One-shot quantum state exchange

The quantum state exchange is a quantum communication task in which two users exchange their respective quantum information in the asymptotic setting. In this work, we consider a one-shot version of the quantum state exchange task, in which the users hold a single copy of the initial state, and they exchange their parts of the initial state by means of entanglement-assisted local operations and classical communication. We first derive lower bounds on the least amount of entanglement required for carrying out this task, and provide conditions on the initial state such that the protocol succeeds with zero entanglement cost. Based on these results, we reveal two counter-intuitive phenomena in this task, which make it different from a conventional SWAP operation. One tells how the users deal with their symmetric information in order to reduce the entanglement cost. The other shows that it is possible for the users to gain extra shared entanglement after this task.

We formally define the OSQSE and its optimal entanglement cost, and derive computable lower bounds on the latter, which in turn yield bounds for the asymptotic quantum state exchange [1,2]. In addition, we provide two useful conditions to decide whether a given initial state enables OSQSE with zero entanglement cost. We then show that there exist counter-intuitive phenomena in the OSQSE task that set it apart from the conventional SWAP operation.
One-shot quantum state exchange.-Consider two users, Alice and Bob, holding parts A and B of the initial state |ψ ≡ |ψ A 1 B 1 A 2 B 2 R with systems A = A 1 A 2 and B = B 1 B 2 , respectively. Alice's and Bob's goal is either to exchange their parts A 1 and B 1 or to exchange their whole parts A and B.
Specifically, let ψ f 1 and ψ f 12 be the final states of the task, where ψ = |ψ ψ|, and the dimension of system X is identical to that of system X. Note that B 1 , B and A 1 , A are Alice's and Bob's systems, respectively. Then three joint operations are called the OSQSE protocols of |ψ , if they are performed by LOCC between Alice and Bob, and satisfy where Ψ and Φ are pure maximally entangled states with Schmidt rank K and L on systems E in A E in B and E out A E out B , respectively. It is possible to generalize the above definitions by arXiv:1905.12332v1 [quant-ph] 29 May 2019 adding errors for approximation to Eq. (2), but it suffices to only consider error-free protocols to obtain our main results.
At this point, it is instructive to inform differences among the three protocols in Eq. (1) as follows: The first two protocols E 1 ψ,K,L and E 1|2 ψ,K,L indicate that only the parts A 1 and B 1 are exchanged, while the whole parts A 1 A 2 and B 1 B 2 are exchanged in the third protocol E 12 ψ,K,L . In addition, the parts A 2 and B 2 can be used for exchanging A 1 and B 1 in the protocol E 1|2 ψ,K,L , while A 2 and B 2 are untouched in the protocol E 1 ψ,K,L . Depending on the types of OSQSE protocols, we define three optimal entanglement costs where the quantity log K − log L is called the entanglement cost of the OSQSE protocol, and the infimums are taken over all joint protocols E 1 ψ,K,L , E 1|2 ψ,K,L , and E 12 ψ,K,L , respectively. By the definitions of the optimal entanglement costs, we obtain the following proposition.
Converse bounds.-A real number r is called a converse bound of the optimal entanglement cost if it is upper bounded by the entanglement cost of any OSQSE protocol. We derive converse bounds of the optimal entanglement costs as follows.
As in the asymptotic scenario [1,2], we consider a oneshot version of the R-assisted quantum state exchange task, in which the reference system R is divided into two systems R A and R B , and then Alice and Bob receive the divided parts R A and R B , respectively, so that the initial state |ψ A 1 B 1 A 2 B 2 R A R B is divided into Alice's parts AR A and Bob's parts BR B . This can be realized by using a quantum channel N : R −→ R A and its complementary channel N c : R −→ R B [20]. Let Eψ ,K,L be an R-assisted OSQSE protocol, Note that Eψ ,K,L is an LOCC protocol by Alice and Bob. Let σ M be the maximally mixed state with rank M. From the majorization condition for LOCC convertibility [21,22], the . Let F be an additive and Schur concave function [23] such that F(σ M ) = log M for any M. From the Schur concavity of the function F, ψ,K,L is also an R-assisted OSQSE protocol for the initial state |ψ , we obtain the following theorem.
Theorem 2. For any input state ψ, the optimal entanglement cost e A 2 B 2 A 1 ↔B 1 (ψ) is lower bounded by where F is an additive and Schur concave function such that F(σ M ) = log M for any M and N(ρ) is a quantum channel from R to R A .
In Theorem 2, if R is directly sent to either Alice or Bob without splitting, and we restrict the function F to the quantum Rényi entropy S α ( ) of order α [23] for a quantum state , then we obtain the following computable converse bounds.
We refer the reader to Appendix A for the proof of Corollary 3. Remark that the converse bound l c 1|2 can be easily computed by means of analytical or numerical methods, since the function f ψ (α) is one-variable and differentiable on (0, ∞). For the different types of the OSQSE protocols, we can also obtain a similar computable converse bound as follows: where the pair (X, Y) can refer either to (A 1 , B 1 ) or to (A, B).
We also remark that in Theorem 2, if F is chosen as the von Neumann entropy [20], then the converse bound l 1|2 recovers a theoretical converse bound in Refs. [1,2]. In addition, a computable converse bound therein is just f ψ (1) in Corollary 3. By virtue of the additivity of F, it is clear that l 1|2 and l c 1|2 are also converse bounds of the optimal entanglement cost for the asymptotic quantum state exchange task. Hence, our converse bounds improve the existing bounds in Refs. [1,2]. For example, if the initial state then we can find a value α 0 ∈ [0, ∞] such that l c 1|2 (ψ 1 ) = f ψ 1 (α 0 ) > f ψ 1 (1) as depicted in Fig. 1. This example shows that our bound l c 1|2 (ψ) is tighter than the existing bound f ψ (1). Conditions for zero entanglement cost.-We now present conditions for OSQSE with zero entanglement cost.
By the converse bound in Eq. (4), it is obvious that if there exist Alice's and Bob's local isometries performing the OS-QSE task, then the optimal entanglement cost is zero. We first characterize this type of strategy. Let (X, Y) be a pair of two systems, which can be either (A 1 , B 1 ) or (A, B), and consider a spectral decomposition of the reduced state ρ XY for |ψ , where {| j } and {|k } indicate the computational bases on Alice's and Bob's systems, respectively. Then we obtain the following sufficient condition.
Here, the isometries U and V indicate Alice's and Bob's local operations exchanging the parts X and Y without shared entanglement. The proof of Theorem 4 is in Appendix B.
From the converse bound in Eq. (4), observe that if the spectrum of Alice's state is different from that of Bob's state, then the optimal entanglement cost cannot be zero. Based on this observation, we obtain the following theorem, whose proof can be found in Appendix C.
We remark that the converse of Theorem 5 is not true in general. Let us consider the following simple initial state from Proposition 1 and Corollary 3. However, the state |ψ 2 satisfies the necessary condition in Theorem 5, since its reduced states ρ A 1 and ρ B 1 are identical.
Counter-intuitive phenomena.-We are now in the position to present two phenomena which show the important differences between the OSQSE task and the SWAP operation.
(1) Symmetric information.-For the initial state |ψ , let us consider a scenario in which Alice and Bob exchange their whole information A and B. Assume that their parts A 2 and B 2 are symmetric, while the remaining parts A 1 and B 1 are not symmetric, i.e., the initial state |ψ satisfies SWAP A 1 ↔B 1 (ψ) ψ and SWAP A 2 ↔B 2 (ψ) = ψ where SWAP X↔Y is the operation swapping quantum states in systems X and Y.
From a viewpoint of the SWAP operation, if Alice and Bob want to exchange A and B, then it suffices for them to exchange A 1 and B 1 , since A 2 is identical to B 2 . This situation can be more easily understood by using a cargo exchange as a FIG. 2: Illustration of the cargo exchange task. The cargoes A 1 A 2 and B 1 B 2 belong to Alice and Bob, respectively. Assume that the cargoes A 2 and B 2 are symmetric, but A 1 and B 1 are not symmetric. When Alice and Bob exchange their whole cargoes A 1 A 2 and B 1 B 2 , it suffices for them to exchange A 1 and B 1 , since A 2 is identical to B 2 . The truck indicates the cost needed for exchanging A 1 and B 1 .
metaphor for the SWAP operation as depicted in Fig. 2. In the cargo exchange, assume that Alice and Bob want to exchange their whole cargoes, and some of the cargoes are symmetric. In terms of efficiency, it is reasonable for them to exchange only A 1 and B 1 in order to reduce the cargo exchange cost, because the cargoes A 2 and B 2 are the same.
On the other hand, in the OSQSE, the proper use of the symmetric parts A 2 and B 2 can more efficiently reduce the entanglement cost compared to exchanging only A 1 and B 1 without using A 2 and B 2 . To be specific, there exists an initial state |ψ such that the parts A 2 and B 2 are symmetric and e A↔B (ψ) = 0 while the rest parts A 1 and B 1 are not symmetric. Consider the specific initial state where A 2 and B 2 are symmetric but A 1 and B 1 are not.
Since Ω 1 AB (φ 1 ) = |00 00| and Ω 2 AB (φ 1 ) = |01 11|, we can show that Ω 1 AB (φ 1 ) and Ω 2 AB (φ 1 ) satisfy the condition in Theorem 4, by setting Thus we obtain that e A↔B (φ 1 ) = 0, which means that A and B can be exchanged by means of LOCC without consuming any non-local resource. As mentioned above, this phenomenon cannot occur when using the SWAP operation.
The above example also shows that the use of the symmetric parts A 2 and B 2 can reduce the entanglement cost for exchanging A 1 and B 1 . Since the initial state |φ 1 does not satisfy the necessary condition in Theorem 5, we obtain e A 1 ↔B 1 (φ 1 ) > 0. Observe that the isometry U (V) in Eq. (6) represents Alice's (Bob's) local operation CNOT A (CNOT B ) whose target and controlled systems are A 1 (B 1 ) and A 2 (B 2 ), respectively. This implies that Alice and Bob can exchange A 1 and B 1 by using local operations. It follows that 0 ≥ e A 2 B 2 A 1 ↔B 1 (φ 1 ). In fact, e A 2 B 2 A 1 ↔B 1 (φ 1 ) = 0 from Corollary 3. Therefore, we obtain e A 1 ↔B 1 (φ 1 ) > e A 2 B 2 A 1 ↔B 1 (φ 1 ). When A 2 and B 2 are symmetric, we can show the following relation between the optimal entanglement costs by definition.  From Proposition 6, we can see that, when Alice and Bob exchange systems A and B of |ψ with symmetric parts A 2 and B 2 , they can achieve the optimal entanglement cost by exchanging only A 1 and B 1 , making the most of this symmetry.
(2) Negative entanglement cost.-As in the asymptotic quantum state exchange task [1,2], there exist initial states to show that the entanglement cost of the OSQSE task can be negative. Assume that Alice and Bob exchange the parts A 1 and B 1 of the initial state where |φ 2 consists of two ebits |e A 1 B 2 and |e B 1 A 2 . To exchange A 1 and B 1 , both Alice and Bob prepare an ebit, respectively, and they locally implement entanglement swapping [24] by performing two Bell measurements on A 2 , B 2 , and the parts of the ebits, as described in Fig. 3. Then they can exchange A 1 and B 1 , and can share two ebits at the same time. This means that the entanglement cost can be negative.
In fact, we have e A 2 B 2 A 1 ↔B 1 (φ 2 ) = −2 from Corollary 3. This is in stark contrast with the SWAP operation, which cannot lead to creation of shared entanglement between Alice and Bob.
From Proposition 6, we can know that if A 2 and B 2 are symmetric, then e A 2 B 2 A 1 ↔B 1 (ψ) cannot be negative. One may ask the question: Is there any condition that implies the nonnegativity of the optimal entanglement cost e A 2 B 2 A 1 ↔B 1 ? To answer this question, we present the following inequalities.
where e A 2 B 2 is the optimal entanglement cost for exchanging B 1 and A 1 when using A 2 and B 2 , and e is the optimal entanglement cost for exchanging A 2 and B 2 when using B 1 and A 1 .
In Proposition 7, the first inequality comes from the fact that Alice and Bob cannot increase the amount of entanglement between them by means of LOCC [25], while the second one is straightforward from the definitions of the optimal entanglement costs. From Proposition 7, we can see that if e A 2 B 2 In particular, let us assume that A 1 and B 1 are symmetric. Then it is obvious that 0 ≥ e A 2 B 2 A 1 ↔B 1 (ψ), from Proposition 1. If 0 > e A 2 B 2 A 1 ↔B 1 (ψ) then it follows from Proposition 7 that e A 2 B 2 B 1 ↔A 1 (ψ f 1 ) > 0. However, since B 1 and A 1 are also symmetric, Proposition 1 implies e A 2 B 2 B 1 ↔A 1 (ψ f 1 ) ≤ 0, which leads to a contradiction. Therefore, we obtain the following corollary.
This tells us that if A 1 and B 1 are symmetric, Alice and Bob cannot increase the amount of shared entanglement after the OSQSE task, even if they make use of the parts A 2 and B 2 .
Conclusion.-In this work, we have considered a one-shot version of the original quantum state exchange task, and have formally defined the OSQSE task and its optimal entanglement costs. We have derived converse bounds on the optimal entanglement costs, and have presented conditions on the initial state to achieve zero entanglement cost. As a related open problem, we can ask the following question: If e A↔B (ψ) = 0, then is it possible to exchange the parts A and B, without classical communication and entanglement, that is, are there local operations L A and L B such that ψ f 12 = (L A ⊗ L B ) (ψ)?
The two counter-intuitive phenomena are the most interesting contribution of this work, showing the major difference between the SWAP operation and the OSQSE. One phenomenon tells us that it is worth using the symmetric parts in order to optimally perform the OSQSE. The other shows that the entanglement cost of the OSQSE can be negative. A further open problem is whether the catalytic use of entanglement [26][27][28] can reduce the optimal entanglement cost for the OSQSE. To be more specific, for the initial state |ψ , do there exist a bipartite entangled state |ψ c A 3 B 3 shared by Alice and Bob and a OSQSE protocol C K,L : Theoretically, the OSQSE is a powerful two-user quantum communication task, which includes quantum teleportation [29] and quantum state merging [3,4] as special cases. Practically, this task can be a fundamental building block for applications involving multiple users, such as distributed quantum computation [30,31] and quantum network [32][33][34].
When X = A and Y = B, consider the Schmidt decompositions of |ψ , For the computational bases {| j } and {|k } on the systems A and B, respectively, we have where [Ω i AB (ψ)] jk = j| A ⊗ k| B |ξ i AB . If the parts A and B are perfectly exchanged, then Alice and Bob hold the final state By the hypothesis, there exist isometries U and V such that for each i, So we have, for each i, [Ω i AB (ψ)] lm j| U |l k| V t |m , which implies that Hence, e A↔B (ψ) = 0. Similarly, we can show that e A 1 ↔B 1 (ψ) = 0 by using isometries U and V such that for each i, for α ∈ (0, 1) ∪ (1, ∞) In addition, if α = 1, then Finally, we have It follows that H α (Z 1 ) = H α (Z 2 ) for all α ∈ [0, ∞]. By the induction hypothesis, there exists a permutation σ ∈ S k−1 such that p i = q σ (i) for all i ∈ [k − 1]. Define σ(1) = 1 and σ(i) = σ (i − 1) with i 1. Then σ ∈ S k and p i = q σ(i) for all i ∈ [k]. Therefore, the statement is true for N = k.
In fact, we can prove Lemma 9 by assuming a weaker condition as follows. Let S be a subset of [0, ∞] including 0, the extended real number ∞, and a sequence {s n } n∈N such that lim n→∞ s n = ∞. Then we can show that if H α (Z) = H α (W) holds for all α ∈ S , then Z and W have the same probability distribution.
The contrapositive of the following lemma proves Theorem 5.
Lemma 10 (Sufficient conditions on the initial state |ψ with e X↔Y (ψ) > 0). Let (X, Y) be the pair of two systems, which can be either (A 1 , B 1 ) or (A, B). Let {λ i } N i=1 and {τ i } M i=1 be nonzero eigenvalues for the reduced states ρ X and ρ Y of |ψ , respectively, which satisfy λ 1 ≥ · · · ≥ λ N , τ 1 ≥ · · · ≥ τ M , and (ii) Suppose that |ψ satisfies N = M and λ i τ i for some i ∈ [N]. Let Z and W be discrete random variables on alphabets Z and W with |Z| = |W| = N, whose probability distributions are {λ i } N i=1 and {τ i } N i=1 , respectively. Let us consider the set then A is a non-empty subset of [N], since i ∈ A. So we can choose the largest element in A, say j. Then λ j τ j and λ i = τ i for all i > j by the definition of the set A. If λ j > τ j (or λ j < τ j ) than λ i > τ j (or λ j < τ i ) for all i ∈ [ j]. Thus λ i τ j (or λ j τ i ) for all i ∈ [ j], which shows that for each σ ∈ S j , there exists i ∈ [ j] such that λ i τ σ(i) . It follows that for each σ ∈ S N , there exists i ∈ [N] such that λ i τ σ(i) . From the contrapositive of Lemma 9, there exists α ∈ [0, ∞] such that H α (X) H α (Y). Therefore, from the converse bound in Eq. (4), we obtain