Erasing correlations, destroying entanglement and other new challenges for quantum information theory Aram Harrow, Bristol Peter Shor, MIT quant-ph/ QIP, 19 Jan 2006

outline General rules for reversing protocols Coherent erasure of classical correlations Disentangling power of quantum operations

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]

(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

[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

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]

The quest for asymmetric unitary gate 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)

et voilà l’asymétrie! 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 U m > m[q ! qq] Meaning: U m ¼ m [q ! qq] and U m y ¼ m [q à qq] (almost worthless w/o ent. assistance!) Upper bound by simulation: m[q ! qq] + O((log m)(log m/  )) ([q ! q] + [q à q]) & U m Similarly, U m y > m[q à qq] and m[q à qq] + O((log m)(log m/  )) ([q ! q] + [q à q]) & U m y

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

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.

more relevant examples Entanglement dilution: |  i AB is partially entangled. E = S(  A ). |  i ­ nE+o(n) ! |  i ­ n OR a size O(n ½ /  ) embezzling state [q-ph/ , Hayden-van Dam] Even |  i ­1 ! |  i ­ n requires  (n ½ ) cbits (in either direction). Quantum Reverse Shannon Theorem for general inputs Input  ­ n requires I(A;B)  [c ! c] + H( N (  )) [qq]. Superpositions of different  ­ n mean consuming superpositions of different amounts of entanglement: we need either extra cbits, embezzling, or another source of entanglement spread. [Bennett, Devetak, Harrow, Shor, Winter]

summary new ideas coherent erasure clean protocols entanglement spread new results asymmetric unitary gate capacities QRST and other converses new directions formalizing entanglement spread clean protocols involving noisy resources (cbits?) READ ALL ABOUT IT! quant-ph/