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

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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

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qubit[q ! q] |0 i A ! |0 i B and |1 i A ! |1 i B ebit[qq] the state (|0 i A |0 i B + |1 i A |1 i B )/ p 2 cbit[c ! c] |0 i A ! |0 i B |0 i E and |1 i A ! |1 i B |1 i E Everything is a resource cobit[q ! qq] |0 i A ! |0 i A |0 i B and |1 i A ! |1 i A |1 i B resource inequalities super-dense coding: [q ! q] + [qq] > 2[c ! c] In fact, [q ! q] + [qq] = 2 [q ! qq] 2 [q ! qq]

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(disentangling power) -[qq] [qq] y = (relation between time-reversal and exchange symmetry) [q Ã q][q ! q] y = |0 i A |0 i B ! |0 i A and |1 i A |1 i B ! |1 i A (coherent erasure??) [q Ã qq] (?)[q ! qq] y = Undoing things is also a resource reversalmeaning

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[qq ! q] > [q ! q] - [q ! qq] = [q ! qq] - [qq] = ([q ! q] - [qq]) / 2 What good is coherent erasure? |0 i A + |1 i A ! |0 i A |0 i B + |1 i A |1 i B (using [q ! qq]) ! |0 i B + |1 i B (using [qq ! q]) [q ! qq] + [qq ! q] > [q ! q] entanglement-assisted communication only In fact, these are all equalities! (Proof: reverse SDC.) = = |x i |y i E (I X x Z y )| i XZ Alice Bob |i|i |x i |y i |x i |y i

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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) = C Ã E (U y ). Time reversal means: U y > C [q Ã qq] = C [qq Ã q] - C [qq]

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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. U m |x i A |0 i B = |x i A |x i B for 0 6 x < 2 m U m |x i A |y i B = |x i A |y-1 i B for 0 < y 6 x < 2 m U m |x i A |y i B = |x i A |y i B for 0 6 x < y < 2 m the construction: (U m acts on 2 m £ 2 m dimensions)

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Upper bound by simulation: 1. Jointly test whether y=0, 0 x. 2. Either send x to Bob using m[q ! qq], map y ! y-1 or do nothing. 3. Test whether y=x, 0 6 y x. m[q ! qq] + O((log m)(log m/ )) ([q ! q] + [q Ã q]) & U m Alice’s input x … 2 m -1 Bob’s input y C ! (U m ) = m À O(log 2 m) > C Ã E (U m ) U m > m[q ! qq] U m |x i A |0 i B = |x i A |x i B for 0 6 x<2 m U m |x i A |y i B = |x i A |y-1 i B for 0

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Alice’s input x … 2 m -1 Bob’s input y C Ã E (U m y ) > m À O(log 2 m) > E(U m y ) U m y |x i A |x i B = |x i A |0 i B for 0 6 x<2 m U m y |x i A |y i B = |x i A |y+1 i B for 0 6 y

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disentanglement clean 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 Example: U m > m[qq], but can only destroy O(log 2 m) [qq] U m y > -m[qq], but can only create O(log 2 m) [qq] clean

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You can’t just throw it away Q: Why not? A: Given unlimited EPR pairs, try creating the state Hayden & Winter [quant-ph/ ] proved that this requires ¼ n bits of communication.

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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 ) := S 0 ( A ) - S 1 ( A ) > 0 is a LU+EPR monotone S 0 ( ) := log rank( ) and S 1 = - log || || 1 Useful because it can increase by at most one per cbit sent. [Hayden & Winter- q-ph/ ] approximate version: ( ) := min { ( ) : || - || 1 6 }

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spread: another resource you didn’t know you needed 1. Entanglement dilution: | i AB is partially entangled. E = S( A ). goal: | i nE+o(n) ! | i n even the weaker | i 1 ! | i n requires (n ½ ) cbits (in either direction). why? concentration requires no communication: | i n ! | i nE+ with probability p( ), where p( ) ¼ Gaussian with mean 0 and std. dev O(n ½ ), meaning | i n LU p p( )| i A | i B | i nE+ Thus (( A ) n ) ~ n ½ and creating | i n means creating O(n ½ ) entanglement spread, which requires communication. (possible [Lo&Popescu, q-ph/ ], necessary [Hayden&Winter, q-ph/ ; Harrow&Lo, q-ph/ ])

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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 U N :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 |0 i 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?

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“Clueless-Eve” Channel on d +1 dimensional Hilbert space conceals information about its input and output from the environment 0 A j j BE d ( j = 1 …d ) j A 0 j BE 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 = 00 m RA + ( j j d ) m RA is not decohered by the Clueless-Eve channel, and so it should not be decohered by a faithful simulation of it R B E A NTP sadly, this appears optimal SLIDE STOLEN FROM CHARLIE BENNETT

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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 L A 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 0 0 R A Sender- receiver entanglement E E’B LALA LBLB Local Environment Output fidelity evaluated here. Non tensor product RA input NTP = ( 00 m + ( j j d ) m ) / to attempted simulation of n instances of Clueless-Eve Channel 0 A j j BE d j A 0 j BE SLIDE STOLEN FROM CHARLIE BENNETT

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

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s-bits?? In some ways, spread smells like a resource. Example: The spread capacity of a unitary gate U is E(U) + E(U y ). 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/ ] ) But it has no canonical instantiation! -cbits are too strong -partially entangled states: | i n scales as p n, 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) À C + (U)…)

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