Limitations of Quantum Advice and One-Way Communication

Scott Aaronson

UC Berkeley IAS

Useful?

What Are Quantum States?To many quantum computing skeptics, they’re exponentially long vectors—and therefore a bad description of Nature

Yet a classical probability distribution over {0,1}n also takes 2(n) bits to specify!“Sure, but each sample is only n bits…”

Distributions over n-bit strings 2n-bit stringsWe give complexity-theoretic evidence that quantum states lie to the left end of this spectrum Supplements information-theoretic evidence (e.g. Holevo)

Quantum Advice

BQP/qpoly: Class of languages decidable by polynomial-size, bounded-error quantum circuits, given a polynomial-size quantum advice state |n that depends only on the input length n

Nielsen & Chuang: “We know that many systems in Nature ‘prefer’ to sit in highly entangled states of many systems; might it be possible to exploit this preference to obtain extra computational power?”

Example (Watrous)

For each n, fix a group Gn and subgroup HnGn

(|Gn|2n, but group operations are polytime)

Given an element xGn as input, is xHn?Solvable in BQP/qpoly using the advice state

1

n

nh Hn

H hH

Idea: Check whether Hn|xHn is 1 or 0

Not known to be in BQP/poly

Maybe BQP/qpoly even contains NP!

Obvious Challenge: Prove an oracle separation between BQP/poly and BQP/qpoly

Buhrman: Hey Scott—why not try for an unrelativized separation? After all, if quantum states are like 2n-bit classical strings, then maybe BQP/qpoly NEEEEE/poly!

Result #1

BQP/qpoly PP/poly

Proof based on new communication result:Given f:{0,1}n{0,1}m{0,1} (partial or total),

D1(f) = O(m Q1(f) logQ1(f))D1(f) = deterministic 1-way communication complexity of fQ1(f) = bounded-error quantum 1-way complexity

Corollary: Can’t show BQP/poly BQP/qpoly

without also showing PP P/poly

Result #2

NPA BQPA/qpolyfor some oracle A (actually, a random oracle)

Proof based on new Direct Product Theorem for quantum search:

Theorem: With few (N) quantum queries, the probability of finding all K marked items is 2-(K)

Fixes a wrong result of Klauck

N items, K of them marked

Result #3(Won’t say any more about this one)

Ambainis: Suppose Alice has x,yFp and Bob has a,bFp. They want to know whether yax+b.1-way quantum communication complexity?

Alice’s point

Bob’s line

Theorem: Alice must send (log p) qubits to Bob

Invented new “trace distance method” to show this

Previously, even randomized complexity was unknown

Then after the measurement, we can recover a ’ such that

'tr

The “Almost As Good As New” Lemma

Suppose a 2-outcome measurement of a

mixed state yields ‘0’ w.p. 1- and ‘1’ w.p.

D1(f) = O(m Q1(f) logQ1(f)) for allf : {0,1}n{0,1}m {0,1}

x y1,y2,… f(x,y)BobAlice

Alice can decrease the error probability to 1/Q1(f)10, by sending K=O(Q1(f)logQ1(f)) qubits

Bob can then compute f(x,y) for Q1(f)2 values of y simultaneously, with probability 0.9

With no communication, he can still do that with probability 0.9/2K, by guessing x=I

1logQ fx f(x,y1)

f(x,y2)

x =

maximally

mixed state?

Alice’s Classical Message

Bob, let p0(y) be the probability you’d guess f(x,y)=1 using I in place of x. Then y1 is the lexicographically

first y for which |p0(y)-f(x,y)|½.

Now let I1 be the reduced state assuming you guessed f(x,y1)

correctly. Let p1(y) be the probability you’d guess f(x,y)=1 using I1 in place of x. Then y2 is the first y after y1 for

which |p1(y)-f(x,y)|½.

y1

y2

Clearly Alice’s message lets Bob compute f(x,y) for any y in his range

Claim: Alice never has to send more than K yi’s—so her total message length is O(mK)

Suppose not. Then Bob would succeed on y1,…,yK+1 simultaneously with probability 1/2K+1

But we already know he succeeds with probability 0.9/2K, contradiction

BQP/qpoly PP/poly

Alice is the “advisor”

Bob is the PP algorithm

Suppose quantum advice has p(n) qubits. Then classical advice consists of K = O(p(n) log p(n)) inputs x1,…,xK{0,1}n, on which algorithm would make the wrong guess using maximally mixed state in place of advice (as before)

Adleman, DeMarrais, Huang: In PP, we can decide which of two sequences of measurement outcomes has greater probability

Improves earlier result:BQP/qpoly EXP/poly

NPA BQPA/qpoly

Claim: If LABQPA/qpoly, then using boosted advice, we can find all 2n/10 elements of S w.h.p. using 2n/10poly(n) quantum queries

Oracle: A(x)=1 iff xS, where S {0,1}n is chosen uniformly at random subject to |S|=2n/10

Language: (y,z)LA iff there exists an xS between y and z lexicographically (clearly LANPA)

Now replace advice by maximally mixed state.Success probability becomes 2-O(poly(n))

Direct Product TheoremGoal: Show that with o(2n/2) quantum queries, the probability of finding all 2n/10 marked items must be doubly exponentially small in n

Beals et al: If a quantum algorithm makes T queries to X{0,1}N, then the probability it accepts a random X with |X|=k is a univariate polynomial p(k) of degree 2T

INTUITIVELY PLAUSIBLE

p(k)0

1

0 1 2 . . . . .k

. . . . . N

Have the algorithm accept iff it finds |S|=2n/10 marked items. Then

(1) p(k)=0 for all k{0,…,|S|–1}

(2) p(|S|) = 2-O(poly(n))

(3) p(k)[0,1] for all k{0,…,2n}

p(k)0

1

0 1 2 . . . k

. . . . . 2n|S|

Theorem: Given the above, 2

degpoly

n Sp

n

(Improved by Klauck et al.)

Idea: Let

Then V.A. Markov (younger brother of A.A. Markov) showed in 1892 that

provided -1p(x)2 for all 0x2n.

0 2max .

n

mm

mx

d pr

dx

2 13deg 2 1 !

2 2 1 !

mm

m

n

p mr

m

On the other hand, one can show by induction on m that r(m) 2-O(poly(n))/m!

Open Questions

Can we show BQP/poly BQP/qpoly relative to an oracle?

What about SZK BQP/qpoly?

Are randomized and quantum 1-way communication complexities polynomially related for all total Boolean functions?(No asymptotic gap is known)

? ?

??