Erasing correlations, destroying entanglement and other new challenges for quantum information

theory

Aram Harrow, Bristol

Peter Shor, MIT

quant-ph/0511219

IHP, 23 Feb 2006

outline

• General rules for reversing protocols

• Coherent erasure of classical correlations

• Disentangling power of quantum operations and entanglement spread as a resource in quantum communication

qubit [q!q] |0iA!|0iB and |1iA!|1iB

ebit [qq] the state (|0iA|0iB + |1iA|1iB)/p2

cbit [c!c] |0iA! |0iB|0iE and |1iA!|1iB|1iE

Everything is a resource

cobit [q!qq] |0iA! |0iA|0iB and |1iA!|1iA|1iB

resource inequalities

super-dense coding:

[q!q] + [qq] > 2[c!c]

In fact, [q!q] + [qq] = 2 [q!qq]

2 [q!qq]

(disentangling power)-[qq][qq]y =

(relation between time-reversal and exchange symmetry)

[qÃq][q!q]y =

|0iA |0iB! |0iA and |1iA |1iB!|1iA

(coherent erasure??)[qÃqq] (?)[q!qq]y =

Undoing things is also a resource

reversal meaning

[qq!q] > [q!q] - [q!qq]

= [q!qq] - [qq]

= ([q!q] - [qq]) / 2

What good is coherent erasure?

|0iA + |1iA ! |0iA|0iB + |1iA|1iB (using [q!qq])

! |0iB + |1iB (using [qq!q])

[q!qq] + [qq!q] > [q!q]

entanglement-assisted communication only

In fact, these are all equalities! (Proof: reverse SDC.)

=

=

|xi

|yi E (I XxZy)|i

X Z

Alice

Bob

|i|xi|yi

|xi|yi

application to unitary gates

U is a bipartite unitary gate (e.g. CNOT)

Known:

U > C[c! c] implies U > C[q!qq]

Corollary: If entanglement is free then C!E(U) = CÃ

E(Uy).

Time reversal means:

Uy > C [qÃqq]

= C [qqÃq] - C [qq]

How to find a gate with asymmetric capacities?

Problem: If U is nonlocal, it has nonzero quantum capacities in both directions. Are they equal?

Yes, if U is 2£2.

No, in general, but for a dramatic separation we will need a gate that violates time-reversal symmetry.

Um|xiA|0iB = |xiA|xiB for 0 6 x < 2m

Um|xiA|yiB = |xiA|y-1iB for 0 < y 6 x < 2m

Um|xiA|yiB = |xiA|yiB for 0 6 x < y < 2m

the construction:

(Um acts on 2m £ 2m dimensions)

Upper bound by simulation:1. Jointly test whether y=0, 0<y6x or y>x.2. Either send x to Bob using m[q!qq], map y!y-1 or do nothing.3. Test whether y=x, 06y<x or y>x.

m[q!qq] + O((log m)(log m/)) ([q!q] + [qÃq]) & Um

Alice’s input x

0 1 2 3 4 … 2m-1

Bob

’s in

put

y

0

1

2

3

4

C!(Um) = m À O(log2m) > CÃE(Um)

Um > m[q!qq]

Um|xiA|0iB = |xiA|xiB for 06x<2m

Um|xiA|yiB = |xiA|y-1iB for 0<y6x<2m

Um|xiA|yiB = |xiA|yiB for 0 6x<y<2m

PICTURE STOLEN FROM CHARLIE BENNETT

Alice’s input x

0 1 2 3 4 … 2m-1

Bob

’s in

put

y

0

1

2

3

4

CÃE(Um

y) > m À O(log2m) > E(Umy)

Umy|xiA|xiB = |xiA|0iB for 06x<2m

Umy|xiA|yiB = |xiA|y+1iB for 06y<x<2m

Umy|xiA|yiB = |xiA|yiB for 0 6x<y<2m

PICTURE STOLEN FROM CHARLIE BENNETT

Meaning: Um ¼ m [q!qq]

and Umy ¼ m [qÃqq] (almost worthless w/o ent. assistance!)

Rerverse everything:Um

y > m[qÃqq] andm[qÃqq] + O((log m)(log m/)) ([q!q] + [qÃq]) & Um

y

disentanglementclean resource inequalities:

means that n can be asymptotically converted to n while discarding only o(n) entanglement.

(equivalently: while generating a sublinear amount of local entropy.)

Example: [q!q] > [qq] and [q!q] > -[qq]clean clean

Example: Um > m[qq], but can only destroy O(log2m) [qq]

Umy > -m[qq], but can only create O(log2m) [qq]

clean

clean

You can’t just throw it away

Q: Why not?

A: Given unlimited EPR pairs, try creating the state

Hayden & Winter [quant-ph/0204092] proved that this requires ¼n bits of communication.

spread: a LU+EPR monotone

Entanglement measures are usually LOCC monotones. e.g. for pure states AB, the entropy of entanglement E() = S(A) cannot increase on average under LOCC.

Under LU + EPR, Alice and Bob have free EPR pairs and local unitaries, but classical communication is expensive.

(A) := S0(A) - S1(A) > 0 is a LU+EPR monotone

S0() := log rank() and S1 = - log ||||1

Useful because it can increase by at most one per cbit sent. [Hayden & Winter- q-ph/0204092]approximate version: () := min { () : || - ||1 6 }

spread: another resource you didn’t know you needed

1. Entanglement dilution:

|iAB is partially entangled. E = S(A).

goal: |i nE+o(n)!|i n

even the weaker |i 1!|i n requires (n½) cbits (in either direction).

why?

concentration requires no communication:

|i n ! |i nE+ with probability p(), where p()¼Gaussian with mean 0 and std. dev O(n½),

meaning |i n LU pp()|iA|iB |i nE+

Thus ((A) n) ~ n½ and creating |i n means creating O(n½) entanglement spread, which requires communication.

(possible [Lo&Popescu, q-ph/9902045], necessary [Hayden&Winter, q-ph/0204092; Harrow&Lo, q-ph/0204096])

spread: another resource you didn’t know you needed

2. Quantum Reverse Shannon Theorem for general inputs

generalizes entanglement dilution.

Goal: Simulate n copies of a channel N:A’!B or its isometric extension UN:A’!AB.

Input n requires I(R;B) [c!c] + H(B) [qq]. note: H(B):=H(N())

[Bennett, Devetak, Harrow, Shor, Winter]

General inputs reduce (via e.g. the Schur transform) to the case of coherent superpositions of different n, which require consuming different amounts of entanglement.

Example: noiseless channel, superposition of |0i and I/2 inputs.

But! Varying changes I(R;B) at the same time as it changes H(B), so we can use max H(B) [qq] + (max H(B) + max H(R)-H(A)) [c!c].

Still terrible! Is this really optimal?

“Clueless-Eve” Channel on d+1 dimensional Hilbert space conceals information about its input and output from the environment

0A j jBEd ( j = 1…d )

jA0 jBE

If source is 0, creates entangled state between Bob and Eve.If source is nonzero, sends the nonzero value to Eve and zero to Bob.

Malicious Non tensor product RA input

NTP = 00mRA

+ ( j jd )mRA

is not decohered by the Clueless-Eve channel, and so it should not be decohered by a faithful simulation of it

R

B

E

ANTP

sadly, this appears optimal

SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT

SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT

m qubits communication

Alice’s Encoder

Bob’s Decoder

Local Environment

Local environments could contain information on whether A was zero or

nonzero. If A is nonzero, entropy of LA will be H(E)+log(md).

Otherwise it will be less, H(E), by conservation of entropy. Because local environments know whether A is zero or not, theynecessarily decohere the 2 terms in the superposition

0

0

R

A

Sender- receiver entanglement

E

E’ B

LA

LBLocal Environment

Output fidelityevaluated here.

Non tensor product RA input NTP = (00m + ( j

jd )m) /to attempted simulation of n instances of Clueless-Eve Channel

0A j jBEd jA 0 jBE

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Spread: where to buy it1. Classical communication in either direction:

|xiA|xiB|iAB n ! |xiA|xiB|i x|00i n-x

uses POVM with elements

2. embezzle it! [q-ph/0205100, Hayden-van Dam]

spread n with error using a size S=n/ embezzling state

s-bits??In some ways, spread smells like a resource.

Example: The spread capacity of a unitary gate U is E(U) + E(Uy). This is a lower bound on the number of cbits needed to simulate U, even given free entanglement.

(Corollary: to prove E(U) À C+(U), we can’t use simulation-based upper bounds.) (Interesting aside: E(U)+E(U^\dag) also upper-bounds C_+^E(U) [Berry&Sanders, q-ph/0207065])But it has no canonical instantiation!

-cbits are too strong

-partially entangled states: |i n scales as pn, and isn’t known to be universal

-embezzling states are non-i.i.d. and give big errors

-unitaries are too strong (unless E(U)ÀC+(U)…)