Quantum Information and the PCP Theorem Ran Raz Weizmann Institute.

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Quantum Information and the PCP Theorem

Ran RazWeizmann Institute

PCP Thm [BFL,FGLSS,AS,ALMSS]:

x 2 SAT can be proved by a poly-

size proof that can be verified by

reading only O(1) of its bits

PCP Thm [BFL,FGLSS,AS,ALMSS]:

x 2 SAT can be proved by poly(n)

blocks of length O(1) that can be

verified by reading only 2 blocks

Same with one block is impossible

(under hardness assumptions)even if each block is of almost linear size

x 2 SAT can be proved by 1) a log-size quantum state |i

and2) a classical proof p of poly(n)

blocks of length polylog eachs.t., after measuring |i the verifier needs to read only oneblock of p

Part I:

The Information of a Quantum State

Information of a Quantum State:

A quantum state |i of n qubits isdescribed by 2n complex numbers.However, a measurement only

gives n bits of information about |i(and the rest is lost)

How much of the information in |i can be used ?

Holevo’s Theorem (1973):

If Bob encodes a1,..,an by |i s.t. Alice can retrieve a1,..,an from |ithen |i is a state of ¸ n qubits.If Alice retrieves each bit ai withprob 1- then |i is a state of ¸ [1-H()]¢n qubits

We can’t communicate n bits by sending less than n qubits

ANTV-Nayak’s Theorem (1999):If Bob encodes a1,..,an by |i s.t. 8 i Alice can retrieve ai from |ithen |i is a state of ¸ n qubits.If Alice can retrieve each bit ai with prob 1- then |i is a state of ¸ [1-H()]¢n qubitsHolevo’s: Alice retrieves a1,..,an Nayak’s: Alice retrieves only one

ai

(of her choice)

Our Result:Bob can encode N=2n bits a1,..,aN

by a state |i of O(n) qubits, s.t. 8 i, ai can be retrieved from |iby a (one round) Arthur-Merlin interactive protocol of size

poly(n)(with a third party, Merlin)

(classical messages) (polynomially small error)

Retrieving ai from |i:

Alice measures |i (gets result e) and sends a question q=q(i,e)Merlin answers by r.Alice computes V(i,e,r) 2 {0,1,err}

Completeness: 8i,q 9 r, V(i,e,r) = ai

Soundness: 8i,q,r, V(i,e,r)2 {ai,err}

(with high probability)

(q,r are poly(n) classical bits)

Retrieving ai from |i:

Alice measures |i (gets result e) and sends a question q=q(i,e)Merlin answers by r.Alice computes V(i,e,r) 2 {0,1,err}

Completeness: 8i,q 9 r, V(i,e,r) = ai

Soundness: 8i,q,r, V(i,e,r)2 {ai,err} (with high probability)

(q,r are poly(n) classical bits)

Bob is trustworthy (|i is correct)

Merlin knows a1,..,aN

More Generally:1) Any constant number of

elements from a1,..,aN can be retrieved in the same way, by a protocol of size poly(n)

2) Any k elements can be retrieved by a protocol of size k¢poly(n)

3) Each ai can be 2 {1,..,N}

A Dequantumized Protocol:|i is not needed:Bob can send a (poly-size)

randomsecret classical string , If Merlin doesn’t know The protocol works as before

Part II:

The Retrieval Protocol

Multilinear Extension:Given a0,..,aN (N=2n-1)

F = field of size n2

A: Fn ! F, s.t.:1) 8 i 2 {0,1}n, A(i) = ai

2) A is multilinear (deg(A) · n)

Quantum Multilinear Extension:A= multilinear extension of

a0,..,aN

1) |i is a state of poly(n) qubits2) When Alice measures |i, she

gets z,A(z) for a random z 2 Fn

(Merlin doesn’t know z)

Retrieving A(i):Alice knows A(z) and wants

A(i)l = the line through i,z (in Fn)Al : l ! F = restriction of A to l

(deg(Al) · n)

The Protocol:Alice sends l, Merlin is required

to give Al : l ! F. Merlin answers byg : l ! F (deg(g) · n)If g(z) Al(z) Alice rejectsOtherwise, Alice assumes

A(i)=g(i)

If g Al then w.h.p. g(z) Al(z) (since both are low degree)Otherwise, A(i) is correct

A Dequantumized Protocol:|i is not needed:Bob can send z,A(z), for a

randomz 2 Fn

(s.t., Merlin doesn’t know z)

The protocol works as before

Part III:

The Exceptional Power of QIP/qpoly

The Class QIP/qpoly:

IP: [B][GMR] x 2 L can be proved by a poly-size interactive proofQIP: [Wat] x 2 L can be proved by

a poly-size quantum interactive proof

QIP/qpoly: x 2 L can be proved by

a poly-size quantum interactive proof with poly-size quantum advice

Quantum Advice:

(captures quantum non-uniformity)A (poly-size) quantum state |L,nigiven to the verifier as an adviceAlternatively, the verifier is aquantum circuit with working spaceinitiated with |L,ni

[NY],[Aar]: Limitations on BQP/qpoly

QIP/qpoly:

QIP/qpoly: x 2 L can be proved bya poly-size interactive proof

wherethe verifier is a poly-size quantumcircuit with working space

initiatedwith an arbitrary state |L,ni

Our Result:QIP/qpoly contains all languages

Proof:

Denote ai 2 {0,1}, ai =1 iff i 2 L |L,ni = the quantum multilinear extension of a0,..,aN (N=2n-1)ai can be retrieved from |L,ni byArthur-Merlin interactive

protocolof size poly(n)(one round, classical

communication)

Randomized Advice:A (poly-size) random string ,chosen from a distribution DL,n,

andgiven to the verifier as an adviceAlternatively, the verifier is a distribution over poly-size

classicalcircuits

Randomized Advice:A (poly-size) random string ,chosen from a distribution DL,n, andgiven to the verifier as an adviceAlternatively, the verifier is a distribution over poly-size classicalcircuits

IP/rpoly: x 2 L can be proved bya poly-size interactive proof

wherethe verifier is a distribution overpoly-size classical circuits

IP/rpoly contains all languages

Part IV:

Quantum Versions of the PCP Theorem

PCP Thm [BFL,FGLSS,AS,ALMSS]:

x 2 SAT can be proved by a poly-

size proof that can be verified by

reading only O(1) of its bits

PCP Thm [BFL,FGLSS,AS,ALMSS]:

x 2 SAT can be proved by poly(n)

blocks of length O(1) that can be

verified by reading only 2 blocks

Same with one block is impossible

(under hardness assumptions)even if each block is of almost linear size

We Show:

x 2 SAT can be proved by 1) a log-size quantum state |i

and2) a classical proof p of poly(n)

blocks of length polylog eachs.t., after measuring |i the verifier needs to read only oneblock of p

We Show:

x 2 SAT can be proved by 1) a log-size quantum state |i and2) a classical proof p of poly(n) blocks of

length polylog eachs.t., after measuring |i the verifier needs to read only oneblock of p

Naive Attempt:a1,..,aN = classical PCP

(N=poly(n))|i = quantum multilinear

extension of a1,..,aN O(log N) qubitsp = Merlin’s answers in the

retrieval protocol

The verifier retrieves a constant number of bits by reading one block

Problem:The verifier can’t trust that |i is a quantum multilinear extension

In the settings of communication orquantum advice, the verifier couldtrust that |i is correct. In the setting of PCP, |i can be anythinge.g. |i is concentrated on a point

Quantum Low Degree Test:The verifier checks that |i is a quantum encoding of a low

degreepolynomial. This is done with

the aid of the classical proof (or equivalently, a classical

prover)

Problem:We are only allowed one query How can we do both: quantum low degree test andretrieval of bits

We combine the two tasks using

ideas from [DFKRS]

Part V:

Scaling up to NEXP

Our Result (for L 2 NEXP):

x 2 L can be proved by 1) a poly-size quantum state |

i 2) a classical proof p of exp(n)

blocks of length poly eachs.t., after measuring |i the verifier needs to read only oneblock of p

Our Result (for L 2 NEXP):

x 2 L can be proved by 1) a poly-size quantum state |i 2) a classical proof p of exp(n) blocks of

length poly eachs.t., after measuring |i the verifier needs to read only oneblock of p

Alternatively (for L 2 NEXP):

x 2 L has a 3 messages (MAM)interactive proof, where the

proveris quantum in round 1 and

classical in round 2:

1) Prover sends |i 2) Verifier sends q3) Prover answers p(q)

Models of 3 Messages Proofs:

IP(3): prover is classicalQIP(3): prover is quantum

The hybrid model:HIP(3): prover is quantum in

first round and classical in second

Models of 3 Messages Proofs:

IP(3): prover is classicalQIP(3): prover is quantum

The hybrid model:HIP(3): prover is quantum in first round and

classical in second

IP(3) µ IP µ PSPACEQIP(3) µ QIP µ EXP [KW]

Our result: HIP(3) = NEXP

Why the prover in our protocol can’t be quantum in both rounds

?

A quantum prover can answer in round 2, based on a

measurementof a state entangled to the stategiven in round 1(fancy version of the EPR

paradox)

The End