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A Short Tour of the Laws of Entanglement (And How to Evade Them)

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A Short Tour of the Laws of Entanglement (And How to Evade Them) A Short Tour of the Laws of Entanglement (And How to Evade Them) Patrick Hayden Stanford University 59 units 60 units Simons Ins<tute UC Berkeley 2014 From the Universal Compendium of Rock Solid Physico-Informa;onal Facts • §1.0.1. – Entanglement cannot be created without interacon – Evasion strategy: embezzlement – Applicaons • §1.0.2. – Entanglement is monogamous – One mathemacal formulaon – No applicaons The Great Laws: Part 1 §1.0.0. Thou shalt not create correla;on without a commensurate investment of interac<on. σAB I(A; B)σ I(A; B)⇢ A B N ⌦ M ⇢AB Embezzlement TheU from a reservoir of wealth sufficiently large that the crime is not no<ced. (Un<l it is.) Embezzling entanglement U U AA0 BB0 The perfect crime: φ AB 00 A B ⌦ φ AB A B | i | i 0 0 ! | i | i 0 0 φ φ A | i B A | i B UAA 0 VBB0 = A’ 00 B’ A’ B’ | i | i Extract the entangled state ψ from the entanglement bank φ without leaving behind a trace in the bank. Trivial solu<on: the infinite reservoir φ =( )⌦1 | iAB | i Drawbacks: * Not even the Federal Reserve has an infinite amount of entanglement * Must keep a separate account for every possible Ψ Embezzling states 1 n 1 φn = j A j B | i pC pj | i | i n j=1 X φn φn A | i B A | i B U AA0 VBB0 A’ 00 B’ A’ B’ | i ' | i Theorem: For every pure state |ψ>A’B’ of Schmidt rank m, there exist unitary transformaons UAA’ and VBB’ such that log m AB φn A B UAA VBB φn AB 00 A B 1 h | 0 0 h | 0 ⌦ 0 | i | i 0 0 ≥ − log n #qubits(φ ) = O( #qubits(ψ)/ε ) for inner product 1-ε n [H-van Dam 2003] Embezzling Bell pairs n 1 1 1 φn = j A j B = ( 00 + 11 ) | i pC pj | i | i | i p2 | i | i n j=1 X Schmidt coefficients: 1 1 1 1 1 φn : | i pCn ⇥ p1 p2 p3 ··· pn 1 1 1 1 1 p1 2 p2 2 p3 2 ··· pn 2 φn : · · · · | i| i pCn ⇥ 1 1 1 1 p1 2 p2 2 p3 2 ··· pn 2 · · · · Embezzling Bell pairs n 1 1 1 φn = j A j B = ( 00 + 11 ) | i pC pj | i | i | i p2 | i | i n j=1 X Schmidt coefficients: 1 1 1 1 1 1 1 φn : | i pC ⇥ p1 p2 p3 p4 ··· pn 1 pn n − 1 1 1 1 1 1 1 φn : | i| i pCn ⇥ p2 p2 p4 p4 ··· pn pn ··· 1 1 1 1 1 1 1 Dot product + + + + + + ≥ C ⇥ 2 2 4 4 ··· n n n ✓ ◆ log(n/2) log 2 =1 ≥ log(n) − log n and much earlier Further developments • [Araki ~1967]: – Exploited (exact) embezzlement in the classificaon of von Neumann algebras • [Leung+Toner+Watrous 2013*]: – Embezzlement for mul<party states • [Dinur+Steurer+Vidick 2013]: – Robust embezzlement • Alice and Bob don’t quite agree on the target state ψ – a.k.a. Quantum correlated sampling lemma – Key step in proving parallel repe<<on for projec<on games • [Leung+Wang 2013]: – Characteris<cs of universal embezzling states • [Haagerup+Scholz+Werner 2014?]: – Universal embezzling algebra • Uniquenessness of universal embezzling “eigenvalue” scaling • Every state in free quantum field theory is embezzling! * Really 2008, but who can be bothered to submit to journals in a <mely fashion these days? Applicaon: Mul<player quantum games j k ⇢ Xj Y k { a} { b } For 2-player games, blow-up in size of shared entangled state is Open queson: polynomial. Best known mul<party universal embezzling states are doubly a b exponen<ally large. Is it possible to do beyer? j k !⇤ =sup sup Ea,b V (j, k a, b)tr[⇢Xa Yb ] dim=:n X,Y,⇢ | ⌦ a,b,j,kX j k =supsup Ea,b V (j, k a, b)tr[⇢nXa Yb ] dim=:n X,Y | ⌦ a,b,j,kX Schmidt rank n embezzling state [Oliveira 2010] Applicaon: Games can require unbounded entanglement 2-player cooperave quantum game DEBBIE LEUNG,BEN TONER, AND JOHN WATROUS Let’s play… φy 1 | i ⇢ RST = ( 0 R 00 ST + 1 R ST ) φ | i p2 | i | i | i | i | ebzi 1 where ST = ( 11 ST + 22 ST ) Alice Bob | i p2 | i | i φ>. Goal: Winning probability approaches 1 as dimension of embezzling state goes to infinity. 1 S T A B win RAB = ( 000 RAB + 111 RAB). Can prove that it is bounded away from 1 for all finite dimensional || i p2 | i | i R Easy if Alice and Bob could apply unitaries r to all of ST since |00> and |ψ> are orthogonal. Figure 1: An illustrationGeneralizes usual model: of a two-player, one-round cooperative quantum game. They can’t because they are separated… quantum messages Definition 3.1 (Two-player, one-round cooperative quantum game). Let R,S,T,A,B be finite-dimensional quantum registers, let r be a quantum state of the registers (R,S,T), and let M be a binary measurementBut since they can coherently embezzle |ψ> from on the registers (R,A,B). Referee performs a projec<on |φebz> they can also coherently unembezzle it! 1. The referee preparesto determine win/loss. (R,S,T) in the state r, and then sends S to Alice and T to Bob. [Leung-Toner-Watrous 2008/13] 2. Alice and Bob transform the registers S and T sent to them however they choose, resulting in registers A and B that are sent back to the referee. 3. The referee applies the binary measurement M on the registers (R,A,B). The outcome 1 means that Alice and Bob win, while the outcome 0 means that they lose. The usual restrictions for provers in a quantum interactive proof system apply to Alice and Bob: they are not permitted to communicate once the game begins, but may agree on a strategy beforehand. Such a strategy may include the sharing of an entangled state y of their own choosing, which they may use | i when transforming the registers sent to them via local quantum operations. Alice’s most general strategy is to apply a quantum channel to S together with her share of the entangled state, outputting A. A general strategy for Bob may be described in a similar way. The complexity of the referee, which corresponds to the verifier in an interactive proof system, is ignored given that we no longer consider an input string. We note that two-player, one-round cooperative quantum games represent a generalization of the non- local games model of [14], where now the referee can send, receive, and process quantum information. CHICAGO JOURNAL OF THEORETICAL COMPUTER SCIENCE 2013, Article 11, pages 1–18 8 Same thing twice, just as nice This idea was originally used as part of the proof of the Quantum Reverse Shannon Theorem: Asympto<cally, every quantum channel can simulate every other using a rate of forward noiseless communicaon given by the rao of their entanglement-assisted capaci<es plus shared entanglement. 26 Embezzlement-assisted QRST in Schur Basis, for general source and channel Purification of a general state input to n instances of quantum channel R E Coherent E Simulation n Trans- A B pE of () form V n acting on to purified Schur input Small Basis Flat sub-channel O(log n) encoding and quantum Alice’s part of pA message Alice’s State Splitting Alice’s Lab part of nQ +o(n) embez- E Random Leftover qubits transfer for state splitting zlement permutation embezzling to symmetrize state Embezzling input. State Variable number of B ebits embezzled. Bob’s part Environment never pB of state learns how many. Bob’s splitting part of embez- Leftover zlement Bob’s Lab embezzling state Fig. 12. QRST on a general input using an entanglement-embezzling state (green). Alice first applies a randomizing permutation ⇡ to the inputs to n instances of her quantum channel, using information shared with Bob (magenta), thereby rendering the overall input approximately permutation-symmetric. * She then uses the ⌧B register to embezzle the correct superposition of (possibly[Benne- very) different amounts of entanglementDevetak needed by her sub-channel-Harrow- encoder, Shor-Winter 2009 ] leaving a negligibly degraded embezzling state behind. At the receiving end (lower right) Bob performs his half of the embezzlement, coherently decodes the sub-channel codes, and undoes the random permutation. The shared randomness needed for the initial randomizing permutation can also be obtained from the embezzling state, and in any case can be made sublinear in n, as shown in Lemma 14. m✏/2 from µ, then with probability 1 D (4/e)− , for any Inefficiencies and errors: Here we briefly tabulate the ≥ − A , various sources of inefficiency and error in our simulation | i m protocols for quantum channels. We will consider allowing 1 2 an inefficiency of O(nδ) in each step of the protocol and will (I U †V˜ †VU) 1 6✏. (87) m |h | ⌦ i i | i| ≥ − i=1 analyze the resulting errors. X Proof: Let ∆ := I V˜ †V . Observe that ∆ 2 and 1) In Lemma 11, an extra communication rate of O(nδ) − k k1 that Eq. (86) implies that ∆ 1 =Tr∆ ✏DA. Define means that the splitting step incurs an error of k k exp( nδ). − Tr ∆ 2) When restricting to typical triples, our definition of ∆0 := EU µ[U∆U †]= I ✏I (88) ⇠ DA T n was chosen so that entropic quantities such as m ,δ 1 HN(Rn)+H(Bn) H(En) would change by nδ + ∆¯ := U ∆U †, (89) − m i i o(n).
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