Download - A MAX Feature Presentation P BQP PSPACE =. Scott Aaronson (IAS) Scotts Grab Bag o Cheap Yuks
A MAX Feature Presentation
P
BQP
PSPACE=
Scott Aaronson (IAS)
Scott’s Grab Bag o’ Cheap Yuks
Scott Aaronson (IAS)
Dr. Scott’s Grab Bag o’ Cheap Yuks
Scott Aaronson (IAS)
Outlook on the Future of Quantum Computing
Scott Aaronson (IAS)
The Remarkable Ubiquity of
Postselection
Scott Aaronson (IAS)
The Stupendous Strength of
Postselection
Scott Aaronson (IAS)
The Hunky, Rippling Manliness of Postselection
Scott Aaronson (IAS)
Lessons Learned in the Gottesman
Institute of Comedy
Scott Aaronson (IAS)
The Amazing Power of Postselection
Learning something about a question by conditioning on the fact that you’re asking it.
What IS Postselection?
BERKELEY CAMBRIDGE
What about the quantum case?
“Anthropic Computing”: A foolproof way to solve NP-complete problems in polynomial time
(1) Accept the many-worlds interpretation
(2) Generate a random truth assignment X
(3) If X doesn’t satisfy , kill yourself
Input: A Boolean formula
In This Talk…
This will lead to:
• An extremely simple proof of the celebrated Beigel-Reingold-Spielman theorem
• Limitations on quantum advice and one-way communication
• Unrelativized quantum circuit lower bounds
• And more!
I’ll study what you could do with a quantum
computer, IF you could measure a qubit and postselect on its being |1
PostBQP
Class of languages decidable by a bounded-error polynomial-time quantum computer, if at any time you can measure a qubit that has a nonzero probability of being |1, and assume the outcome will be |1
I hereby define a newcomplexity class…
(Postselected BQP)
Another Important Animal: PP
Class of languages decidable by a nondeterministic poly-time Turing machine that accepts iff the majority of its paths do
NP
PP
P#P=PPP
PSPACE
P
Theorem: PostBQP = PP
Easy half: PostBQP PP
Adleman, DeMarrais, and Huang (1997) showed BQP PP using “Feynman sum-over-histories”
Idea: Write acceptance and rejection probabilities as sums of exponentially many easy-to-compute terms; then use PP to decide which sum is greater
For PostBQP, just sum over postselected outcomes only
To Show PP PostBQP…Given a Boolean function f:{0,1}n{0,1},let s=|{x : f(x)=1}|. Need to decide if s>2n-1
From
/ 2
0,1
2n
n
x
x f x
2 22 2
2 0 1 1/ 2 2 0 1/ 2 2 2 1,
2 2
n n n
n n
s s sH
s s s s
we can easily prepare
Goal: Decide if these amplitudes have the same or opposite signs
Prepare |0|+|1H| for some ,.Then postselect on second qubit being |1
/ 22 2 2
0 1/ 2 2 2 1:
/ 2 2 2
n
n
s s
s s
Yields in first qubit
To Show PP PostBQP…
/ 22 2 2
0 1/ 2 2 2 1:
/ 2 2 2
n
n
s s
s s
Yields in first qubit
1
0
Suppose s and 2n-2s are both positive
Then by trying / = 2i for all i{-n,…,n}, we’ll eventually get close to
0 1
2
On the other hand, if 2n-2s is negative, then we won’t. QED
Totally unexpectedly, the PostBQP=PP theorem turned out to have implications for classical complexity theory…
Beigel, Reingold, Spielman 1990: PP is “closed under intersection”Solved a problem that was open for 18 years…
Other Classical Results Proved With Quantum Techniques: Kerenidis & de Wolf, A., Aharonov & Regev, …
Observation: PostBQP is trivially closed under intersection PP is too
Given L1,L2PostBQP, to decide if xL1 and xL2, postselect on both computations succeeding, and accept iff they both accept
Other Results that PostBQP=PP Makes Simpler
(Fortnow and Reingold)
(Fortnow and Rogers)
(Kitaev and Watrous)
PPPPP ||
PPPPBQP PPQMA
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
Mike & Ike: “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?”
How powerful is quantum advice?
Could it let us solve problems that are not even recursively enumerable given classical advice of similar size?!
Limitations of Quantum Advice
NP BQP/qpoly relative to an oracle(Uses direct product theorem for quantum search)
BQP/qpoly PostBQP/poly( = PP/poly)
.log 111 fQfmQOfD
Closely related: for all (partial or total) Boolean functions f : {0,1}n {0,1}m {0,1},
Alice’s Classical Advice
Bob, suppose you used the maximally mixed state in place of your
quantum advice. Then x1 is the lexicographically first input for which you’d output the right answer with
probability less than ½.But suppose you succeeded on x1,
and used the resulting reduced state as your advice. Then x2 is the
lexicographically first input after x1 for which you’d output the right answer
with probability less than ½...
x1
x2
Given an input x, clearly lets Bob
decide in PostBQP whether xL
But how many inputs must Alice specify?
We can boost a quantum advice state so that the error probability on any input is at most (say) 2-100n; then Bob can reuse the advice on as many inputs as he likes
We can decompose the maximally mixed state on p(n) qubits as the boosted advice plus 2p(n)-1 orthogonal states
Alice needs to specify at most p(n) inputs x1,x2,…, since each one cuts Bob’s total success probability by least half, but the probability must be (2-p(n)) by the end
PPP Does Not Have Quantum Circuits of Size nk
Does U accept x0 w.p. ½?If yes, set x0LIf no, set x0L
U: Picks a size-nk quantum circuit uniformly at random
and runs it
x0
x1
x2
x3
x4
x5
Conditioned on deciding x0 correctly, does U accept x1 w.p. ½?If yes, set x1LIf no, set x1L
Conditioned on deciding x0 and x1 correctly, does U accept x2 w.p. ½?If yes, set x2LIf no, set x2L
For any k, defines a language L that does not have quantum circuits of size nk
Why? Intuitively, each iteration cuts the number of potential circuits in half, but there were at most circuits to begin with
kn2~
On the other hand, clearly L PPP
Even works for quantum circuits
with quantum advice!
Also…If a function f:{0,1}n{0,1} has a polynomial-size quantum circuit, then a PostBQP machine can find such a circuit using queries to f
Reminiscent of a classical learning theory result of Bshouty, Cleve, et al.
Intuition: Guess random inputs xt and quantum circuits Ct. Repeatedly use postselection to find
• An input xt on which Ct fails
• A circuit Ct+1 that succeeds on x1,…,xt
Even works for quantum circuits
with quantum advice!
And now for a grand finale…0-1-NPC - #AC0 - #L - #L/poly - #P - #W[t] - +EXP - +L - +L/poly - +P - +SAC1 - A0PP - AC - AC0 - AC0[m] - ACC0 - AH - AL - AlgP/poly - AM - AM-EXP - AM intersect coAM - AM[polylog] - AmpMP - AmpP-BQP - AP - AP - APP - APP - APX - AUC-SPACE(f(n)) - AVBPP - AvE - AvP - AW[P] - AWPP - AW[SAT] - AW[*] - AW[t] - βP - BH - BPE - BPEE - BPHSPACE(f(n)) - BPL - BP•NP - BPP - BPPcc - BPPKT - BPP-OBDD - BPPpath - BPQP - BPSPACE(f(n)) - BPTIME(f(n)) - BQNC - BQNP - BQP - BQP/log - BQP/poly - BQP/qlog - BQP/qpoly - BQP-OBDD - BQPtt/poly - BQTIME(f(n)) - k-BWBP - C=AC0 - C=L - C=P - CFL - CLOG - CH - Check - CkP - CNP - coAM - coC=P - cofrIP - Coh - coMA - coModkP - compIP - compNP - coNE - coNEXP - coNL - coNP - coNPcc - coNP/poly - coNQP - coRE - coRNC - coRP - coSL - coUCC - coUP - CP - CSIZE(f(n)) - CSL - CZK - D#P - Δ2P - δ-BPP - δ-RP - DET - DiffAC0 - DisNP - DistNP - DP - DQP - DSPACE(f(n)) - DTIME(f(n)) - DTISP(t(n),s(n)) - Dyn-FO - Dyn-ThC0 - E - EE - EEE - EESPACE - EEXP - EH - ELEMENTARY - ELkP - EPTAS - k-EQBP - EQP - EQTIME(f(n)) - ESPACE - BPP - NISZK - EXP - EXP/poly - EXPSPACE - FBQP - Few - FewP - FH - FNL - FNL/poly - FNP - FO(t(n)) - FOLL - FP - FPNP[log] - FPR - FPRAS - FPT - FPTnu - FPTsu - FPTAS - FQMA - frIP - F-TAPE(f(n)) - F-TIME(f(n)) - GA - GAN-SPACE(f(n)) - GapAC0 - GapL - GapP - GC(s(n),C) - GI - GPCD(r(n),q(n)) - G[t] - HeurBPP - HeurBPTIME(f(n)) - HkP - HVSZK - IC[log,poly] - IP - IPP - L - LIN - LkP - LOGCFL - LogFew - LogFewNL - LOGNP - LOGSNP - L/poly - LWPP - MA - MA' - MAC0 - MA-E - MA-EXP - mAL - MaxNP - MaxPB - MaxSNP - MaxSNP0 - mcoNL - MinPB - MIP - MIP*[2,1] - MIPEXP - (Mk)P - mL - mNC1 - mNL - mNP - ModkL - ModkP - ModP - ModZkL - mP - MP - MPC - mP/poly - mTC0 - NC - NC0 - NC1 - NC2 - NE - NE/poly - NEE - NEEE - NEEXP - NEXP - NEXP/poly - NIQSZK - NISZK - NISZKh - NL - NL/poly - NLIN - NLOG - NP - NPC - NPcc - NPC - NPI - NPcoNP - (NPcoNP)/poly - NP/log - NPMV - NPMV-sel - NPMVt - NPMVt-sel - NPO - NPOPB - NP/poly - (NP,P-samplable) - NPR - NPSPACE - NPSV - NPSV-sel - NPSVt - NPSVt-sel - NQP - NSPACE(f(n)) - NT - NTIME(f(n)) - OCQ - OptP - P - P/log - P/poly - P#P - P#P[1] - PAC0 - PBP - k-PBP - PC - Pcc - PCD(r(n),q(n)) - P-close - PCP(r(n),q(n)) - PermUP - PEXP - PF - PFCHK(t(n)) - PH - PHcc - Φ2P - PhP - Π2P - PINC - PIO - PK - PKC - PL - PL1 - PLinfinity - PLF - PLL - PLS - PNP - PNP[k] - PNP[log] - PNP[log^2] - P-OBDD - PODN - polyL - PostBQP - PP - PP/poly - PPA - PPAD - PPADS - PPP - PPP - PPSPACE - PQUERY - PR - PR - PrHSPACE(f(n)) - PromiseBPP - PromiseBQP - PromiseP - PromiseRP - PrSPACE(f(n)) - P-Sel - PSK - PSPACE - PT1 - PTAPE - PTAS - PT/WK(f(n),g(n)) - PZK - QAC0 - QAC0[m] - QACC0 - QAM - QCFL - QCMA - QH - QIP - QIP(2) - QMA - QMA+ - QMA(2) - QMAlog - QMAM - QMIP - QMIPle - QMIPne - QNC0 - QNCf0 - QNC1 - QP - QPLIN - QPSPACE - QSZK - R - RE - REG - RevSPACE(f(n)) - RHL - RL - RNC - RP - RPP - RSPACE(f(n)) - S2P - S2-EXP•PNP - SAC - SAC0 - SAC1 - SAPTIME - SBP - SC - SEH - SelfNP - SFk - Σ2P - SKC - SL - SLICEWISE PSPACE - SNP - SO-E - SP - SP - span-P - SPARSE - SPL - SPP - SUBEXP - symP - SZK - SZKh - TALLY - TC0 - TFNP - Θ2P - TreeBQP - TREE-REGULAR - UAP - UCC - UE - UL - UL/poly - UP - US - VNCk - VNPk - VPk - VQPk - W[1] - WAPP - W[P] - WPP - W[SAT] - W[*] - W[t] - W*[t] - XOR-MIP*[2,1] - XP - XPuniform - YACC - ZPE - ZPP - ZPTIME(f(n))
Quantum Karp-Lipton Theorem
If PP BQP/qpoly, then the counting hierarchy—consisting ofetc.—collapses to PP
,,,PPPPPP PPPPPP
But there’s more: With no assumptions, PP does not have quantum circuits of size nk
And more: PEXP requires quantum circuits of size f(n), where f(f(n))2n
Even Stronger Separations
QMAEXP (a subclass of PEXP) is not in BQP/qpoly
QCMAEXP (a subclass of QMAEXP) is not in BQP/poly
A0PP (a subclass of PP) does not have quantum circuits of size nk
NONRELATIVIZING
Conclusions
I started out with a weird philosophical question
Try it—it works!
I ended up with seven or eight results