Entanglement teleportation through GHZ-class states

We first consider teleportation of entangled states shared between Claire and Alice to Bob1 and Bob2 when Alice and the two Bobs share a single copy of a GHZ state and where all the four parties are at distant locations. We extend this result to a more general state than GHZ-state and show that still a class of pure entangled states can be teleported, where the entanglement of this class ranges from 0 to e (⩽1), depending on the entanglement (defined in the text) of the channel state. We then generalize this situation to the case of teleportation of entangled states shared between Claire1, Claire2, ..., Claire(N - 1) and Alice to Bob1, Bob2, ..., BobN when Alice and the N Bobs share a single copy of a GHZ-class state and where again all the 2N parties are at distant locations.


48.2
state is reconstructed). This seemingly impossible phenomenon is made possible (in the case of a two level system) by allowing Alice and Bob to a priori share a maximally entangled state of two qubits say, It is important to note that the entanglement in the channel state vanishes completely after it has been used to send the qubit by using the BBCJPW protocol. Now if the sent qubit is a priori entangled with another qubit (possessed by Claire), it will remain so after teleportation. That is, the previously shared entanglement between Alice and Claire would now be shared between Bob and Claire. Consider now a different situation. A source delivers an arbitrary two-qubit entangled state to Alice which must finally be shared between Bob1 and Bob2. Instead of state teleportation, Alice therefore has the task of entanglement teleportation. It would be sufficient if Alice shares a maximally entangled state with Bob1 and another with Bob2. Alice would then just teleport the two qubits using the BBCJPW protocol [2] †.
But what if Alice shares with the Bob1-Bob2 system, less than two ebits of entanglement? Suppose for example that instead of the two maximally entangled states, Alice, Bob1 and Bob2 share the GHZ state [3] |GHZ = 1 √ 2 (|000 + |111 ).
Would the same feat be possible now? Gorbachev and Trubilko [4] have considered this case and shown that if Alice knows that the state has been prepared in the Schmidt basis {|0 0 , |1 1 }, i.e., if Alice knows that the state is of the form |χ = α|0 0 + β|1 1 with known |0 0 and |1 1 but unknown Schmidt coefficients α, β, then this state can be made to share between Bob1 and Bob2. In this paper we simplify their protocol. Shi et al [5] generalized this situation to the case in which the state |χ is shared between two separated parties Alice and Claire ‡. The protocol of Shi et al is probabilistic as in their case Alice and the two Bobs share the state instead of a GHZ state. Here we show that even if Alice and the two Bobs share the state where |φ and |φ are not necessarily orthogonal, it is possible for Alice and Claire to make the two Bobs share the state Our protocol is independent of the ones in [5]. More important is the fact that our protocol is generalizable to the N -party situation. We also touch the probabilistic case in both situations.

48.3
Suppose a source prepares the state |χ (with known |0 0 and |1 1 but unknown α, β) and delivers it to Alice who wants to make it shared between Bob1 and Bob2 through a GHZ state which she shares with the Bobs. This situation has been considered in [4]. We simplify their protocol and show that |χ can be made to share between the two Bobs by simply using the BBCJPW protocol. Indeed |χ is essentially a qubit. What is important is that there is no nonlocal operation involved between the two Bobs in the protocol. Let us elaborate.
First Alice transforms |χ to which is possible as the Schmidt basis {|0 0 , |1 1 } is known. The combined state |ξ 12 |GHZ AB 1 B 2 may be written as Alice now performs a projection-valued measurement on her three qubits (1, 2 and A) with the projection operators and communicates the result to Bob1 and Bob2. If P GHZ 1 clicks, the Bobs are to do nothing. They would then already share the state |ξ . If P GHZ 2 clicks, Bob2 does nothing but Bob1 applies σ z on his particle. If P GHZ 3 clicks, both of them apply σ x i.e. they apply σ x ⊗ σ x . And if P GHZ 4 clicks they apply σ x ⊗ iσ y . P GHZ 5 would never click as probability of occurance of this projector in the input state considered above is zero. Finally, Bob1 and Bob2 share the state |ξ on which they apply U 1 ⊗ U 2 to transform it to |χ where U 1 is the unitary operator that transforms |0 → |0 and |1 → |1 and U 2 the unitary operator that transforms |0 → |0 and |1 → |1 . Note that throughout the process there is no nonlocal operation involved between Bob1 and Bob2.
Interestingly Shi et al [5] have generalized this situation to the case in which two separated parties Alice and Claire share the state |χ , and they cannot perform joint operation on their two particles. Instead of describing that protocol where Alice, Bob1 and Bob2 share a GHZ state, we generalize this result where Alice and the two Bobs share the state where |φ and |φ are not necessarily orthogonal, it is possible for Alice and Claire to make the two Bobs share the state |χ = α|φ0 + β|φ 1 (see figure 1).
Note that |ghz is a state of the 'GHZ-class' [6]. The initial combined state is where the particles 1 and 2 belong to Claire and Alice respectively. First of all Alice performs the unitary operation U , on the qubit 2, that transforms |0 → |0 and |1 → e −iε |1 where φ|φ = re iε so that |χ 12 transforms to where |φ = e −iε |φ . Alice also applies the unitary operator, on qubit A, that transforms |0 → |0 and |1 → e −iε |1 so that |ghz transforms to The state of the five particles is now which may be rewritten as

48.5
Alice now conducts a projection measurement (the Bell measurement) on her two particles with the projection operators After that she sends two bits of classical message to each of Bob1 and Bob2 to tell them the result of the Bell measurement.
As φ|φ (=r) belongs to [0, 1], there exists a unique orthonormal basis {|a , |a } such that Note that θ, |a , |a are all known. Claire performs a projective measurement on just the basis {|a , |a } and communicates the result to the Bobs.
It is now straightforward to see that there always exists a product-unitary operation between the two Bobs, depending upon the results communicated by Alice and Claire, so that they (the Bobs) end up sharing the state |ξ . If P 1 clicks in Alice's measurement, then Claire and the two Bobs share the state If after her measurement, Claire obtains the result |a , then the Bobs are to do nothing. On the other hand if she obtains the result |a , only Bob2 is to perform a unitary operation that transforms |0 → |0 and |1 → −|1 , which is just σ z . If P 2 clicks, then Claire and the Bobs share the state In this case, the Bobs are to do nothing if Claire obtains |a and only Bob2 is to apply σ z if Claire obtains |a . If P 3 clicks, then the shared state is Since the inner product of |φ and |φ is real, there exists a unitary operator U that transforms |φ → |φ and |φ → |φ . Irrespective of what Claire obtained, Bob1 is to apply just this operator. However, Bob2 is to do nothing if Claire obtains |a but to apply σ z if Claire obtains |a . If P 4 clicks in Alice's measurement, Bob1 again applies U irrespective of Claire's result. And Bob2 is to do nothing if Claire obtains |a and to apply σ z if she obtains |a . At the end of 48.6 all these the Bobs are left with the state |ξ on which Bob2 has to perform a rotation to transform it to |χ . Precisely Bob2 has to apply (U ) −1 .
Looking at the protocol described above, which is essentially a generalization of the BBCJPW protocol, it may seem that the measurement of Claire must be preceded by that of Alice, and so Alice must communicate to Claire that her (Alice's) measurement has been performed. But that is not true. The protocol would go through irrespective of whether Alice or Claire performed the first measurement. Indeed the measurements are to be performed in two different Hilbert spaces and the corresponding projection operators would therefore commute. For example, if Alice obtains |φ + and Claire obtains |a , the Bobs would share the state α|φ0 + β|φ 1 irrespective of who performed the first measurement. Now we define entanglement Q n (|ϕ ) of a n-qubit pure state |ϕ = x∈S n u x |x (where S n = {0, 1} × {0, 1} × · · · n times, and x∈S n |u x | 2 = 1), provided by Meyer and Wallach [7]: where the linear map l j (b) (for b = 0, 1) is given by Using this definition, we have Q 3 (|ghz ) = 1 3 (3 − | φ|φ | 2 ), while Q 2 (|χ ) = 4|αβ| 2 (1 − | φ|φ | 2 ). Thus, for all α and β, the entanglement (Q 2 ) of |χ varies from 0 to (1−| φ|φ | 2 ) = e (say). And the entanglement (Q 3 ) of the channel state |ghz varies from 2 3 to 1 3 (2 + e). Thus here we see that for exactly teleporting a two qubit pure state, whose entanglement (Q 2 ) is <1, a three qubit pure state, whose entanglement (Q 3 ) is <1, is sufficient as a channel state.
For completeness, note that if Alice and two Bobs share the state |ghz = a|0φ0 + b|1φ 1 the above entanglement teleportation is possible in a probabilistic manner where Alice has to change her operations in the same way as exact teleportation was changed to probabilistic teleportation in [8]. Suffice it to mention that the combined state may be written as We now go over to the N -party case. Suppose there are N Bobs, Bob1, Bob2, . . . , BobN and they share with Alice a GHZ-class state where |φ i and |φ i (i = 1, 2, . . . , N − 1) are not necessarily orthogonal. It would then be possible to make the N Bobs share the state initially shared between Claire1, Claire2, . . . , Claire(N − 1) and Alice where |φ 1 φ 2 . . . φ N −1 0 and |φ 1 φ 2 . . . φ N −1 1 are known but α, β are unknown. The particles 1, 2, . . . , N belong respectively to Claire1, Claire2, . . . , Claire(N − 1) and Alice. Alice first transforms |χ N to 1, 2, . . . , N − 1). To effect this transformation, Alice has to apply the unitary operator, on qubit N , that transforms Alice also applies the unitary operator, on qubit A, that transforms |0 → |0 and |1 → e −i N −1 i=1 ε i |1 so that |ghz transforms to The combined state of the 2N + 1 particles is now ..B N which may be written as As before, Alice now performs a Bell measurement on her two qubits. And Claire(i) performs a projection measurement in the orthonormal basis {|a i , |a i } determined uniquely by where θ i ∈ [0, π/2] (i = 1, 2, . . . , (N − 1)). Alice and the Claires now communicate their results to the Bobs. By now it is obvious that whatever the result at Alice and the Claires is, there would always exist a product-unitary operator between the N Bobs so that they (the Bobs) are left with the state |ξ N which may henceforth be transformed locally to |χ N . Let us add that if Alice and the N Bobs share the state the above entanglement teleportation of |χ N would be possible in a probabilistic manner.

48.8
In all the above cases of entanglement teleportation considered above, the teleported entangled state is essentially a qubit as each of them is a superposition of two different states. And in the deterministic cases, Alice shares 1 ebit of entanglement with the Bobs. Now 1 ebit may be used to teleport at most 1 qubit. For if it were possible to teleport two qubits, these could a priori be separately entangled maximally with two other qubits resulting in the creation of two ebits of entanglement using a single ebit †.
To summarize, we have considered here set-dependent entanglement teleportation when the available channel resource is less than what is needed for universal entanglement teleportation. In the case of teleportation of entangled states through the channel state 1 √ 2 (|000 + |111 ) AB 1 B 2 , the pure entangled states from the plane spanned by the known states |0 0 and |1 1 can be exactly teleported by Alice to Bob1-Bob2 [4]. Note that the amount of entanglement of these pure states vary from 0 to 1. Shi et al [5] have considered the probabilistic case when the channel is a|000 AB 1 B 2 +b|111 AB 1 B 2 and when the states to be teleported are themselves shared between Alice and a distant party Claire.
We have shown that an arbitrary pure entangled state which is a linear combination of the two known states |φ0 and |φ 1 (where |φ and |φ are arbitrary but fixed, in general, non-orthogonal states) can be deterministically or probabilistically teleported (from Alice-Claire to Bob1-Bob2) by using the channel as the GHZ-class states [6] 1 √ 2 (|0φ0 + |1φ 1 ) or a|0φ0 + b|1φ 1 respectively. Note that as the states |φ and |φ are in general non-orthogonal, the amount of the entanglement of the final states after teleportation vary from 0 to e, where e ≤ 1. And our protocol is deterministic or probabilistic depending on whether Alice shares 1 ebit with the Bob1-Bob2 system or less than that. It is interesting to note that although for both |GHZ = 1 √ 2 (|000 + |111 ) AB 1 B 2 and |ghz = 1 √ 2 (|0φ0 + |1φ 1 ) AB 1 B 2 , Alice shares 1 ebit with the Bobs, the channel state between Alice and Bob2 is distillable for |ghz (when |φ and |φ are non-orthogonal) and separable for |GHZ , while the channel states between Alice and Bob1 are separable for both ‡. This could somehow be the reason as to why 0 to 1 entanglement is not transferred through |ghz (when |φ and |φ are non-orthogonal) although the same is possible through |GHZ . Interestingly one can check that if Alice, Bob1 and Bob2 share the W-class state |W = 1 √ 2 |0 ⊗ |00 + 1 √ 2 |1 ⊗ 1 √ 2 (|01 + |10 ) [6] (where there is 1 ebit of entanglement between Alice and Bob1-Bob2), and uses the same teleportation protocol as described above, then no pure entangled state (shared between Claire and Alice) can be teleported exactly to Bob1 and Bob2. This situation can easily be generalized to the N -party case.