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### Transcript of Appendix A Solutions of Exercises - Springer 978-3-319-45241... Appendix A Solutions of Exercises...

• Appendix A Solutions of Exercises

Exercise 1.1 PρM(1) = 916 , PρM(2) = 116 , and PρM(3) = 38 .

Exercise 1.2

A = E1 − E2, where E1 = 12 ( 1 1 1 1

) and E2 = 12

( 1 −1

−1 1 ) .

Exercise 1.3 We can calculate it in two ways. (1) 1 × 12 + (−1) × 12 = 0. (2) Tr ρA = 0.

Exercise 1.4 We can calculate it in two ways. (1) 12 × 12 + (−1)2 × 12 = 1. (2) Tr ρA2 = 1.

Exercise 1.5

Since

( 1 2

1 4

1 4

1 2

) = 34

( 1 2

1 2

1 2

1 2

) + 14

( 1 2 − 12

− 12 12

) , we have

( 1 2

1 4

1 4

1 2

)−1/2 = (3

4 )−1/2

( 1 2

1 2

1 2

1 2

) + (1

4 )−1/2

( 1 2 − 12

− 12 12

)

= 2√ 3

( 1 2

1 2

1 2

1 2

) + 2

( 1 2 − 12

− 12 12

) =

( 1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

) .

Hence, the POVM {Mi }3i=1 is given as

M1 = (

1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

) 1

2

( 1 2

1 2

1 2

1 2

) ( 1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

)

= ( 1

3 1 3

1 3

1 3

) ,

© Springer International Publishing AG 2017 M. Hayashi, A Group Theoretic Approach to Quantum Information, DOI 10.1007/978-3-319-45241-8

205

• 206 Appendix A: Solutions of Exercises

M2 = (

1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

) 1

4

( 1 0 0 0

) ( 1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

)

= (

1 3 + 12√3 − 16

− 16 13 − 12√3

) ,

M3 = (

1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

) 1

4

( 0 0 0 1

) ( 1√ 3

+ 1 1√ 3

− 1 1√ 3

− 1 1√ 3

+ 1

)

= (

1 3 − 12√3 − 16

− 16 13 + 12√3

) .

Exercise 1.6

Since XX† = 12 ( 1 0 0 1

) , the Schmidt rank is 1. The Schmidt coefficients are

( 1√ 2 , 1√

2 ).

Exercise 1.7

Y ⊗ Z |X〉〉A,B = ∑ k, j

∑ k ′, j ′

yk,k ′ z j, j ′xk ′, j ′ |k〉 ⊗ | j〉 = |Y X ZT 〉〉A,B . (A.1)

Exercise 1.8

TrB |X〉〉A,B A,B〈〈Y | = TrB ∑ k, j

xk, j |k〉 ⊗ | j〉 ∑ k ′, j ′

yk ′, j ′ 〈k ′| ⊗ 〈 j ′| (A.2)

= ∑ k ′,k, j

xk, j yk ′, j |k〉〈k ′| = XY †. (A.3)

Exercise 1.9√ p|0〉|0〉 + √1 − p|1〉|1〉.

Exercise 1.10 Define the POVM Mi := X piρTi (XT )−1. Using Exercise 1.8, we have

TrB Mi |X〉〉〈〈X | = TrB X piρTi (XT )−1|X〉〉〈〈X | = TrB |XX−1 piρi (X†)−1〉〉〈〈X | = XX−1 piρi (X†)−1X† = piρi .

http://dx.doi.org/10.1007/978-3-319-45241-8_1

• Appendix A: Solutions of Exercises 207

Exercise 1.11 Choose the system HB as the space spanned by the CONSs {|i〉}i . Then, we define |X〉〉 as ∑i √pi |xi 〉|i〉. Define the PVM E as Ei := |i〉〈i |. So, we can check the desired condition.

Exercise 1.12 Since TrA 1d |Ui 〉〉〈〈Ui ||ρA ⊗ I = TrA(ρA ⊗ I ) 1d |Ui 〉〉〈〈Ui | = TrA 1d |ρAUi 〉〉〈〈Ui | = 1 dU

† i ρAUi , we have Tr

1 d |Ui 〉〉〈〈Ui |ρA ⊗ ρmix,B = TrB 1dU †i ρAUiρmix,B = 1d2 .

Exercise 2.1 Let x1 and x2 be two elements of the domain of f . For an arbitrary number λ ∈ [0, 1], the linearity of f implies that f (λx1(1−λ)x2) = λ f (x1)+(1−λ) f (x2). Since g is a concave function,wehave g(λ f (x1)+(1−λ) f (x2)) ≤ λg( f (x1))+(1−λ)g( f (x2)). Combining these two inequalities, we obtain g( f (λx1(1 − λ)x2)) ≤ λg( f (x1)) + (1 − λ)g( f (x2)).

Exercise 2.2 For a square matrix X , we choose two Hermitian matrices X1 and X2 such that X = X1 + X2i . Then, we have |Tr Xρ|2 = (Tr X1ρ)2 + (Tr X2ρ)2. Exercise 2.1 guarantees the convexity of the functions (Tr X1ρ)2 and (Tr X2ρ)2. Since a sum of convex functions is a convex function, we obtain the desired statement.

Exercise 2.3 We fix two elements x1, x2 ∈ X and λ ∈ (0, 1). For any � > 0, we choose a′ such that fa′(λx1 + (1 − λ)x2) ≥ supa fa(λx1 + (1 − λ)x2) − �. Since

fa′(λx1 + (1 − λ)x2) ≤ λ fa′(x1) + (1 − λ) fa′(x2) ≤ λ(sup

a fa(x1)) + (1 − λ)(sup

a fa(x2)),

we have

sup a

fa(λx1 + (1 − λ)x2) − � ≤ λ(sup a

fa(x1)) + (1 − λ)(sup a

fa(x2)).

Since � > 0 is arbitrary, we have

sup a

fa(λx1 + (1 − λ)x2) ≤ λ(sup a

fa(x1)) + (1 − λ)(sup a

fa(x2)),

which is the desired statement.

Exercise 2.4 Let {pi } be an arbitrary distribution on {1, . . . , d}. So, we have∑ki=1 p↓i ≥ kd , which implies the desired statement.

http://dx.doi.org/10.1007/978-3-319-45241-8_2

• 208 Appendix A: Solutions of Exercises

Exercise 2.5

Iρ(A : B) = H(ρA) + H(ρB) − H(ρ) = − Tr ρ(log ρA) ⊗ IB − Tr ρIA ⊗ (log ρB) + Tr ρ log ρ = − Tr ρ log(ρA ⊗ ρB) + Tr ρ log ρ = D(ρ‖ρA ⊗ ρB).

Exercise 2.6 We focus on the eigenvectors of the density matrix ρ whose eigenvalue is not λ. We move such eigenvectors cyclically and obtain another density matrix. We take the average among such density matrices with equal weight. Such the averaged density matrix has von Neumann entropy (1 − λ) log dimH + h(λ). Hence, the concavity of von Neumann entropy yields the desired statement.

Exercise 2.7 The (2.44) can be shown as follows.

D(ρ‖σA ⊗ σB) = Tr ρ(log ρ − logσA ⊗ σB) = Tr ρ(log ρ − (logσA) ⊗ IB − IA ⊗ (logσB)) = Tr ρ(log ρ − (logσA) ⊗ IB − IA ⊗ (log ρB))

+ Tr ρ(IA ⊗ (log ρB) − IA ⊗ (logσB)) = Tr ρ(log ρ − log(σA ⊗ ρB)) + Tr ρB(log ρB − logσB) = D(ρ‖σA ⊗ ρB) + D(ρB‖σB). (A.4)

Exercise 2.8 We denote the spectral decomposition E by {Ei }. We define a unitary representation f(θ1, . . . , θn) := ∑nj=1 eiθ j E j of the group Rn . The pinching ΛE with respect to the PVM E satisfies that ΛE (ρ) = ρ. Due to Theorem 2.9, the image ΛE (ρ) of the pinching is the point closest to ρ among the invariant density matrices with respect to this representation in the sense of relative entropy. That is, Theorem 2.9 yields (2.45).

Exercise 2.9 Since 〈Φ|Fm ⊗ I |Φ〉 = Tr Fmρ, we have

F2e (ρ,Λ) = 〈Φ|(Λ ⊗ id)(|Φ〉〈Φ|)|Φ〉 = 〈Φ|

∑ m

(Fm ⊗ I )|Φ〉〈Φ|(Fm ⊗ I )†|Φ〉

= ∑ m

|〈Φ|Fm ⊗ I |Φ〉|2 = ∑ m

|Tr Fmρ|2. (A.5)

http://dx.doi.org/10.1007/978-3-319-45241-8_2 http://dx.doi.org/10.1007/978-3-319-45241-8_2 http://dx.doi.org/10.1007/978-3-319-45241-8_2 http://dx.doi.org/10.1007/978-3-319-45241-8_2

• Appendix A: Solutions of Exercises 209

Exercise 2.10 Since 〈Φ|(Fm ⊗ I )|Φ〉 = Tr Fmρ, we have

F2e (ρ,Λ) = 〈Φ|(Λ ⊗ id)(|Φ〉〈Φ|)|Φ〉 = 〈Φ|

∑ m

(Fm ⊗ I )|Φ〉〈Φ|(Fm ⊗ I )†|Φ〉

= ∑ m

〈Φ|(Fm ⊗ I )|Φ〉〈Φ|(Fm ⊗ I )†|Φ〉 = ∑ m

|Tr Fmρ|2.

Exercise 2.11 Based on Exercise 1.11, we choose the reference systemHR , a purification |X〉〉 of ρ onHB , and a PVM E = {Ei }i onHR such that pi |xi 〉〈xi = TrB Ei |X〉〉〈〈X |. Hence,

F2e (ρ,Λ) = 〈Φ|(Λ ⊗ id)(|Φ〉〈Φ|)|Φ〉 ≤ F2(ΛE (|Φ〉〈Φ|),ΛE ((id ⊗ Mi )(Λ ⊗ id)(|Φ〉〈Φ|))) =

∑ j

p j 〈x j |Λ(|x j 〉〈x j |)|x j 〉. (A.6)

Exercise 2.12 H1+s(ρx) is calculated to be 1s log((

1+‖x‖ 2 )

1+s + ( 1−‖x‖2 )1+s). Exercise 2.13 To show 2.82, it is enough to show that

D1+s(ρ‖ρA ⊗ σB) ≥ D1+s(ρ‖ρA ⊗ σ∗B(1 + s)) (A.7)

for any state σB . This inequality is equivalent to

esD1+s (ρ‖ρA⊗σB ) ≤ esD1+s (ρ‖ρA⊗σ∗B (1+s)) (A.8)

for s ∈ (−1, 0) and

esD1+s (ρ‖ρA⊗σB ) ≥ esD1+s (ρ‖ρA⊗σ∗B (1+s)) (A.9)

for s ∈ (0,∞). (A.8) is equivalent to the following inequality

Tr ρ1+sρ−sA ⊗ σ−sB = TrB(TrA ρ1+sρ−sA )σ−sB ≤ (TrB(TrA ρ1+sρ−sA ) 1

1+s )1+s . (A.10)

We apply matrix Hölder inequality (2.21) to the case when A = (TrA ρ1+sρ−sA ), B = σ−sB , p = 11+s > 0 and q = 1−s > 0 with s ∈ (−1, 0). Since the RHS of (2.21) equals the RHS of (A.10), we obtain (A.10).

http://dx.doi.org/10.1007/978-3-319-45241-8_1 http://dx.doi.org/10.1007/978-3-319-45241-8_2 http://dx.doi.org/10.1007/978-3-319-45241-8_2 http://dx.doi.org/10.1007/978-3-319-45241-8_2

• 210 Appendix A: Solutions of Exercises

Similarly, (A.9) is equivalent to the following inequality

TrB(TrA ρ 1+sρ−sA )σ

−s B ≥ (TrB(TrA ρ1+sρ−sA )

1 1+s )1+s . (A.11)

We obtain (A.11) by applying matrix reverse Hölder inequality (2.22) to the case when A = (TrA ρ1+sρ−sA ), B = σ−sB , p = 11+s > 0 and q = 1−s > 0 with s ∈ (0,∞).

Exercise 3.1 Due to Theorem 3.1, it is sufficient to show that TrB |Ψ 〉〈Ψ | � TrB |Φ〉〈Φ|. This relation is trivial from the definition of majorization