1. Coherent Classical Communication Aram Harrow (MIT) quant-ph/0307091

2. Outline What is coherent classical communication (CCC)? Where does CCC come from? What is CCC good for? Remote state preparation with CCC Noisy CCC and applications

3. beyond qubits and cbits Let {|x i } x=0,1 be a basis for C 2. [q ! q]: |x i A ! |x i B (qubit) [c ! c]: |x i A ! |x i B |x i E (cbit) [qq]: |  i =2 -1/2  x |x i A |x i B (ebit) [cc]: 2 -1/2  x |x i A |x i B |x i E (rbit) [[c ! c]]: |x i A ! |x i A |x i B (coherent cbit) (notation due to Devetak and Winter, quant- ph/0304196)

4. simple resource relations Trivial relations: [q ! q] > [[c ! c]] > [c ! c] > [cc] [q ! q] > [[c ! c]] > [qq] > [cc] Teleportation (TP): 2[c ! c] + [qq] > [q ! q] Super-dense coding (SDC): [q ! q] + [qq ] > 2[[c ! c]] (coherent output!)

5. distributed unitary gates Theorem: If U is a unitary gate on H A ­H B such that U + e [qq] > C ! [c ! c] + C à [c à c] (A) then U + e [qq] > C ! [[c ! c]] + C à [[c à c]] (A). Examples: CNOT AB |x i A |0 i B =|x i A |x i B (HZ a ­ I)CNOT AB (X a ­ Z b )2 -1/2  x |x i A |x i B =|b i A |a i B Note: 1. The proof requires careful accounting of ancillas. 2. It also holds for isometries (e.g. |x i A ! |x i A |x i B )

6. Teleportation H XZ 2 [c ! c] + 1 [qq] > 1 [q ! q] + 2 [cc] uniformly random Before measuring, the state is 2 - 1  ab |a i |b i A Z a X b |  i B.

7. Teleportation with coherent communication H XZ 2 [[c ! c]] + 1 [qq] > 1 [q ! q]+ 2 [qq] coherent classical comm. 2 -1  ab |ab i A Z a X b |  i B 2 -1  ab |ab i A |ab i B Z a X b |  i B

8. the power of coherent cbits Teleportation with recycling: 2 [[c ! c]] + 1 [qq] > 1 [q ! q]+ 2 [qq] 2 [[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.

9. Recycling in the remote CNOT H = [c ! c] + [c à c] + [qq] > CNOT [Gottesman, quant-ph/9807006] [[c ! c]] + [[c à c]] + [qq] > CNOT + 2 [qq] [[c ! c]] + [[c à c]] > CNOT + [qq] (C)

10. the power of a CNOT Making a remote CNOT coherent: [[c ! c]] + [[c à c]] > CNOT + [qq] (C) Using a CNOT for bidirectional communication: (HZ a ­ I)CNOT AB (X a ­ Z b )2 -1/2  x |x i A |x i B =|b i A |a i B CNOT + [qq] > [[c ! c]] + [[c à c]] Combined: CNOT + [qq] = [[c ! c]] + [[c à c]] (C) 2 CNOT = 2 [[c ! c]] + 2 [[c à c]] – 2 [qq] = [q ! q] + [q à q] = SWAP (C)

11. Remote State Preparation 1 cbit + 1 ebit > 1 remote qubit Given |  d i and a description of  2 C d, Alice can prepare  in Bob’s lab with error  by sending him log d + O(log (log d)/  2 ) bits. [Bennett, Hayden, Leung, Shor and Winter, quant-ph/0307100]

12. definitions of remote qubits What does it mean for Alice to send Bob n remote qubits? She can remotely prepare one of

13. RSP lemma For any d and any  >0, there exists n=O(d log d/  2 ) and a set of d x d unitary gates R 1,…,R n such that for any , Use this to define a POVM:

14. RSP protocol k

15. Neumark’s theorem: any measurement can be made unitary k UMUM

16. Entanglement recycling in RSP UAUA discard coherent classical communication of log n bits

17. 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.)

18. RSP of entangled states (eRSP) Let E ={p i,  i } be an ensemble of bipartite pure states. Define S( E )=S(  i p i Tr A  i ), E( E )=  i p i S(Tr A  i ),  ( E )=S( E )-E( E ). eRSP:  ( E ) [c ! c] + S( E ) [qq] > E (A) [BHLSW] make it coherent:  ( E ) [[c ! c]] + E( E ) [qq] > E (A) use super-dense coding:  ( E )/2 [q ! q] + (E( E )+  ( E )/2) [qq] > E (A)

19. Unitary gate capacities Define C e to be the forward classical capacity of U assisted by e ebits of entanglement per use, so that 1 use of U + e [qq] > C e [c ! c] (A) (In [BHLS; quant-ph/0205057], this was proved for e= 1.) Solution: C e =sup E {  (U E ) -  ( E ) : E( E ) - E(U E ) 6 e }

20. Warmup: entanglement capacity Define E(U) to be the largest number satisfying U > E(U) [qq] (A). Claim: E(U) = sup |  i E(U|  i ) – E(|  i ) Proof: [BHLS; quant-ph/0205057] |  i + U > U|  i > E(U|  i ) [qq] (concentration) > |  i + E(U|  i )-E(|  i ) [qq] (dilution) Thus: U > E(U|  i )-E(|  i ) [qq] (A)

21. Coherent HSW coding Lemma: Let E ={p i,  i } be an ensemble of bipartite pure states that Alice can prepare in superposition. Then E >  ( E ) [[c ! c]] + E( E ) [qq] (A) Proof: Choose a good code on E ­ n. Bob’s measurement obtains ¼ n  ( E ) bits of Alice’s 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 ¼ nE( E ) ebits.

22. Protocol achieving C e E + U > U E >  (U E ) [[c ! c]] + E(U E ) [qq](coherent HSW) > E + (  (U E )-  ( E )) [[c ! c]] + (E(U E )-E( E )) [qq] (coherent RSP) Thus, U + (E( E )-E(U E )) [qq] > (  (U E )-  ( E )) [[c ! c]] (A)

23. Quantum capacities of unitary gates Define Q e (U) to be the largest number satisfying U + e [qq] > Q e [q ! q]. Using 2[[c ! c] = 1[q ! q] + 1[qq], we find

24. Summary 2 coherent cbits = 1 qubit + 1 ebit 2 CNOT = SWAP (catalysis) 1 qubit + 1 ebit > 2 remote qubits (catalysis) eSDCQS using  /2 qubits and S-  /2 ebits. Single-letter expressions for C e and Q e. Remote state capacities and classical capacities are equal for unitary gates.

25. Noisy CCC [Devetak, 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.

26. Noisy CCC: definitions Let  AB be a bipartite state and |  i ABE its purification. I(A:B) = H(A) + H(B) – H(E) I(A:E) = H(A) + H(E) – H(B) I c = H(B) – H(E) = ½ (I(A:B) – I(A:E)) If N is a noisy channel, then evaluate the above quantities on (I ­N )|   i, where |   i is a purification of Alice’s input . {qq} = one copy of  AB {q ! q} = one use of N

27. Noisy CCC: applications Old results: S(A) [qq] + {q ! q} > I(A:B) [[c ! c]] [BSST; q- ph/0106052] {q ! q} > I c [q ! q] [Shor; unpublished] S(A) [q ! q] + {qq} > I(A:B) [[c ! c]] [HHHLT; q- ph/0106080] I(A:E) [c ! c] + {qq} > I c [qq] [DW; q- ph/0306078] New results: I(A:E)/2 [qq] + {q ! q} > I(A:B)/2 [q ! q][father] I(A:E)/2 [q ! q] + {qq} > I(A:B)/2 [qq][mother]

28. A family of quantum protocols father mother hashing inequality [DW] I(A:B) [c ! c] + {qq} > I c [q ! q] [HDW/Burkard] C E [BSST] noisy SDC [HHHLT] {q ! q} > I c [q ! q] [Shor] TPTP TPTP TPTP SD C [q ! q] > [qq] SD C TP= teleportation SDC = super-dense coding

29. A gate with asymmetric capacities? x=0,…,d-1, U 2 C d ­ C d U |x0 i = |xx i U |xx i = |x0 i U |xy i = |xy i for x  y  0. C 1 = log d C 2 > (log d)/2