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

The Power of Unentanglement

Scott Aaronson (MIT)

Salman Beigi (MIT)

Andrew Drucker (MIT)

|

Bill Fefferman (Caltech)

Peter Shor (MIT)

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“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 ResultsProving 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:

• xL exists a witness | with poly(n) qubits, causing polytime quantum verifier Arthur to accept w.p. 2/3

• xL 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]

Classically, this is completely uninteresting: MA(k)=MA

But quantumly, a single Merlin could cheat by entangling the k proofs!

Class of languages L such that for all inputs x:

• xL there exist witnesses |1,…,|k causing Arthur to accept w.p. 2/3

• xL for all |1,…,|k, Arthur accepts w.p. 1/3

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 3SATWe’ll work not with 3SAT but with “2-out-of-4-SAT”:

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.

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)

xi + xj + xk + xl = 2 (mod 4)

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

Then Arthur can measure | in a basis corresponding to the clauses of , obtaining the outcome

n

i

x in

i

1

11

where x1,…,xn is a claimed satisfying assignment for . (I.e., a proper state.)

lkji lkji xxxx 1111

for some clause C=(xi,xj,xk,xl). A further measurement reveals whether C is satisfied with (1) probability.

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)log(n)log(n)

n

i

x in

i

1

11

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

The Uniformity TestPick a random matching M on [n]

for each (i,j)M.

Measure each witness | in a basis containing

2,

2

jiji

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).

|3-|4 |8+|10 |2+|9 |8-|10

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

AmplificationFor QMA, it’s easy to amplify success probability, even if Merlin cheats by entangling the witnesses

“Entanglement Swapping”

So then what’s the problem with amplifying QMA(2)?

Witness1 Witness2 Witness3

Witness1 Witness2 Witness3

Witness1 Witness2 Witness3

Merlin1:

Merlin2:

Uh-oh!

Accept

Yet it seems possible to defend against this bizarre behavior…

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…

W1 W2 W3 W4 W5 W6 W7

W1 W2 W3 W4 W5 W6 W7

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

Is EF superadditive?

k

i

BAF

BBAAF

iikk EE1

,, 11

Shor 2003: Equivalent to proving the “additivity of quantum channel capacity,” a famous open problem

Good for us

Good for us

Assuming the Additivity Conjecture, we show that…

QMA(2) protocols can be amplified to exponentially small error

QMA(2)=QMA(k) for all 2kpoly(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

Upper Bounds for QMA(2)It’s obvious that QMA(2)NEXP. Embarrassingly, we still don’t have a better upper bound!

On the other hand: If amplification that reuses both witnesses is possible, then PSPACE=NEXP!

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

EXP

NEXP

PP

QMA(2

)

?

Does QMA(2)=QMA?Right now, even proving an oracle separation between them seems way beyond reach!

We show: If you want a perfect disentangler, then the input Hilbert space needs to be infinite-dimensional.

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

More Open ProblemsIn 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

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?

Can we improve on Õ(n), or get evidence against this?

Remove additivity/amplification assumptions!