coherent classical communication and entanglement recycling 经典通信和量子纠缠相干循环

CoherentClassicalCommunication,Aram Harrow (MIT)quant-ph/0307091,Outline,What is coherent classical communication (CCC)?Where does CCC come from?What is CCC good for? Remote state preparation with CCCNoisy CCC and applications,beyond qubits and cbits,Let |xix=0,1 be a basis for C2.q!q: |xiA!|xiB (qubit)c!c: |xiA!|xiB|xiE (cbit)qq: |Fi=2-1/2x|xiA|xiB (ebit)cc: 2-1/2x|xiA|xiB|xiE (rbit)c!c: |xiA!|xiA|xiB (coherent cbit)(notation due to Devetak and Winter, quant-ph/0304196),simple resource relations,Trivial relations:q!q c!c c!c ccq!q c!c qq ccTeleportation (TP):2c!c + qq q!qSuper-dense coding (SDC):q!q + qq 2c!c (coherent output!),distributed unitary gates,Theorem:If U is a unitary gate on HAHB such that U + e qq C!c!c + Ccc (A)then U + e qq C!c!c + Ccc (A).Examples:CNOTAB|xiA|0iB=|xiA|xiB(HZaI)CNOTAB(XaZb)2-1/2x|xiA|xiB=|biA|aiBNote:1. The proof requires careful accounting of ancillas.2. It also holds for isometries (e.g. |xiA!|xiA|xiB),Teleportation,H,X,Z,2 c!c + 1 qq 1 q!q,Before measuring, the state is 2-1ab|ai|biAZaXb|yiB.,Teleportation with coherent communication,H,X,Z,2 c!c + 1 qq 1 q!q+ 2 qq,coherentclassicalcomm.,the power of coherent cbits,Teleportation with recycling:2 c!c + 1 qq 1 q!q+ 2 qq2 c!c 1 q!q+ 1 qq (C)Super-dense coding:1 q!q+ 1 qq 2 c!c (C)Therefore:2 c!c = 1 q!q+ 1 qq (C)Teleportation and super-dense coding are no longer irreversible.,Recycling in the remote CNOT,H,=,c!c + cc + qq CNOT Gottesman, quant-ph/9807006c!c + cc + qq CNOT + 2 qqc!c + cc CNOT + qq (C),the power of a CNOT,Making a remote CNOT coherent:c!c + cc CNOT + qq (C)Using a CNOT for bidirectional communication:(HZaI)CNOTAB(XaZb)2-1/2x|xiA|xiB=|biA|aiBCNOT + qq c!c + ccCombined: CNOT + qq = c!c + cc (C)2 CNOT = 2 c!c + 2 cc 2 qq = q!q + qq = SWAP (C),Remote State Preparation,1 cbit + 1 ebit 1 remote qubitGiven |Fdi and a description of y2Cd, Alice can prepare y in Bobs lab with error e by sending him log d + O(log (log d)/e2) bits.Bennett, Hayden, Leung, Shor and Winter, quant-ph/0307100,definitions of remote qubits,What does it mean for Alice to send Bob n remote qubits?She can remotely prepare one of,RSP lemma,For any d and any e0, there exists n=O(d log d/e2) and a set of d x d unitary gates R1,Rn such that for any y,Use this to define a POVM:,RSP protocol,k,Neumarks theorem:any measurement can be made unitary,k,Entanglement recycling in RSP,UA,discard,coherentclassicalcommunicationof log n bits,Implications of recycled RSP,1 coherent cbit 1 remote qubit (with catalysis),Corollary 1: The remote state capacity of a unitary gate equals its classical capacity.Corollary 2: Super-dense coding of quantum states (SDCQS)1 qubit + 1 ebit 2 remote qubits (with catalysis)(Note: Harrow, Hayden, Leung; quant-ph/0307221 have a direct proof of SDCQS.),RSP of entangled states (eRSP),Let E=pi,yi be an ensemble of bipartite pure states. Define S(E)=S(ipiTrAyi), E(E)=ipiS(TrAyi), c(E)=S(E)-E(E).eRSP: c(E) c!c + S(E) qq E (A) BHLSWmake it coherent: c(E) c!c + E(E) qq E (A)use super-dense coding: c(E)/2 q!q + (E(E)+ c(E)/2) qq E (A),Unitary gate capacities,Define Ce to be the forward classical capacity of U assisted by e ebits of entanglement per use, so that1 use of U + e qq Ce c!c (A),(In BHLS; quant-ph/0205057, this was proved for e=1.),Solution:Ce=supE c(UE) - c(E) : E(E) - E(UE)6e,Warmup: entanglement capacity,Define E(U) to be the largest number satisfyingU E(U) qq (A).Claim: E(U) = sup|yi E(U|yi) E(|yi)Proof: BHLS; quant-ph/0205057 |yi + U U|yi E(U|yi) qq (concentration)|yi + E(U|yi)-E(|yi) qq (dilution)Thus: U E(U|yi)-E(|yi) qq (A),Coherent HSW coding,Lemma: Let E=pi,yi be an ensemble of bipartite pure states that Alice can prepare in superposition. ThenE c(E) c!c + E(E) qq (A)Proof: Choose a good code on En. Bobs measurement obtains nc(E) bits of Alices message and determines the codeword with high probability, causing little disturbance. Thus, this measurement can be made coherent. Since Alice and Bob know the codeword, they can then do entanglement concentration to get nE(E) ebits.,Protocol achieving Ce,E + U UE c(UE) c!c + E(UE) qq(coherent HSW) E + (c(UE)-c(E) c!c + (E(UE)-E(E) qq (coherent RSP)Thus, U + (E(E)-E(UE) qq (c(UE)-c(E) c!c (A),Quantum capacities of unitary gates,Define Qe(U) to be the largest number satisfyingU + e qq Qe q!q.Using 2c!c = 1q!q + 1qq, we find,Summary,2 coherent cbits = 1 qubit + 1 ebit2 CNOT = SWAP (catalysis)1 qubit + 1 ebit 2 remote qubits (catalysis)eSDCQS using c/2 qubits and S-c/2 ebits.Single-letter expressions for Ce and Qe.Remote state capacities and classical capacities are equal for unitary gates.,Noisy CCCDevetak, Harrow, Winter; quant-ph/0308044,Two minute proofs of the hashing inequality and the quantum channel capacity.Generalizations of these protocols to obtain the full trade-off curves for quantum channels assisted by a limited amount of entanglement and entanglement distillation with a limited amount of communication.,Noisy CCC: definitions,Let rAB be a bipartite state and |yiABE its purification.I(A:B) = H(A) + H(B) H(E)I(A:E) = H(A) + H(E) H(B)Ic = H(B) H(E) = (I(A:B) I(A:E)If N is a noisy channel, then evaluate the above quantities on (IN)|Fri, where |Fri is a purification of Alices input r.qq = one copy of rABq!q = one use of N,Noisy CCC: applications,Old results:S(A) qq + q!q I(A:B) c!c BSST; q-ph/0106052q!q Ic q!q Shor; unpublishedS(A) q!q + qq I(A:B) c!c HHHLT; q-ph/0106080I(A:E) c!c + qq Ic qq DW; q-ph/0306078New results:I(A:E)/2 qq + q!q I(A:B)/2 q!qfatherI(A:E)/2 q!q + qq I(A:B)/2 qqmother,A family of quantum protocols,fa