We are now in a position to construct an interactive proof for any language inBQP. To this end, letLbe a language inBQP. Then fromTheorem 2.1and the definition ofBQPfor any inputxthere exists an adaptive measurement6on a polynomially sized triangular graph state such that
• Ifx∈Lthen the measurement outputs “ACCEPT” with probabilityccalc≥2/3. • Ifx∈/Lthen the measurement outputs “ACCEPT” with probabilityscalc≤1/3.
The adaptive measurement supplies the measurements required for each vertex via anglesθv. It also supplies the functions required for the adaptation. The interactive proof is given by the following procedure.
Procedure 3.
1. Randomly choose “CALCULATE” with probabilityqor “TEST” with probability 1−q. 2. If “CALCULATE”:
(a) Query provers according to the measurement-based computation. (b) Accept if the computation accepts.
3. if “TEST”:
(a) Perform the test for honesty given inProcedure 2. (b) Accept if the test accepts.
Now we calculate the optimal value ofq.
Lemma 4.1. Let L be a language and x in input. Suppose we are given an adaptive measurement on a
triangular cluster state on n vertices which implements a measurement-based computation which decides whether x∈L with error at most1/3(i. e., ccalc≥2/3and scalc≤1/3). Let0<δ <1/6be given and set
q= ctest−stest
1+ctest−scalc−stest−δ . (4.1)
Then
• if x∈L then for honest proversProcedure 3accepts with probability at least cip, • if x∈/L then for any set of proversProcedure 3accepts with probability at most sip, where
cip−sip≥
δ8
1018.8n11. (4.2)
Proof. Letctestbe the probability of honest provers passing the test, and letstestbe the probability of
dishonest provers passing the test, given inCorollary 3.9. Here, by dishonest we mean that the probability of some outcome of an adaptive measurement made using the provers differs from the honest case by more than the givenδ.
Then we have two cases:
• The input is in the language: then we only care about the honest case, in which the probability of accepting iscip≥qccalc+ (1−q)ctest.
• The input is not in the language: then there are two subcases:
– The provers pass the test with probability at leaststest. Then byCorollary 3.9the probability
of accepting on the calculation is at mostscalc+δ and the probability of accepting on the test
is at mostctestfor an overall probability of at mostq(scalc+δ) + (1−q)ctest.
– The provers pass the test with probability less thanstest. Then the probability of accepting on
the calculation could be as high as 1, since we gain no information from the test. The overall probability of accepting is then less thanq+ (1−q)stest.
The two different cases in thex∈/Lcase give two different gaps which are, in the first case
qccalc+ (1−q)ctest−q(scalc+δ)−(1−q)ctest=q(ccalc−scalc−δ) (4.3)
and in the second case
qccalc+ (1−q)ctest−q−(1−q)stest=q(ccalc−1) + (1−q)(ctest−stest). (4.4)
The overall gap is the minimum of these two. We wish to findqwhich maximizes the minimum. The two equations are just lines inqwhich cross each other. At the point where they are equal we find the maximum overall gap. The crossing point is easily found to be at
q= ctest−stest
1+ctest−stest−scalc−δ (4.5)
which gives a gap of
cip−sip = (ccalc−scalc−δ)(ctest−stest)
1+ctest−stest−scalc−δ (4.6)
≥ 1
12
δ8
1017.7n11. (4.7)
On the first line, we see that the denominator can be no larger than 2, and the first factor in the numerator is at least 1/6 (whenδ =1/6 andccalc−scalc=1/3). So we can lower bound the gap by
1
12(ctest−stest), which is estimated inCorollary 3.9.
We are now in a position to prove our main claim. This is obtained by applying a standard gap amplification procedure.
Procedure 4.
1. PerformProcedure 3Ntimes and let the number of times the procedure accepts beM. 2. IfM>Ncip+sip
2 then accept. 3. Otherwise reject.
Note that there is ambiguity inProcedure 4. In particular, it does not mention whether the repetition
ofProcedure 3should be done serially or in parallel. In fact this does not matter, as we shall see in the
proof below. We can addNsets ofnprovers and query each prover once, or use one set ofnprovers and query each oneNtimes.
Proof ofTheorem 1.1. LetL∈BQPbe given along with inputx. Theorem 2.1tells us that we can find an adaptive measurement on a triangular cluster state onn=poly(|x|)vertices whose outcome tells us whetherx∈Lwith error less than 1/3. FromLemma 4.1we have an interactive proof such that ifx∈L
then we accept with probability at leastcipfor honest provers, and ifx∈/Lwe accept with probability at mostsip.
Now we amplify usingProcedure 4. Let us first consider the casex∈L. Then we are interested in the case of honest provers, in which case we haveN independent and identical Bernoulli trials with some probability of acceptingp≥cip. Using Hoeffding’s inequality, the probability that we mistakenly reject
is bounded by P M≤Ncip+sip 2 ≤ exp −2 N p−Ncip+2sip 2 N (4.8) ≤ exp −N(cip+sip) 2 2 . (4.9)
Setting this equal to 1/3 we solve forNto find the minimum number of trials to achieve our desired error rate.
N≥ 2 ln 3
(cip+sip)2
. (4.10)
Substituting in forcip−sipas estimated inLemma 4.1we obtain N≥10
37.9n22
δ16 . (4.11)
The same number of repetitions also suffices to bound the probability of accepting whenx∈/Lto below 1/3. The analysis is similar, however if the provers are not honest they may vary their behavior on each trial, so the trials are not necessarily identically distributed. However, since the probabilitypof accepting satisfiesp≤sipfor every trial we can still use Hoeffding’s inequality.
To make this argument formal, letAj be the random variable corresponding to the verifier accepting on the jth trial with 1 indicating acceptance and 0 indicating rejection. Further, letebe some event which may depend on trials 1, . . . ,j−1. Now consider the probabilityP(Aj=1|e).
First note that all operations that contribute toAj commute with operations on all other trials (parallel repetition with fresh provers), or happen before the jth trial (sequential repetition with a single set of provers). Also, the verifier’s actions are independent for each trial. Therefore we may assume without loss of generality that the jth trial happens after everything else has completed. Because everything else in independent from the verifier’s actions on the jth trial, the provers can simulate their actions for all other trials by choosing randomness themselves. Hence one option for the provers’ strategy on a single trial is to perform theNtrial protocol for trials 1, . . . ,j−1, repeating until the eventeis observed, and then proceeding. This is effectively a state preparation for the jth trial. The probability of acceptance is then bounded by the single trial upper bound,P(Aj=1|e)≤sip.
LetMj =∑kj=1Ak. We will apply induction on jand show that P(Mj ≥T)is upper bounded by
the case of a binomial distribution on jtrials with probabilitysip. The case for j=1 is trivial. For the inductive step let 0≤T be given.
P(Mj≥T) =P(Aj=1|Mj−1≥T−1)P(Mj−1≥T−1) +P(Aj=0|Mj−1≥T)P(Mj−1≥T). (4.12)
Applying the upper bound onP(Aj|e)we find
P(Mj≥T)≤sipP(Mj−1≥T−1) + (1−sip)P(Mj−1≥T) (4.13)
and with the induction hypothesis we see that this is further upper bounded by assuming the binomial distribution with jtrials and probability of a 1 given bysip. We may thus apply an argument similar
that forx∈L, finding that the same number of trials suffices to bound the probability of error to within 1/3.