Electronic Colloquium on Computational Complexity, Report No. 131 (2019) A Simple Proof of Vyalyi’s Theorem and some Generalizations Lieuwe Vinkhuijzen Andre´ Deutz Universiteit Leiden Universiteit Leiden Email: [email protected] Email: [email protected] Abstract—In quantum computational complexity the- Vyalyi’s proof of Theorem 1 introduces a new ory, the class QMA models the set of problems effi- complexity class called A0 PP and uses Gap func- ciently verifiable by a quantum computer the same tions to show that QMA ⊆ A0 PP ⊆ PP; then way that NP models this for classical computation. it uses Gap functions and a strong version of Vyalyi proved that if QMA = PP then PH ⊆ QMA. Toda’s Theorem to show that if A0 PP = PP then In this note, we give a simple, self-contained proof PH ⊆ PP. Specifically, it uses the version of #P[1] of the theorem, using only the closure properties Toda’s Theorem which states that PH ⊆ P : of the complexity classes in the theorem statement. all languages in the polynomial hierarchy can be We then extend the theorem in two directions: solved with only one query to a counting oracle. (i) we strengthen the consequent, proving that if PP Our new proof is, in our view, simpler. It has QMA = PP then QMA = PH , and (ii) we weaken three ingredients: (i) the usual version of Toda’s the hypothesis, proving that if QMA = CO-QMA Theorem [4] (namely PH ⊆ PPP), (ii) that PP is then PH ⊆ QMA. Lastly, we show that all the closed under complement [3] and (iii) that QMA\ above results hold, without loss of generality, for the CO-QMA is closed under Turing reductions. The class QAM instead of QMA. We also formulate a third ingredient is, to the best of our knowledge, “Quantum Toda’s Conjecture”. new. 1. Introduction We anticipate the objection that our proof still uses Toda’s Theorem and that therefore the com- A major open question in quantum computa- plexity of the original proof is not eliminated, tional complexity theory is to find the relationships but merely outsourced. Our response is to give between quantum complexity classes and classical a second, wholly self-contained elementary proof, ones. In particular, we do not know how QMA whose ideas we immediately use to improve The- relates to the polynomial hierarchy or to PP. For orem 1 in two ways. First, we strengthen the the first problem, no containment is known in consequent, as follows: either direction. While it is known that QMA is PP contained in PP, it is open whether the inclusion is Theorem 2. If QMA = PP then QMA = PH . strict. Progress in this direction was made in 2003 when Vyalyi showed that the two questions are in That is, the hypothesis implies that the poly- fact related: nomial hierarchy collapses relative to a count- ing oracle. Second, we weaken the hypothesis of Theorem 1 (Vyalyi [1]). If QMA = PP then Vyalyi’s Theorem from QMA = PP to merely PH ⊆ PP. QMA = CO-QMA: For the proof, see page 6. Vyalyi took this as evidence that QMA 6= PP because otherwise Theorem 3. If QMA = CO-QMA then PH ⊆ PH ⊆ QMA, which seems unlikely. PP. ISSN 1433-8092 A Simple Proof of Vyalyi’s Theorem and some Generalizations 2. Preliminaries The class PP, or Probabilistic Polynomial time, was defined by Gill [3], who showed that PP is We assume that the reader is familiar with the closed under complement. Toda’s Theorem states basics of Computational Complexity, in particular that PH ⊆ PPP [4]. the polynomial hierarchy (see Arora and Barak [5]) and with quantum computing (see Nielsen and Chuang [6] or Kitaev, Shin and Vyalyi [7]). 3. Closure properties of QMA We work with languages over the binary alphabet f0; 1g. In this section, we show that the class QMA \ A Turing reduction from a language L to a CO-QMA is closed under polynomial-time Turing language K is an algorithm with oracle access to reductions. That is, K which solves L. If L Turing-reduces to K, we \CO Theorem 4. QMA \ CO-QMA = PQMA -QMA write L ≤T K. If the reduction algorithm runs in polynomial time, we say that there is a polynomial- For the proof, see page 4. The ideas in the time deterministic Turing reduction from L to K proof are best illustrated by recalling two other p K and write L ≤T K or L 2 P . A class C is closed theorems: (i) the class NP\CO-NP is closed under under polynomial-time Turing reductions, written Turing reductions (Theorem 5), and (ii) the class C p C = P , if L ≤T K implies L 2 C for every QMA is closed under intersection (Theorem 7). To language K 2 C and L ⊆ f0; 1g∗. A class is closed streamline the proof of Theorem 7, we introduce under complement, i.e. C = CO−C, if L 2 C () the Entanglement Independence Lemma, (Lemma L 2 C for all L 2 C, where L = f0; 1g∗ n L. 6). \CO The class QMA was defined by Kitaev et al. The result NP \ CO-NP = PNP -NP is clas- [7] (they called it BQNP): sic, and the idea of the Entanglement Independence Definition 1 (The class QMA: Quantum Mer- Lemma is simply the technique Kitaev et al. used lin-Arthur games). A language L ⊆ f0; 1g∗ is in for error amplification when they defined QMA QMA if there are polynomials m(n); w(n) and [7]. a polynomial-time constructible family of quantum Theorem 5. NP \ CONP is closed under circuits fUxgx2f0;1g∗ receiving an m(n)-qubit in- polynomial-time deterministic Turing reductions: \CO put and using w(n) qubits of workspace, possess- PNP NP = NP \ CONP. ing completeness and soundness: • Completeness: If x 2 L then the circuit Proof. The trivial direction is NP \ CO-NP ⊆ NP\CO-NP Ux accepts some input state j i with P , so we will only show the other direc- m(n) NP\CO-NP probability at least 1 − 2−n tion, P ⊆ NP \ CO-NP. To this end, it is sufficient to show that PNP\CO-NP ⊆ NP because • Soundness: If x 62 L then the circuit Ux rejects all input states with probability at P is closed under complement. NP\CO-NP least 1 − 2−n Let K 2 P be a language decided by the polynomial-time (say O(t(n))-time) Turing The circuit family is called a protocol for L. Machine M L, with access to an oracle language The class QMA is often studied as a set of L 2 NP \ CO-NP. Because L 2 NP \ CO-NP, promise problems (which are pairs (Lyes;Lno), there are non-deterministic Turing Machines Y and where the algorithm is allowed to behave arbitrar- N recognizing the languages L and L, respectively, 0 ily on inputs outside of Lyes[Lno), because in that both running in time O(t (n)). If we manage to context it allows for complete problems, notably simulate M L with a non-deterministic machine, the Local Hamiltonian Problem [7]. For us it will we will have proved the theorem. be more natural to consider QMA simply as a set Clearly if we manage to obtain the answers to of languages, because in this context the operation all the queries M L makes, then we can faithfully of using a language as an oracle is cleaner, but simulate M L. The central insight is that we can we stress that this decision is without loss of obtain the answers by guessing and then verifying generality. them. Therefore the algorithm will be as follows. 2 A Simple Proof of Vyalyi’s Theorem and some Generalizations Before we compute anything, (i) we non- j i deterministically guess that the queries that M L is m U going to make are the strings s1; : : : ; st(n), (ii) we j0i guess the answers a1; : : : ; at(n) to all the queries w that M L makes, and lastly (iii) we guess certifi- cate strings y1; : : : ; yt(n) and z1; : : : ; zt(n) for the jφim 1 machines Y and N, respectively. The remainder V of the computation is deterministic. The algorithm j0i checks that its guesses were correct: for each i, it w checks that Y (s ; y ) = a and N(s ; z ) = :a i i i i i i Figure 1. The circuits U and V receive inputs that may be (meaning that, for example, if a2 = 1 then Y entangled. accepts the certificate y2 we guessed for s2 and N rejects z2). If any of these checks fail, we have evidently guessed incorrectly, and we reject on this Lemma 6 (Entanglement Independence Lemma). computation path. Let U; V be two quantum circuits as in Figure 1, Lastly, we simulate M L. When it makes the i- both receiving an m-qubit input and a w-qubit ? workspace, with measurement operators ΠU ; ΠV . th query, we check that it queries si 2 L; if so, we Suppose that U and V accept with probability at feed it the answer ai and continue the simulation, most a and b, respectively, regardless of their m- but if not, we immediately reject, because we qubit input. Then the probability that both U and have incorrectly guessed which strings M L would L V accept when they are implemented jointly, and query. When M halts and accepts, we accept; when their inputs may be entangled, is at most a·b. otherwise we reject.