Homework 3 - Solutions

Homework 3 - Solutions The Transpose an Partial Transpose. 1 Let { 1, 2,, } be an orthonormal basis for C. The transpose map efine with respect to this basis is a superoperator Γ that acts on an operator M LC by M m ij i j ΓM m ij j i. 1 i,j1 i,j1 As emphasize before, this is a basis-epenent operation in the sense that if { 1, 2,, } is another orthonormal basis, then the transpose map efine with respect to this other basis is the superoperator Γ with action given by M m ij i j Γ M m ij j i. 2 i,j1 i,j1 a By proviing an explicit example, show that ΓM Γ M in general. b Nevertheless, show that ΓM an Γ M will always have the same eigenvalue spectrum. a Consier the two bases B { 0, 1 } an B { 0, 1 }, where ± 1/2 0 ± i 1. Take the operator M 1/2 0 + i 1 0 1/2 + + +. Then ΓM 1/2 0 0 + i 1 Γ M 1/2 + + + since + 1/2 0 i 1. Part b follows from the two facts that i any two orthonormal bases can be relate by a unitary transformation, an ii eigenvalue spectrums remain invariant uner unitary transformations. In more etail, let U be the unitary that rotates i i U i. Then Γ M UΓMU so the eigenvalues of ΓM are the same as ΓM. 1

2 Now consier two systems, A an B, with bipartite state space C C an computational prouct basis { i j } i,j1. The partial transpose of system B with respect to the computational basis is the superoperator I A Λ B acting on LC C, where Λ B is the transpose map on B an I A is the trivial map on A; i.e. I A T T for all T LC. For an arbitrary M LC C, the action of I A Γ B is given by M m ij,kl i k A j l B I A Γ B M m ij,kl i k A Γ j l B i,j,k,l1 i,j,k,l1 m ij,kl i k A l j B. 3 i,j,k,l1 The partial transpose of system A with respect to the computational basis is similarly efine as the map Γ A I B. For a bipartite operator M, its partial transpose on B is enote as M Γ B : I A Γ B M, an its partial transpose on A is likewise enote by M Γ A : Γ A I B M. a Let M be an operator on C 2 C 2. Then it is represente as a 4 4 complex matrix in the computational basis. Give the matrix representations for both M Γ A an M Γ B. b For C C state space, prove that taking a partial transpose oes not affect the partial trace: Tr A M Tr A M Γ A an Tr B M Tr B M Γ B. c If M is hermitian, show that both M Γ A an M Γ B are also hermitian. If M is hermitian, show that M Γ A an M Γ B have the same eigenvalue spectrum. a Write a general two-qubit operator as M 1 i,j,k,l0 m ij,kl i k j l. Then M Γ A 1 i,j,k,l0 m kj,il i k j l an M Γ B 1 i,j,k,l0 m il,kj i k j l. In matrix form, the three matrices are m 00,00 m 00,01 m 00,10 m 00,11 M. m 01,00 m 01,01 m 01,10 m 01,11 m 10,00 m 10,01 m 10,10 m 10,11, m 11,00 m 11,01 m 11,10 m 11,11 m 00,00 m 00,01 m 10,00 m 10,01 m 00,00 m 01,00 m 00,10 m 01,10 M Γ. A m 01,00 m 01,01 m 11,00 m 11,01 m 00,10 m 00,11 m 10,10 m. 10,11 MΓ B m 00,01 m 01,01 m 00,11 m 01,11 m 10,00 m 11,00 m 10,10 m 11,10. m 01,10 m 01,11 m 11,10 m 11,11 m 10,01 m 11,01 m 10,11 m 11,11 2

b For M i,j,k,l1 m ij,kl i k A j l B an partial transpose M Γ B i,j,k,l1 m il,kj i k A j l B, their partial transposes are given by Tr B M Tr B M Γ B n1 n1 i,j,k,l1 i,l,k,j1 m ij,kl i k A n j l B n m il,kj i k A n j l B n n1 n1 i,k1 i,k1 m in,kn i k A ; m in,kn i k A. So the two are equal. The same argument works for the partial trace on system A. M Γ B c The assumption that M M means that m ij,kl m kl,ij. The components of are m il,kj an so M Γ B has components m kj,il, which are equal to m il,kj since M M. The same argument shows that M Γ A is hermitian. Suppose that M Γ A ψ λ ψ an write this as the matrix equation M Γ A ψ ψ λ ψ ψ. Take the transpose of both sies to obtain ψ ψ M Γ A T λ ψ ψ, where we use the fact that ψ ψ T ψ ψ for complex conjugation taken in the same basis that the transpose is performe. Since M Γ A T M Γ B, we have ψ ψ M Γ B λ ψ ψ. Now take the hermitian conjugate of both sies to obtain M Γ B ψ ψ λ ψ ψ λ ψ ψ, where we have use the facts that M Γ B is hermitian an that λ is real since M Γ A is also hermitian. Therefore, λ is an eigenvalue of M Γ B with corresponing eigenvector ψ. Positive an Completely Positive Maps. 3 Consier the map ΛX TrXI X for X LC. Is Λ a positive map? Is it completely positive? Let ψ C be an normalize vector. Then ψ ΛX ψ TrX ψ X ψ 0. 4 The reason for this inequality is because TrX ψ X ψ + i ψ i X ψ i, where the { ψ, ψ i } 1 i1 form an orthonormal basis for C. Hence Λ is positive. To check for complete positivity, we compute Thus Λ is also completely positive. Λ IΦ I I Φ Φ 0. 5 4 3

The fully ephasing map is a completely positive superoperator that acts on a linear operator X LC accoring to X i i X i i. 6 i1 a Restrict attention to C 3, an consier the superoperator ΛX 2 X + ΠXΠ X for X LC 3, 7 where Π 1 2 + 2 3 + 3 1 is a permutation. Is Λ completely positive? b* Is Λ given in Eq. 7 a positive map? Compute the Choi matrix Ω Λ Λ IΦ 3 2 3 3 i1 ii ii + 3 i1 i, i 1 i, i 1 Φ 3. Thus Φ 3 Λ IΦ 3 Φ 3 2 3 1 < 0. The map is not CP. For b, let x ij be the elements of X when expresse in the computational basis. Then it s straightforwar to compute ΛX. 2x 11 + x 22 x 12 x 13 x12 2x 22 + x 33 x 23. x13 x23 2x 33 + x 11 If X 0, then x ii 0 an x ii x jj x ij 2. For the positivity of ΛX, by Sylvester s criterion it suffices to check that the three leaing principal minors of ΛX are nonnegative. Clearly 2x 11 + x 22 0 an et 0. Thus it remains to check the et ΛX. We have 2x11 +x 22 x 12 x12 2x 22 +x 33 etλx 2x 11 + x 22 [4x 22 + x 33 x 33 + x 11 x 23 2 + x 12 [ 2x 12x 33 + x 11 x 23 x 13] x 13 [x 12 x 23 + 2x 13x 22 + x 33 ] 8x 11 + x 22 x 22 + x 33 x 33 + x 11 2x 11 + x 22 x 22 x 33 2x 11 x 22 x 33 + x 11 2x 11 x 33 x 22 + x 33 2Rex 12 x 23 x 13 8x 11 + x 22 x 22 + x 33 x 33 + x 11 2x 11 + x 22 x 22 x 33 2x 11 x 22 x 33 + x 11 2x 11 x 33 x 22 + x 33 2x 11 x 22 x 33 0. 8 Entanglement Fielity. 4

For a quantum channel E : LC LC, its entanglement fielity is the quantity given by Φ + I EΦ+ Φ+, where Φ+ 1 i1 ii an Φ+ Φ+ Φ+. The entanglement fielity essentially measures how well entanglement is preserve when sening one half of the entangle state through the channel. 4 A Pauli channel is any qubit channel of the form Eρ 1 p x p y p z ρ + p x σ x ρσ x + p y σ y ρσ y + p z σ z ρσ z 9 for nonnegative numbers p x, p y, p z calle channel parameters. For a Paul channel E, compute the ensity matrix I EΦ + an the entanglement fielity as a function of the channel parameters p x, p y, p z. For two qubits, recall that Φ b0 b 1 I σ b 0 z σ b 1 x Φ +. The ensity matrix I EΦ + 2 is then given by I EΦ + 2 I E Φ 00 Φ 00 1 p 1 p 2 p 3 Φ 00 Φ 00 + p 1 I σ x Φ 00 Φ 00 I σ x + p 2 I σ y Φ 00 Φ 00 I σ y + p 3 I σ z Φ 00 Φ 00 I σ z p 0 Φ + Φ + + p x Ψ + Ψ + + p y Ψ Ψ + p z Φ Φ, 10 where p 0 1 p x p y p z. Such states are calle Bell iagonal since their eigenstates are the Bell states. Since the Bell states are orthogonal, the entanglement fielity is easily compute to be Φ + 2 I EΦ+ 2 Φ+ 2 p 0 1 p x p y p z. 5 For any integer N, consier again the unitary U e iπσy/n, an now efine the qubit channel E N ρ 1 N Uk ρu k. Compute the ensity matrix I E N Φ + an the entanglement fielity. Recall from problem 5 in Homework 2 that we compute the matrix representation U k coskπ/n sinkπ/n sinkπ/n coskπ/n U k 0 coskπ/n 0 + sinkπ/n 1 U k 1 coskπ/n 0 sinkπ/n 1. 5

Now when E N acts on half of Φ +, we obtain the state I E N Φ + 1 2 [ 0 0 E N 0 0 + 0 1 E N 0 1 + 1 0 E N 1 0 + 1 1 E N 1 1 ]. Using our previous calculations, we compute Bob s part on each term: E N 0 0 1 N 1 N E N 0 1 1 N 1 N E N 1 0 1 N 1 N E N 1 1 1 N Therefore, 1 N U k 0 0 U k cos 2 kπ/n coskπ/n sinkπ/n U k 0 1 U k cos 2 kπ/n coskπ/n sinkπ/n U k 1 0 U k cos 2 kπ/n coskπ/n sinkπ/n U k 1 1 U k cos 2 kπ/n coskπ/n sinkπ/n coskπ/n sinkπ/n sin 2 1 kπ/n 2 I, coskπ/n sinkπ/n 1 sin 2 2 σ z, kπ/n coskπ/n sinkπ/n sin 2 1 kπ/n 2 σ z, coskπ/n sinkπ/n sin 2 1 kπ/n 2 I. I E N Φ + 1 4 [ 0 0 I + 0 1 + 1 0 σ z + 1 1 I] The entanglement fielity is compute to be 1 4 [I I + σ x σ z ]. 11 1 4 [1 + Φ+ σ x σ z Φ + ] 1 4 [1 + 1 2 00 + 11 + 0 1 ] 1 4 1 + 1. 2 12 Teleportation Fielity. The stanar teleportation fielity of a bipartite state ρ is the quantity given by f ρ ˆn ˆn Λ T0,ρ ˆn ˆn ˆn, 13 6

where Λ T0,ρ is the qubit channel inuce by performing the stanar teleportation protocol T 0 on the bipartite state ρ. 6 As a function of the channel parameters p 1, p 2, p 3, compute the stanar teleportation fielity of the bipartite state obtaine by sening half of the maximally entangle state through a Pauli channel. That is, compute the stanar teleportation fielity of the state ρ I EΦ +, where E is the channel given by Eq. 9. In Exercise 5, the ensity matrix obtaine by sening Φ + 2 through a Pauli channel is ρ I EΦ + 2 p 0 Φ 00 Φ 00 + p x Ψ 01 Ψ 01 + p z Ψ 10 Ψ 10 + p y Φ 11 Φ 11, with p 0 1 p x p z p y. Suppose now that Alice an Bob share this state ρ AB an they use it to perform the stanar teleportation protocol T 0. Let ˆn A be the state that Alice wants to teleport. The effect of the stanar teleportation protocol is to perform a generalize measurement {M A AB b 0 b 1 } b0,b 1 {0,1} on systems A AB with Kraus operators M A AB b 0 b 1 Φ b0 b 1 Φ b0 b 1 A A σ b 0 z σ b 1 x. Using the fact that Φ b0 b 1 AB I A σ b 0 z σ b 1 x Φ 00 AB, we can compute M b0 b 1 ˆn A Φ 00 AB Φ b0 b 1 A A ˆn B M b0 b 1 ˆn A Φ 01 AB M b0 b 1 I A A σ B x ˆn A Φ 00 AB Φ b0 b 1 A A σ x ˆn B M b0 b 1 ˆn A Φ 10 AB M b0 b 1 I A A σ B z ˆn A Φ 00 AB Φ b0 b 1 A A σ z ˆn B M b0 b 1 ˆn A Φ 11 AB M b0 b 1 I A A σ B y ˆn A Φ 00 AB Φ b0 b 1 A A σ y ˆn B. 14 Thus for a given input state ˆn A, the average output state for Bob is Tr A A M b0 b 1 ˆn ˆn A ρ AB M b 0 b 1 b 0 b 1 {0,1} p 0 ˆn ˆn + p x σ x ˆn ˆn σ x + p z σ z ˆn ˆn σ z + p y σ y ˆn ˆn σ y. 15 We now average over ˆn cosθ/2 0 + e iφ sinθ/2 1. Clearly the first term correspons to perfect fielity. The secon term has average fielity ˆn ˆn σ x ˆn 2 1 φθ sin θ cosθ/2 sinθ/2e iφ + e iφ 2 4π 1 φθ sin θ cos 2 θ/2 sin 2 θ/24 cos 2 φ 4π θ sin θ cos 2 θ/2 sin 2 θ/2 1/3. 16 Similar calculations show that ˆn ˆn σ y ˆn 2 1/3 an ˆn ˆn σ z ˆn 2 0. Therefore in total, f ρ ˆn ˆn Λ T0,ρ ˆn ˆn ˆn p 0 + p x + p y /3. 17 7

7 How well can Alice an Bob simulate the teleportation of a ranom qubit state using no share entanglement? This is the question you will explore in this exercise. a Consier first the scenario where Alice oes not sen Bob any classical information. She ranomly generates a state ˆn with ˆn istribute uniformly on the Bloch sphere, an Bob s goal is to prouce a state ˆn that maximizes the average fielity ˆn ˆn 2. Uner these restrictions show that the maximum average fielity ˆn ˆn ˆn 2 is 1/2. b Next we consier strategies that allow Alice to classically communicate with Bob, just like in stanar teleportation. Suppose that Alice measures her generate state ˆn in the { 0, 1 } basis an obtains outcome a, for a {0, 1}. She computes to Bob the outcome of her measurement, an then Bob prouces the state ˆn a. What is average fielity ˆn ˆn 2 obtaine using such a strategy? a Without any aitional information from Alice, Bob s strategy involves ranomly choosing some point ˆn on the Bloch sphere an preparing the state ˆn. On average, the fielity will be the same no matter what irection he chooses, so without loss of generality we can assume that Bob always prepares 0. The average fielity is then ˆn ˆn 0 2 θφ sin θ 4π cos2 θ/2 1 2. 18 b For a given state ˆn cosθ/2 0 + e iφ sinθ/2 1, Alice obtains outcome 0 with probability cos 2 θ/2 an outcome 1 with probability sin 2 θ/2. Thus following the protocol where Bob prepares a when Alice measures a, for a {0, 1}, the average fielity for state ˆn is cos 2 θ/2 ˆn 0 2 + sin 2 θ/2 ˆn 1 2 cos 4 θ/2 + sin 4 θ/2. Therefore, the average fielity is given by θφ sin θ 4π [cos4 θ/2 + sin 4 θ/2] 2 3. 19 References 8