Aram Harrow, Bristol Peter Shor, MIT

Erasing correlations, destroying entanglement and other new challenges for quantum information theory quant-ph/0511219 Aram Harrow, Bristol Peter Shor, MIT IHP, 23 Feb 2006

outline • General rules forreversing protocols • Coherent erasureof classical correlations • Disentangling powerof quantum operations and entanglement spread as a resource in quantum communication

Everything is a resource resource inequalities super-dense coding: [q!q] + [qq] > 2[c!c] 2 [q!qq] In fact, [q!q] + [qq] = 2 [q!qq]

Undoing things is also a resource [q!q]y = [qÃq] (relation between time-reversal and exchange symmetry) [qq]y = -[qq] (disentangling power) [q!qq]y = [qÃqq] (?) |0iA |0iB! |0iA and |1iA |1iB!|1iA (coherent erasure??) reversal meaning

[qq!q] > [q!q] - [q!qq] = [q!qq] - [qq] = ([q!q] - [qq]) / 2 What good is coherent erasure? = = entanglement-assisted communication only In fact, these are all equalities! (Proof: reverse SDC.) |xi E Alice (I XxZy)|Fi |Fi |yi X Z |xi |xi Bob |yi |yi a|0iA + b|1iA! a|0iA|0iB + b|1iA|1iB (using [q!qq]) !a|0iB + b|1iB (using [qq!q]) [q!qq] + [qq!q] > [q!q]

application to unitary gates U is a bipartite unitary gate (e.g. CNOT) Known: U > C[c! c] implies U > C[q!qq] Time reversal means: Uy> C [qÃqq] = C [qqÃq] - C [qq] Corollary: If entanglement is free then C!E(U) = CÃE(Uy).

How to find a gate with asymmetric capacities? the construction: (Um acts on 2m £ 2m dimensions) 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 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.

C!(Um) = m À O(log2m) > CÃE(Um) Alice’s input x 0 1 2 3 4 … 2m-1 0 PICTURE STOLEN FROM CHARLIE BENNETT Bob’s input y 1 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 2 3 4 Um> m[q!qq] 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/e)) ([q!q] + [qÃq]) & Um

CÃE(Umy) > m À O(log2m) > E(Umy) Alice’s input x 0 1 2 3 4 … 2m-1 0 PICTURE STOLEN FROM CHARLIE BENNETT Bob’s input y 1 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 2 3 4 Rerverse everything: Umy> m[qÃqq] and m[qÃqq] + O((log m)(log m/e)) ([q!q] + [qÃq]) & Umy Meaning: Um¼ m [q!qq] and Umy¼ m [qÃqq] (almost worthless w/o ent. assistance!)

disentanglement clean clean Example: [q!q] > [qq] and [q!q] > -[qq] clean Example: Um> m[qq], but can only destroy O(log2m) [qq] Umy> -m[qq], but can only create O(log2m) [qq] clean clean resource inequalities: means that an can be asymptotically converted to bn while discarding only o(n) entanglement. (equivalently: while generating a sublinear amount of local entropy.)

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 yAB, the entropy of entanglement E(y) = S(yA) cannot increase on average under LOCC. Under LU + EPR, Alice and Bob have free EPR pairs and local unitaries, but classical communication is expensive. D(yA) := S0(yA) - S1(yA) > 0 is a LU+EPR monotone S0(r) := log rank(r) and S1 = - log ||r||1 Useful because it can increase by at most one per cbit sent. [Hayden & Winter- q-ph/0204092] approximate version:De(r) := min { D(s) : || r-s ||16e }

spread: another resource you didn’t know you needed 1. Entanglement dilution: |yiAB is partially entangled. E = S(yA). goal: |FinE+o(n)!|yin even the weaker |Fi1!|yin requires W(n½) cbits (in either direction). why? concentration requires no communication: |yin! |FinE+d with probability p(d), where p(d)¼Gaussian with mean 0 and std. dev O(n½), meaning |yin @LUådpp(d)|diA|diB |FinE+d Thus D((yA)n) ~ n½ and creating |yin 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 [Bennett, Devetak, Harrow, Shor, Winter] 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 rn requires I(R;B)r [c!c] + H(B) [qq]. note: H(B):=H(N(r)) General inputs reduce (via e.g. the Schur transform) to the case of coherent superpositions of different rn, which require consuming different amounts of entanglement. Example: noiseless channel, superposition of |0i and I/2 inputs. But! Varying r changes I(R;B) at the same time as it changes H(B), so we can use maxr H(B) [qq] + (maxr H(B) + maxr H(R)-H(A)) [c!c]. Still terrible! Is this really optimal?

sadly, this appears optimal SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT “Clueless-Eve” Channel on d+1 dimensional Hilbert space conceals information about its input and output from the environment |0ñA => S | j jñBE/Öd ( j =1…d ) |jñA => |0 jñBE 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. R Malicious Non tensor product RA input YNTP = |00ñÄmRA+ ( S | j jñ /Öd)ÄmRA is not decohered by the Clueless-Eve channel, and so it should not be decohered by a faithful simulation of it YNTP A B E SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT

Non tensor product RA input YNTP = (|00ñÄm + ( S | j jñ /Öd)Äm) /Ö2 to attempted simulation of n instances of Clueless-Eve Channel |0ñA => S | j jñBE/Öd , | jñA => |0 jñBE SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT SLIDE STOLEN FROM CHARLIE BENNETT R A Alice’s Encoder Output fidelity evaluated here. E m qubits communication 0 Bob’s Decoder Sender- receiver entanglement E’ B 0 Local Environment LA Local Environment LB 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, they necessarily decohere the 2 terms in the superposition

Spread: where to buy it 2. embezzle it![q-ph/0205100, Hayden-van Dam] spread n with error e using a size S=n/e embezzling state 1. Classical communication in either direction: |xiA|xiB|FiABn ! |xiA|xiB|Fix|00in-x uses POVM with elements

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: |yin 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)…)

Aram Harrow, Bristol Peter Shor, MIT