The Power of Quantum Advice Scott Aaronson Andrew Drucker.

The Power of Quantum Advice

Scott AaronsonAndrew Drucker

Freeze-Dried Computation

Motivating Question: How much useful computational work can one “store” in a quantum state, for later retrieval?

If quantum states are exponentially large objects, then possibly a huge amount!

Yet we also know that quantum states have no more “general-purpose storage capacity” than classical strings of the same size

Cast of CharactersBQP/qpoly is the class of problems solvable in quantum polynomial time, with the help of polynomial-size “quantum advice states”

Formally: a language L is in BQP/qpoly if there exists a polynomial time quantum algorithm A, as well as quantum advice states {|n}n on poly(n) qubits, such that for every input x of size n, A(x,|n) decides whether or not xL with error probability at most 1/3

YQP (“Yoda Quantum Polynomial-Time”) is the same, except we also require that for every alleged advice state , A(x,) outputs either the right answer or “FAIL” with probability at least 2/3

BQP YQP QMA BQP/qpoly

Watrous 2000: For any fixed, finite black-box group Gn and subgroup Hn≤Gn, deciding membership in Hn is in BQP/qpoly

The quantum advice state is just an equal superposition |Hn over the elements of Hn We don’t know how to solve the same problem in BQP/poly

A. 2004: BQP/qpoly PP/poly = PostBQP/poly Quantum advice can be simulated by classical advice, combined with postselection on unlikely measurement outcomes

A. 2006: HeurBQP/qpoly = HeurYQP/polyTrusted quantum advice can be simulated on most inputs by trusted classical advice combined with untrusted quantum advice

A.-Kuperberg 2007: There exists a “quantum oracle” separating BQP/qpoly from BQP/poly

QUANTUM ADVICE IS POWERFUL

NO IT ISN’T

New Result: BQP/qpoly = YQP/polyTrusted quantum advice is equivalent in power to trusted classical advice combined with untrusted quantum advice.

(“Quantum states never need to be trusted”)

Given an n-qubit state and parameters m,, there exists a local Hamiltonian H on poly(n,m,1/) qubits (e.g., a sum of 2-qubit interactions) for which the following holds:

For any ground state | of H, and any binary measurement E on performed by a circuit with ≤m gates, there’s an efficient measurement f(E) that we can perform on | such that

.Tr EEf

FOR THE PHYSICISTS

What Does It Mean?Preparing quantum advice states is no harder than preparing ground states of local Hamiltonians

This explains a once-mysterious relationship between quantum proofs and quantum advice: efficient preparability of ground states would imply both QMA=QCMA and BQP/qpoly=BQP/poly

“Quantum Karp-Lipton Theorem”: NP-complete problems are not efficiently solvable using quantum advice, unless some uniform complexity classes collapse unexpectedly

QCMA/qpoly QMA/poly: classical proofs and quantum advice can be simulated with quantum proofs and classical advice

BQP

YQP QCMABQP/poly

BQP/qpoly=YQP/poly QCMA/poly QMA

QCMA/qpoly

QMA/poly PP

PP/polyQMA/qpoly

PSPACE/polyA.’06

This work

Majority-Certificates

Lemma

Real Majority-Certificates Lemma

Circuit Learning (Bshouty et al.)

Minimax Theorem

Safe Winnowing

Lemma

Holevo’s Theorem

Random Access Code Lower

Bound (Ambainis et al.)

BQP/qpoly=YQP/poly

HeurBQP/qpoly=HeurYQP/poly(A.’06)

Quantum advice no harder than ground state preparation

Fat-Shattering Bound (A.’06)

Covering Lemma (Alon et al.)

Learning of p-Concept Classes (Bartlett & Long)

LOCAL HAMILTONIANS is QMA-complete

(Kitaev)

Cook-Levin Theorem

QMA=QMA+(Aharonov & Regev)

Used as lemma

Generalizes

that computes some Boolean function f:{0,1}n{0,1} belonging to a “small” set S (meaning, of size 2poly(n)). Someone wants to prove to us that f equals (say) the all-0 function, by having us check a polynomial number of outputs f(x1),…,f(xm).

Intuition: We’re given a black box (think: quantum state)

fx f(x)

This is trivially impossible!f0 f1 f2 f3 f4 f5

x1 0 1 0 0 0 0

x2 0 0 1 0 0 0

x3 0 0 0 1 0 0

x4 0 0 0 0 1 0

x5 0 0 0 0 0 1

But … what if we get 3 black boxes, and are allowed to simulate f=f0 by taking the point-wise MAJORITY of their outputs?

Majority-Certificates Lemma

Lemma: Let S be a set of Boolean functions f:{0,1}n{0,1}, and let f*S. Then there exist m=O(n) certificates C1,…,Cm, each of size k=O(log|S|), such that

(i)Some fiS is consistent with each Ci, and

(ii)If fiS is consistent with Ci for all i, then MAJ(f1(x),…,fm(x))=f*(x) for all x{0,1}n.

Definitions: A certificate is a partial Boolean function C:{0,1}n{0,1,*}. A Boolean function f:{0,1}n{0,1} is consistent with C, if f(x)=C(x) whenever C(x){0,1}. The size of C is the number of inputs x such that C(x){0,1}.

Proof IdeaBy symmetry, we can assume f* is the all-0 function. Consider a two-player, zero-sum matrix game:

Alice picks a certificate C of size k consistent

with some fS

Bob picks an input x{0,1}n

Alice wins this game if f(x)=0 for all fS consistent with C.

Crucial Claim: Alice has a mixed strategy that lets her win >90% of the time.

The lemma follows from this claim! Just choose certificates C1,…,Cm independently from Alice’s winning

distribution. Then by a Chernoff bound, almost certainly MAJ(f1(x),…,fm(x))=0 for all f1,…,fm consistent with C1,…,Cm respectively and all inputs x{0,1}n. So clearly there exist

C1,…,Cm with this property.

Proof of ClaimUse the Minimax Theorem! Given a distribution D over x, it’s enough to create a fixed certificate C such that

.10

11 s.t. with consistent Pr

xfCf

Dx

Stage I: Choose x1,…,xt independently from D, for some t=O(log|S|). Then with high probability, requiring f(x1)=…=f(xt)=0 kills off every fS such that

.10

11Pr

xf

Dx

Stage II: Repeatedly add a constraint f(xi)=bi that kills at least half the remaining functions. After ≤ log2|S| iterations, we’ll have winnowed S down to just a single function fS.

“Lifting” the Lemma to QuantumlandBoolean Majority-Certificates BQP/qpoly=YQP/poly Proof

Set S of Boolean functions Set S of p(n)-qubit mixed states

“True” function f*S “True” advice state |n

Other functions f1,…,fm Other states 1,…,m

Certificate Ci to isolate fi Measurement Ei to isolate I

New Difficulty Solution

The class of p(n)-qubit quantum states is infinitely large! And even if we discretize it, it’s still doubly-exponentially large

Result of A.’06 on learnability of quantum states (building on Ambainis et al. 1999)

Instead of Boolean functions f:{0,1}n{0,1}, now we have real functions f:{0,1}n[0,1] representing the expectation values

Learning theory has tools to deal with this: fat-shattering dimension, -covers… (Alon et al. 1997)

How do we verify a quantum witness without destroying it?

QMA=QMA+ (Aharonov & Regev 2003)

What if a certificate asks us to verify Tr(E)≤a, but Tr(E) is “right at the knife-edge”?

“Safe Winnowing Lemma”

Theorem: BQP/qpoly = YQP/poly.Proof Sketch: YQP/poly BQP/qpoly is immediate. For the other direction, let LBQP/qpoly. Let M be a quantum algorithm that decides L using advice state |n. Define

accepts ,Pr: xMxf

Let S = {f : }. Then S has “fat-shattering dimension” at most poly(n), by A.’06. So we can apply a real analogue of the Majority-Certificates Lemma to S. This yields certificates C1,…,Cm (for some m=poly(n)), such that any states 1,…,m consistent with C1,…,Cm respectively satisfy

xfxfxfm nnm

1

1

for all x{0,1}n (regardless of entanglement). To check the Ci’s, we use the “QMA+ super-verifier” of Aharonov & Regev.

Promised Application to “Physics”

Furthermore, in their reduction, the witness is a “history state”

So given any language LBQP/qpoly=YQP/poly, we can use the Kitaev et al. reduction to get a local Hamiltonian H whose unique ground state is |’. We can then use |’ to recover the YQP witness |, and thereby decide L

By Kitaev et al., we know LOCAL HAMILTONIANS is QMA-complete.

T

ttt

T 1

1:

Measuring this state yields the original QMA witness |1 with (1/poly(n)) probability. Hence |1 can be recovered from

npoly:'

Quantum Karp-Lipton Theorem

Our quantum analogue:

If NP BQP/qpoly, then coNPNP QMAPromiseQMA.

Karp-Lipton 1982: If NP P/poly, then coNPNP = NPNP.

Proof Idea: A coNPNP statement has the form x y R(x,y).

By the hypothesis and BQP/qpoly = YQP/poly, there exists an advice string s, such that any quantum state consistent with s lets us solve NP problems (and some such is consistent).

In QMAPromiseQMA, first guess an s that’s consistent with some state . Then use the oracle to search for an x and such that, if is consistent with s, then R(x,Q(x,)) holds, where Q is a quantum algorithm that searches for a y such that R(x,y).

A Theory of Isolatability

Which classes of functions C are “isolatable”—in the sense that for any fC, one can give a small number of conditions such that any f1,…,fmC satisfying the conditions can be used to compute f efficiently on all inputs?

We can generalize the majority-certificates idea well beyond what we have any application for

Another application of the Majority-Certificates Lemma: it substantially simplifies the proof that

BQPSPACE/coin = PSPACE/poly

We study the following abstract question, inspired by computational learning theory:

Open Problems

Improve QMA/qpoly PSPACE/poly to QMA/qpoly P#P/poly

Find other applications of the majority-certificates technique

Circuit complexity? Communication complexity? Learning theory? Quantum information?

Is the dependence on n, log|S|, and 1/ optimal?

Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly

Although this work closes off a chapter in the quantum advice story, there are still