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

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Presentation on theme: "Erasing correlations, destroying entanglement and other new challenges for quantum information theory Aram Harrow, Bristol Peter Shor, MIT quant-ph/0511219."— Presentation transcript:

1 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

2 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

3 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]

4 (disentangling power) -[qq] [qq] y = (relation between time-reversal and exchange symmetry) [q à 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 à qq] (?)[q ! qq] y = Undoing things is also a resource reversalmeaning

5 [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

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

7 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)

8 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 à q]) & U m Alice’s input x 0 1 2 3 4 … 2 m -1 Bob’s input y 0 1 2 3 4 C ! (U m ) = m À O(log 2 m) > C à 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4240993/slides/slide_8.jpg", "name": "Upper bound by simulation: 1. Jointly test whether y=0, 0 x.", "description": "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 à q]) & U m Alice’s input x 0 1 2 3 4 … 2 m -1 Bob’s input y 0 1 2 3 4 C . (U m ) = m À O(log 2 m) > C à 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

9 Alice’s input x 0 1 2 3 4 … 2 m -1 Bob’s input y 0 1 2 3 4 C à 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 m[q à qq] and m[q à qq] + O((log m)(log m/  )) ([q ! q] + [q à q]) & U m y

10 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

11 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 ¼ n bits of communication.

12 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/0204092] approximate version:   (  ) := min {  (  ) : ||  -  || 1 6  }

13 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/9902045], necessary [Hayden&Winter, q-ph/0204092; Harrow&Lo, q-ph/0204096])

14 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?

15 “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

16 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

17 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/0205100, Hayden-van Dam] spread n with error  using a size S=n/  embezzling state

18 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/0207065] ) 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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