One-shot quantum state exchange
Yonghae Lee, Hayata Yamasaki, Gerardo Adesso, Soojoon Lee

TL;DR
This paper investigates a one-shot quantum state exchange protocol, establishing entanglement bounds, conditions for zero entanglement cost, and revealing counter-intuitive phenomena that differ from standard SWAP operations.
Contribution
It introduces the first analysis of one-shot quantum state exchange, deriving entanglement bounds and uncovering novel phenomena in entanglement management.
Findings
Lower bounds on entanglement needed for one-shot exchange
Conditions for zero entanglement cost in the protocol
Discovery of counter-intuitive phenomena in entanglement handling
Abstract
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…
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One-Shot Quantum State Exchange
Yonghae Lee
Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University, Seoul 02447, Korea
Hayata Yamasaki
Photon Science Center, Graduate School of Engineering, The University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan
Gerardo Adesso
School of Mathematical Sciences and Centre for the Mathematics and Theoretical Physics of Quantum Non-Equilibrium Systems, University of Nottingham, University Park, Nottingham NG7 2RD, United Kingdom
Soojoon Lee
Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University, Seoul 02447, Korea
School of Mathematical Sciences and Centre for the Mathematics and Theoretical Physics of Quantum Non-Equilibrium Systems, University of Nottingham, University Park, Nottingham NG7 2RD, United Kingdom
Abstract
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.
pacs:
03.67.Hk, 89.70.Cf, 03.67.Mn
Introduction.— In quantum information theory, the quantum state exchange Oppenheim and Winter ; Lee et al. (2019) is a quantum communication task, in which two users, Alice and Bob, exchange their quantum information by means of local operations and classical communication (LOCC) assisted by shared entanglement. A main research aim in the study of the quantum state exchange is to evaluate the least amount of entanglement needed for the task, as in other quantum communication tasks, such as quantum state merging Horodecki et al. (2005, 2006) and quantum state redistribution Devetak and Yard (2008); Yard and Devetak (2009).
Most quantum communication tasks Schumacher (1995); Horodecki et al. (2005, 2006); Devetak and Yard (2008); Yard and Devetak (2009); Abeyesinghe et al. (2009) including the quantum state exchange usually assume the asymptotic scenario, in which users can have an unbounded number of independent and identically distributed copies of an initial state, and they carry out their task with the copies. On the other hand, it is not easy in a realistic situation to prepare a sufficiently large number of state copies, and the amount of non-local resources available for the users is limited. To reflect these practical difficulties, quantum information research has focused more recently on the one-shot scenario Berta et al. (2011); Renes and Renner (2012); Wang and Renner (2012); Datta and Hsieh (2013); Berta et al. (2016); Zhao et al. (2018); Anshu et al. (2018); Regula et al. (2018); Yamasaki and Murao .
In this work, we introduce and study the one-shot quantum state exchange (OSQSE) task. This is not only a useful quantum communication task, but can also have a potential application in quantum computation. Let us consider a specific situation as follows. Alice and Bob want to carry out the SWAP gate Nielsen and Chuang (2010), which plays an important role in universal quantum computation Jozsa and Miyake (2008). The problem is that they cannot directly apply the SWAP gate, because they are far apart. If Alice and Bob are sharing prior entanglement, then the OSQSE can be a method to non-locally perform the SWAP gate, as both operationally provide the same result. However, the OSQSE has unique properties which we reveal in this work.
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 Oppenheim and Winter ; Lee et al. (2019). 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 and of the initial state with systems and , respectively. Alice’s and Bob’s goal is either to exchange their parts and or to exchange their whole parts and .
Specifically, let and be the final states of the task,
[TABLE]
where , and the dimension of system is identical to that of system . Note that , and , are Alice’s and Bob’s systems, respectively. Then three joint operations
[TABLE]
are called the OSQSE protocols of , if they are performed by LOCC between Alice and Bob, and satisfy
[TABLE]
where and are pure maximally entangled states with Schmidt rank and on systems and , respectively. It is possible to generalize the above definitions by 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. (One-Shot Quantum State Exchange) as follows: The first two protocols and indicate that only the parts and are exchanged, while the whole parts and are exchanged in the third protocol . In addition, the parts and can be used for exchanging and in the protocol , while and are untouched in the protocol .
Depending on the types of OSQSE protocols, we define three optimal entanglement costs
[TABLE]
where the quantity is called the entanglement cost of the OSQSE protocol, and the infimums are taken over all joint protocols , , and , respectively.
By the definitions of the optimal entanglement costs, we obtain the following proposition.
Proposition 1**.**
For any input state , .
Converse bounds.— A real number 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 Oppenheim and Winter ; Lee et al. (2019), we consider a one-shot version of the -assisted quantum state exchange task, in which the reference system is divided into two systems and , and then Alice and Bob receive the divided parts and , respectively, so that the initial state is divided into Alice’s parts and Bob’s parts . This can be realized by using a quantum channel and its complementary channel Wilde (2013). Let be an -assisted OSQSE protocol,
[TABLE]
with the entanglement cost such that , where . Note that is an LOCC protocol by Alice and Bob. Let be the maximally mixed state with rank . From the majorization condition for LOCC convertibility Nielsen (1999); Mari et al. (2014), the state majorizes the state . Let be an additive and Schur concave function van Dam and Hayden such that for any . From the Schur concavity of the function , the inequality holds. Since is additive and , it follows that . Since any protocol is also an -assisted OSQSE protocol for the initial state , we obtain the following theorem.
Theorem 2**.**
For any input state , the optimal entanglement cost is lower bounded by
[TABLE]
where is an additive and Schur concave function such that for any and is a quantum channel from to .
In Theorem 2, if is directly sent to either Alice or Bob without splitting, and we restrict the function to the quantum Rényi entropy of order van Dam and Hayden for a quantum state , then we obtain the following computable converse bounds.
Corollary 3**.**
For any input state , ,where is a function of and defined by .
We refer the reader to Appendix A for the proof of Corollary 3. Remark that the converse bound can be easily computed by means of analytical or numerical methods, since the function is one-variable and differentiable on . For the different types of the OSQSE protocols, we can also obtain a similar computable converse bound as follows:
[TABLE]
where the pair can refer either to or to .
We also remark that in Theorem 2, if is chosen as the von Neumann entropy Wilde (2013), then the converse bound recovers a theoretical converse bound in Refs. Oppenheim and Winter ; Lee et al. (2019). In addition, a computable converse bound therein is just in Corollary 3. By virtue of the additivity of , it is clear that and 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. Oppenheim and Winter ; Lee et al. (2019). For example, if the initial state is
[TABLE]
then we can find a value such that as depicted in Fig. 1. This example shows that our bound is tighter than the existing bound .
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 OSQSE task, then the optimal entanglement cost is zero. We first characterize this type of strategy. Let be a pair of two systems, which can be either or , and consider a spectral decomposition of the reduced state for , , where with . For each , we define the matrix by
[TABLE]
where and indicate the computational bases on Alice’s and Bob’s systems, respectively. Then we obtain the following sufficient condition.
Theorem 4**.**
Let be either or . If there exist isometries and such that, for each , , then .
Here, the isometries and indicate Alice’s and Bob’s local operations exchanging the parts and 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.
Theorem 5**.**
*Let be either or . If , then there exists an isometry such that . *
We remark that the converse of Theorem 5 is not true in general. Let us consider the following simple initial state
[TABLE]
then , from Proposition 1 and Corollary 3. However, the state satisfies the necessary condition in Theorem 5, since its reduced states and 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 , let us consider a scenario in which Alice and Bob exchange their whole information and . Assume that their parts and are symmetric, while the remaining parts and are not symmetric, i.e., the initial state satisfies and where is the operation swapping quantum states in systems and .
From a viewpoint of the SWAP operation, if Alice and Bob want to exchange and , then it suffices for them to exchange and , since is identical to . This situation can be more easily understood by using a cargo exchange as a 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 and in order to reduce the cargo exchange cost, because the cargoes and are the same.
On the other hand, in the OSQSE, the proper use of the symmetric parts and can more efficiently reduce the entanglement cost compared to exchanging only and without using and . To be specific, there exists an initial state such that the parts and are symmetric and while the rest parts and are not symmetric. Consider the specific initial state
[TABLE]
where and are symmetric but and are not. Since and , we can show that and satisfy the condition in Theorem 4, by setting
[TABLE]
Thus we obtain that , which means that and 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 and can reduce the entanglement cost for exchanging and . Since the initial state does not satisfy the necessary condition in Theorem 5, we obtain . Observe that the isometry () in Eq. (6) represents Alice’s (Bob’s) local operation () whose target and controlled systems are () and (), respectively. This implies that Alice and Bob can exchange and by using local operations. It follows that . In fact, from Corollary 3. Therefore, we obtain .
When and are symmetric, we can show the following relation between the optimal entanglement costs by definition.
Proposition 6**.**
*, if the parts and of are symmetric. *
From Proposition 6, we can see that, when Alice and Bob exchange systems and of with symmetric parts and , they can achieve the optimal entanglement cost by exchanging only and , making the most of this symmetry.
(2) Negative entanglement cost.— As in the asymptotic quantum state exchange task Oppenheim and Winter ; Lee et al. (2019), 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 and of the initial state
[TABLE]
where consists of two ebits and . To exchange and , both Alice and Bob prepare an ebit, respectively, and they locally implement entanglement swapping Żukowski et al. (1993) by performing two Bell measurements on , , and the parts of the ebits, as described in Fig. 3. Then they can exchange and , and can share two ebits at the same time. This means that the entanglement cost can be negative. In fact, we have 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 and are symmetric, then cannot be negative. One may ask the question: Is there any condition that implies the non-negativity of the optimal entanglement cost ? To answer this question, we present the following inequalities.
Proposition 7**.**
[TABLE]
where is the optimal entanglement cost for exchanging and when using and , and is the optimal entanglement cost for exchanging and when using and .
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 Vedral et al. (1997), while the second one is straightforward from the definitions of the optimal entanglement costs. From Proposition 7, we can see that if or is non-positive, then cannot be negative. Moreover, if the condition holds, then Proposition 7 implies .
In particular, let us assume that and are symmetric. Then it is obvious that , from Proposition 1. If then it follows from Proposition 7 that . However, since and are also symmetric, Proposition 1 implies , which leads to a contradiction. Therefore, we obtain the following corollary.
Corollary 8**.**
, if and are symmetric.
This tells us that if and are symmetric, Alice and Bob cannot increase the amount of shared entanglement after the OSQSE task, even if they make use of the parts and .
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 , then is it possible to exchange the parts and , without classical communication and entanglement, that is, are there local operations and such that ?
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. By observing the aforementioned examples involving the phenomena, we can provide another interesting open problem: If , do there exist Alice’s and Bob’s local operations and such that ?
A further open problem is whether the catalytic use of entanglement Jonathan and Plenio (1999); Eisert and Wilkens (2000); Majenz et al. (2017) can reduce the optimal entanglement cost for the OSQSE. To be more specific, for the initial state , do there exist a bipartite entangled state shared by Alice and Bob and a OSQSE protocol such that and ?
Theoretically, the OSQSE is a powerful two-user quantum communication task, which includes quantum teleportation Bennett et al. (1993) and quantum state merging Horodecki et al. (2005, 2006) as special cases. Practically, this task can be a fundamental building block for applications involving multiple users, such as distributed quantum computation Cirac et al. (1999); Bruß et al. (2004) and quantum network Cirac et al. (1997); Azuma et al. (2016); Pirker et al. (2018).
Acknowledgements.
We would like to thank Ryuji Takagi and Bartosz Regula for fruitful discussion. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science and ICT (NRF-2019R1A2C1006337) and the MSIT (Ministry of Science and ICT), Korea, under the ITRC (Information Technology Research Center) support program (IITP-2019-2018-0-01402) supervised by the IITP (Institute for Information & communications Technology Promotion). H. Y. acknowledges Grant-in-Aid for JSPS Research Fellow, JSPS KAKENHI Grant No. 18J10192, Cross-ministerial Strategic Innovation Promotion Program (SIP) (Council for Science, Technology and Innovation (CSTI)), and CREST (Japan Science and Technology Agency) JPMJCR1671. G. A. acknowledges support from the ERC Starting Grant GQCOP (Grant Agreement No. 637352).
Appendix A Proof of Corollary 3
We show that there exists a number such that . Note that the function is continuous on the compact set . So the extreme value theorem implies that there exists a number such that for all . Let us consider the function on the interval defined as
[TABLE]
then is continuous on . By using the extreme value theorem again, there exists a number such that for all . It follows that there exists a number such that for all . By setting , we obtain that for all .
Appendix B Proof of Theorem 4
When and , consider the Schmidt decompositions of ,
[TABLE]
where with . For the computational bases and on the systems and , respectively, we have
[TABLE]
where . If the parts and are perfectly exchanged, then Alice and Bob hold the final state
[TABLE]
By the hypothesis, there exist isometries and such that for each ,
[TABLE]
So we have, for each ,
[TABLE]
which implies that
[TABLE]
Hence, .
Similarly, we can show that by using isometries and such that for each , .
Appendix C Proof of Theorem 5
We use the following lemma in order to prove Theorem 5.
Lemma 9**.**
Let and be any discrete random variables on alphabets and with and . Let and be probability distributions for and , respectively. If the following equality holds for all ,
[TABLE]
where is the Rényi entropy of classical random variables, then and there exists a permutation such that for all , where is the set of all permutations on .
Note that and for each .
Proof.
Suppose that for all . Since , it holds that . For convenience, we assume that any probability distribution satisfies for all .
We now prove the statement by using mathematical induction on .
(i) If , then implies and so . Thus the statement is true.
(ii) Suppose that the statement is true for . Let and be discrete random variables on alphabets and with . Let and be probability distributions for and , respectively. Since , . By setting and for each , we can construct random variables and on alphabets and whose probability distributions are and , respectively. Obviously, , and so . Observe that for
[TABLE]
In addition, if , then
[TABLE]
Finally, we have
[TABLE]
It follows that for all . By the induction hypothesis, there exists a permutation such that for all . Define and with . Then and for all . Therefore, the statement is true for . ∎
In fact, we can prove Lemma 9 by assuming a weaker condition as follows. Let be a subset of including 0, the extended real number , and a sequence such that . Then we can show that if holds for all , then and have the same probability distribution.
The contrapositive of the following lemma proves Theorem 5.
Lemma 10** (Sufficient conditions on the initial state with ).**
Let be the pair of two systems, which can be either or . Let and be non-zero eigenvalues for the reduced states and of , respectively, which satisfy , , and . Then , if one of the following conditions holds:
(i) .
(ii) and for some .
Proof.
(i) If , then , which means
[TABLE]
by the converse bound in Eq. (4).
(ii) Suppose that satisfies and for some . Let and be discrete random variables on alphabets and with , whose probability distributions are and , respectively. Let us consider the set
[TABLE]
then is a non-empty subset of , since . So we can choose the largest element in , say . Then and for all by the definition of the set . If (or ) than (or ) for all . Thus (or ) for all , which shows that for each , there exists such that . It follows that for each , there exists such that . From the contrapositive of Lemma 9, there exists such that . Therefore, from the converse bound in Eq. (4), we obtain
[TABLE]
∎
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