Equilibrium Value Method For the proof of QIP=PSPACE
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Transcript of Equilibrium Value Method For the proof of QIP=PSPACE
Equilibrium Value Method For the proof of
QIP=PSPACEXiaodi Wu
EECS, University of Michigan, Ann ArborJanuary, 2010
•The work was conducted while the author was visiting the Institute for Quantum •Computing, University of Waterloo, Ontario, Canada.
Interactive Proof System : Intuitive Picture
Prover
Verifier
input x
input x
Goal : determine whether x is in a language L
Computationally unbounded power.
Try to convince the verifier that x is in language L.
Probabilistic Polynomial time
power. To determine
whether x is in L.
Polynomial Rounds interactions of classical messages
Quantum Interactive Proof System
Quantum Prover
Quantum
Verifier
input x, still classical
input x
Goal : determine whether x is in a language L
Computationally unbounded
power = anything allowed
by quantum physics
Probabilistic Polynomial
time on quantum
computer.
Polynomial Rounds interactions of quantum messages
Previous Result:
IP=PSPACE [Lund, Fortnow, Karloff, and Nisan 1992; Shamir 1992]
Easy Direction : IP PSPACE, since all the messages are of polynomial size.
Hard Direction: PSPACE IP
µ
µ
Solved by a method called Arithmetizationwhich constructs a polynomial round interaction protocols for PSPACE-complete problem. Polynomial rounds is necessary !
What about Quantum Case
IP QIP, and thus we know PSPACE QIP The easier direction becomes hard in
quantum caseIt is an open problem for almost a decade to
show QIP=PSPACE.However, we do know something non-trivial
about the QIP in the very beginning. QIP=QIP(3)[KW00], 3 rounds are sufficient
for quantum caseQIP(3) EXP[KW00], by formulated as a SDP.
µ µ
µ
3 rounds Quantum Interactive Proof System
1 2 3
Qubits
efficient quantum circuits
all- power , any quantum circuits
Notations: linear algebra : complex Euclidean spaces : space of operators acting on : set of positive semidefinite operators
(or matrices) acting on
X ;Y ;ZL(X ) XPos(X )
X
D(X ) := f½2 Pos(X ) Tr(½) = 1gQuantum State:
One part of the whole state: Given
½2 D(X Y )The X part of the state is and Y part of the state is TrY (½) TrX (½)
This is called Partial Trace
More precisely, we have partial trace be the unique linearmapping such that
TrY : L (X Y ) ! L (X )
TrY (A B) = Tr(B)A 8;A 2 L(X );B 2 L(Y )
Notations: continuedQuantum Operations: a operation maps a quantum state to a quantum state.
More precisely:
©: L(Y ) ! L (Z)
auxiliary space
(Complete Positivity) (Trace Preserving)
½ 0) ¾= (I X ª )½ 0Tr(½) = Tr(¾) = 1
Quantum Measurement : an irreversible quantum operations defined by a set of positive operators such that .M = fM i 2 Pos(X )g
Pi M i = I X
The outcome k occurs with probability hMk;½i
3 rounds Quantum Interactive Proof System(revisit)
1 2 3
P M VQuantum State on
efficient quantum operations
all- power , any quantum operation
SDP?
RoadmapMatch from both sides
[KW00]
General Model for interaction
[GW07]
SDP formulation but with exponential size
Polynomial algorithm for SDP (IPM,
Ellipsoid) QIP in EXP [KW00]
PSPACE ?
How to parallelize ?
Matrix Multiplicative Weights Update
method[AHK05]
Parallelizable for some SDP and Equilibrium
Value ?
PSPACE =NC(poly)[Bord77]
One Possible Way
Roadmap(continued)
Bad News: old formulation of QIP still open
Good News: reformulation becomes solvable
SDP reformulation
[JJUW09], August 09
QIP=QMAM[MW05]
1 2 3Only one classical
bit is sent in the second
step
Simpler solvable SDP
QMAM in NC(poly) too complicated and technical
Starts with definition, simpler SDP but not that simple.Using MMW to solve SDP involves more complicated
steps.
Roadmap(continued)A Neater Proof is available [Wu09]
August, 09
Starts with QIP-Complete problem
[RW05]
Resulted in a simple Equilibrium Value
Problem
Quantum Circuit Distinguishability:
Given two short quantum circuits, distinguish their
distance between two promises.
Solved by Matrix Multiplicative Weight Update Method
Two Key Ingredients
Matrix Multiplicative Weight Update Method
w1
w2
.
.
.
wn
n agents weights
Update weights according toperformance:
wit+1 Ã wi
t (1 + ¢ performance of i)
1$ for correct prediction
0$ for incorrect
Long History(cited from Sanjeev Arora)
N “experts” on TVCan we perform as good as the best
expert ?
The answer is Yes by using multiplicative weight update
Matrix Multiplicative Weight Update Method
Density operator Observation
hM (t);½(t) i = Tr(M (t);½(t))cost of round t
updated in this way
my performance any agent’s performance
some small gap
Proof Hint: use potential function Tr(W (t)) and matrix
inequality.
Equilibrium Value
C1 £ C2Consider C1, C2 are convex compact sets, function f is a bilinear function on
mina2C1 maxb2C2 f (a;b) =maxb2C2 mina2C1 f (a;b)
Moreover, there exists an equilibrium point (a¤;b¤)maxb2C2 f (a¤;b) = f (a¤;b¤) =mina2C1 f (a;b¤)equilibrium value ¤
Question: How to compute the equilibrium value ?
Pick a random series of points from C1, and then get the maximum over C2
a1a2....at
b1b2....bt
¸¤ = mina2C1
maxb2C2
f (a;b)· min
tmaxb2C2
f (at;b) =mint
f (at;bt)
a upper bound is obtained easily
How about the lower bound ? choose a better series
Equilibrium ValueIn our settings, C1 is the set of density operators, C2 is the set
The bilinear function is h¦ ;¥(½)i ;½2 C1; ¦ 2 C2 linear mapping
a1a2....at
b1b2....bt
a1
a2....at
b1
b2....bt
MMW helps to generate the
series
Intuitively thinking : Why this works?
Problem with the upper bound is it can be any large. MMW helps to make the value less than the “best agent”
plus small gap.
equilibrium point
Equilibrium Value
max¦ h¦ ;¥(½(t))i
make own “observation”use MMW to get
Get for the round t½(t)
½(t+1)substitute
Equilibrium Value
use equilibrium point (½¤; ¦ ¤)Considermaxb2C2 f (a¤;b) = f (a¤;b¤) =mina2C1 f (a;b¤)
approximated value
Conclusion: with precision , need rounds! ±
Convert QIP-Complete to Equilibrium Value ProblemGiven any two quantum mixed state circuits Q1, Q2, wants to distinguish between
This norm measures the distance between two circuitsor channel. It is called diamond norm.
Induced from L1 norm for super-operators:k©k1 =maxk½k· 1k©(½)k1k©k¦ =maxk½k· 1k© I (½)k1
To:
better representation of the distinguishing powerby using entanglement with auxiliary space.
Proved to be QIP-Complete when
a+b=2, 0<b<1<a<2 [RW05]
A formulation of equilibrium value simply follows!
The Converted Problem
two promises with constant gap!
constant precision will do the job!
two promises!
1.9
0.1
converted from Q0,Q1
equilibrium value
kf©A ¡ f©B k1min
The Conversion : sketch
diamond norm to fidelity
fidelity to L1 norm
max
“ – “ sign, min implied
k½k1 =max¦ h½;2I ¡ ¦ ithen we have a min-max form
Simulation by NC(poly)constant polynomial
polynomial time matrix operations
matrix operations in NC
Finally, polynomial compositions of NC(poly), still NC(poly) !
thus in PSPACE
ConclusionsCorollary: QIP=PSPACE
SDP reformulation
[JJUW09], August 09
A Neater Proof is available [Wu09]
August, 09
Use QIP=QMAMUse definition to
formulateSolve SDP by MMW
Use QIP-Complete Problem
Formulated as equilibrium value
Solved by MMW
Open Questions
How to make more applications of MMW method?
For quantum, QRG(2), QRG, QMA(2) candidates
In other fields, like algorithmic game theory…
MMW method for Convex Programming, under KKT conditions, or …
Thank You!Q&A