A Theory of Isolatability Scott Aaronson Andrew Drucker MIT.

A Theory of Isolatability

Scott Aaronson Andrew DruckerMIT

Freeze-Dried Computation

Motivating Question: How much useful computational work can one “store” in (say) an n-qubit quantum state, or a coin whose bias is an arbitrary real number?

Potentially a huge amount!

We give a new tool—called “isolatability”—for ruling out the possibility of such extravagant encodings.

Idea: Take some advice resource (such as a coin or a quantum state), and simulate it using a short classical string, together with a polynomial number of untrusted advice resources.

In other words, all the relevant information in the advice resource gets packed into an ordinary string (which we say “isolates” the resource), and we're left with just a computational search problem—of finding coins, quantum states, etc. that are consistent with the string.

Part I: The Majority-Certificates Lemma

Our basic tool

Part II: Application to Quantum Advice

BQP/qpoly QMA/poly

Part III: Application to Advice Coins

PSPACE/coin PSPACE/poly

Part I: The Majority-Certificates Lemma

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

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 O(log|S|), such that

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

(ii)If f1S is consistent with C1, f2S is consistent with C2, and so on, then MAJ(f1,…,fm)=f*, where MAJ denotes pointwise majority.

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.

Part II: Application to Quantum Advice

BQP/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”)

Let ρ be any quantum state on n qubits. Then for all m,ε, there exists a 2-local Hamiltonian H=H1+ +H⋯ L on poly(n,m,1/ε) qubits, such that any ground state |φ of H can be used to simulate ρ (with error ε) on all quantum circuits of size at most m. In other words, there exists an efficient mapping C→C such ′that for all circuits C of size m,

.accepts Praccepts 'Pr CC

“PHYSICS” IMPLICATION

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

“Lifting” the Majority-Certificates LemmaBoolean 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”

Majority-Certificates Lemma, Real CaseLemma: Let S be a set of functions f:{0,1}ⁿ→[0,1], let f∗ S, and ∈let ε>0. Then we can find m=O(n/ε²) functions f1,…,fm S, sets ∈X1,…,Xm {0,1}ⁿ each of size⊆

and

for which the following holds. All functions g1,…,gm S that ∈satisfy for all i[m] also satisfy

where

,fat 48/3

Sn

Ok

Sn 48/

2

fat

xfxg iiXx i

max

,max *

1,0

xfxg

nx

.1: 1 xgxgm

xg m

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 the real version 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.

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).

Part III: Application to Advice Coins

Erik Demaine (motivated by a computational genetics problem): “Suppose a PSPACE machine can flip a coin with Bernoulli probability p an unlimited number of times. Can it extract an exponential amount of information (or even more) about p?”

Me: “I’m sure whatever the answer is, it’s obvious...”

Didn’t seem too likely there could be superpowerful “Advice Coins”

Indeed, Hellman & Cover proved the following in 1970...

Suppose a finite automaton M is trying to decide whether a coin has p=½ or p=½+. Then even if it can flip the coin an unlimited number of times, M needs (1/) states to succeed with probability (say) 2/3.

This result seems to imply PSPACE/coinPSPACE/poly, since we could take the first poly(n) bits of p as the advice. But it breaks down if p is close to 0 or 1!

Furthermore, quantum mechanics nullifies the Hellman-Cover Theorem!

Theorem: For any >0, it’s possible to distinguish a coin with p=½ from a coin with p=½+ using a single qubit of memory, with error probability independent of .

0

10 1

2

Keep flipping the coin.Whenever the coin lands heads, rotate /100 radians counterclockwise. Whenever it lands tails, rotate /100 radians clockwise.

Halt with probability ~2/100 at each time stepExpected difference in final angle after halting, in p=½ vs. p=½+ cases: 1 radianStandard deviation in angle:

Theorem: Despite these obstacles,BQPSPACE/coin = PSPACE/coin = PSPACE/poly.

We’re interested in a fixed-point of p: a mixed state p such that

Proof: Suffices to show BQPSPACE/coin PSPACE/poly.Let 0 = quantum operation applied to our memory

qubits whenever coin lands heads,1 = operation applied when it lands tails

Then induced operation at each time step:

Fixed-Points of Superoperators

Already studied by [A.-Watrous 2008], in the context of quantum computing with closed timelike curves

Our result there: BQPCTC = PCTC = PSPACE

Quantum computers with CTCs have exactly the same power as classical computers with CTCs, namely PSPACE (or: “CTCs make time and space equivalent as computational resources”)

Key Lemma: Let a(p) be the probability that an S-state quantum finite automaton accepts, if each input bit is 1 with independent probability p. Then a(p) is a degree-S2 rational function of p.

Proof Idea: We can write a(p) as

Furthermore, each entry of p can be written as a degree-S2 rational function of p, by using Cramer’s Rule on S2S2 matrices.

.AccAccAccAcc1

lim1

0 p

T

t

tp

T T

Now, by calculus, a degree-S2 rational function can cross the line y=½ at most 2S2 times…

p

ax(p)

Hence a(p) is itself a degree-S2 rational function of p, except possibly at a finite number of singularities:

A further continuity argument rules out the singularities, except possibly at p=0 and p=1.

p

ax(p)

Given a BQPSPACE/coin machine M, let ax(p) be its acceptance probability on input x{0,1}n and a coin with Bernoulli probability p.Challenge: How can we describe p well enough to compute ax(p) for every x, using only poly(n) bits?

Finishing the Argument (Sketch)We’ve upper-bounded how many distinct Boolean functions f:{0,1}n{0,1} can be expressed, as we vary p from 0 to 1.So by the Majority-Certificates Lemma, we can simulate PSPACE/coin using a PSPACE/poly machine, combined with poly(n) untrusted advice coins.We then get a computational search problem: finding coin biases p1,…,pm that are consistent with the /poly advice string.We can solve that problem in PSPACE, using NC algorithms for root-finding developed by Neff and Pan

When Can An O(1)-State Finite Automaton Detect an Change to

the Bias of a Coin?

Open Problems

Quantum finite automata: are their limiting acceptance probabilities continuous functions of p, for p(0,1)?

Find other applications of isolatability

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

Optimality of the Majority-Certificates Lemma?

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

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:'