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Simulating MIP(2,1) protocols by QMIP(2,1)

Hirotada Kobayashi and Keiji Matsumoto

December 3rd, 2006

1 Introduction

This note shows how to prevent quantum provers cheat by quantum strategy in 1 round protocol. Namely,we are interested in simulating (1) classical 2-prover 1-round proof systems. Here, quantum provers shareunlimitted amount of entanglement.

2 Notations and conventions U(H): the totality of unitary operations actiong onH. Projection ontoHis also denoted byH. Hilbert spaces with the same dimensionality are often identied with each other. jjAjj1 := 12 tr

pAyA= 12

P(singular values)

jjthe dierence of probability distributionsjj1 means the total variation (statistical distance). d(; ) := jj jj1, d(ji ; ji) := jj ji hj ji hj jj1 P1,P2: the 1st and 2nd prover.

V: the verier. P1,P2: the private space of the 1st and 2nd prover. V: Vs private space. M1: the message space between V andP1 (M2 is dened analogously). Mq01: the register in the message space between V andP1, whereVwrites his query. Mq1: the subspace ofMq01, which is sppannded byfjqi1g, whereqruns over all possible query used in the

simulated claasical protocol.

Ma1 : the register in the message space between V andP1, where honestP1is supposed to write his answer.

K= dim

M1 = dim

M2

Ka = dim Mq1 = dim Mq2, Kq= dim Mq1 = dim Mq2 fjaijg: a computational basis ofMaj . jeq;ai1 := 1 H logKa jq; ai1. Letting i = (q; a), we often denote this byjeii1. fjiiVg: the computational basis ofV, respectively. ji: the initial state ofP1 P2. U1 2 U(P1 M1), U2 2 U(P2 M2): the rst operation ofP1 andP2, respectively. U1ij := 1 heij Uk jeji1, U2ij := 2 heij Uk jeji2 .

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3 A protocol (+Some more symbols are dened)

We do one of the following tests uniformly randomly, without telling the provers which test is performed.

(A) We modify a classical protocol which asksqand q0 to the rst and the second provers respectively obeying

the joint probabilitypq(q; q0). We dene

p((q; a); (q0; a0)) := 1

Kapq(q; q

0):

As is explained in the next section, we can assume p(i; j) = p(j; i)without loss of generality: any protocolcan be symmetrized by randomly exchanging the role ofP1andP2. Let us denote the marginal distributionP

q0pq(q; q0)and

Pjp(i; j) bypq(q)and p(i). Based on a classical protocol, we do as follows.

(A.1)V sets the registerM1 M2 V toXi;j

pp(i; j) jii1 jji2 ji; jiV =

1

Ka

Xq;q0;a;a0

qpq(q; q0) jqi1 jai1 jq0i2 ja0i2 j(q; a); (q0; a0)iV

= Xi;j pp(i; j) jeii1 jeji2 jeiiV;1 jejiV;2(A.2)P1 andP2 performs their operations.

(A.3)V measures M1M2Vby standard basis, and decode the answers from the provers by the operationjai1 ja0i2 j(a00; a000)iV! ja a00i1 ja0 a000i2 j(a00; a000)iV.

(A.4)V follows the verication procedure of the given classical protocol.

(B) The algorithm of the test depends on the input probability distribution p(i; j).

(B.1)V sets the registerM1 M2 tojeii1 jeji2 , wherei,j are chosen randomly according to the probabilitydistributionp(i; j). Before sending o the message, V records i; j in his register.

(B.2)P1 andP2 performs their operations.

(B.3)V measuresM1 M2 to check whether the state is not changed by the provers. If the state is notchanged, he accepts the input.

(C) (C.1) Let R be a register in the Vs private space. Vsets the register M1M2R toPK

i=1

pp(i) jeii1 jeii2 jiiR

.

(C.2)P1 andP2 performs their operations.

(C.3)V measuresM1 M2 R to check whether the state is not changed by the provers. If the state is notchanged, he accepts the input.

(D) (D.1) Let R be a register in the Vs private space. Vsets the register M1M2R to 1p2 jeii1 jeji2 ji; jiR+ jeji1 je

wherei; j is chosen randomly, according to the probability distribution p(i)p(j)

(D.2)P1 andP2 performs their operations.

(D.3)V measuresM1 M2 to check whether the state is not changed by the provers. If the state is notchanged, he accepts the input.

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4 Some properties and some more denitions

First, we can assume that U1 = U2(=: U, hereafter), andji is symmetric, without loss of generality. Moreprecicely, for any given provers, we can nd symmetric provers which corresponds to the same acceptanceprobability. This is due to the symmetry in our protocol. Hence, we often drop the subsript which identies

the prover. For example,jeii1 are often denoted byjeii, and so on. Also, we assume thatP1 andP2 , arehomogeneous. Dene a hilbert spacePwhich is homogenious toP1 andP2, for notational convention.Second, in doing the each test, the provers are not told which test is undrgoing. Therefore, the provers are

forced to perform U irrespective to which tests are undergoing. Note that this does not prohibit the proversto modify their strategies depending on the messages. Such strategy will be descriibed by a control unitaryoperation with (a part of )M1 andM2 being the control bits.

Observe that our protocol is designed so that Tests (A)-(E) are indistinguishable by looking atP1 (, orP2)only, to disable provers strategies to discriminate each test.

5 Some useful facts

Lemma 1 Given vectorsjui andjvi withkjuik = andkjvik =;

kjui huj jvi hvjk1 12 kjui jvik ( + ) :

Proof. Letcos = hu jvikjuikkjvik , and we have

kjui huj jvi hvjk1=

2 cos2 2 2 sin cos 2 sin cos 2 sin2

1

=2

"

cos2 22

sin cos

sin cos sin2

#1

= 2

2

s(1

2

2)2 + 4

2

2sin2 =

1

2

q(2 2)2 + 422 sin2 = 1

2

q(2 + 2)2 422 cos2

= 1

2rkjuik2 +

kjv

ik2

2

4

jhu

jv

ij2 =

1

2rkjuik2 +

kjv

ik2

2

jhu

jv

ijkjuik2 +

kjv

ik2 + 2

jhu

jv

ij 1

2

rkjuik2 + kjvik2 2

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.To dene Ti, we consider Hilbert spacesP0' P01' P02 with dimP

0

dimP = K, and regardsP,P1, andP2 is asubspace ofP0,P01, andP02, respectively. Also, U is extended toP0 so that it acts as identity operator in thesubspace orthogonal to P. Obviously,

P 1M UP 1M = P 1MU =UP 1M = U;P1 P2 ji = ji :

OurTi is a unitary operator inP0 withPTiP= Uii. An example of such a unitary matrix is

Ti=

266666664

Uii Ui1 Ui2 UiKU1i U11 U12

. . . U1K

U2i U21. . .

. . . U2K...

. . . . . .

. . . ...

UK i UK1 UKK

377777775;

where the rst column acts on

Pandn takes values from 1 to Kexcepti.

Tests (C)-(D) allows us to replace Ti by some other operator, sayeTi, which has some good property. Thisreplacement can change the acceptance probability of Test (A) at most

d

0@Xi;j

pp(i; j) eTi eTj ji jeii1 jeii1 jeji2 jeji2 ;X

i;j

pp(i; j)U Uji jeii1 jeii1 jeji2 jeji2

1A d

0@Xi;j

pp(i; j) eTi eTj ji jeii1 jeii1 jeji2 jeji2 ;X

i;j

pp(i; j)Ti Tjji jeii1 jeii1 jeji2 jeji2

1A+d

0

@Xi;j

pp(i; j)U Uji jeii1 jeii1 jeji2 jeji2 ;

Xi;j

pp(i; j)Ti Tjji jeii1 jeii1 jeji2 jeji2

1

A:

The second term of the RHS is upperbounded by (B), and the rst term is upperbounded as follows. First,

X

i;j

pp(i; j) eTi eTj ji jeii1 jeii1 jeji2 jeji2 X

i;j

pp(i; j)Ti Tjji jeii1 jeii1 jeji2 jeji2

=

Xi;j

p(i; j) eTi eTj ji Ti Tjji

Xi;j

p(i; j)eTi 1 ji Ti 1ji + eTj 1 ji Tj 1ji

= 2Xi p(i) eTi 1 ji Ti 1ji= 2Ei

eTi 1 ji Ti 1ji :Recall thatd(; ) k k :Therefore, we have

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d

0

@Xi;j

pp(i; j)

eTi

eTj ji jeii1 jeii1 jeji2 jeji2 ;

Xi;j

pp(i; j)Ti Tjji jeii1 jeii1 jeji2 jeji2

1

A Xi;j pp(i; j) eTi eTj ji jeii1 jeii1 jeji2 jeji2 Xi;j pp(i; j)Ti Tjji jeii1 jeii1 jeji2 jeji2=

sXi;j

p(i; j) eTi eTj ji Ti Tjji2

sX

i;j

p(i; j) eTi eTj ji Ti Tjji

=

sXi

p(i) eTi 1 ji Ti 1ji +X

j

p(j)1 eTj ji 1 Tjji

r2Ei eTi 1 ji Ti 1ji (1)Therefore,

Ei

eTi 1 ji Ti 1jiis upperbounded by Test (C) and the follwoing tests.

7 Analysis of the test (B)

Theorem 3 Test (B) passes with probability not more than

1

1

2d0@

K

Xi;j=1pp(i; j)OU OUji jeii1 jeii1 jeji2 jeji2 ;K

Xi;j=1pp(i; j)U Uji jeii1 jeii1 jeji2 jeji21A2

Proof. Let

d:= d

0@ KXi;j=1

pp(i; j)OU OUji jeii1 jeii1 jeji2 jeji2 ;

KXi;j=1

pp(i; j)U Uji jeii1 jeii1 jeji2 jeji2

1A ;and we have

1 d2

=

*

K

Xi;j=1pp(i; j)OU OUji jeii1 jeii1 jeji2 jeji2 ;K

Xi;j=1pp(i; j)U Uji jeii1 jeii1 jeji2 jeji2+2

=

KX

i;j=1

p(i; j)

Ti Tj ji jeii1 jeji2 ; U Uji jeii1 jeji22

=

KX

i;j=1

p(i; j) hj TiUii Tj Ujj ji2

=

KXi;j=1

p(i; j)Uii Ujj ji jeii1 jeji22

2

;

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where the last equality is due to

hTi Tj ji Uii Ujj ji ; Uii Ujj jii = 0:Therefore, the acceptance probability equals

p1 d2 1 12 d2:8 Analysis of the test (C)

Lemma 4 Let p0(i; j) be an arbitrary probability distribution, where i = 1; ; N, j = 1; ; M . Given asystem of vectors

ij with ij2 = 1 , there is a sequence of vectors j such that -balls B(; j) (j = 1; ; M) centered at

j satisesMX

j=1

NXi=1

p0(i; j)i;j

2

1 2Prj ; Pr i ; i;j

=2 B(;

j

) :

Proof. Observe that

Ep0

ijj

i;j Ep0i0jj i0j2 = Ep0ijj NXi=1

ij2 Ep0ijj ij2 = 1 Ep0ijj ij2 :On the other hand,

Ep0

j Ep0

ijji;j Ei0jj i0j2 2Ep0j Prn i ; i;j =2 B(; Ep0i0jj i0j)o

2Prn

j; Prn

i ;i;j 2 B(; Ep0i0jj i0j)o o

Therefore, letting

j

= Ep

0

i0jj

i0j

, we have the lemma.

Theorem 5 The test (C) passes at most1 2 min

jiPr fi ; Ti Ti ji =2 B(; ji)g :

Proof. The probability that the test (C) passes is given byKX

i;j=1

p(i)p(j) 1 heij 2 heijR hij U Uji jeji1 jeji2 jjiR

2

=

KX

i=1

p(i)Uii Uii ji2

SinceKX

i=1

p(i)Uii Uii ji =KX

i=1

p(i)P1TiP1 P2TiP2 ji

= P1 P2 KX

i=1

p(i)Ti Ti ji ;

KX

i=1

p(i)Ti Ti ji2

KXi=1

p(i)Uii Uii ji2

:

Therefore, the test (C) passes at most with the probabilityKX

i=1

p(i)Ti Ti ji2

1 2 minji

Pr fi ; Ti Ti ji =2 B(; ji)g :

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9 Analysis of the test (D)

Theorem 6 The test (D) passes with the probability at most

1 4

Pr i; Prj ; k(Ti Tj Tj Ti) jik p

:Proof. Ifi; j with i 6=j are chosen in the step (D1), the test passes with the probability12(Uii Ujj + Ujj Uii) ji

2 12(Ti Tj+ Tj Ti) ji2

= 1

2

kTi Tj jik2 + kTj Ti jik2

12(Ti Tj Tj Ti) ji

2= 1 1

4k(Ti Tj Tj Ti) jik2 :

Therefore, the average probability of the acceptance is upperbounded by

1 1

4

K

Xi;j=1p(i)p(j) k(Ti Tj Tj Ti) jik

2

:

(In the derivation, the probability of acceptance in case of i = j is upperbounded by unity.) This quantity isupperbounded by

1 4

Pr

i ; Pr

j ; k(Ti Tj Tj Ti) jik p

:

10 What is tested by test (B)-(D) (outline)

In this section, we roughly describe what the all the tests are about. Before going into detailed analysis, roughargument is described.

First, test (B) assures us that the Provers unitary is of the form ctrl-Ti. the test (C) assures us that

Ti Ti ji jiholds for the majority ofi0s for some stateji :Pick up Ti with Ti Ti ji ji, and dene

T0i :=TiT1i

:

Then, we have that T0i T0iji ji : The test (D) assures us Ti Tj ji Tj Ti ji, which, with j = i,leads to

T0i Iji 1 T0iji :It is also easy to derive

T0i T0j ji T0j T0iji :

Combination of these two relations yields

T0i T0j 1 ji T0j T0i 1 ji :

Observe that

ji = T0i T0iji = (T0i 1) (1 T0i ) ji (T0i 1) (T0i 1) ji= T0

2

i Iji :As a result, T0j are approximately commute with T

0i , and the eigenvalues of T

0j , T

0i are either 1 or -1,

approximately.

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11 What is tested by the test (B)- (D) (detail)

Suppose Test (C) and (D) passes with the probability at least1 1f(n)2 , and1 1f(n) 52 . Then, letting = 1f(n)

in Theorem5, we have

minji Pr(i ; Ti Ti ji =2 B 1pf(n) ; ji!) 1f(n) ;and letting = 4f(n) and=

1pf(n)

in Theorem 6, we have

Pr

(i ; PrJi 1 1p

f(n)

) 1

f(n); (2)

where

Ji :=

(j ; k(Ti Tj Tj Ti) jik < 2p

f(n)

):

Choose i so that Ti Ti ji 2 U(1=f(n); ji) and Pr Ji >1 2pf(n) , and let

Ii :=(

j ; Tj Tj ji 2 B 1pf(n)

; ji!) \ Ji:The probability measure of this set satises

Pr Ii 1 1

f(n) 2p

f(n) 1 3p

f(n): (3)

Denejiand T0i byji =Ti Ti ji ; T0i :=TiT1i ;

respectively. Ifi 2 Ii, the relationskT0i T0iji jik = kTi Tiji Ti Ti jik

kTi

Ti

j

i jik

+kj

i

Ti

Tij

ik 1 1p

f(n)(9)

holds for all a. Suppose that (q0; a0) with q0

2Q is not a memeber ofJ(q;a) with q

2Q. Then, obviously (9)

cannot be true. Therefore, for all q02 Q and for all a0,J(q0;a0) f(q; a); q2 Qg:

Therefore, relations (4)-(7) hold for all the elements of

f(q; a); q2 Qg:Suppose the provers performs ctrl-Ti. ( This is not the case usually. However, the dierence of acceptanceprobability due to this imperfection had already evaluated in the previous section. ) Then, in the sep (A.3),the verier obtain the classical bits a, a0 with the probability

1

K2a Xb;b0(1)aba0b0Tqb Tq0b0 ji

2

1

K2a Xb;b0(1)aba0b0T00qb T00q0b0 ji

2

;

and the dierence between the both ends can be evaluated using (5). Let us dene

Wqa := 1

Ka

Xb

(1)abT00qb:

Then, a system of CP mapsfqaga with qa() := Wqa Wqya denes an instrument acting onP, and the mapq() :=

Xa

Wqa Wqya =

Xa

qa()

is a CPTP map. Then, for any q2Q, we havekq 1 (ji hj) 1 q (ji hj)k1

Xa kqa 1 (ji hj) 1 qa(ji hj)k1

Xa

k(Wqa 1 1 Wqa ) jik

1Ka

Xa;b

T00qb 1 1 T00qb ji=

Xb

T00qb 1 1 T00qb ji 8Kap

f(n):

With the help of the inequality (7), for any q2Q and q02Q, we have

qq0a0 1 (ji hj) q0a0q 1 (ji hj)1

X

a

qaq0a0 1 (ji hj) q0a0qa 1 (ji hj)1

1Ka

Xa;b

T00qbT00q0b0 T00q0b0T00qb 1 ji=

Xb

T00qbT00q0b0 T00q0b0T00qb 1 ji 24K

2ap

f(n):

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Therefore,

q0

a0

1Y=N

q() 1 (ji hj) q

(N)

q0

a0

1Y=N1

q() 1 (ji hj)

1

q0a02Y

=N

q() q(1) (ji hj) q(N)q0

a0

2Y=N1

q() q(1) (ji hj) +2 8Kapf(n)

q0a0q(N) N1Y=1

q()

(ji hj) q(N)

q0

a0N1Y=1

q()

(ji hj) +(N 1) 2 8Kapf(n)

24K2ap

f(n)+

(N 1) 2 8Kapf(n)

;

which leads to

q0

a0

1Y=N

q() 1 (ji hj)

1Y=N

q()

q0

a0 1 (ji hj)1

24K2apf(n)

+(N 1) 2 8Kapf(n)

! N 24Np

f(n)

K2a + N Ka

:

Let us denote the probability distribution of the answers from the provers by

(q(1;1); ; q(1;N); q(2;1); ; q(2;M))(a(1;1); ; a(1;N); a(2;1); ; a(2;M))

: = tr

0@ 1Y=N

q(1;)

a(1;)

1Y=M

q(2;)

a(2;)

1A (ji hj) ;and its marginal distribution is denoted by (q(1;1);

; q(1;N); q(2;1);

; q(2;M))q(1;);q(2;)(a(1;); a(2;)). Animportant observation is that

(q(1;1); ; q(1;N); q(2;1); ; q(2;M))

q(1;1);q(2;2)=(q(1;1); q(2;1))

We number all the possible questions of the veriers so that q(1); ; q(jQj) 2 Q, and q(jQj+1); ; q(Kq) =2Q. Suppose that the provers share not an entangled state but the randomness which obeys joint probabilitydistribution(q(1); q(Kq); q(1); q(Kq)). Then, obviously, they cannot cheat Test (A). Below, we show thatthe real distribution of the answers to the questions q(0) andq(0) is close to the one obtaind as the marginalof this imaginary shared randomness, provided that q(0); q(0) 2 Q:

(q(1); ; q(0); q(Kq); q(1); ; q(0); ; q(Kq))

q(0);q(0) (q(0); q(0))

1

= (q(1); ; q(0); q(1); ; q(0)q(0);q(0) (q(0); q(0))1=

Xa;a0

tr

q(0)

a

1Y=01

q()

!0@q(0)a0 1Y

=01q

()

1A (ji hj) trq(0)a q(0)a0 (ji hj)

=Xa;a0

tr

q(0)

a

Q1=01

q()

q(0)

a0Q1

=01q()

(ji hj)

tr

q(0)

a

Q1=01

q()

q(0)

a0 (ji hj)

+Xa;a0

tr

q(0)

a

1Y=01

q()

! q

(0)

a0 (ji hj) trq(0)

a q(0)

a0 (ji hj) :

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The rst term is evaluated as follows.

1

2

Xa;a0

tr

q(0)

a

Q1=01

q()

q(0)

a0Q1

=01q()

(ji hj)

tr

q(0)

a

Q1=01

q()

q

(0)

a0 (ji hj)

= 1

2

Xa;a0

trq(0)a Q1=01q() q(0)a0 Q1=01q() (ji hj)

tr

q(0)

a

Q1=01

q()

Q1

=01q()

q(0)

a0

(ji hj)

Xa;a0

q(0)

a

Q1=01

q()

q(0)

a0Q1

=01q()

(ji hj)

q(0)

a

Q1=01

q()

Q1

=01q()

q(0)

a0

(ji hj)

1

Xa;a0

1 0@q(0)a0 1Y

=01q

()

1A (ji hj) 1 0@ 1Y

=01q

()

q(0)

a0

1A (ji hj)1

:

where the inequality in the third line and the forth line are due to the monotonicity of the trace norm with

respect to tr, q()

, and q(0)

a (recall Lemma 2) .The second term is evaluated almost prallelly, and(q(1); ; q(0); q(Kq); q(1); ; q(0); ; q(Kq))q(0);q(0) (q(0); q(0))1

Xa;a0

1 0@q(0)a0 1Y

=01q

()

1A (ji hj) 1 0@ 1Y

=01q

()

q(0)

a0

1A (ji hj)1

+Xa;a0

q(0)

a

1Y=01

q()

! 1 (ji hj)

1Y

=01q

()

q(0)

a

! 1 (ji hj)

1

If q(1); ; q(0) 2 Q ( q(1); ; q(0) 2 Q), the rst term (the second term, respectively) of the RHS isupperbounded by

2Xa;a0

240pf(n)

K2a + 0Ka

2 Ka 24Kqpf(n)

K2a + KqKa

100K5pf(n)

:

Therefore, in average, the dierence is

Xq();q()

pq(q(); q())

(q(1); ; q(); q(Kq); q(1); ; q(); ; q(Kq))

q();q()

(q(); q())

1

The sumP

q();q()is decomposed into four parts,P

q()2Q;q()2Q,P

q()2Q;q() =2Q,P

q() =2Q;q()2Q,P

q() =2Q;q() =2Q,and

2 100K5pf(n) + KaKqf(n)1=3 2 + 100K

5pf(n)!+

100K5p

f(n)+

KaKqf(n)1=3

2!

+

KaKqf(n)1=3

2 + KaKqf(n)1=3

2

= 400K5p

f(n)+ 8 KaKq

f(n)1=3 K

5

f(n)1=3:

Let us denote by ~(q(1); ; q(0); q(Kq); q(1); ; q(0); ; q(Kq))the probability distribution of the answer

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when the provers sharesjiand measures their particles byn

1Ka

Pb(1)ab Tqb

o. Due to (5) and (8), we have

Xq();q()

pq(q(); q())

~(q(); q()) (q(); q())1

12

Xq();q();a;a0

pq(q(); q())

1K2a Pb;b0(1)aba0b T00q()b T00q()b0 ji2 1K2a Pb;b0(1)aba0b Tq()b Tq()b0 ji2

Xq();q();a;a0

pq(q(); q())

1K2a Pb;b0(1)aba0b T00q()b T00q()b0 ji 1K2a Pb;b0(1)aba0b Tq()b Tq()b0 ji

Xq();q();a;a0

pq(q(); q())

1K2aP

b;b0(1)aba0b

T00q()b

T00q()b0

ji 1K2a

Pb;b0(1)aba

0bTq()b Tq()b0 ji

1

K2a

Xq();q();a;a0;b;b0

pq(q(); q())

T00q()b T00q()b0 ji Tq()b Tq()b0 ji= X

q();q();b;b0

pq(q(); q())T00q()b T00q()b0 ji Tq()b Tq()b0 ji

Xq();;b;b0

pq(q())

T00q()b 1 ji Tq()b 1 ji+

Xq();b;b0

pq(q())

1 T00q()b0 ji 1 Tq()b0 ji 2 K2a

2

p2p

f(n)+

KaKqf(n)1=3

2!

5K3aKq

f(n)1=3:

Therefore,

Xq();q()

pq(q(); q())

(q(1); ; q(); q(Kq); q(1); ; q(); ; q(Kq))q();q()~(q(); q()) 1

K5

f(n)1=3+

5K3aKqf(n)1=3

2K5

f(n)1=3:

Note ~(q(); q()) is not yet the true probability distirbution of q(), q(), and the dierence from the trueprobability is, due to Thorem 3, not larger than

2 (1 (prob. ofacceptaceof (B))) = 2f(n)

52

K5

f(n)1=3:

The following theorem is the direct consequence of the above lemma.

Theorem 8 Suppose there is a classical two provers one round interactive proof system which accepts a languageL with the probability1 andps for completeness side and soundness side, respectively. Based on such a classicalprotocol, we can construct QMIP(2,1)-protocol whose acceptance probability for completeness side is 1. Its acceptance probabilitypquantums for the soundeness side is upperbounded as follows:

pquantums 1 1

4min

1

ps+

3K5

f(n)1=3

;

1

f(n)52

;

wheref(n) is an arbitrary function of the input sizen.

13

8/9/2019 2-prover

14/14

The best known classical protocols can accept NP languages using log-bit communications, meaning thatKis plynomial. Therefore, if we set f(n)to be a polynomial function, pquantums is1 1polynomial . On the otherhand, the best known classical protocols for NEXP languages uses polynomially many message bits, meaningthatKis exponentially large. Therefore, by setting f(n)an exponential function, we can prove that pquantums 1 1exp(cn) . Therefore,

Corollary 9 NP QMIP(2; 1; 1; 1 1polynomial ), NEXP QMIP(2; 1; 1; 1 1exp(cn) ).

With trivial modication of Kitaev-Watrous argument, we can prove QIP(1; 1 1exp(cn)) EXP. This showsgap (??) between QMIP and QIP.

QIP(1; 1 1exp(cn)

) EXP NEXP QMIP(2; 1; 1; 1 1exp(cn)

):

(this is not so impressive ...)