Xiaodi Wu
EECS, University of Michigan, Ann Arbor
January, 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
Goal : determine whether x is in a language L
input x
Prover
Polynomial
Rounds interactions of classical messages
input x
Verifier
Quantum Interactive Proof System
Goal : determine whether x is in a language L
input x, still classical
Polynomial
Rounds interactions of quantum messages
input x
Previous Result:
 IP=PSPACE [Lund, Fortnow, Karloff, and Nisan 1992;
Shamir 1992]
 Easy Direction : IPµ PSPACE, since all the messages
are of polynomial size.
 Hard Direction: PSPACE µ IP
Solved by a method called
which constructs a polynomial round interaction protocols for
PSPACE-complete problem. Polynomial rounds is necessary !
What about Quantum Case
 IP µ QIP, and thus we know PSPACE µ QIP
 The easier direction becomes hard in quantum case
 It 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 case
 QIP(3) µ EXP[KW00], by formulated as a SDP.
3 rounds Quantum Interactive Proof System
Qubits
1
efficient quantum circuits
2
3
all- power
, any
quantum
circuits
Notations: linear algebra
 X ; Y; Z : complex Euclidean spaces
 L (X ) : space of operators acting on X
Pos(X ) : set of positive semidefinite operators (or
matrices) acting on X
Quantum State: D(X ) := f ½2 Pos(X )
Tr (½) = 1g
One part of the whole state: Given ½2 D(X - Y)
The X part of the state is Tr Y (½) and Y part of the state is Tr X (½)
More precisely, we have partial trace Tr Y : L(X - Y) !
mapping such that
This is called
Partial Trace
L(X ) be the unique linear
Tr Y (A - B) = Tr (B)A 8; A 2 L(X ); B 2 L(Y)
Notations: continued
Quantum Operations: a operation maps a quantum state to a quantum state.
More precisely:
auxiliary space
© : L(Y) ! L(Z )


½¸ 0 ) ¾= (I X - ª )½¸ 0
Tr (½) = Tr (¾) = 1
(Complete Positivity)
(Trace Preserving)
Quantum Measurement : an irreversible quantum operations
defined by a set of
P
positive operators M = f M i 2 Pos(X )g such that i M i = I X .
The outcome k occurs with probability hM k ; ½i
3 rounds Quantum Interactive Proof System(revisit)
Quantum State on P - M - V
1
efficient quantum operations
2
3
all- power
, any
quantum
operation
Roadmap
Match from both
sides [KW00]
General Model for
interaction [GW07]
SDP formulation but
with exponential size
One Possible Way
PSPACE
PSPACE ?
=NC(poly)
[Bord77]
Polynomial algorithm for
SDP (IPM, Ellipsoid)
QIP in EXP [KW00]
How to parallelize ?
Matrix Multiplicative
Weights Update method
[AHK05]
Parallelizable for some SDP
and Equilibrium Value
Roadmap(continued)
Bad News: old formulation of QIP
still open 
Good News: reformulation becomes
solvable 
SDP reformulation
[JJUW09], August 09
QIP=QMAM
[MW05]
Simpler solvable SDP
QMAM in NC(poly)
Only one
Starts with definition,
simpler SDP
classical bit
but not
thatin
simple.
is sent
the
Using MMWsecond
to solvestep
SDP involves
more complicated steps.
1
2
too complicated and technical 
3
Roadmap(continued)
A Neater Proof is available
[Wu09] August, 09 
Starts with QIP-Complete
problem [RW05]
Quantum Circuit Distinguishability:
Given two short quantum circuits,
distinguish their distance between
two promises.
Resulted in a simple
Equilibrium Value Problem
Solved by Matrix Multiplicative
Weight Update Method
Matrix Multiplicative Weight Update Method
n agents
weights
w1
w2
Long History
(cited from
Sanjeev Arora)
.
.
Update weights according to
performance:
wit+1 Ã wit (1 +  ¢ performance of i)
.
wn
The answer is Yes by using
multiplicative weight update
1$ for correct prediction
 N “experts” on TV
 Can we perform as good as the best expert ?
0$ for incorrect
Matrix Multiplicative Weight Update Method
Density operator
Proof Hint: use potential function
Tr(W (t)) and matrix inequality.
hM (t ) ; ½(t ) i = Tr (M (t ) ; ½(t ) )
cost of round t
my performance
Observation
updated in this way
some small gap
any agent’s
performance
Equilibrium Value
Consider C1, C2 are convex compact sets, function f is a bilinear function on C1 £ C2
mina2 C 1 maxb2 C 2 f (a; b) = maxb2 C 2 mina2 C 1 f (a; b)
¤ ¤
Moreover, there exists an equilibrium point (a ; b )
equilibrium value ¸ ¤
maxb2 C 2 f (a¤ ; b) = f (a¤ ; b¤ ) = mina2 C 1 f (a; b¤ )
Question: How to compute the equilibrium value ?
Pick a random series of points from C1, and then get the maximum over C2
a1
a2
..
..
at
b1
b2
..
..
bt
¸¤
=
·
min max f (a; b)
a2 C 1 b2 C 2
min max f (at ; b) = min f (at ; bt )
t
t
b2 C 2
a upper bound is obtained easily
How about the lower bound ?
choose a better series
Equilibrium Value
In our settings, C1 is the set of density operators, C2 is the set
The bilinear function is h¦ ; ¥(½)i ; ½2 C1 ; ¦ 2 C2
a1
a2
..
..
at
b1
b2
..
..
bt
a1
b1
a2
..
..
at
b2
..
..
bt
linear mapping
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
Get ½( t ) for the round t
max¦ h¦ ; ¥(½(t ) )i
make own “observation”
substitute
use MMW to get
½( t + 1)
Equilibrium Value
approximated value
use equilibrium point (½¤ ; ¦ ¤ )
Consider maxb2 C 2 f (a¤ ; b) = f (a¤ ; b¤ ) = mina2 C 1 f (a; b¤ )
Conclusion: with precision ± , need
rounds!
Convert QIP-Complete to Equilibrium Value Problem
Given any two quantum mixed state circuits Q1, Q2, wants to distinguish between
This norm measures the distance between two circuits
or channel. It is called diamond norm.
Proved to be QIP-Complete
when a+b=2, 0<b<1<a<2
[RW05]
A formulation of equilibrium
value simply follows!
Induced from L1 norm for super-operators:
k©k1 = maxk½k· 1 k©(½)k1
To:
k©k¦ = maxk½k· 1 k© - I (½)k1
better representation of the distinguishing power
by using entanglement with auxiliary space.
The Converted Problem
two promises!
converted from Q0,Q1
1.9
0.1
fA ¡ ©
f B k1
min k©
equilibrium value
two promises with constant gap!
constant precision will do the job!
The Conversion : sketch
max
diamond norm to fidelity
fidelity to L1 norm
“ – “ sign, min implied
k½k1 = max¦ h½; 2I ¡ ¦ i
then 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
Conclusions
Corollary: QIP=PSPACE
SDP reformulation
[JJUW09], August 09
A Neater Proof is available
[Wu09] August, 09 
 Use QIP=QMAM
 Use QIP-Complete Problem
 Use definition to formulate
 Formulated as equilibrium
 Solve SDP by MMW
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 …