue of the cyclic property of the trace. (b) The simplest way is to note that F is symmetric in σ̂ and ρ̂ (see the following exercise), which means that i2 h p |ψihψ|ρ̂|ψihψ| F = Tr = hψ|ρ̂|ψiTr(|ψihψ|) = hψ|ρ̂|ψi .

(8.25) Recall that the eigenvalues {λi } are the solutions of the characteristic equation:

det ρ̂1/2 σ̂ ρ̂1/2 − λi Î = 0 .

The determinant of a product of matrices is the product of the corresponding determinants and therefore:

det ρ̂1/2 σ̂ ρ̂1/2 − λi Î = det ρ̂1/2 σ̂ ρ̂1/2 − λi Î det(Î)

= det ρ̂1/2 σ̂ ρ̂1/2 − λi Î det σ̂ −1/2 ρ̂−1/2 ρ̂1/2 σ̂ 1/2

= det σ̂ −1/2 ρ̂−1/2 ρ̂1/2 σ̂ ρ̂1/2 ρ̂1/2 σ̂ 1/2 − λi Î

= det σ̂ 1/2 ρ̂σ̂ 1/2 − λi Î .

It follows that ρ̂1/2 σ̂ ρ̂1/2 and σ̂ 1/2 ρ̂σ̂ 1/2 have the same eigenvalues and hence so do their square roots, and the fidelities are therefore the same.

Quantum information theory 37

(8.26) If the density operators commute then it necessarily follows that they have a complete set of common eigenvectors and we can write X ρ̂ = pi |iihi| σ̂ =

i X

qi |iihi|

i

⇒ ρ̂1/2 σ̂ ρ̂1/2 =

X

pi qi |iihi|

i

⇒

p

ρ̂1/2 σ̂ ρ̂1/2 =

X√

pi qi |iihi|

i

⇒ F (P, Q) =

X√

! pi qi

.

i

Clearly the minimum value is 0, which occurs when the sets of non-zero probabilities {pi } and {qi } are mutually exclusive. The maximum value occurs when pi = √ qi . The quickest way to prove this is to use Lagrange’s method. Let f = F . !

X√ X f˜ = pi qi + µ pi − 1 i

⇒ df˜ =

X 1 r qi i

2

p1

i + µ dpi .

Hence the extremum (maximum) occurs for pi ∝ qi and normalization then fixes pi = qi .

(8.27)

We start by noting that for any 2 × 2 matrix A we have

1 (TrA)2 − Tr(A2 ) 2 ⇒ (TrA)2 = Tr(A2 ) + 2detA . detA =

If we set A =

p

ρ̂1/2 σ̂ ρ̂1/2 then we find

p F (ρ̂, σ̂) = Tr ρ̂1/2 σ̂ ρ̂1/2 + 2det ρ̂1/2 σ̂ ρ̂1/2 p = Tr (ρ̂σ̂) + 2 detρ̂1/2 σ̂ ρ̂1/2 p = Tr (ρ̂σ̂) + 2 det(ρ̂)det(σ̂) .

(8.28) 1 + αβ ∗ + α∗ β 2 1 det (ρ̂) = − |α|2 4 det (σ̂) = p(1 − p) − |β|2

Tr (ρ̂σ̂) =

1 ⇒ F = + αβ ∗ + α∗ β + 2 2

s

1 − |α|2 (p(1 − p) − |β|2 ) .

4

38 Quantum information theory

(a) (i) To minimize the fidelity we want p(1 − p) to be as small as possible. The smallest value this can take, which with is consistent σ̂ p 1 2 2 being a positive operator is |β| for which p = 2 1 ± 1 − 4|β| . (ii) To maximize the fidelity we want p(1 − p) to be as large as possible, which occurs for p = 12 . (b) (i) To minimize the fidelity, first we choose the argument of β such that αβ ∗ is real and negative. Next we set |β|2 = p(1 − p). (ii) To maximize the fidelity we choose the argument of β such that αβ ∗ is real and positive. This leaves the fidelity in the form r 1 1 F = + 2|α||β| + 2 ( − |α|2 )(p(1 − p) − |β|2 ) .

2 4 Differentiating with respect to |β| and setting this derivative to zero gives

1 2 p(1 − p) 2 4 − |α| , |β| = 1 + 14 − |α|2 which is physically acceptable as |β|2 < p(1 − p).

(8.29) We start by expressing our density operators in terms of the Pauli operators: 1 ρ̂ = Î + α0 σ̂x − α00 σ̂y 2

1 1 σ̂ = Î + + p − σ̂z + β 0 σ̂x − β 00 σ̂y , 2 2 where α = α0 + iα00 (and similarly for β). Hence, from eqn 8.63 p D(ρ̂, σ̂) = (p − 1/2)2 + |α − β|2 . (a) (i) The distance is clearly minimized for p = 21 . (ii) The distance is maximized when either p or 1 − p takes its maximum possible value, which is |β|. (b) (i) The minimum distance occurs when β = α. If this is not possible (because this value does not correspond to a positive density operator) then the distance is minimzed for arg α = arg β and |β|2 = p(1 − p). (ii) The maximum distance occurs for arg α = arg β + π and |β|2 = p(1 − p).

(8.30) Let P̂ be the projector on to the space of positive-eigenvalue eigenstates of ρ̂ − σ̂. It follows that 1 D(ρ̂, σ̂) = Tr |ρ̂ − σ̂| 2 1 1 = Tr[P̂ (ρ̂ − σ̂)] − Tr[(Î − P̂ )(ρ̂ − σ̂)] 2 2 = Tr[P̂ (ρ̂ − σ̂)] .

Quantum information theory 39

It follows that (with a different projector) we can write " " ! # # X X X X D pi ρ̂i , qi σ̂i = Tr P̂ pi ρ̂i − Tr P̂ qi σ̂i i

i

= ≤

X

i i h i X pi Tr P̂ (ρ̂i − σ̂i ) + (pi − qi )Tr[P̂ σ̂i ]

i X

i X pi D(ρ̂i , σ̂i ) + (pi − qi )Tr[P̂ σ̂i ] ,

i

i

where the equality holds only if P̂ is the projector onto the positiveeigenvalue eigenstates of each of the ρ̂i − σ̂i . To complete the proof we write

X X 1 (pi − qi )Tr[P̂ σ̂i ] = Tr (pi − qi )P̂ + (qi − pi ) σ̂i 2 i i

X 1 |pi − qi |σ̂i ≤ Tr 2 i = D(P, Q) , where the equality holds only if the space of states projected by P̂ corresponds to those situations in which pi ≥ qi . Putting these two inequalities together completes the proof.

(8.31)

The transformation changes the density operators as: 1 ˆ ) → 1 (Î + ~r0 · ~σ ˆ) ρ̂ = (Î + ~r · ~σ 2 2 1 ˆ ) → 1 (Î + ~s0 · ~σ ˆ) , σ̂ = (Î + ~r · ~σ 2 2

where we must be careful which σ corresponds to a density operator and which to a Pauli matrix! Here rx0 s0 = x = (1 − 2py − 2pz ) = ax rx sx ry0 s0y = = (1 − 2pz − 2px ) = ay ry sy rz0 s0 = z = (1 − 2px − 2py ) = az . rz sz It follows that the operation changes the trace distance from q 1 1 |~r − ~s| = (rx − sx )2 + (ry − sy )2 + (rz − sz )2 2 2 to 1 0 1q 2 |~r − ~s0 | = ax (rx − sx )2 + a2y (ry − sy )2 + a2z (rz − sz )2 2 2 1 ≤ |~r − ~s| 2

40 Quantum information theory

as the factors ai are constrained to have modulus less than or equal to 1 in order to represent a physical operation.

(8.32) Any two pure states span a 2-dimensional state space and hence we can write them, in general, in the form |ψi = cos θ|Ai + sin θ|Bi |φi = cos θ|Ai − sin θ|Bi , where |Ai and |Bi are a pair of orthonormal state vectors. It follows that |ψihψ| − |φihφ| = sin(2θ)(|BihA| + |AihB|) ⇒ D(|ψi, |φi) = | sin(2θ)| .

For the fidelity we have F (|ψi, |φi) = |hψ|φi|2 = cos2 (2θ) √ ⇒D= 1−F .

(8.33) The probability for mistakenly identifying a probability distribution P as Q is approximately P (Q|P ) ≈ 2−N H(Q||P ) .

Hence we require the relative entropies for the two probability distribu3 3 3 3 3 1 tions P1 = { 16 , 16 , 61 , 16 , 16 , 16 } and P2 = { 20 , 20 , 20 , 20 , 20 , 4 }: For these probability distributions we find 1 5 3 1 1 − log − log 6 6 20 6 4 = 0.029 3 3 1 1 1 H(P2 ||P1 ) = log + log − log 4 20 4 4 6 = 0.322 .

H(P1 ||P2 ) = log

(a) To be confident with the given probability that the die is fair we require roughly N rolls: 2−N H(P2 ||P1 ) ≈ 10−6 log 106 ⇒N≥ = 619 .

H(P2 ||P1 ) (b) To have the same confidence that the die is loaded we require roughly N rolls where now: 2−N H(P1 ||P2 ) ≈ 10−6 log 106 = 690 . ⇒N≥ H(P1 ||P2 )

Quantum information theory 41

(8.34) (a) S(ρ̂A ⊗ ρ̂B ||σ̂A ⊗ σ̂B ) = TrAB [ρ̂A ⊗ ρ̂B (log(ρ̂A ⊗ ρ̂B )) − log(σ̂A ⊗ σ̂B )] = TrAB [ρ̂A ⊗ ρ̂B (log(ρ̂A ) ⊗ IB + ÎA ⊗ log(ρ̂B ) − log(σ̂A ) ⊗ IB + ÎA ⊗ log(σ̂B ))] = TrA [ρ̂A (log ρ̂A − log σ̂A )] + TrB [ρ̂B (log ρ̂B − log σ̂B )] = S(ρ̂A ||σ̂A ) + S(ρ̂B ||σ̂B ) . (b) S(pσ̂1 + (1 − p)σ̂2 ||ρ̂) = Tr[(pσ̂1 + (1 − p)σ̂2 )(log(pσ̂1 + (1 − p)σ̂2 ) − log ρ̂)] = −Tr[(pσ̂1 + (1 − p)σ̂2 ) log ρ̂] − S(pσ̂1 + (1 − p)σ̂2 ) ≤ −Tr[(pσ̂1 + (1 − p)σ̂2 ) log ρ̂] − pS(σ̂1 ) − (1 − p)S(σ̂2 ) = pS(σ̂1 ||ρ̂) + (1 − p)S(σ̂2 ||ρ̂) , where we have used eqn 8.16. (c) X

pi S(ρ̂i ||σ̂) =

i

X

pi Tr[ρ̂i (log ρ̂i − σ̂)]

i

=

X

pi Tr[ρ̂i (log ρ̂i − log ρ̂)] + Tr[ρ̂(log ρ̂ − log σ̂)]

i

=

X

pi S(ρ̂i ||ρ̂) + S(ρ̂||σ̂) .

i

(8.35) h i F = Inf |ψi |hψ|σ̂z (1 − p − q)1/2 σ̂z |ψi|2 + p|hψ|σ̂z σ̂x |ψi|2 + q|hψ|σ̂z σ̂y |ψi|2

= Inf |ψi (1 − p − q) + p|hψ|σ̂y |ψi|2 + q|hψ|σ̂x |ψi|2 =1−p−q, which occurs when |ψi is an eigenstate of σ̂z .

(8.36) We can write any single-qubit unitary operator in the form ˆ , where ~a is a unit vector. If we write the unitary cos θÎ + i sin θ~a · ~σ † 0 operator Û Û in this form and insert this into our expression for the gate fidelity then we find the necessary minimum value by choosing as state ˆ is zero. The corresponding value for which the expectation value of ~a · ~σ 2 of the gate fidelity is cos θ. Consider a general maximally entangled state, which we can write in the form |φi = 2−1/2 (|0i|bi) + |1i|bi, where |ai and |bi are a pair of orthonormal states. For this state we find hφ|Û † Û 0 ⊗ Î|φi

2

1 = Tr Û † Û 0 2 = cos2 θ = F .

42 Quantum information theory

(8.37)

The eigenvalues of |ψihψ| − ρ̂ are q λ± = ± sin2 θp + cos2 θ(1 − p) − p(1 − p) q ⇒ D = sin2 θp + cos2 θ(1 − p) − p(1 − p) .

Differentiating to find the minimum gives p = cos2 θ as required. Note that this is reasonable as for this value the diagonal elements of both states (in the computational basis) are the same.

(8.38) F = hψ|ρ̂|ψi = p cos2 θ + (1 − p) sin2 θ .

Differentiating with respect to p does not give a maximum therefore the maximum value lies at the edge of the range, either p = 0 or p = 1, with the value depending on the value of θ. The maximum occurs for p = 0 if sin2 θ > cos2 θ and for p = 1 if cos2 θ > sin2 θ.

(8.39) We start by noting that the state |Ψ− i is orthogonal to all product pure states in which the two qubits are in the same state, such as |00i. Hence we star by considering product pure states for which the two qubits are in orthogonal states, such as |01i. (It doesn’t matter wihich of these we choose as the Bell state |Ψ− i is invariant (up to a global phase) under the action of Û ⊗ Û . It remains only to determine whether the minimum distance and maximum fidelity occur only for a pure state or occur, also, for a mixture of such states. (a) Let us consider a mixed state (chosen by considering the Schmidt decomposition of the entangled state) of the form ρ̂ = p|0ih0| ⊗ |1ih1| + (1 − p)|1ih1| ⊗ |0ih0| .

We can obtain the distance as half the sum of the moduli of eigenvalues of ρ̂ − |Ψ− ihΨ− |. This simple procedure gives r 1 D = p(p − 1) + , 2 so that the minimum distance occurs for p = 12 corresponding to an equally weighted mixture of |01i and |10i. Note that an equally weighted mixture of any pair of states in which the two qubits are in orthogonal stats will give the same distance of 21 . (b) The fidelity will clearly be maximized if ρ̂ is a pure product state in which the two qubits are prepared in mutually orthogonal states, for example |01i. Because the fidelity is linear on ρ̂ and any pure state in which the two qubits are in orthogonal states will maximize the fidelity, it follows that the fidelity is maximized for ρ̂ being any mixture of such states.

Quantum information theory 43

(8.40) To some extent this is open to discussion, but there is one clear justification that can be made. Recall form the discussion on p.

209 that H(P ||Q) is associated with the probability for identifying the probability distribution as P when it was, in fact, Q. Hence we can associate S(ρ̂unent ||ρ̂) with the probabiity that we mistakenly interpret a genuinely entangled state as an unentangled one. More important, of course, is whether a close unentangled state can pass as the desired entangled state and the probability for this is related to (ρ̂||ρ̂unent ).

(8.41) There is a formal proof of this in the paper “Entanglement measures and purification processes” by Vedral and Plenio, Phys. Rev. A 57, 1619 (1998). A simpler (if perhaps less rigorous) approach is as follows. We start by writing our entangled pure state in its Schmidt basis: X√ pn |n, ni |Ψi = n

⇒ ERE (|Ψi) = −Inf ρ̂unent |hΨ| log ρ̂unent |Ψi X√ pn pm hm, m| log ρ̂unent |n, ni . = m,n

The only basis of states specified in this expression is the Schmidt basis and for this reason we write ρ̂unent in terms of this basis: X ˆunent ρ̂unent = q` |`, `ih`, `| + ρ̃ ` unent

ˆ where ρ̃ is orthogonal to the first part of this expression. From this we note two features: (i) if any of the qn = 0 for which pn 6= 0 then the logarithm diverges and hence this is not possible in the required minˆunent is present then there must be a corresponding imum and (ii) If ρ̃ decrease in the q` so that Trρ̂unent = 1, and this decrease must correspondingly increase the logarithm of the probabilities. This reasoning leads us to the form: X ρ̂unent = q` |`, `ih`, `| `

⇒⇒ ERE (|Ψi) = −Inf {qn }

X

pn log qn

n

=−

X

pn log pn = S(A) = S(B) ,

n

where we have used the positivity of the relative entropy.

(8.42) (a) 2 1 P (ai , bi ) = hψi |π̂i |ψi = 3 9 1 1 P (ai , bj )i6=j = hψi |π̂j |ψi = 3 18

44 Quantum information theory

⇒ H(A, B) = −3 ×

2 2 1 1 log − 6 × log 9 9 18 18

1 = − log 2 + 2 log 3 3 H(A) = log 3 = H(B) 1 ⇒ H(A : B) = ≈ 0.333 .

3 (b) P (ai , bi ) = 0 1 P (ai , bj )i6=j = 6 1 1 log = log 3 + 1 6 6 H(A) = log 3 = H(B)

⇒ H(A, B) = −6 ×

⇒ H(A : B) = log 3 − 1 ≈ 0.585 . (c) Let us choose, for example, |ψ1 i and |ψ2 i. As noted in the text, the measurement that maximizes the mutual information in this case is simply the one that also minimizes the probability of error. For the two states selected this corresponds to the two measurement operators: 1 (|0i + |1i) (h0| + h1|) 2 1 π̂2 = (|0i − |1i) (h0| − h1|) .

2 π̂1 =

Hence we have

√ 1 2+ 3 P (ai , bi ) = hψ1 |π̂1 |ψ1 i = 2 √ 8 2− 3 P (ai , bj )i6=j = 8 √ √ √ √ 2+ 3 2+ 3 2− 3 2− 3 ⇒ H(A, B) = − log − log 4 √ 8 4 √ 8 √ √ 2− 3 2+ 3 log(2 + 3) − log(2 − 3) =3− 4 4 H(A) = 1 = H(B) √ √ √ √ 2+ 3 2− 3 ⇒ H(A : B) = log(2 + 3) + log(2 − 3) − 1 4 4 ≈ 0.671 .

It is believed that this is the maximum allowed value for signally based on the trine ensemble.

(8.43)

For the correlated state given in eqn 8.115 we have   X X S(A, B) = −Tr  P (ai )|ai ihai | ⊗ ρ̂Bi log P (aj )|aj ihaj | ⊗ ρ̂Bj  i

j

Quantum information theory 45

! =− =−

X

X

TrB

P (ai )ρ̂Bi log P (ai )ρ̂Bi

i

i

X

P (ai ) log P (ai ) −

X

i

ρ̂A =

X

P (ai )TrB (ρ̂Bi log ρ̂Bi )

i

P (ai )|ai ihai |

i

⇒ S(A) = −

X

P (ai ) log P (ai )

i

ρ̂B =

X

P (ai )ρ̂Bj

i

  X X ⇒ S(B) = −Tr  P (ai )ρ̂Bi log P (aj )ρ̂Bj  i

j

  X X ⇒ S(A : B) = −Tr  P (ai )ρ̂Bi log P (aj )ρ̂Bj  i

j

+

X

P (ai )TrB (ρ̂Bi log ρ̂Bi )

i

! =S

X i

P (ai )ρ̂Bi

−

X

P (ai )S (ρ̂Bi ) .

i

(8.44) Apologies, the two states should, of course, have been ρ̂1 = q|0ih0| + (1 − q)|1ih1| ρ̂2 = (1 − q)|0ih0| + q|1ih1| .

Let us select these with probabilities p1 = p and p2 = 1 − p respectively.

This gives !

X S pi ρ̂i = −(1 − p − q + 2pq) log(1 − p − q + 2pq) i

−(p + q − 2pq) log(p + q − 2pq) S(ρ̂1 ) = −q log q − (1 − q) log(1 − q) = S(ρ̂2 ) It follows that !

X X S pi ρ̂i − pi S (ρ̂i ) = −(1 − p − q + 2pq) log(1 − p − q + 2pq) i

i

−(p + q − 2pq) log(p + q − 2pq) +q log q + (1 − q) log(1 − q) .

The maximum value of this occurs for p = 21 ; substituting this value into this expression leads the sum of the first two terms to be 1 bit, which is the global maximum. This gives, finally, the channel capacity C = χ = 1 − H(q) ,

46 Quantum information theory

which is the channel capacity found in eqn 1.82. This should not surprise as our two signal states commute and form, in this sense, a classical channel.

(8.45) This might make an interesting group discussion. The key point is that in quantum dense coding, the state available to Alice and Bob for their communication is not simply the state of the qubit sent from Alice to Bob but rather the combined state of this qubit with its entangled partner retained by Bob. Another point is that we should also consider in the communication the process of the initial sharing the ebit.

(8.46) (a) If she has only one bit at her disposal then the best she can do is to select one of the bits at random and send this to Bob. The probability that this gives the correct result is 1 if the correct bit is sent and 12 if it is not, as Bob then has to guess: PC =

1 1 1 2 +2× × = .

3 3 2 3

Purely for interest, if Bob has to choose among n bits to send, the 1 probability Bob will get the correct bit is 12 + 2n . (b) To encode the possible values for three bits, we need eight different states. The natural way to construct these is to choose the so-called tetrad states (which form a regular tetrahedron on the Bloch sphere): √ 1 |ψ1 i = √ (|0i + 2|1i) 3 √ 1 |ψ2 i = √ (|0i + 2ei2π/3 |1i) 3 √ 1 |ψ3 i = √ (|0i + 2ei4π/3 |1i) 3 |ψ4 i = |0i , and the states orthogonal to these: 1 √ |ψ1⊥ i = √ ( 2|0i − |1i) 3 1 √ ⊥ |ψ2 i = √ ( 2|0i − ei2π/3 |1i) 3 √ 1 √ ⊥ |ψ3 i = √ ( 2|0i − 2ei4π/3 |1i) 3 ⊥ |ψ4 i = |1i .

Note that the combination of these eight states forms a cube on the surface of the Bloch sphere. The best way to encode the bit values onto these (and indeed the natural way) is to make associate the numbers that are most different with the pairs of orthogonal states. With this in mind we set: 000 → |ψ4 i

Quantum information theory 47

111 → |ψ4⊥ i 001 → |ψ1 i 110 → |ψ1⊥ i 010 → |ψ2 i 101 → |ψ2⊥ i 100 → |ψ3 i 011 → |ψ3⊥ i .

Let us suppose that Bob needs to recover the bit vale A. In this case he needs to perform the minimum-error measurement that distinguished between the four equiprobable states that correspond to the value 0 and the remaining 4 that correspond to the value 1. We can write the associated density operators in the form: 1 ρ̂0 = (|ψ4 ihψ4 | + |ψ1 ihψ1 | + |ψ2 ihψ2 | + |ψ3⊥ ihψ3⊥ |) 4 1 1 = Î + (|ψ3⊥ ihψ3⊥ | − |ψ3 ihψ3 |) 2 4 1 1 ρ̂1 = Î + (|ψ3 ihψ3 | − |ψ3⊥ ihψ3⊥ |) , 2 4 P where we have used the fact that i |ψi ihψ1 | = 2Î. Clearly the optimal measurement to perform in this case is to determine whether the qubit is found in the state |ψ3⊥ i which we then associate with the bit value 0 or in the state |ψ3 i which we assign the value 1. The probability of correctly identifying the bit value in this case is hψ3⊥ |ρ̂0 |ψ3⊥ i = 34 . The symmetry of the problem then implies that this is true also for the other bit values so that the overall probability for correctly identifying the state is 34 , which is clearly greater than the classical, single-bit, probability of 32 .

(8.47) The simplest (if brute force) way to do this is to introduce two ancilla qubits prepared in the state |0i and use these to mark the states to be interchanged: |110i and |001i. This is depicted in Fig. 8.1.

There is a simpler transformation as if the third qubit is found in the state |1i then the compression has failed and the states of the other two qubits are not relevant. Hence a combination of two Toffoli and two CNOT gates suffices, as in Fig. 8.2.

(8.48) (a) The unitary transformation performed in the compression step leaves the three qubits in the state

Û |ψj i ⊗ |ψk i ⊗ |ψl i = cos3 θ|000i + cos2 θ sin θ (−1)j−1 |100i

+(−1)k−1 |010i + (−1)l−1 |110i

+ cos θ sin θ (−1)j−k |001i + (−1)j−l |101i

+(−1)k−l |011i + sin3 θ(−1)j+k+l−1 |111i .

The compression success probability is simply the probability that the

48 Quantum information theory

Fig. 8.1

Fig. 8.2

third qubit is found in the state |0i. This is Psucc = cos6 θ + 3 cos4 θ sin2 θ = 1 − 3 sin4 θ ≈ 1 − 3θ4 . (b) In the decompression step, Bob supplies a third qubit in the state |0i and applies the unitary operator Û † (= Û ) to the produce the state 1 p

2

cos θ|000i + sin θ (−1)j−1 |110i

1 + 2 sin θ

+(−1)k−1 |010i + (−1)l−1 |001i .

The fidelity is simply the modulus squared of the overlap between this state and the initial state: F= p

1

2 4

2

2

(cos θ + 3 cos θ sin θ) 1 + 2 sin2 θ = (1 + 2 sin2 θ) cos4 θ ≈ 1 − 3θ4 .

(8.49) In each case we need to find the a priori von Neumann entropy for each single qubit. (a)

For the three trine states we have ρ̂ =

1 1 1 1 |ψ1 ihψ1 | + |ψ2 ihψ2 | + |ψ3 ihψ3 | = Î .

3 3 3 2

This is already the maximum value and no compression is possible.

Quantum information theory 49

(b)

For the states indicated the density operator is

ρ̂ = p|ψ1 ihψ1 | + (1 − p)|ψ2 ihψ2 | = cos2 θ|0ih0| + sin2 θ|1ih1| + (1 − 2p) sin θ cos θ(|0ih1| + |1ih0|) .

The eigenvalues of this density operator are

q 1 λ± = 1 ± 1 − 16p(1 − p) cos2 θ sin2 θ 2 ⇒ S = −λ+ log λ+ − λ− log λ− , and the N qubits can be compressed into N S qubits.

(8.50) Each qubit pair is in one of two orthogonal qubit states, |Ψ− i or |Ψ+ i, hence we can compress by a factor of 2 by mapping each state onto a corresponding single qubit state, |0i or |1i. A suitable transformation is |Ψ− i → |0i|0i

|Ψ+ i → |1i|0i ,

at which stage the second qubit can be discarded. Note that this transformation can readily be performed by a combination of a CNOT and a Hadamard gate.

(8.51)

The channel capacity is bounded by " ! # X X 0 0 C ≤ Sup S P (ai )ρ̂i − P (ai )S(ρ̂i ) . i

i

We can achieve the maximum possible value of 1 bit by choosing our signal states to be |0i and |1i, which are unaffected by the channel transformation and selecting these two with equal probability. ˆ ) then (8.52) If the initial qubit state is ρ̂i = 1 (Î + ~r · ~σ 2

1 4p ˆ . ρ̂0i = Î + 1 − ~r · ~σ 2 3 This is a mixture of the original state and the flipped state (with the direction of the Bloch vector reversed). The HSW bound is achieved if we start with a set of pure states for which the ensemble density operator is 12 Î:

2p 2p 2p 2p C = 1+ log + 1− log 1 − .

3 3 3 3 The simplest way to achieve this bound is to select, with equal probability, from a pair of orthogonal states for each qubit.

(8.53) If Alice’s measurement results show the her two qubits to be in the same state then the resulting state is |ψ2 i →

cos2 θ|00iA |00iB + sin2 θ|11iA |11iB p . cos4 θ + sin4 θ

50 Quantum information theory

This state is less entangled than the original p state, as the probability p amplitudes cos2 θ/ cos4 θ + sin4 θ and sin2 θ/ cos4 θ + sin4 θ differ by more than the original amplitudes, cos θ and sin θ.

Determining whether the entanglement increases, decreases or stays the same requires some care. If we rely on Fent then we find that the average entanglement is unchanged. This is not conclusive, however. For Dent , however, we find that this is reduced on average, from | cos θ sin θ| to 2 cos2 θ sin2 θ = | sin(2θ)|| cos θ sin θ| so the resulting state is, on average, less strongly entangled.

(8.54) As the state is clearly unentangled, the concurrence should be zero. We can check this by explicit calculation. First note that for the given state

1 ˆ ⊗ 1 Î − ~s · ~σ ˆ . ˆ= Î − ~r · ~σ ρ̄ 2 2 Note that this commutes with ρ̂ and hence R̂2 = ρ̂ρ̂0 = so the eigenvalues of R̂ are all zero.

√

1 (1 − r)Î 4

1 − r/2 and hence the concurrence is

(8.55) It follows directly from the form of the Werner state that ρ̂0W = ρ̂W and hence R̂ = ρ̂W and the four eigenvalues are p/4, p/4, p/4 and 1 − 3p/4. Hence the concurrence is 0 if p ≥ 2/3 and is C =1−

3p 2

if p < 2/3. It follows that the entanglement of formation is zero if p ≥ 2/3 and otherwise is ! ! p p 1 + 3p(1 − 3p/4) 1 + 3p(1 − 3p/4) EF = − log 2 2 ! ! p p 1 − 3p(1 − 3p/4) 1 − 3p(1 − 3p/4) − log .

2 2