BQP and the Polynomial Hierarchy 1 Introduction

BQP and The Polynomial Hierarchy based on ‘BQP and The Polynomial Hierarchy’ by Scott Aaronson

Deepak Sirone J., 17111013 Hemant Kumar, 17111018 Dept. of Computer Science and Engineering Dept. of Computer Science and Engineering Indian Institute of Technology Kanpur Indian Institute of Technology Kanpur

Abstract

The problem of comparing two complexity classes boils down to either finding a problem which can be solved using the resources of one class but cannot be solved in the other thereby showing they are different or showing that the resources needed by one class can be simulated within the resources of the other class and hence implying a containment. When the relation between the resources provided by two classes such as the classes BQP and PH is not well known, researchers try to separate the classes in the oracle query model as the first step. This paper tries to break the ice about the relationship between BQP and PH, which has been open for a long time by presenting evidence that quantum computers can solve problems outside of the polynomial hierarchy. The first result shows that there exists a relation problem which is solvable in BQP, but not in PH, with respect to an oracle. Thus gives evidence of separation between BQP and PH. The second result shows an oracle relation problem separating BQP from BPPpath and SZK.

1 Introduction

The problem of comparing the complexity classes BQP and the polynomial heirarchy has been identiﬁed as one of the grand challenges of the ﬁeld. The paper “BQP and the Polynomial Heirarchy” by Scott Aaronson proves an oracle separation result for BQP and the polynomial heirarchy in the form of two main results:

A 1. There exists an oracle A relative to which FBQPA 6⊂ FBPPPH , where FBQP and FBPP are the relation versions of BQP and BPP respectively. 2. There exists an oracle A relative to which BQPA 6⊂ BPPpathA and BQPA 6⊂ SZKA.

The problem of comparing BQP and the Polynomial Hierarchy is interesting for the following reasons:

1. The question of what classical resources are needed to simulate quantum computers is still not known. Questions like whether quantum amplitudes can be simulated using approximate counting (which would imply that BQP ⊂ BPPNP are still open). 2. BQP 6⊂ PH would imply that quantum computers can solve problems other than the NP interme- diate problems like graph isomorphism. It might be fruitful to search for quantum algorithms not even in PH. 3. The question of whether BQP could provide exponential speedups even if P = NP is another reason why the question of whether or not BQP 6⊂ PH is interesting.

We will give a brief overview of the proofs of the two theorems in proceeding sections.

1 Complexity Classes

A brief overview of the various complexity classes presented in this report are as follows: NP A language L ∈ NP if there exists a polynomial time Turing machine M such that

x ∈ L iﬀ ∃yM(x, y) = 1

P PH For i ≥ 1, a language L is in Σi if there exists a polynomial time Turing machine M and a polynomial q such that

q(|x|) q(|x|) q(|x|) x ∈ L iﬀ ∃u1 ∈ {0, 1} ∀u2 ∈ {0, 1} ...Qiui ∈ {0, 1} M(x, u1, . . . ui) = 1

where Qi denotes ∀ or ∃ depending on whether i is even or odd, respectively. The polynomial S P hierarchy is the set PH= Σi i FC The class of functions computable in class C whose output is a bit string rather than a Boolean value.

BPP The language L is said to be in class BPP, if there exists a polynomial time probabilistic Turing machine P such that P r[P (x) = L(x)] ≥ 2/3, r where probability is over random strings r chosen by P . [5] BQP A language L is in BQP if and only if there exists a polynomial-time uniform family of quantum circuits {Qn : n ∈ N} such that

• For all n ∈ N, Qn takes n qubits as input and outputs 1 bit. 2 • For all x ∈ L, P r[Q|x|(x) = 1] ≥ 3 2 • For all x 6∈ L, P r[Q|x|(x) = 0] ≥ 3 FBQP FBQP is the class of relations R ⊂ {0, 1}∗ × {0, 1}∗ for which there exists a quantum polynomial time algorithm A, such that given an x ∈ {0, 1}n produces an output y such that

P r[(x, y) ∈ R] = 1 − o(1)

where the probability is over A’s internal randomness. FBPP FBPP is deﬁned the same way as above except that now A is a classical probabilistic polynomial time algorithm. CA The set of languages solvable by a class C machine when given an oracle access to language A. This means that the C can ask the oracle A membership queries without incurring any additional cost in time. AC0 The class of languages computable by circuit families of constant depth and polynomial size, whose gates have unbounded fanin. [5] BPPpath The class of languages L for which there exists a pair of polynomial time Turing machines A and B such that • if x ∈ L, then P r[A(x, r)|B(x, r)] ≥ 2 r 3 • if x 6∈ L, then P r[A(x, r)|B(x, r)] < 1 r 3 • P r[B(x, r)] > 0. [4] r

2 IP A language L is in IP, if there exists a polynomial time probabilistic Turing machine M such that on c input x and random sting r ∈ {0, 1}|x| , x ∈ L =⇒ ∃PP r[M(x, r, P ) = 1] ≥ 2/3, r x 6∈ L =⇒ ∀PP r[M(x, r, P ) = 1] ≤ 1/3, r where P is proof provided by all powerful prover having infinite computation power. [5] SZK Zero-knowledge protocol is a method by which the prover can prove to the verifier that a given statement is true, without conveying any information apart from the fact that the statement is indeed true. In the IP protocol, along with prover and verifier, their exists an expected probabilistic simulator S such that it can predict the conversation between the verifier and the prover. We say the protocol is statistical zero knowledge if the distribution of simulator and the distribution of verifier-prover conversation is statistically close. [3]

2 Preliminaries

The paper discusses two problems on boolean functions namely Fourier Fishing and Fourier Check- ing. The boolean functions considered in the following sections will have the form f : {0, 1}n → {−1, 1}. n ˆ The Fourier transform of f in Z2 , denoted as f is 1 X fˆ = √ (−1)x.zf(x) N x∈{0,1}n where N = 2n.

2.1 Fourier Fishing n Given n Boolean functions f1, . . . , fn : {0, 1} → {−1, 1} as input. We have to output n strings n ˆ z1, . . . , zn ∈ {0, 1} , such that at least 75% of which satisfy |fi(zi)| ≥ 1 and at least 25% of which ˆ satisfy |fi(zi)| ≥ 2.

Call an n-tuple hf1, . . . , fni of Boolean functions good if n X X ˆ 2 fi(zi) ≥ 0.8Nn, i=1 zi:|fˆi(zi)|≥1 n X X ˆ 2 fi(zi) ≥ 0.26Nn i=1 zi:|fˆi(zi)|≥2 The promise version of Fourier Fishing assumes that we are given n Boolean functions as input which n are promised to be good and we have to output n strings z1, . . . , zn ∈ {0, 1} , such that at least 75% of ˆ ˆ which satisfy |fi(zi)| ≥ 1 and at least 25% of which satisfy |fi(zi)| ≥ 2.

2.2 Fourier Checking Given two Boolean function f, g : {0, 1}n → {−1, 1} as input, it is promised that either 1. hf, gi was drawn from a uniform distribution U, which sets each f(x) and g(y) by a fair, independent coin toss. 2. hf, gi was drawn from a forrelated distribution F, deﬁned as the distribution generated by the following process:

N A random real vector is chosen v = (vx)x∈{0,1}n ∈ R , by drawing each entry from a Gaussian distribution N (0, 1). Then f(x) and g(x) are set as follows: v vˆ f(x) = x , g(x) = x |vx| |vˆx| n wherev ˆ is Fourier transform of v in Z2 . As g is highly correlated to f’s Fourier transform, it is called a forrelated distribution.

3 The problem is to accept if hf, gi was drawn from F and reject if drawn from U. The promise version of this problem says that the quantity

 2 1 X p(f, g) := f(x)(−1)x.yg(y) N 3   x,y∈{0,1}n

is at least 0.05 or at most 0.01. Accept hf, gi in the ﬁrst case and reject in the second case.

3 Oracle Separation Between FBQP and FBPP

The oracle separation between FBQP and FBPP results from a series of reductions using AC0 circuit n lower bounds. First, we get a circuit lower bound for ε-Bias Detection by reducing the 2 -Hamming Weight problem to it and secondly, we reduce ε-Bias Detection to Fourier Fishing and thereby derive a circuit lower bound for Fourier Fishing.

n 2 -Hamming Weight → ε-Bias Detection → Fourier Fishing exp Ω n1/(d−1) exp Ω 1/ε1/(d+2) exp Ω N 1/(2d+8)

3.1 Quantum Algorithm

The quantum algorithm for Fourier Fishing FF-ALG with oracle access to hf1, f2, . . . fni is as follows: 1. for i = 1 . . . n (a) Create an n-qubit state |0i 1 P (b) Apply Hadamard gate and create the equal superposition √ n |xi N x∈{0,1} 1 P (c) Query the f oracle and get the state √ n f (x) |xi i N x∈{0,1} i (d) Apply Hadamard gate again and measure in the computational basis

(e) Output the measurement as zi

As shown in the paper, the lemma 7 and lemma 8 claims that the FF-ALG will output the required zi’s with probability 1 − 1/exp(n).

Lemma 7 Assuming hf1, f2, . . . fni is good, FF-ALG succeeds with probability 1 − 1/exp(n).

Lemma 8 hf1, f2, . . . fni is good with probability 1−1/exp(n), if the fi’s are chosen uniformly at random. Skipping the proof discussed in the paper, the end result of lemma 8 is as follows:

n X X ˆ 2 fi(zi) ≥ 0.8Nn i=1 ˆ zi:|fi(zi)|≥1

n X X ˆ 2 fi(zi) ≥ 0.26Nn i=1 ˆ zi:|fi(zi)|≥2 with probability 1 − 1/exp(n).

The FF-ALG makes one query to each fi oracle, and there are n of them, thus, clearly, FF-ALG is in FBQPA where A is the oracle outputting random boolean functions. The classical algorithm for this problem have to do exponentially many queries to each oracle fi so, it certainly not in FBPP. The next section on constant circuit lower bounds proves even a stronger result that Fourier Fishing is not even in FBPPPH.

4 3.2 Preparation of Circuit Lower Bound n In this part, we will present how the reduction of Fourier Fishing to ε-Bias Detection to 2 - Hamming Weight is done. The theorem 11 is directly stated from Hastad’s book. Theorem 11[2] Any depth-d circuit that accepts all n-bit strings of Hamming weight n/2 + 1, and rejects all strings of Hamming weight n/2, has size exp Ω n1/(d−1). Corollary 12 Let U[ε] be the distribution over {0, 1}m where each bit is 1 with independent probability 1/2 + ε. Then any depth-d circuit such that

P r [C(x)] − P r [C(x)] = Ω(1) x∼U[ε] x∼U[0]

has size exp Ω 1/ε1/(d+2). n The corollary 12 probabilistically reduces the 2 -Hamming Weight to the ε-Bias Detection. The reduction is based on the analogy that if the input string has hamming weight n/2 then the distribution induced by sampling bits from it will have the independent probability of 1 occurring equal to 1/2, otherwise the independent probability of 1 occurring will be 1/2 + ε. The sampling is done by taking m bits from the n-bit input string, m < n, and checking the bias of the sampled bits. The reduction repeats the procedure for O(n) times, collects the results and uses an AC0 circuit of depth 2 and size poly(O(n)) for the Approximate Majority problem to take a decision about the Hamming weight of the input string.

3.3 Secretly Biased Fourier Coeﬃcients

In this section we show that the ε-Bias Detection problem can be reduced to fourier ﬁshing using an AC0 circuit. For doing this consider two parties Alice and Bob. Fix a string s ∈ {0, 1}n. Deﬁne distributions A[s], B[s] and D[s] over functions f : {0, 1}n → {−1, 1} where each f(x) is 1 with independent probability pa, pb as follows:

• A[s]: p = 1 + (−1)s.x √1 a 2 2 N

• B[s]: p = 1 + (−1)s.x √1 b 2 2 N • D[s] : A coin is tossed and it’s outcome decides whether A[s] is used or B[s] is used. Alice now chooses a string s ∈ {0, 1}n uniformly at random. She then draws f according to D[s], keeps s a secret and sends the truth table of f to Bob. After examining the entire truth table Bob can output any string z ∈ {0, 1}n such that |fˆ(z)| ≥ β. Bob’s goal is to avoid outputting s. The following lemma says that regardless of Bob’s strategy, if β is sufficiently greater than 1, then Bob must output s with 1 probability significantly greater than N . Lemma 13 eβ + e−β P r[s = z] ≥ √ 2 eN As this probability is significantly greater than 1/N, Alice can use Bob to solve the ε-Bias Detection problem. Alice is an AC0 reduction from the ε-Bias Detection problem to fourier fishing. This will be detailed in the coming sections. Lemma 14 1 kD − Uk = O(√ ) N Lemma 14 can be used to conclude that if a Fourier Fishing algorithm succeeds with probability p on hf1 . . . fni, then it also succeeds with n p − kDn − U nk ≥ p − O(√ ) N n on hf1 . . . fni drawn from D

5 Theorem 15 Any depth-d circuit that solves the Fourier Fishing problem with probability 0.99 at 1/(2d+8) least over f1, . . . , fn chosen uniformly at random, has size exp Ω N . 0 Let C be the AC circuit for Fourier Checking having depth d and size S. Let G is set of all hf1, . . . , fni for which C succeeds. As asserted in theorem 15

P r[hf1, . . . , fni ∈ G] ≥ 0.99. U n From Lemma 14, we have, for large n,

n P r[hf1, . . . , fni ∈ G] ≥ 0.99 − O √ ≥ 0.98. Dn N √ Theorem 15 describes a reduction from ε-Bias Detection to Fourier Fishing, where ε := 1/2 N. Let C0 be a circuit for ε-Bias Detection having depth d0 and size S0. C0, ﬁrst, samples a M = N 2n bit M string R = r1 . . . rM ∈ {0, 1} where each ri is 1 with independent probability p. Here we have to decide whether p = 1/2 or p = 1/2 + ε, analogous to whether the distribution was U[0] or U[ε], respectively. The circuit C0 works as follows:

M 2 1. samples a string R = r1 . . . rM ∈ {0, 1} , M = N n where each ri is 1 with independent probability p 2. for each j = 1 ...N do

n (a) sample strings sj1, . . . , sjn ∈ {0, 1} and bits bj1, . . . , bjn ∈ {0, 1} uniformly at random

(b) sample kj ∈ [n] uniformly at random n (c) deﬁne a set of functions fj1, . . . , fjn : {0, 1} → {−1, 1} using (j − 1)Nn + 1 to jNn bits of 0 0 0 R, call it R = r1 . . . rNn, where each fji is deﬁned as

0 r +si.x+bi fji(x) = (−1) (i−1)N+x

(d) feed fj1, . . . , fjn as input to C, and call the output of C as zj1, . . . , zjn, and output of this iteration zj = zjk

3. check whether there exists a j ∈ [N] such tha zj = sjkj

If p = 1/2, then fj1, . . . , fjn are independent and uniformly random, and C does not get any information

about sj1, . . . , sjn, so P r[zj = sjkj ] = 1/N.

If p = 1/2 + ε then, as discussed in secretly biased Fourier coeﬃcients, each fij is sampled independently from biased distribution D[s]. If C succeeds then,

1 e2 + e−2 3 1 e1 + e−1 1.038 P r [zj = sjkj ] ≥ √ + − √ ≥ fj1,...,fjn,kj 4 2 eN 4 4 2 eN N n So, for a random fj1, . . . , fjn drawn from D

1.038 1.017 P r [zj = sjk ] ≥ 0.98 ≥ n j fj1,...,fjn,kj ,D N N

For each j = 1,...,N, we have 1 1 p = =⇒ P r[z = s ] = 2 j jkj N 1 1.017 p = + ε =⇒ P r[z = s ] ≥ 2 j jkj N

Let E be the event that there exists a j ∈ [N] such that zj = sjkj .

1 1 N 1 P r[C0(R)] : p = =⇒ P r[E] = 1 − 1 − ' 1 − ≤ 0.633 U[0] 2 N e 1 1.017N 1 P r[C0(R)] : p = + ε =⇒ P r[E] = 1 − 1 − ' 1 − ≥ 0.638 U[ε] 2 N e

6 P r[C0(R)] − P r[C0(R)] ≥ 0.638 − 0.633 = 0.005 = Ω(1) U[ε] U[0] The above result matches with the requirement of ε-Bias Detection. Thus C0 is a circuit for ε-Bias Detection. Step 3 of C0 can be done with depth-2 circuit of size O(Nn). This gives d0 = d + 2, and S0 = O(NS) + O(Nn). Performing the reverse calculation for S with the circuit lower bound given by corollary 12 gives S = exp Ω N 1/(2d+8). Thus, we have achieved the series of reduction presented in the start of the section.

3.4 Giving Separation between FBQP and FBPP This separation follows from the standard conversion between PH and AC0, for every n there exists 0 poly(n) poly log(N) probabilistic AC circuit CM,n of size 2 = 2 .

A Theorem 16 FBQPA 6⊂ FBPPPH with probability 1 for a random oracle A. n Here, oracle A is the encoding of n random Boolean functions fn1, . . . , fnn : {0, 1} → {−1, 1}, for each integer n. Let R be the relation problem, with input 0n such that we succeed iﬀ we output n z1, . . . , zn ∈ {0, 1} satisfying the requirements of the Fourier Fishing problem. By lemma 7 and lemma 8, there exists an FBQPA machine deciding R with success probability 1 − 1/exp(n).

A Let M be FBPPPH machine deciding R. Then, by standard conversion from PH to AC0, for each n, AC0 poly log(N) n circuit CM,n have size 2 which takes A as input and simulates M(0 ). Whereas, from theorem 15 we have circuit size as exp Ω N 1/(2d+8). Therefore, by theorem 15,

P r[CM,n succeeds ] < 0.99 A for suﬃciently large n. As all the fni’s are independent, this would be true for if we condition on all CM,i succeeds. P r[CM,1,CM,2,... all succeed] = 0. A So by the union bound, P r[∃M : CM,1,CM,2,... all succeed] = 0. A

A It follows that FBQPA 6⊂ FBPPPH with probability 1 over A.

4 Oracle Separation Between BQP, BPPpath and SZK

The oracle separation between BQP, BPPpath and SZK comes from the ability to distinguish between a uniform distribution to a highly correlated distribution. The Fourier Checking is one natural problem in which we are given oracle access to the functions f and g and we have to distinguish whether they are sampled from a uniform distribution or a forrelated1 distribution.

4.1 Quantum Algorithm The quantum algorithm for Fourier Checking,FC-ALG, is as follows:

1. Create an n-qubit state |0i

1 P 2. Apply Hadamard gate to all qubits and create equal superposition √ n |xi N x∈{0,1} 1 P 3. Query f and get the state √ n f(x) |xi N x∈{0,1} 1 P x.y 4. Apply Hadamard gate to all qubits to create the state N x,y∈{0,1}n f(x)(−1) |yi

1 P x.y 5. Query g and get the state N x,y∈{0,1}n f(x)(−1) g(y) |yi

1 P x.y y.z 6. Apply Hadamard gate to all qubits to create the state N 3/2 x,y,z∈{0,1}n f(x)(−1) g(y)(−1) |zi

1A distribution which is closely related to the Fourier transform of a ﬁxed string.

7 7. Measure in computational basis and accept only when measured n-qubit state |0i

FC-ALG makes only one query to f oracle and one query to g oracle. By calculating the probability of success of FC-ALG in the coming section will show that given√ oracle access to f and g Fourier Checking is indeed in BQP. The classical algorithm takes O( N) = O(2n/2) queries to solve the Fourier Checking. Theorem 9 If hf, gi is drawn from uniform distribution U, then 1 E[p(f, g)] = U N If hf, gi is drawn from forrelated distribution F, then

E[p(f, g)] > 0.07. F The proof follows from the way the two Boolean functions f and g are defined in section 2.2. The ratio for U is simply the probability of measuring√|0i in the final state. The√ ratio for F is derived from approximating the angle between flat(w) = f/ N and flat(Hw) = g/ N, where H is the n-qubit Hadamard gate, and w is defined as wx = vx/kvxk.

2 p(f, g) = flat(w)T Hflat(Hw)

By using triangle inequality law,

arccos flat(w)T Hflat(Hw) ≤ arccos flat(w)T w + arccos wT Hflat(Hw) .

By performing some statistical calculations, with 1 − 1/exp(N) probability we have, ! r 2 arccos flat(w)T w ≤ arccos + ε π

! r 2 arccos wT Hflat(Hw) ≤ arccos + ε π ! r 2 arccos flat(w)T Hflat(Hw) ≤ 2 arccos + 2ε = 1.3 π

Thus p(f, g) ≥ (cos1.3)2 ' 0.072. Deﬁnition 18 A distribution D over N-bit strings is ε-almost k-wise independent, if for every k-term C,

P r [C(X)] 1 − ε ≤ X∼D ≤ 1 + ε P r [C(X)] X∼U √ √ Theorem 19 F is O k2/ N -almost k-wise independent for all k ≤ 4 N. First, the proof shows the analogous statement for the real valued functions F and G drawn from F, then generalises the notion for discrete functions f and g where each C term is conjunction of K terms of form F (x) ≤ 0 or F (x) ≥ 0 and L terms of form G(x) ≤ 0 or G(x) ≥ 0.

P r[C] (K + L)2 k2 F = 1 ± O √ = 1 ± O √ P r[C] N N U Lemma 20 Suppose a probability distribution D over oracle strings is 1/t(n)-almost poly(n)-wise independent, for some superpolynomial function t. Then no BPPpath machine or SZK protocol can distinguish D from the uniform distribution U with non-negligible bias.

8 2 Let M be a BPPpath machine, let pD is the probability that M accepts an oracle string drawn from D. Then pD must be of form aD/sD, where sD is fraction of strings postselected and aD is fraction of strings that are accepted and postselected. As D is 1/t(n)-almost poly(n)-wise independent, 1 a 1 1 s 1 1 − ≤ D ≤ 1 + and 1 − ≤ D ≤ 1 + t(n) aU t(n) t(n) sU t(n)

1 2 a /s 1 2 1 − ≤ D D ≤ 1 + t(n) aU /sU t(n) Let’s discuss a lemma about detecting the closeness in distributions. As proved in [1], Statistical Differ- ence, detecting the closeness between two distributions, is in SZK and it is complete problem in SZK. Lemma (Polarization Lemma)[1] There is polynomial time computable function that takes a triple k (C0,C1, 1 ), where C0 and C1 are circuits and outputs a pair of circuits (D0,D1) such that 1 kC − C k < =⇒ kD − D k < 2−k 0 1 3 0 1 2 kC − C k > =⇒ kD − D k > 1 − 2−k 0 1 3 0 1 The first case is the reject case for SD, and second case is the accept case for SD. By above stated lemma, there exists polynomial time computable distributions A and A0 such that an SZK protocol P accepts, then kA − A0k ≤ 1/3, while if P rejects, then kA − A0k ≥ 2/3 (since, SZK is closed under complementation). But since each computation can examine at most poly(n) oracle bits 0 0 and D is 1/t(n)-almost poly(n)-wise independent, we have kAD −AU k ≤ 1/t(n) and kAD −AU k ≤ 1/t(n). Hence, 2 | kA − A0 k − kA − A0 k | ≤ kA − A k + kA0 − A0 k ≤ . D D U U D U D U t(n) As the statistical difference between D and U is at most 2/t(n), P will always reject and never be able to tell the difference between D and U, thus no SZK protocol exists. Theorem 21 There exists an oracle A realtive to which BQPA 6⊂ BPPpathA and BQPA 6⊂ SZKA. n The oracle A encodes the truth table of Boolean functions f1, f2,... and g1, g2,... where fn, gn : {0, 1} → {−1, 1}. For each n, with 1/2 probability hfn, gni is sampled from uniform distribution U and with 1/2 probability hfn, gni is distributed from forrelated distribution F. Let L be the unary language consisting n of all 0 for which hfn, gni is sampled from F. By theorem 9, there exists BQPA machine M which decides L with probability 1 over A.

By lemma 20, no BPPpath machine can distinguish between F and U. Let En(M) be the event that the BPPpath machine M correctly decides 0n ∈ L. Then

P r[En(M)] ≤ 1/2 A

and by conditioning on Ei(m), i ∈ [n] and by union bound,

P r[∃M : E1(M) ∧ E2(M) ∧ ...] = 0. A

Thus it follows that BQPA 6⊂ BPPpathA with probability 1 over A. By following the similar lines of proof, it follows BQPA 6⊂ SZKA with probability 1 over A. The proceeding sections of the original paper discuss relations among BQP, PH and AM, which requires deeper knowledge in computational complexity theory and mathematics, which is currently out of our reach. So we will be restricting ourselves till here, and we will not include those sections in this report.

2 A language L is in class BPPpath, if there exists two machines A and B such that if x ∈ L then P rr[A(x, r)|B(x, r)] > 2/3, and if x 6∈ L then P rr[A(x, r)|B(x, r)] < 1/3. The machine A is called acceptor, and the machine B is called post-selector.

9 5 Conclusion

The discussion in section 3 gives an oracle separation between BQP and PH relative to a random oracle. The section 4 gives oracle separation between BQP and BPPpath, and BQP and SZK. This provides subtle evidence that the BQP is not contained in any of those classes which suggests that quantum algorithm designers must look for solutions to problems not in NP, or in general in PH.

Acknowledgement

We would like to thanks Prof. Rajat Mittal to give us opportunity to work in this topic.

References

[1] Amit Sahai and Salil Vadhan, A Complete Problem for Statistical Zero Knowledge, Journal of the ACM, Vol. 50, No. 2, March 2003, pp. 196–249. [2] J. H˚astad.Computational Limitations for Small Depth Circuits. MIT Press, 1987.

[3] https://en.wikipedia.org/wiki/Zero-knowledge proof

[4] https://complexityzoo.uwaterloo.ca/Complexity Zoo [5] Sanjeev Arora and Boaz Barak, Computational Complexity: A Modern Approach, Cambridge Uni- versity Press.