Network quantum information theory:

Not a jungle anymore

Patrick Hayden (McGill University)

♀

A 1 A 2 E |  i n ABE B With Anura Abeyesinghe, Igor Devetak, Andreas Winter, and Jon Yard

Overview

 Distributed compression  Haar mother, Slepian-Wolf and all that  Quantum multiple access channels  Including stern words about non-additive capacity formulas  Quantum broadcast channels  Preview

♀ Haar Mother: Overview

 Who is she?

 What does she do for us?

 The simplest possible protocol  Polytime encodings for Alice  Application to distributed compression

♀ Who is she?

A 1 A A 2 E |  i n ABE B 1. Alice sends A 1 to Bob. A 1 is ~n ¢ I(A;E)  /2 qubits Devetak, Harrow, Winter [2003]

♀ Who is she?

A 2 E |  i n ABE A 1 B B 2 1. Alice sends A 1 to Bob. A 1 is ~n ¢ I(A;E)  /2 qubits 2. Bob performs a local operation At end: a) Alice and Bob share ~n ¢ I(A;B)  /2 ebits b) Bob holds a purification of Eve’s state Devetak, Harrow, Winter [2003]

♀ What does she do for us?

Quantum communication transfer state from A to B is ½ ¢ Ebits [I(A;E)-I(A;B)] = -H(A|E) = H(A|B).

1. Teleporting the [q !

q] gives entanglement distillation (hashing <>) 2. Further teleporting gives superdense coding from Alice to Bob using  AB 3. Teleporting over output entanglement proves “state merging” [HOW’05] Known proofs are complicated. Example: Perform mixed state superdense coding using  AB by invoking the HSW theorem. Prove that the data transmitted is “private”. Argue the protocol can be made coherent.

Devetak, Harrow, Winter [2003]

♀ The simplest possible protocol

Before: A 1 A 2 After: A 2 E |  i n ABE E |  i n ABE B B 1 B 2 0. Alice projects onto the typical subspace for  A 0.5. Alice applies a random unitary U A to her system 1. Alice sends A 1 to Bob. A 1 is ~n ¢ I(A;E)  /2 qubits The entanglement must be somewhere principle: it suffices to check that  A2 E = tr A1 [(U A I E )  n AE (U A y I E )] is nearly a product state on A 2 E.

The calculation

Hinges on purity of  A2E : Recall I(A 2 ;E) = H(A 2 ) + H(E) – H(A 2 E) SWAP The entanglement must be somewhere principle: it suffices to check that  A2 E = tr A1 [(U A I E )  n AE (U A y I E )] is nearly a product state on A 2 E.

The condition

The entanglement must be somewhere principle: it suffices to check that  A2 E =tr A1 [(U A I E )  AE (U A y I E )] is nearly a product state on A 2 E.

A 1 A 2 In our case: d d A1 A2 tr(   =  n ~ 2 n[I(A;E)  /2] ~ 2 AE 2 n[I(A;B) ) ~ 2  /2] -n[H(AE)  ] E |  i n ABE B Substituting gives near product state provided log d A1 >> nI(A;E)/2.

Belabouring the point

▪ ▪ ▪ The protocol couldn’t really be any simpler No conditional typical subspaces No concentration estimates required (just a single lowly average purity)

Bonus: Efficient encoding

   Replace the U A U A group twirl.

“twirl” by a Clifford Elements of the Clifford group can be drawn from the uniform distribution in polynomial time [DLT01].

Alice’s encoding can be implemented using a Clifford group element, so can be implemented in polynomial time.

Interim summary

 QIT coding theorems can be radically simplified – just Schumacher compression and averages over the unitary group required  Resulting proof easily modified to yield polytime quantum algorithm for Alice’s encoding

E

Distributed compression

A |  i n ABE B Alice and Bob transfer their shares of a pure tripartite entangled state to Charlie using as few qubits of communication as possible.

Distributed compression

A 1 A 2 E |  i n ABE A B B 1 B 2 Alice and Bob transfer their shares of a pure tripartite entangled state to Charlie using as few qubits of communication as possible.

Mother provides a way to do this, provided: R A ¸ ½ I(A;E) R B ¸ ½ I(B;E) R A + R B ¸ ½ [H(A)+H(B)+H(AB)]

Distributed compression

E |  i n ABE A 1 A 2 A’ Bob to Charlie: H(B) Alice to Charlie: I(A;E)/2 A B B Alice and Bob transfer their shares of a pure tripartite entangled state to Charlie using as few qubits of communication as possible.

Mother provides a way to do this, provided: R A ¸ ½ I(A;E) R B ¸ ½ I(B;E) R A + R B ¸ ½ [H(A)+H(B)+H(AB)] H(B)+ ½ I (A;E) = ½[H(A)+H(B)+H(AB)]

Rate region

A + R B > H(AB) Mother provides a way to do this, provided: R A ¸ ½ I(A;E) R B ¸ ½ I(B;E) R A + R B ¸ ½ [H(A)+H(B)+H(AB)] Conditions always necessary Condition necessary if  AB is separable and has a decomposition into irreducible ensemble of product states: [ADHW quant-ph/0403042.] ●If the decoder is restricted to isometric operations, a much stronger bound holds: R R A ¸ A ¸ H(A) and R H(A|B) R B B ¸ ¸ H(B).

●Doesn’t apply here: Charlie must discard extra entanglement with Alice.

Complete exploitation of correlation : R A + R B = H(AB)

Quantum multiple access channels

N : A’ B’ !

C

N

Classical MAC: capacity region found by Ahlswede ’71 and Liao ’72.

Classical capacity of a QMAC: Winter ‘01 Classical-quantum and quantum-quantum capacities: Yard, Devetak, H ‘05 quant-ph/0501045

QMAC: Quantum-quantum capacity

A A’ B’ B n n N

n

D

A* B* ´ S R Alice and Bob prepare inputs to n copies of the channel and Charlie decodes. The output should approximate two maximally entangled states, one with Alice of about nR+o(n) ebits and one with Bob of nS+o(n) ebits. In the limit n is achievable.

!1

, the fidelity should go to 1.

The capacity region consists of the (R,S) rate pairs for which this A A* B* B

Solution circa QIP 2005

Interpretation: Alice and Bob treat each others actions as noise. Independent decoding.

No-go theorem for use of quantum side information.

Solution post-QIP 2005

Interpretation: Charlie decodes Alice’s quantum data first and uses it to help him decode Bob’s. (Or vice-versa.) Go theorem for use of quantum side information.

Stern warning

   Can’t infer very much from capacity formulas that require a limit n ! 1 Classical information theorists knew this: classical MAC capacity has similar inequivalent characterizations Capacity formulas should (at least) be validated by finding non-trivial single-letter examples before we start believing in them

Money where my mouth is

1 S Bob and Charlie each get one qubit: R,S · 1 0 1 Q( N ) = 2 – H(p) ¸ R+S R

Solution post-QIP 2005

Interpretation: Charlie decodes Alice’s quantum data first and uses it to help him decode Bob’s. (Or vice-versa.) Go theorem for use of quantum side information.

Proof idea

Let N :A’B’ !

C and  ABC Show that the rate pair R ~ I c (A i = (1 AB N )(  C), S ~ I c (B i 1 AA’  2 BB’ ) mixed AC) is achievable.

Alice uses random LSD codes for the channel N 1 Bob uses random LSD codes for the channel N 2 Charlie uses a fancy decoder: = N ( ¢  2 B’ = N (  1 AA’ ¢ ) ) pure

Notions of quantum capacity

    Amount of entanglement that can be created using N Amount of entanglement that can be sent through N Dimension of Hilbert space all of whose states can be sent reliably through N Dimension of Hilbert space all of whose

entangled states

can be sent reliably through N: strong subspace transmission

From entanglement transmission to strong subspace transmission

“Pruning” works for 1-sender/1-receiver channels but not for QMAC.

Suppose this circuit sends maximally entangled states with high fidelity

E 1

N

n

D E 2

Shared Alice-Charlie and Bob-Charlie randomness Arbitrary inputs can be made to look like halves of maximally entangled states by randomizing.

To eliminate shared randomness, imagine purifying it. After decoding, shared randomness and decoded entanglement are decoupled. Re-use the shared randomness!

Quantum broadcast channels

N : A’ !

BC

N

In general, the classical capacity of a classical broadcast channel is still an open problem. Will focus on a solvable case.

Degraded broadcast channels

a p(b,c|a) c b ´ a p(b|a) b p(c|b) c Consider the capacity to send a common message at rate R to Bob and Charlie simultaneously while also sending an additional message of rate R B to Bob.

The capacity region is given by the union of rectangles 0 · R · I(T;C) 0 · R B · I(A;B|T) where the union is over distributions p(t,a).

[Cover, Bergmans, Gallagher]

Isometric extensions of dephasing channels

U

BC

●Let N g (|x ih x|) = |  x ih ● Note that Tr B U BC = N g  x |. ± Tr C U BC. Like degradability.

Let R be the rate at which Alice, Bob and Charlie can establish GHZ states using U BC while simultaneously establishing EPR pairs between Alice and Bob at a rate R B .

The capacity region is given by the union of rectangles 0 · R · I(T;C) 0 · R B · H(X|T) – H(C|T) evaluated for the state  TXC =  t,x p(t,x) |t,x ih t,x| TX  x C

Summary

 After pessimistic missteps corrected, network quantum Shannon theory is making rapid progress  Distributed compression  Quantum multiple access channels  Quantum broadcast channels  Open problems: mix and match!