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# The Power of Unentanglement

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The Power of Unentanglement
| | | Scott Aaronson (MIT) Salman Beigi (MIT) Andrew Drucker (MIT) Bill Fefferman (Caltech) Peter Shor (MIT)

Not what I’ll be talking about today
“It is not yet entirely clear what advances in our understanding of quantum computation and quantum information can be expected as a result of the study of quantitative measures of entanglement.” —Nielsen & Chuang (2000) In this work, we connect quantum complexity theory to entanglement theory—ironically, by studying the power of lack of entanglement! Previous 3 talks, 3 talks at upcoming FOCS: Quantum multi-prover interactive proof systems where provers share entanglement Not what I’ll be talking about today

Main Results Proving 3SAT With Õ(n) Qubits
Let  be a 3SAT instance of size n. Someone can prove to you that  is satisfiable by giving you only O(n polylog n) qubits—provided you know certain subsets of the qubits are unentangled with the rest Proof is nonrelativizing, and requires a tight PCP theorem Additivity  Amplification and Collapse Multi-prover quantum MA can be amplified to exponentially small error, and three or more Merlins can be simulated with two, assuming the famous Additivity Conjecture from quantum information theory

QMA: Quantum Merlin-Arthur [Kitaev and Watrous, 2000]
Class of languages L such that for all inputs x: xL  exists a witness | with poly(n) qubits, causing polytime quantum verifier Arthur to accept w.p.  2/3 xL  for all witnesses |, Arthur accepts w.p.  1/3 We know a reasonable amount about QMA: it’s contained in PP, allows amplification, has natural complete promise problems…

QMA(k) [Kobayashi, Matsumoto, Yamakami 2003]
Class of languages L such that for all inputs x: xL  there exist witnesses |1,…,|k causing Arthur to accept w.p. 2/3 xL  for all |1,…,|k, Arthur accepts w.p. 1/3 Classically, this is completely uninteresting: MA(k)=MA But quantumly, a single Merlin could cheat by entangling the k proofs!

Do Multiple Quantum Proofs Ever Actually Help?
Liu, Christandl, Verstraete: Natural problem from quantum chemistry, pure state N-representability, which is in QMA(2) but not known to be in QMA Blier and Tapp (independent of us): 3-COLORING admits a 2-prover QMA protocol with witnesses of size log(n), and a (1/n6) probability of catching cheating provers This work: A protocol for 3SAT with Õ(n) quantum witnesses of size log(n), and constant soundness

Our Protocol for 3SAT We’ll work not with 3SAT but with “2-out-of-4-SAT”: xi + xj + xk + xl = 2 (mod 4) We need our 2-out-of-4-SAT instance to be balanced (each variable occurs in O(1) clauses), as well as a PCP (either satisfiable or -far from satisfiable) Theorem: We can get all of this using known classical reductions from 3SAT (including Dinur’s gap amplification procedure), incurring a polylog(n) blowup in the number of variables and clauses. 7

So suppose Arthur has done all this, to obtain a 2-out-of-4-SAT instance . And suppose Merlin sends him a log(n)-qubit state of the form where x1,…,xn is a claimed satisfying assignment for . (I.e., a proper state.) Then Arthur can measure | in a basis corresponding to the clauses of , obtaining the outcome for some clause C=(xi,xj,xk,xl). A further measurement reveals whether C is satisfied with (1) probability. 8

Problem: How can Arthur force Merlin to send him a proper state. (E. g
Problem: How can Arthur force Merlin to send him a proper state? (E.g., what if Merlin cheats by putting all amplitude on a few computational basis states?) Solution: More Merlins! | n log(n) 9

The Protocol With 1/3 prob.
Pick a random |k and do the Satisfiability Test described earlier With 1/3 prob. Pick two random |k’s and do a Swap-Test (Ensures most |k’s are pretty much identical) With 1/3 prob. Do a Uniformity Test to make sure the |k’s are close to proper states 10

The Uniformity Test |3-|4 |8+|10 |2+|9 |8-|10
Pick a random matching M on [n] Measure each witness | in a basis containing for each (i,j)M. |3-|4 |8+|10 |2+|9 |8-|10 Since there are n witnesses, by the Birthday Paradox, with constant probability we’ll see a collision: two outcomes involving the same edge (i,j). 11

 If both outcomes are |i+|j or both |i-|j, accept.
If one outcome is |i+|j and the other is |i-|j, reject. Accepts with certainty if the witnesses are identical and proper Theorem: Rejects with (1) probability if witnesses are close to each other but far from proper Proof: So intuitively obvious, it takes 14 pages to prove Why doesn’t our protocol work with entangled witnesses? Because the Merlins could send a state that passes all Swap-Tests, yet doesn’t produce collisions 12

“Entanglement Swapping”
Amplification For QMA, it’s easy to amplify success probability, even if Merlin cheats by entangling the witnesses Witness1 Witness2 Witness3 So then what’s the problem with amplifying QMA(2)? Witness1 Witness2 Witness3 Merlin1: Merlin2: Uh-oh! Accept “Entanglement Swapping” 13

Yet it seems possible to defend against this bizarre behavior…
W1 W2 W3 W4 W5 W6 W7 n100 pairs of witnesses, of which we only measure a random n Does any tiny amount of entanglement that’s created during this protocol “spread itself thinly” across the registers in a reasonable way? To answer this question, we need a way to measure entanglement… 14

Entanglement of Formation EF(AB)
Intuitively, minimum # of EPR pairs needed to prepare AB Fun Facts: EF can only increase by  2K when we act on a K-qubit register If EF(AB)0, then AB is close to a separable state in trace distance Good for us Good for us Is EF superadditive? Shor 2003: Equivalent to proving the “additivity of quantum channel capacity,” a famous open problem 15

Assuming the Additivity Conjecture, we show that…
QMA(2) protocols can be amplified to exponentially small error QMA(2)=QMA(k) for all 2kpoly(n) (building on [KMY]) SymQMA(k)=QMA(k) (SymQMA(k): All k Merlins send the same state) For every fixed polynomial p, p(n) entanglement gives the Merlins no extra power to cheat 16

? Upper Bounds for QMA(2) QMA(2)
It’s obvious that QMA(2)NEXP. Embarrassingly, we still don’t have a better upper bound! NEXP QMA(2) ? EXP Our Result: QMA(2)PSPACE, assuming “Strong Amplification” of QMA(2) protocols (Amplification that reuses one of the Merlin’s witnesses over and over) PSPACE On the other hand: If amplification that reuses both witnesses is possible, then PSPACE=NEXP! PP 17

Does QMA(2)=QMA? Right now, even proving an oracle separation between them seems way beyond reach! Conjecture (Watrous): There’s no way to simulate QMA(2) in QMA by taking an arbitrary polynomial-size witness, and “disentangling” it to produce an arbitrary roughly-separable witness All separable states All states We show: If you want a perfect disentangler, then the input Hilbert space needs to be infinite-dimensional. 18

More Open Problems In our 3SAT protocol, can the assumption of unentanglement be removed? If so, then we get a 2Õ(n) quantum algorithm for 3SAT! Conjecture: Our protocol can be modified to require only two provers sending Õ(n) qubits each Can we improve on Õ(n), or get evidence against this? In defining QMA(2), does it matter whether amplitudes are real or complex? Are there natural group-theoretic problems in QMA(2)? Does QMA(2) have natural complete promise problems? Remove additivity/amplification assumptions!

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