%% ToC vol 3 no 7 \documentclass{toc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \setcounter{MaxMatrixCols}{30} \providecommand{\U}[1]{\protect\rule{.1in}{.1in}} \tocdetails{% volume=3, number=7, year=2007, firstpage=129, received={July 27, 2006}, revised={August 23, 2007}, published={September 14, 2007} } \runningauthor{S. Aaronson and G. Kuperberg} \runningtitle{Quantum vs. classical proofs and advice} \copyrightauthor{Scott Aaronson and Greg Kuperberg} \bibliographystyle{tocplain} \begin{document} \begin{frontmatter}[classification=text] \title{Quantum Versus Classical Proofs and Advice} \author[scott]{Scott Aaronson\thanks{MIT. \ Email: aaronson@csail.mit.edu.}} \author[greg]{Greg Kuperberg\thanks{UC Davis. \ Email: greg@math.ucdavis.edu.}} \tockeywords{quantum computing, QMA, black-box groups, oracles, hidden subgroup problem} \begin{abstract} This paper studies whether quantum proofs are more powerful than classical proofs, or in complexity terms, whether $\mathsf{QMA}=\mathsf{QCMA}$. We prove three results about this question. First, we give a ``quantum oracle separation'' between $\mathsf{QMA}$ and $\mathsf{QCMA}$. More concretely, we show that any quantum algorithm needs $\Omega\left( \sqrt{\frac{2^{n}}{m+1}}\right) $ queries to find an $n$-qubit ``marked state'' $\left\vert \psi \right\rangle $, even if given an $m$-bit classical description of $\left\vert \psi\right\rangle $ together with a quantum black box that recognizes $\left\vert \psi\right\rangle $. Second, we give an explicit $\mathsf{QCMA}$ protocol that nearly achieves this lower bound. Third, we show that, in the one previously-known case where quantum proofs seemed to provide an exponential advantage, \emph{classical} proofs are basically just as powerful. In particular, Watrous gave a $\mathsf{QMA}$ protocol for verifying non-membership in finite groups. Under plausible group-theoretic assumptions, we give a $\mathsf{QCMA}$ protocol for the same problem. Even with no assumptions, our protocol makes only polynomially many queries to the group oracle. We end with some conjectures about quantum versus classical oracles, and about the possibility of a \emph{classical} oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$. \end{abstract} \tocams{03F20, 68Q10, 68Q15, 81P68} \tocacm{F.1.2, F.1.3} \end{frontmatter} \section{Introduction\label{INTRO}} If someone hands you a quantum state, is that more ``useful'' than being handed a classical string with a comparable number of bits? In particular, are there truths that you can efficiently verify, and are there problems that you can efficiently solve, using the quantum state but not using the string? These are the questions that this paper addresses, and that it answers in several contexts. Recall that $\mathsf{QMA}$, or Quantum Merlin-Arthur, is the class of decision problems for which a ``yes'' answer can be verified in quantum polynomial time, with help from a polynomial-size quantum witness state $\left\vert \psi\right\rangle $. Many results are known about $\mathsf{QMA}$: for example, it has natural complete promise problems~\cite{kkr}, allows amplification of success probabilities~\cite{mw}, and is contained in $\mathsf{PP}$~\cite{mw}. Raz and Shpilka~\cite{razshpilka} have also studied communication complexity variants of $\mathsf{QMA}$. Yet, as Aharonov and Naveh~\cite{an} pointed out in 2002, the very definition of $\mathsf{QMA}$ raises a fundamental question. Namely: is it really essential that the witness be quantum, or does it suffice for the\emph{ }algorithm \emph{verifying }the witness to be quantum? To address this question, Aharonov and Naveh defined the class $\mathsf{QCMA}$, or ``Quantum Classical Merlin-Arthur,'' to be the same as $\mathsf{QMA}$ except that now the witness is classical.\footnote{Some say that this class would more accurately be called $\mathsf{CMQA}$, for ``Classical Merlin Quantum Arthur.'' But $\mathsf{QCMA}$ has stuck.} We can then ask whether $\mathsf{QMA}=\mathsf{QCMA}$. Not surprisingly, the answer is that we don't know. If we can't decide whether two complexity classes are equal, the usual next step is to construct a relativized world that separates them. This would provide at least some evidence that the classes are different. But in the case of $\mathsf{QMA}$ versus $\mathsf{QCMA}$, even this limited goal has remained elusive. Closely related to the question of quantum versus classical proofs is that of quantum versus classical \emph{advice}. Compared to a proof, advice has the advantage that it can be trusted, but the disadvantage that it can't be tailored to a particular input. More formally, let $\mathsf{BQP/qpoly}$ be the class of problems solvable in quantum polynomial time, with help from a polynomial-size ``quantum advice state'' $\left\vert \psi_{n}\right\rangle$ that depends only on the input length $n$. Then the question is whether $\mathsf{BQP/qpoly}=\mathsf{BQP/poly}$, where $\mathsf{BQP/poly}$ is the class of problems solvable in quantum polynomial time with help from polynomial-size \emph{classical} advice. Aaronson~\cite{aar:adv} showed that $\mathsf{BQP/qpoly}\subseteq \mathsf{PP/poly}$, which at least tells us that quantum advice is not ``infinitely'' more powerful than classical advice. But, like the $\mathsf{QMA}$ versus $\mathsf{QCMA}$ question, the $\mathsf{BQP/qpoly}$ versus $\mathsf{BQP/poly}$ question has remained open, with not even an oracle separation known. \subsection{Our results\label{RESULTS}} This paper introduces new tools with which to attack $\mathsf{QMA}$ versus $\mathsf{QCMA}$ and related questions. First, we achieve an oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$, but only by broadening the definition of ``oracle.'' In particular, we introduce the notion of a \emph{quantum oracle}, which is just an infinite sequence of unitaries $U=\left\{ U_{n}\right\} _{n\geq1}$ that a quantum algorithm can apply in a black-box fashion. Just as a classical oracle models a subroutine to which an algorithm has black-box access, so a quantum oracle models a quantum subroutine, which can take quantum input and produce quantum output. We are able to give a quantum oracle that separates $\mathsf{QMA}$ from $\mathsf{QCMA}$: \begin{theorem} \label{qosepthm}There exists a quantum oracle $U$ such that $\mathsf{QMA} ^{U}\neq\mathsf{QCMA}^{U}$. \end{theorem} Similarly, there exists a quantum oracle $V$ such that $\mathsf{BQP} ^{V}\mathsf{/qpoly}\neq\mathsf{BQP}^{V}\mathsf{/poly}$. \expref{Theorem}{qosepthm} implies that if $\mathsf{QMA}=\mathsf{QCMA}$, then any proof of this fact will require ``quantumly nonrelativizing techniques'': techniques that are sensitive to the presence of quantum oracles. Currently, we do not know of \emph{any} quantumly nonrelativizing techniques that are not also classically nonrelativizing. For this reason, we believe that quantum oracle separations merit the same informal interpretation as classical oracle separations: almost any argument that one might advance against the former, is also an argument against the latter! The difference is that quantum oracle results are sometimes much easier to prove than classical ones. To our knowledge, this paper provides the first example of this phenomenon, but other examples have since emerged~\cite{aar:copy,moscastebila}. It might be objected that, even if quantum oracle separations are no less trustworthy than classical ones, they certainly aren't \emph{more} trustworthy, and complexity theorists have known since the celebrated $\mathsf{IP}=\mathsf{PSPACE}$ theorem~\cite{shamir} that oracle results sometimes ``point in the wrong direction.'' We wish to stress two points in response. First, oracle results provide at least \emph{some} understanding, thereby opening the way to further progress. This is particularly true in quantum computing, where even the oracle results tend to be much less intuitively obvious than they are in the classical world. Second, complexity theorists do not currently have any nonrelativizing technique for ``non-interactive'' classes such as $\mathsf{QMA}$ and $\mathsf{QCMA}$ even remotely analogous to the arithmetization technique that Shamir~\cite{shamir} used to show $\mathsf{IP}=\mathsf{PSPACE}$. We hope such a technique will someday be discovered. Underlying \expref{Theorem}{qosepthm} is the following lower bound. Suppose a unitary oracle $U_{n}$ acts on $n$ qubits, and suppose that either (i) $U_{n}$ is the identity, or (ii) there exists a secret $n$-qubit ``marked state'' $\left\vert \psi _{n}\right\rangle $ such that $U_{n}\left\vert \psi_{n}\right\rangle =-\left\vert \psi_{n}\right\rangle $, but $U_{n}\left\vert \varphi \right\rangle =\left\vert \varphi\right\rangle $ whenever $\left\vert \varphi\right\rangle $ is orthogonal to $\left\vert \psi_{n}\right\rangle $. Then even if a quantum algorithm is given $m$ bits of classical advice about $\left\vert \psi_{n}\right\rangle$, the algorithm still needs $\Omega\bigl( \sqrt{{2^{n}}/(m+1)}\bigr)$ queries to $U_{n}$ to distinguish these cases. Note that when $m=0$, we recover the usual $\Omega\bigl( \sqrt {2^{n}}\bigr) $ lower bound for Grover search as a special case. At the other extreme, if $m\approx2^{n}$ then our bound gives nothing---not surprisingly, since the classical advice might contain explicit instructions for preparing $\left\vert \psi_{n}\right\rangle$. The point is that, if $m$ is \emph{not} exponentially large, then exponentially many queries are needed. Since $\left\vert \psi_{n}\right\rangle $ is an arbitrary $2^{n}$-dimensional unit vector, it might be thought obvious that $2^{\Omega\left( n\right)}$\ bits are needed to describe that vector. The key point, however, is that the $\mathsf{QCMA}$ verifier is given not only a classical description of $\left\vert \psi_{n}\right\rangle $, but also oracle access to $U_{n}$. So the question is whether some \emph{combination} of these resources might be exponentially more powerful than either one alone. We prove that the answer is no, using the hybrid argument of Bennett et al.~\cite{bbbv} together with geometric results about partitionings of the unit sphere. In \expref{Section}{UPPER}, we show that our lower bound is basically tight, by giving an algorithm that finds $\left\vert \psi_{n}\right\rangle $ using $O\bigl( \sqrt{2^{n}/m}\bigr) $ queries when $m\geq2n$. This algorithm has the drawback of being \emph{computationally} inefficient. To fix this, we give another algorithm that finds $\left\vert \psi_{n}\right\rangle $ using $O\bigl( n\sqrt{2^{n}/m}\bigr) $ queries together with $O\bigl( n^{2}\sqrt{2^{n}/m}+\operatorname*{poly}\left( m\right) \bigr) $ computational steps. Having separated $\mathsf{QMA}$ from $\mathsf{QCMA}$ by a quantum oracle, we next revisit the question of whether these classes can be separated by a \emph{classical} oracle. Right now, we know of only one candidate problem for such a separation in the literature: the Group Non-Membership ($\operatorname*{GNM}$) problem, which Watrous~\cite{watrous} placed in $\mathsf{QMA}$ even though Babai~\cite{babai:am} showed it is not in $\mathsf{MA}$ as an oracle problem.\footnote{Interestingly, the classes $\mathsf{MA}$ and $\mathsf{AM}$ were originally defined by Babai in connection with GNM~\cite{babai:am2}.} In Group Non-Membership, Arthur is given black-box access to a finite group $G$, together with a subgroup $H\leq G$ specified by its generators and an element $x\in G$. \ Arthur's goal is to verify that $x\notin H$, using a number of group operations polynomial in $\log\left\vert G\right\vert $. (Note that the group \emph{membership} problem is in $\mathsf{NP}$ by a result of Babai and Szemer\'{e}di~\cite{bs:matrix}.) In Watrous's protocol, the quantum witness is simply an equal superposition $\left\vert H\right\rangle $ over the elements of $H$. Given such a witness, Arthur can check non-membership by comparing the states $\left\vert H\right\rangle $ and $\left\vert xH\right\rangle $, and can similarly check the veracity of $\left\vert H\right\rangle $ by comparing it to $\left\vert hH\right\rangle $, where $h$ is an almost-uniformly random element of $H$. Evidently a classical proof of non-membership would have to be completely different. Nevertheless, in \expref{Section}{GNM} we show the following: \begin{theorem} \label{hspthm}$\operatorname*{GNM}$ has polynomially-bounded $\mathsf{QCMA}$ query complexity. \end{theorem} \expref{Theorem}{hspthm} implies that it is pointless to try to prove a classical oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$ by proving a lower bound on the quantum query complexity of Group Non-Membership. If such a separation is possible, then a new approach will be needed. The idea of the proof of \expref{Theorem}{hspthm} is that Merlin can ``pull the group out of the black box.'' In other words, he can claim an embedding of a model group $\Gamma$ into $G$. This claim is entirely classical, but verifying it requires solving the Normal Hidden Subgroup Problem ($\operatorname*{NHSP}$) in $\Gamma$. This problem has low query complexity by a result of Ettinger, H\o yer, and Knill~\cite{ehk}, but is not known to be in $\mathsf{BQP}$. In addition, analyzing the description of $\Gamma$ is not known to be computationally efficient. Nonetheless, in \expref{Section}{COMP} we discuss evidence that $\operatorname*{NHSP}$ is in $\mathsf{BQP}$ and that non-membership for $\Gamma$ is in $\mathsf{NP}$. Based on this evidence, we conjecture the following: \begin{conjecture} \label{gnminqcma}$\operatorname*{GNM}$ is in $\mathsf{QCMA}$. \end{conjecture} Given our results in \expref{Section}{GNM}, the question remains of whether there is some other way to prove a classical oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$. In \expref{Section}{SQO}, we conjecture that the answer is yes: \begin{conjecture} \label{sepconj}There exists a classical oracle $A$ such that $\mathsf{QMA}^{A}\neq\mathsf{QCMA}^{A}$. Furthermore, this can be proven by exhibiting an oracle problem with polynomial $\mathsf{QMA}$ query complexity but exponential $\mathsf{QCMA}$ query complexity. \end{conjecture} The reason we believe \expref{Conjecture}{sepconj} is that it seems possible, for many purposes, to ``encode'' a quantum oracle into a classical one. In \expref{Section}{SQO} we explain more concretely what we mean by that, and present some preliminary results. For example, we show that there exists a $\mathsf{BQP}$ algorithm that maps an oracle string $A$ to an $n$-qubit pure state $\left\vert \psi_{A}\right\rangle $, such that if $A$ is uniformly random, then $\left\vert \psi_{A}\right\rangle $ is (under a suitable metric) close to uniformly random under the Haar measure. We also study the question of applying a random $N\times N$ unitary matrix using a random classical oracle in the same way. We do not know how to do this, but we show that one quantum query will not suffice for this purpose. To prove this, we show that a quantum algorithm that uses just one query can apply at most $4^{N}$ different $N\times N$ unitaries, whereas the number of unitaries required to approximate the uniform distribution grows like $2^{\Theta\left( N^{2}\right) }$. We end in \expref{Section}{OPEN} with some open problems. \section{Preliminaries\label{PRELIM}} Throughout this paper, we refer to the set of $N$-dimensional pure states as $\mathbb{CP}^{N-1}$ (that is, complex projective space with $N-1$ dimensions). We use $\Pr$ to denote probability, and $\operatorname*{E}$ to denote expectation. We assume familiarity with standard complexity classes such as $\mathsf{BQP}$ and $\mathsf{MA}$. For completeness, we now define $\mathsf{QMA}$, $\mathsf{QCMA}$, $\mathsf{BQP/qpoly}$, and $\mathsf{BQP/poly}$. \begin{definition} $\mathsf{QMA}$ is the class of languages $L\subseteq\left\{ 0,1\right\} ^{n}$ for which there exists a polynomial-time quantum verifier $\mathcal{Q}$ and a polynomial $p$ such that, for all $x\in\left\{ 0,1\right\} ^{n}$: \begin{enumerate} \item[(i)] If $x\in L$ then there exists a $p\left( n\right) $-qubit quantum proof $\left\vert \varphi\right\rangle $ such that $\mathcal{Q}$ accepts with probability at least $2/3$ given $\left\vert x\right\rangle \left\vert \varphi\right\rangle $ as input. \item[(ii)] If $x\notin L$ then $\mathcal{Q}$ accepts with probability at most $1/3$ given $\left\vert x\right\rangle \left\vert \varphi\right\rangle $ as input, for all purported proofs $\left\vert \varphi\right\rangle $. \end{enumerate} The class $\mathsf{QCMA}$ is defined similarly, except that $\left\vert \varphi\right\rangle $ is replaced by a classical string $z\in\left\{ 0,1\right\} ^{p\left( n\right) }$. \end{definition} \begin{definition} $\mathsf{BQP/qpoly}$ is the class of languages $L\subseteq\left\{ 0,1\right\} ^{n}$ for which there exists a polynomial-time quantum algorithm $\mathcal{Q}$, together with a set of states $\left\{ \left\vert \psi _{n}\right\rangle \right\} _{n\geq1}$ (where $\left\vert \psi_{n}\right\rangle $ has size $p\left( n\right) $ for some polynomial $p$), such that for all $x\in\left\{ 0,1\right\} ^{n}$: \begin{enumerate} \item[(i)] If $x\in L$ then $\mathcal{Q}$ accepts with probability at least $2/3$ given $\left\vert x\right\rangle \left\vert \psi_{n}\right\rangle $ as input. \item[(ii)] If $x\notin L$ then $\mathcal{Q}$ accepts with probability at most $1/3$ given $\left\vert x\right\rangle \left\vert \psi_{n}\right\rangle $ as input. \end{enumerate} The class $\mathsf{BQP/poly}$ is defined similarly, except that $\left\vert \psi_{n}\right\rangle $ is replaced by a classical string $a_{n}\in\left\{ 0,1\right\} ^{p\left( n\right) }$. \end{definition} Let us now explain what we mean by a ``quantum oracle.'' For us, a quantum oracle is simply an infinite sequence of unitary transformations, $U=\left\{ U_{n}\right\} _{n\geq1}$. We assume that each $U_{n}$ acts on $p\left( n\right) $ qubits for some known polynomial $p$. We also assume that given an $n$-bit string as input, a quantum algorithm calls only $U_{n}$, not $U_{m}$ for any $m\neq n$. This assumption is only made for simplicity; our results would go through without it.\footnote{If one made the analogous assumption in \emph{classical} complexity---that given an input of length $n$, an algorithm can query the oracle only on strings of length $n$---one could simplify a great many oracle results without any loss of conceptual content.} When there is no danger of confusion, we will refer to $U_{n}$ simply as $U$. Formally, the oracle access mechanism is as follows. Assume a quantum computer's state has the form \[ \left\vert \Phi\right\rangle =\sum_{z}\alpha_{z}\left\vert z\right\rangle \left\vert \phi_{z}\right\rangle\, , \] where $\left\vert z\right\rangle $ is a workspace register and $\left\vert \phi_{b,z}\right\rangle $ is a $p\left( n\right) $-qubit answer register. Then to ``query $U_{n}$'' means to apply the $p\left( n\right) $-qubit unitary transformation that maps $\left\vert \Phi\right\rangle $ to \[ \left\vert \Phi^{\prime}\right\rangle =\sum_{z}\alpha_{z}\left\vert z\right\rangle U_{n}\left\vert \phi_{z}\right\rangle\, . \] Let $\mathcal{C}$ be a quantum complexity class, and let $U=\left\{ U_{n}\right\} _{n\geq1}$ be a quantum oracle. Then by $\mathcal{C}^{U}$, we will mean the class of problems solvable by a $\mathcal{C}$ machine that, given an input of length $n$, can query $U_{n}$ at unit cost as many times as it likes. In defining the notion of quantum oracle, at least two choices present themselves that have no counterpart for classical oracles: \begin{enumerate} \item[(1)] If we can apply a quantum oracle $U$, then can we also apply controlled-$U$ (that is, $U$ conditioned on a control qubit $\left\vert b\right\rangle $)? \item[(2)] If we can apply $U$, then can we also apply $U^{-1}$? \end{enumerate} At least for the present paper, the answers to these questions will not matter, for the following reasons. First, all of the quantum oracles $U$ that we consider will be self-inverse (that is, $U=U^{-1}$). Second, while our algorithms \emph{will} need to apply controlled-$U$, that is only for the technical reason that we will define $U$ so that $U\left\vert \psi\right\rangle =-\left\vert \psi\right\rangle $ if $\left\vert \psi\right\rangle $ is the marked state, and $U\left\vert \varphi \right\rangle =\left\vert \varphi\right\rangle $ whenever $\left\langle \varphi|\psi\right\rangle =0$. If we stipulated instead that $U\left\vert \psi\right\rangle \left\vert b\right\rangle =\left\vert \psi\right\rangle \left\vert b\oplus1\right\rangle $ and $U\left\vert \varphi\right\rangle \left\vert b\right\rangle =\left\vert \varphi\right\rangle \left\vert b\right\rangle $ whenever $\left\langle \varphi|\psi\right\rangle =0$, then $U$ alone would suffice. Yet even though these choices will not matter for our results, it still seems worthwhile to discuss them a bit, since they might arise in future work involving quantum oracles. One could argue that (i) the purpose of an oracle is to model a \emph{subroutine} that an algorithm can call without understanding its internal structure, and that (ii) given a quantum circuit for applying some unitary operation $U$, one can easily produce a circuit for applying controlled-$U$ or $U^{-1}$, without understanding anything about the original circuit's structure. In particular, to produce a circuit for controlled-$U$, one simply conditions each gate on the control qubit; while to produce a circuit for $U^{-1}$, one simply inverts all the gates and reverses their order. These considerations suggest that the answers to questions (1) and (2) should both be `yes.' On the other hand, it would still be interesting to know whether disallowing controlled-$U$ or $U^{-1}$ would let us prove more quantum oracle separations. (Note that if we disallow these operations, then the set of inequivalent quantum oracles becomes \emph{larger}.) \section{Quantum oracle separations\label{QORACLE}} The aim of this section is to prove \expref{Theorem}{qosepthm}: that there exists a quantum oracle $U$ such that $\mathsf{QMA}^{U}\neq\mathsf{QCMA}^{U}$. The same ideas will also yield a quantum oracle $V$ such that $\mathsf{BQP}^{V} \mathsf{/qpoly}\neq\mathsf{BQP}^{V}\mathsf{/poly}$. To prove these oracle separations, we first need a geometric lemma about probability measures on quantum states. Let $\mu$ be the uniform probability measure over $N$-dimensional pure states (that is, over $\mathbb{CP}^{N-1}$). The following notion will play a key role in our argument. \begin{definition} \label{puniform}For all $p\in\left[ 0,1\right] $, a probability measure $\sigma$ over $\mathbb{CP}^{N-1}$ is called $p$\emph{-uniform} if $p\sigma\leq\mu$. Equivalently, $\sigma$ is $p$-uniform if it can be obtained by starting from $\mu$, and then conditioning on an event that occurs with probability at least $p$. \end{definition} So for example, we obtain a $p$-uniform measure if we start from $\mu$ and then condition on $\log_{2}1/p$ bits of classical information about $\left\vert \psi\right\rangle $. Our geometric lemma says that if $\left\vert \psi\right\rangle $ is drawn from a $p$-uniform measure, then for every mixed state $\rho$, the squared fidelity between $\left\vert \psi\right\rangle $ and $\rho$ has small expectation. More precisely: \begin{lemma} \label{sumip}Let $\sigma$ be a $p$-uniform probability measure over $\mathbb{CP}^{N-1}$. Then for all $\rho$, \[ \operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\langle \psi|\rho|\psi\right\rangle \right] =O\left( \frac{1+\log1/p}{N}\right)\,. \] \end{lemma} The proof of \expref{Lemma}{sumip} is deferred to \expref{Section}{LEMMA}. In this section we assume the lemma, and show how to use it to prove our main result. In particular, we show that any quantum algorithm needs $\Omega\bigl( \sqrt{{2^{n}}/{(m+1)}}\bigr) $ queries to find an $n$-qubit marked state $\left\vert \psi\right\rangle $, even if given $m$ bits of classical advice about $\left\vert \psi\right\rangle $. \begin{theorem} \label{qosep}Suppose we are given oracle access to an $n$-qubit unitary $U$, and want to decide which of the following holds: \begin{itemize} \item[(i)] There exists an $n$-qubit ``quantum marked state'' $\left\vert \psi\right\rangle$ such that $U\left\vert \psi\right\rangle =-\left\vert \psi\right\rangle$, but $U\left\vert \phi\right\rangle =\left\vert \phi\right\rangle$ whenever $\left\langle \phi|\psi\right\rangle =0$; or \item[(ii)] $U=I$ is the identity operator. \end{itemize} Then even if we have an $m$-bit classical witness $w$ in support of case (i), we still need $\Omega\left( \sqrt{\frac{2^{n}}{m+1}}\right) $ queries to verify the witness, with bounded probability of error. \end{theorem} \begin{proof} If $m=\Omega\left( 2^{n}\right) $ then the theorem is certainly true, so suppose $m=o\left( 2^{n}\right) $. Let $A$ be a quantum algorithm that queries $U$. Also, let $U_{\psi}$ be an $n$-qubit unitary such that $U_{\psi}\left\vert \psi\right\rangle =-\left\vert \psi\right\rangle $, but $U_{\psi}\left\vert \phi\right\rangle =\left\vert \phi\right\rangle $ whenever $\left\langle \phi|\psi\right\rangle =0$. Then $A$'s goal is to accept if and only if $U=U_{\psi}$ for some $\left\vert \psi\right\rangle $. For each $n$-qubit pure state $\left\vert \psi\right\rangle $, let us fix a classical witness $w\in\left\{ 0,1\right\} ^{m}$ that maximizes the probability that $A$ accepts, given $U_{\psi}$ as oracle. Let $S\left( w\right) $ be the set of $\left\vert \psi\right\rangle $'s associated with a given witness $w$. Since the $S\left( w\right) $'s form a partition of $\mathbb{CP}^{2^{n}-1}$, clearly there exists a witness, call it $w^{\ast}$, such that \[ \Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \psi\right\rangle \in S\left( w^{\ast}\right) \right] \geq\frac{1}{2^{m}}\,. \] Fix that $w^{\ast}$ (or in other words, hardwire $w^{\ast}$ into $A$). Then to prove the theorem, it suffices to establish the following claim: $A$ cannot distinguish the case $U=U_{\psi}$ from the case $U=I$ by making $o\bigl( \sqrt{{2^{n}}/(m+1)}\bigr)$ queries to $U$, with high probability if $\left\vert \psi\right\rangle $ is chosen uniformly at random from $S\left( w^{\ast}\right)$. To prove the claim, we use a generalization of the hybrid argument of Bennett et al.~\cite{bbbv}. Suppose that $A$ makes $T$ queries to $U$. (Technically speaking, we should also allow queries to controlled-$U$, but this will make no difference in our analysis.) Then for all $0\leq t\leq T$, let $\left\vert \Phi_{t}\right\rangle $ be the final state of $A$, assuming that $U=I$ for the first $t$ queries, and $U=U_{\psi}$ for the remaining $T-t$ queries. Thus $\left\vert \Phi_{0}\right\rangle $ is the final state in case (i), while $\left\vert \Phi_{T}\right\rangle $ is the final state in case (ii). We will argue that $\left\vert \Phi_{t}\right\rangle $ cannot be very far from $\left\vert \Phi_{t-1}\right\rangle $, with high probability over the choice of marked state $\left\vert \psi\right\rangle $. Intuitively, this is because the computations of $\left\vert \Phi _{t}\right\rangle $ and $\left\vert \Phi_{t-1}\right\rangle $ differ in only a single query, and with high probability that query cannot have much overlap with $\left\vert \psi\right\rangle $. We will then conclude, by the triangle inequality, that $\left\vert \Phi_{0}\right\rangle $ cannot be far from $\left\vert \Phi_{T}\right\rangle $ unless $T$ is large. More formally, let $\rho_{t}$ be the marginal state of the query register just before the $t^{th}$ query, assuming the ``control case'' $U=I$. Also, let $\rho_{t}=\sum p_{i}\left\vert \varphi_{i}\right\rangle \left\langle \varphi_{i}\right\vert $ be an arbitrary decomposition of $\rho_{t}$ into pure states. Then for every $i$, the component of $\left\vert \varphi_{i}\right\rangle $ orthogonal to $\left\vert \psi\right\rangle $ is unaffected by the $t^{th}$ query. Therefore \begin{align*} \left\Vert \left\vert \Phi_{t}\right\rangle -\left\vert \Phi_{t-1} \right\rangle \right\Vert _{2} & \leq\sum_{i}p_{i}\cdot2\left\vert \left\langle \varphi_{i}|\psi\right\rangle \right\vert =2\sum_{i}p_{i}\sqrt{\left\langle \psi|\varphi_{i}\right\rangle \left\langle \varphi_{i}|\psi\right\rangle }\\ & \leq2\sqrt{\sum_{i}p_{i}\left\langle \psi|\varphi_{i}\right\rangle \left\langle \varphi_{i}|\psi\right\rangle } =2\sqrt{\left\langle \psi|\rho_{t}|\psi\right\rangle }\,, \end{align*} where the third line uses the Cauchy-Schwarz inequality (the average of the square root is at most the square root of the average). Now let $\sigma$ be the uniform probability measure over $S\left( w^{\ast}\right) $, and observe that $\sigma$ is $2^{-m}$-uniform. So by \expref{Lemma}{sumip}, \begin{align*} \operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\Vert \left\vert \Phi_{t}\right\rangle -\left\vert \Phi_{t-1}\right\rangle \right\Vert _{2}\right] & \leq2\operatorname*{E}_{\left\vert \psi \right\rangle \in\sigma}\left[ \sqrt{\left\langle \psi|\rho_{t} |\psi\right\rangle }\right] \leq2\sqrt{\operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma }\left[ \left\langle \psi|\rho_{t}|\psi\right\rangle \right] }\\ & \leq2\sqrt{\frac{1+\ln\left( 1/2^{-m}\right) }{2^{n}}} =O\left( \sqrt{\frac{m+1}{2^{n}}}\right) \, , \end{align*} where the second line again uses the Cauchy-Schwarz inequality. Finally, \[ \operatorname*{E}_{\left\vert \psi\right\rangle \in S\left( w^{\ast}\right) }\left[ \left\Vert \left\vert \Phi_{T}\right\rangle -\left\vert \Phi _{0}\right\rangle \right\Vert _{2}\right] \leq\sum_{t=1}^{T}\operatorname*{E}_{\left\vert \psi\right\rangle \in S\left( w^{\ast}\right) }\left[ \left\Vert \left\vert \Phi_{t}\right\rangle -\left\vert \Phi_{t-1}% \right\rangle \right\Vert _{2}\right] =O\left( T\sqrt{\frac{m+1}{2^{n}}}\right) \] by the triangle inequality. This implies that, for $\left\vert \Phi _{T}\right\rangle $ and $\left\vert \Phi_{0}\right\rangle $ to be distinguishable with $\Omega\left( 1\right)$ bias, we must have $T=\Omega\bigl( \sqrt{{2^{n}}/(m+1)}\bigr)$. \end{proof} Using \expref{Theorem}{qosep}, we can straightforwardly show a quantum oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$. \begin{proof} [Proof of \expref{Theorem}{qosepthm}] Let $L$ be a unary language chosen uniformly at random. The oracle $U=\left\{ U_{n}\right\} _{n\geq1}$ is as follows: if $0^{n}\in L$, then $U_{n}\left\vert \psi_{n}\right\rangle =-\left\vert \psi_{n}\right\rangle $ for some $n$-qubit marked state $\left\vert \psi _{n}\right\rangle $ chosen uniformly at random, while $U_{n}\left\vert \varphi\right\rangle =\left\vert \varphi\right\rangle $ whenever $\left\langle \varphi|\psi_{n}\right\rangle =0$. Otherwise, if $0^{n}\notin L$, then $U_{n}$ is the $n$-qubit identity operation. Almost by definition, $L\in\mathsf{QMA}^{U}$. For given a quantum witness $\left\vert \varphi\right\rangle $, the $\mathsf{QMA}$ verifier first prepares the state $({1}/{\sqrt{2}}) \left( \left\vert 0\right\rangle \left\vert \varphi\right\rangle +\left\vert 1\right\rangle \left\vert \varphi\right\rangle \right) $, then applies $U_{n}$ to the second register conditioned on the first register being $\left\vert 1\right\rangle $. Next the verifier applies a Hadamard gate to the first register, measures it, and accepts if and only if $\left\vert 1\right\rangle $ is observed. If $0^{n}\in L$, then there exists a witness---namely $\left\vert \varphi \right\rangle =\left\vert \psi_{n}\right\rangle $---that causes the verifier to accept with probability $1$. On the other hand, if $0^{n}\notin L$, then \emph{no} witness causes the verifier to accept with nonzero probability. On the other hand, we claim that $L\notin\mathsf{QCMA}^{U}$ with probability $1$ over the choice of $L$ and $U$. This can be seen as follows. Fix a $\mathsf{QCMA}$ machine $M$, and let $S_{M}\left( n\right) $ be the event that $M^{U}$ \emph{succeeds} on $0^{n}$: that is, either $0^{n}\in L$ and there exists a string $w$ such that $M^{U}$ accepts $\left\vert 0^{n}\right\rangle \left\vert w\right\rangle $ with probability at least $2/3$, or $0^{n}\notin L$ and $M^{U}$ accepts $\left\vert 0^{n}\right\rangle \left\vert w\right\rangle $ with probability at most $1/3$ for all $w$. Then \expref{Theorem}{qosep} readily implies that there exists a positive integer $N$ such that for all $n\geq N$, \[ \Pr_{L,U}\left[ S_{M}\left( n\right) \mid S_{M}\left( 1\right) ,\ldots,S_{M}\left( n-1\right) \right] \leq\frac{2}{3}\,. \] Hence \[ \Pr_{L,U}\left[ S_{M}\left( 1\right) \land S_{M}\left( 2\right) \land\cdots\right] =0\,. \] Now, because of the Solovay-Kitaev Theorem~\cite{kitaev:ec}, the number of possible $\mathsf{QCMA}$ machines is only countably infinite. So by the union bound, \[ \Pr_{L,U}\left[ \exists M:S_{M}\left( 1\right) \land S_{M}\left( 2\right) \land\cdots\right] =0 \] as well. \end{proof} We can similarly show a quantum oracle separation between $\mathsf{BQP/qpoly}$ and $\mathsf{BQP/poly}$. \begin{theorem} \label{qpoly}There exists a quantum oracle $U$ such that $\mathsf{BQP}^{U}\mathsf{/qpoly}\neq\mathsf{BQP}^{U}\mathsf{/poly}$. \end{theorem} \begin{proof} In this case $U_{n}$ will act on $2n$ qubits. Let $L$ be a binary language chosen uniformly at random, and let $L\left( x\right) =1$ if $x\in L$ and $L\left( x\right) =0$ otherwise. Also, for all $n$, let $\left\vert \psi_{n}\right\rangle $ be an $n$-qubit state chosen uniformly at random. Then $U_{n}$ acts as follows: for all $x\in\left\{ 0,1\right\}^{n}$, \[ U_{n}\left( \left\vert \psi_{n}\right\rangle \left\vert x\right\rangle \right) =\left( -1\right) ^{L\left( x\right) }\left\vert \psi _{n}\right\rangle \left\vert x\right\rangle\, , \] but $U_{n}\left( \left\vert \phi\right\rangle \left\vert x\right\rangle \right) =\left\vert \phi\right\rangle \left\vert x\right\rangle $ whenever $\left\langle \phi|\psi_{n}\right\rangle =0$. Clearly $L\in\mathsf{BQP}^{U}\mathsf{/qpoly}$; we just take $\left\vert \psi_{n}\right\rangle $ as the advice. On the other hand, by essentially the same argument as for \expref{Theorem}{qosepthm}, one can show that $L\notin\mathsf{BQP}^{U}\mathsf{/poly}$ with probability $1$ over $L$ and $U$. \end{proof} \subsection{Proof of geometric lemma\label{LEMMA}} In this section we fill in the proof of \expref{Lemma}{sumip}, thereby completing the oracle separation theorems. In proving \expref{Lemma}{sumip}, the first step is to ask the following question: among all $p$-uniform probability measures $\sigma$, which is the one that maximizes $\operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\bigl[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\bigr]$? We can think of the set of quantum states $\mathbb{CP}^{N-1}$ as a container, which contains a fluid $\sigma$ that is gravitationally attracted to the state $\left\vert 0\right\rangle $. Then intuitively, the answer is clear: the way to maximize $\operatorname*{E}_{\left\vert \psi\right\rangle \in \sigma}\bigl[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\bigr]$ is to ``fill the container from the bottom,'' subject to the density constraint $p\sigma\leq\mu$. In other words, the optimal $\sigma$ should be the uniform measure over the region $\mathcal{R}\left( p\right) $ given by $\left\vert \left\langle \psi|0\right\rangle \right\vert \geq h\left( p\right) $, where $h\left( p\right) $ is chosen so that the volume of $\mathcal{R}\left( p\right) $ is a $p$ fraction of the total volume of $\mathbb{CP}^{N-1}$. The following lemma makes this intuition rigorous. \begin{lemma} \label{gravity}Among all $p$-uniform probability measures $\sigma$ over $\mathbb{CP}^{N-1}$, the one that maximizes \[ \operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\vert \left\langle \psi |0\right\rangle \right\vert ^{2}\right] \] is $\tau\left( p\right)$, the uniform measure over the region $\mathcal{R}\left( p\right) $ defined above. \end{lemma} \begin{proof} Since $\left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}$ is nonnegative, we can write \[ \operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\right] =\int_{0}^{\infty }\Pr_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] dy\,. \] We claim that setting $\sigma:=\tau\left( p\right) $ maximizes the integrand for every value of $y$. Certainly, then, setting $\sigma :=\tau\left( p\right) $ maximizes the integral itself as well. To prove the claim, we consider two cases. First, if $y\leq h\left( p\right) ^{2}$, then \[ \Pr_{\left\vert \psi\right\rangle \in\tau\left( p\right) }\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] =1\,, \] which is certainly maximal. Second, if $y>h\left( p\right) ^{2}$, then \[ \Pr_{\left\vert \psi\right\rangle \in\tau\left( p\right) }\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] =\frac{1} {p}\cdot\Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] \, . \] This is maximal as well, since \[ \Pr_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] \leq\frac{1}{p}\cdot \Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\geq y\right] \] for all $p$-uniform probability measures $\sigma$. \end{proof} \expref{Lemma}{gravity} completely describes the probability measure that maximizes $\operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\bigl[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\bigr]$, except for one detail: the value of $h\left( p\right) $ (or equivalently, the radius of $\mathcal{R}\left( p\right)$). The next lemma completes the picture. \begin{lemma} \label{hball}For all $p$, \[ h\left( p\right) =\sqrt{1-p^{1/\left( N-1\right) }}=\Theta\left( \sqrt{\frac{\log1/p}{N}}\right) \, . \] \end{lemma} \begin{proof} We will show that for all $h$, \[ \Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert \geq h\right] =\left( 1-h^{2}\right) ^{N-1}\,, \] where $\mu$ is the uniform probability measure over $\mathbb{CP}^{N-1}$. Setting $p:=\Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert \geq h\right] $ and solving for $h$ then yields the lemma. Let $\vec{z}=\left( z_{0},\ldots,z_{N-1}\right) $ be a complex vector; then let $\vec{r}=\left( r_{0},\ldots,r_{N-1}\right) $ and $\vec{\theta}=\left( \theta_{0},\ldots,\theta_{N-1}\right) $ be real vectors such that $z_{k}=r_{k}e^{i\theta_{k}}$ for each coordinate $k$. Also, let $\mathcal{D}$ be a Gaussian probability measure on $\mathbb{C}^{N}$, with density function \[ P\left( \vec{z}\right) =P\left( \vec{r}\right) =\frac{1}{\pi^{N}}e^{-\left\Vert \vec{r}\right\Vert _{2}^{2}}\,. \] Let $d\vec{r}$ be shorthand for $dr_{0}\cdots dr_{N-1}$. Then we can express the probability that $\left\vert \left\langle \psi|0\right\rangle \right\vert \geq h$ as \begin{align*} \Pr_{\left\vert \psi\right\rangle \in\mu}\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert \geq h\right] & =\Pr_{\vec{z} \in\mathcal{D}}\left[ \left\vert z_{0}\right\vert \geq h\left\Vert \vec{z}\right\Vert _{2}\right] \\ & =\Pr_{\vec{r},\vec{\theta}}\left[ r_{0}\geq h\left\Vert \vec{r}\right\Vert _{2}\right] \\ & =\int_{\vec{r},\vec{\theta}~:~r_{0}\geq h\left\Vert \vec{r}\right\Vert _{2}}P\left( \vec{r}\right) ~r_{0}\cdots r_{N-1}~d\vec{r}d\vec{\theta}\\ & =\left( 2\pi\right) ^{N}\int_{\vec{r}~:~r_{0}\geq h\left\Vert \vec{r}\right\Vert _{2}}\frac{1}{\pi^{N}}e^{-\left\Vert \vec{r}\right\Vert _{2}^{2}}~r_{0}\cdots r_{N-1}~d\vec {r}\\ & =\int_{r_{1},\ldots,r_{N-1}=0}^{\infty}\left( \int_{r_{0}=h\sqrt {\frac{r_{1}^{2}+\cdots+r_{N-1}^{2}}{1-h^{2}}}}^{\infty}2e^{-r_{0}^{2}} r_{0}dr_{0}\right) 2^{N-1}e^{-r_{1}^{2}-\cdots-r_{N-1}^{2}}~r_{1}dr_{1}\cdots r_{N-1}dr_{N-1}\\ & =\int_{r_{1},\ldots,r_{N-1}=0}^{\infty}e^{-\left( r_{1}^{2}+\cdots +r_{N-1}^{2}\right) \cdot h^{2}/\left( 1-h^{2}\right) }2^{N-1}e^{-r_{1}^{2}-\cdots-r_{N-1}^{2}}~r_{1}dr_{1}\cdots r_{N-1}dr_{N-1}\\ & =\int_{r_{1},\ldots,r_{N-1}=0}^{\infty}2^{N-1}e^{-\left( r_{1}^{2} +\cdots+r_{N-1}^{2}\right) /\left( 1-h^{2}\right) }~r_{1}dr_{1}\cdots r_{N-1}dr_{N-1}\\ & =\left( \int_{r=0}^{\infty}2e^{-r^{2}/\left( 1-h^{2}\right) }rdr\right) ^{N-1}\\ & =\left( 1-h^{2}\right) ^{N-1}. \end{align*} \end{proof} By combining Lemmas~\ref{gravity} and~\ref{hball}, we can now prove \expref{Lemma}{sumip}: that if $\sigma$ is $p$-uniform, then for all mixed states $\rho$, \[ \operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma}\left[ \left\langle \psi|\rho|\psi\right\rangle \right] =O\left( \frac{1+\log1/p}{N}\right) \, . \] \begin{proof} [Proof of \expref{Lemma}{sumip}]If $p\leq e^{-\Omega\left( N\right) }$ then the lemma is certainly true, so suppose $p\geq e^{-O\left( N\right) }$. Since the concluding inequality is linear in $\rho$, we can assume without loss of generality that $\rho$ is a pure state. Indeed, by symmetry we can assume that $\rho=\left\vert 0\right\rangle \left\langle 0\right\vert $. So our aim is to upper-bound $\operatorname*{E}_{\left\vert \psi\right\rangle \in\sigma }\bigl[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\bigr] $, where $\sigma$ is any $p$-uniform probability measure. By \expref{Lemma}{gravity}, we can assume without loss of generality that $\sigma =\tau\left( p\right) $ is the uniform measure over all $\left\vert \psi\right\rangle $ such that $\left\vert \left\langle \psi|0\right\rangle \right\vert \geq h\left( p\right) $. Then letting \begin{align*} \left\vert \psi\right\rangle & =\alpha_{0}\left\vert 0\right\rangle +\cdots+\alpha_{N-1}\left\vert N-1\right\rangle ,\\ r & =\sqrt{\left\vert \alpha_{1}\right\vert ^{2}+\cdots+\left\vert \alpha_{N-1}\right\vert ^{2}}, \end{align*} we have \begin{align*} \operatorname*{E}_{\left\vert \psi\right\rangle \in\tau\left( p\right) }\left[ \left\vert \left\langle \psi|0\right\rangle \right\vert ^{2}\right] & =\operatorname*{E}_{\left\vert \psi\right\rangle ~:~\left\vert \alpha _{0}\right\vert \geq h\left( p\right) }\left[ \left\vert \alpha _{0}\right\vert ^{2}\right] \\ & =\operatorname*{E}_{\left\vert \psi\right\rangle ~:~r^{2}\leq1-h\left( p\right) ^{2}}\left[ 1-r^{2}\right] \\ & =\frac{\int_{0}^{\sqrt{1-h\left( p\right) ^{2}}}r^{2N-3}\left( 1-r^{2}\right) dr}{\int_{0}^{\sqrt{1-h\left( p\right) ^{2}}}r^{2N-3}dr}\\ & =\frac{\left[ \frac{r^{2N-2}}{2N-2}-\frac{r^{2N}}{2N}\right] _{0}^{\sqrt{1-h\left( p\right) ^{2}}}}{\left[ \frac{r^{2N}}{2N}\right] _{0}^{\sqrt{1-h\left( p\right) ^{2}}}}\\ & =\frac{1-\left( 1-\frac{1}{N}\right) \left( 1-h\left( p\right) ^{2}\right) }{\left( 1-\frac{1}{N}\right) \left( 1-h\left( p\right) ^{2}\right) }\\ & =O\left( \frac{1}{N}+h\left( p\right) ^{2}\right) \\ & =O\left( \frac{1+\log1/p}{N}\right) , \end{align*} where the last line follows from \expref{Lemma}{hball}. \end{proof} \section{Upper bound\label{UPPER}} In this section we show that the lower bound of \expref{Theorem}{qosep} is basically tight. In particular, let $U$ be an $n$-qubit quantum oracle, and suppose we are given an $m$-bit classical proof that $U$ is not the identity, but instead conceals a marked state $\left\vert \psi\right\rangle $ such that $U\left\vert \psi\right\rangle =-\left\vert \psi\right\rangle $. Then provided $2n\leq m\leq2^{n}$, a quantum algorithm can verify the proof by making $O\bigl( \sqrt{2^{n}/m}\bigr) $ oracle calls to $U$. This matches our lower bound when $m\geq2n$.\footnote{When $m\ll2n$, the best upper bound we know is the trivial $O\left( \sqrt{2^{n}}\right)$. However, we conjecture that $O\left( \sqrt{2^{n}/m}\right) $ is achievable in this case as well.} Let $N=2^{n}$ be the dimension of $U$'s Hilbert space. Then the idea of our algorithm is to use a ``mesh'' of states $\left\vert \phi_{1}\right\rangle ,\ldots,\left\vert \phi_{M}\right\rangle \in\mathbb{CP}^{N-1}$, at least one of which has nontrivial overlap with every pure state in $\mathbb{CP}^{N-1}$. A classical proof can then help the algorithm by telling it the $\left\vert \phi_{i}\right\rangle $ that is closest to $\left\vert \psi\right\rangle $. More formally, define the $h$\emph{-ball} about $\left\vert \phi\right\rangle $ to be the set of $\left\vert \varphi\right\rangle $ such that $\left\vert \left\langle \phi|\varphi\right\rangle \right\vert \geq h$. Then define an $h$ \emph{net for} $\mathbb{CP}^{N-1}$ \emph{of size} $M$ to be a set of states $\left\vert \phi_{1}\right\rangle ,\ldots,\left\vert \phi _{M}\right\rangle $ such that every $\left\vert \psi\right\rangle \in\mathbb{CP}^{N-1}$ is contained in the $h$-ball about $\left\vert \phi _{i}\right\rangle $ for some $i$.\footnote{These objects are often called $\varepsilon$-nets, with the obvious relation $h=\cos\varepsilon$.} We will use the following theorem, which follows from Corollary 1.2 of B\"{o}r\"{o}czky and Wintsche~\cite{boroczky}. \begin{theorem} [\cite{boroczky}]\label{hnet}For all $0N$. Then \[ s_{1}+\cdots+s_{L}=x_{1}^{2}+\cdots+x_{N}^{2}=1\,, \] so certainly there exists an $i\in\left\{ 1,\ldots,L\right\} $ such that $s_{i}\geq1/L$. Fix that $i$. Then since the $x_{j}$'s are arranged in nonincreasing order, we have \[ x_{2^{i-1}}\geq\sqrt{\frac{s_{i}}{2^{i-1}}}\geq\sqrt{\frac{1}{2^{i-1}L}}\,. \] There are now two cases. First, if $k\leq2^{i-1}$ then \[ \max_{1\leq t\leq k}\left[ \frac{x_{1}+\cdots+x_{t}}{\sqrt{t}}\right] \geq\frac{x_{1}+\cdots+x_{k}}{\sqrt{k}}\geq\frac{k}{\sqrt{k}}x_{2^{i-1}} \geq\sqrt{\frac{k}{2^{i-1}L}}\geq\sqrt{\frac{k}{N\left\lceil \log _{2}N\right\rceil }}\,. \] Second, if $2^{i-1}\leq k$ then \[ \max_{1\leq t\leq k}\left[ \frac{x_{1}+\cdots+x_{t}}{\sqrt{t}}\right] \geq\frac{x_{1}+\cdots+x_{2^{i-1}}}{\sqrt{2^{i-1}}}\geq\frac{2^{i-1}}{\sqrt{2^{i-1}}}x_{2^{i-1}}\geq\sqrt{\frac{1}{L}}\geq\sqrt{\frac {k}{N\left\lceil \log_{2}N\right\rceil }}\,. \] This completes the proof.\footnote{One might wonder whether the $\sqrt {1/\left\lceil \log_{2}N\right\rceil }$ factor can be eliminated. However, a simple example shows that it can be improved by at most a constant factor. Suppose $x_{j}:=\sqrt{\frac{1}{jw}}$, where $w=\sum_{j=1}^{n}\frac{1}{j}\approx\ln N$. Then for all $t\in\left\{ 1,\ldots,N\right\} $, we have \[ \frac{x_{1}+\cdots+x_{t}}{\sqrt{t}}\approx\frac{2}{\sqrt{\ln N}}\,. \] } \end{proof} We now use \expref{Lemma}{ineq} to construct an $h$-net. \begin{theorem} \label{explicit}For all $00$, then $f=g$. \end{proposition} \begin{proof} Fix $x\in\Gamma$; then for all $y\in\Gamma$, we have $f\left( x\right) =f\left( y\right) f\left( y^{-1}x\right) $ and $g\left( x\right) =g\left( y\right) g\left( y^{-1}x\right) $. By the union bound, \[ \Pr_{y\in\Gamma}\left[ f\left( y\right) =g\left( y\right) \land f\left( y^{-1}x\right) =g\left( y^{-1}x\right) \right] \geq1-\Pr_{y\in\Gamma }\left[ f\left( y\right) \neq g\left( y\right) \right] -\Pr_{y\in\Gamma }\left[ f\left( y^{-1}x\right) \neq g\left( y^{-1}x\right) \right] >0\,. \] Hence there exists a $y$ such that $f\left( y\right) =g\left( y\right) $ and $f\left( y^{-1}x\right) =g\left( y^{-1}x\right) $. But this implies that $f\left( x\right) =g\left( x\right) $. \end{proof} In particular, \expref{Proposition}{homclose} implies that if a function $f$ is $1/5$-close to a homomorphism, then it is $1/5$-close to a \emph{unique} homomorphism ($1/5$ being an arbitrary constant less than $1/4$). \begin{proposition} [Ben-Or et al.~\cite{bclr}]\label{bclrprop}Given finite groups $\Gamma$ and $G$, a function $f:\Gamma\to G$, and a real number $\varepsilon>0$, if \[ \Pr_{x,y\in\Gamma}\left[ f\left( xy\right) \neq f\left( x\right) f\left( y\right) \right] \leq\varepsilon \] then $f$ is $\varepsilon$-close to a homomorphism. \end{proposition} Together, Propositions~\ref{homclose} and~\ref{bclrprop} have the following easy corollary. \begin{corollary} \label{homtest}There is a randomized algorithm which, given finite groups $\Gamma$ and $G$ and a function $f:\Gamma\to G$ as input, makes $O\left(1\right)$ oracle queries to $f$, accepts with probability $1$ if $f$ is a homomorphism, and rejects with probability at least $2/3$ if $f$ is not $1/5$-close to a homomorphism. Also, if $f$ is $1/5$-close to some homomorphism $\widetilde{f}$, then there exists a randomized algorithm that, given an input $x\in\Gamma$, makes $O\left(r\right)$ oracle queries to $f$, and outputs $\widetilde{f}\left(x\right)$ with probability at least $1-1/2^{r}$. \end{corollary} In the present context, our algorithms are not limited in space or time, and we can say for simplicity that $\Gamma$ is represented by its entire multiplication table. It is then easy, as the proof will require, to pick elements of $\Gamma$ uniformly at random. By contrast, $G$ is represented by oracle access, but there will be no need to choose its elements at random. %% But note that even if access %% to $\Gamma$ were restricted by an oracle model, it is still possible %% to pick elements at random using an algorithm of Babai %% \cite{babai:walk}; this is relevant to the discussion in %% \expref{Section}{COMP}. \begin{proof} The first algorithm simply chooses $O\left( 1\right) $ pairs $x,y\in\Gamma $ uniformly at random, accepts if $f\left( xy\right) =f\left( x\right) f\left( y\right) $ for all of them, and rejects otherwise. Let $k=O\left( r\right) $. Then the second algorithm chooses $z_{1}, \ldots,z_{k}\in\Gamma$ uniformly at random, and outputs the plurality answer among \[ f\left( z_{1}\right) f\left( z_{1}^{-1}x\right) ,\ldots,f\left( z_{k}\right) f\left( z_{k}^{-1}x\right)\, \] breaking ties arbitrarily. \end{proof} It follows from the Classification of Finite Simple Groups that there are at most two finite simple groups of any particular order (see~\cite{atlas} for example). The following well-known result is a combination of that fact and of a theorem due to Neumann~\cite{neumann}. \begin{theorem} \label{iso}There are $N^{O\left(\left(\log_{2}N\right)^2\right)}$ groups of order $N$ up to isomorphism.\footnote{The most accurate asymptotic result on the number of groups of order $N$, in terms of the prime factorization of $N$, appears in a paper by Pyber~\cite{pyber}.} \end{theorem} Finally, recall that the Hidden Subgroup Problem ($\operatorname*{HSP}$) is defined as follows. We are given a finite group $G$, and oracle access to a function $f:G\to\mathbb{Z}$. We are promised that there exists a ``hidden subgroup'' $H\leq G$ such that $f\left( x\right) =f\left( y\right) $ if and only if $x$ and $y$ belong to the same left coset of $H$. The problem is then to output a set of generators for $H$. Whether $\operatorname*{HSP}$ can be solved in quantum polynomial time, for various non-abelian groups $G$, is one of the most actively studied questions in quantum computing. However, if we only care about query complexity, then Ettinger, H\o yer, and Knill~\cite{ehk} proved the following useful result. \begin{theorem} [\cite{ehk}]\label{ehkthm}There is a quantum algorithm such that, given any finite group $G$ as oracular input, solves $\operatorname*{HSP}$ using only $\operatorname*{polylog}\left( \left\vert G\right\vert \right) $ quantum queries to $f$ (together with a possibly exponential amount of postprocessing).\footnote{Indeed, for \emph{Normal} $\operatorname*{HSP}$ (which is the special case we care about), Hallgren, Russell, and Ta-Shma~\cite{hrt} improved this result, showing how to find a hidden subgroup using only $O\left(\log\left\vert G\right\vert\right)$ queries to $f$ (again, with exponential postprocessing).} \end{theorem} We can now prove \expref{Theorem}{hspthm}: that $\operatorname*{GNM}$ has polynomially-bounded $\mathsf{QCMA}$ query complexity. \begin{proof} [Proof of \expref{Theorem}{hspthm}]Let $G$ be a group of order at most $2^{n}$, and let $\mathcal{O}$ be a group oracle that maps each element of $G$ to an $n$-bit label. Also, given (the labels of) group elements $x,h_{1}, \ldots,h_{m}\in G$, let $H$ be the subgroup of $G$ generated by $\left\langle h_{1},\ldots,h_{m}\right\rangle $. Then the problem is to decide if $x\notin H$. In our $\mathsf{QCMA}$ protocol for this problem, Merlin's witness will consist of the following: \begin{itemize} \item An explicit ``model group'' $\Gamma$, of order at most $2^{n}$. \item A list of elements $\gamma_{1},\ldots,\gamma_{k}\in\Gamma$, where $k=O\left(\log\left\vert \Gamma\right\vert \right)$. \item A corresponding list $g_{1},\ldots,g_{k}\in G$. \item Another list $z,\lambda_{1},\ldots,\lambda_{m}\in\Gamma$. \end{itemize} We should be more explicit about the notion of an ``explicit'' group $\Gamma$, and about the syntax of this witness. By \expref{Theorem}{iso}, there are at most $2^{\operatorname*{poly}\left( n\right)}$ groups of order $\left\vert \Gamma\right\vert \leq2^{n}$ up to isomorphism. Since Arthur is allowed unlimited computation and is only restricted in queries, he can construct a full multiplication table for $\Gamma$ using only the name of its isomorphism type. The multiplication table is not unique, because the elements of $\Gamma$ can be permuted; but for instance Arthur could construct the lexicographically first such table. Since Merlin can anticipate Arthur's construction of $\Gamma$, he can then refer to elements of $\Gamma$ using the same construction. He can also refer to elements of $G$ since he understands the oracle. In conclusion, Merlin can specify the witness using only $\operatorname*{poly}\left(n\right)$ bits. If Merlin is honest, then the witness will have the following three properties: \begin{enumerate} \item[(1)] $\gamma_{1},\ldots,\gamma_{k}$ is an efficient generating set for $\Gamma$. \item[(2)] $z\notin\Lambda$, where $\Lambda$ is the subgroup of $\Gamma $ generated by $\left\langle \lambda_{1},\ldots,\lambda_{m}\right\rangle $. \item[(3)] There exists an embedding $\widetilde{f}:\Gamma\to G$, such that (i) $\widetilde{f}\left( \gamma_{i}\right) =g_{i}$ for all $i\in\left\{ 1,\ldots,k\right\} $, (ii) $\widetilde{f}\left( \lambda _{j}\right) =h_{j}$ for all $j\in\left\{ 1,\ldots,m\right\} $, and (iii) $\widetilde{f}\left( z\right) =x$. \end{enumerate} Suppose for the moment that (1)-(3) all hold. Then there exists an embedding $\widetilde{f}:\Gamma\to G$, which maps the set $\left\langle \gamma_{1},\ldots,\gamma_{k}\right\rangle $ in $\Gamma$ to the set $\left\langle g_{1},\ldots,g_{k}\right\rangle $ in $G$. Furthermore, this embedding satisfies $\widetilde{f}\left( \Lambda\right) =H$ and $\widetilde{f}\left( z\right) =x$. Since $z\notin\Lambda$ by (2), it follows that $x\notin H$ as well, which is what Arthur wanted to check. So it suffices to verify (1)-(3). In the remainder of the proof, we will explain how to do this using a possibly exponential amount of computation, but only $\operatorname*{poly}\left( n\right) $ quantum queries to the group oracle $\mathcal{O}$. First, since properties (1) and (2) only involve the explicit group $\Gamma$, not the black-box group $G$, Arthur can verify these properties ``free of cost.'' In other words, regardless of how much computation he needs, he never has to query the group oracle. The nontrivial part is to verify (3). It will be convenient to split (3) into the following sub-claims: \begin{enumerate} \item[(3a)] There exists a homomorphism $\widetilde{f}:\Gamma\to G$ such that $\widetilde{f}\left( \gamma_{i}\right) =g_{i}$ for all $i\in\left\{ 1,\ldots,k\right\} $. \item[(3b)] $\widetilde{f}$ satisfies $\widetilde{f}\left( z\right) =x$ and $\widetilde{f}\left( \lambda_{j}\right) =h_{j}$ for all $j\in\left\{ 1,\ldots,m\right\} $. \item[(3c)] $\widetilde{f}$ is injective (\ie\ is an embedding into $G$). \end{enumerate} To verify (3a), first Arthur fixes a ``canonical representation'' of each element $\gamma\in\Gamma$. This representation has the form \[ \gamma=\gamma_{1}^{e_{1}}\cdots\gamma_{k}^{e_{k}}\,, \] where $\left\langle \gamma_{1},\ldots,\gamma_{k}\right\rangle $ is the efficient generating set for $\Gamma$, and $e_{1},\ldots,e_{k}\in\left\{ 0,1\right\} $ are bits depending on $\gamma$. Next he defines a function $f:\Gamma\to G$ by \[ f\left( \gamma\right) :=g_{1}^{e_{1}}\cdots g_{k}^{e_{k}} \] for all $\gamma\in\Gamma$. By using the canonical representation of $\gamma $, Arthur can evaluate $f\left( \gamma\right) $ using at most $k-1$ queries to the group oracle $\mathcal{O}$. Finally Arthur appeals to \expref{Corollary}{homtest}. If $f$ is not $1/5$-close to a homomorphism, then by using $O\left( 1\right) $ queries to $f$, with high probability Arthur can detect that $f$ is not a homomorphism. In that case Merlin has been caught cheating, so Arthur rejects. On the other hand, if $f$ is $1/5$-close to some homomorphism $\widetilde{f}$, then by using $O\left( \log\left\vert \Gamma\right\vert \right) $ queries to $f$, with high probability Arthur can ``correct'' $f$ to $\widetilde{f}$. In that case it remains only to check that $\widetilde{f}\left( \gamma_{i}\right) =g_{i}$ for all $i\in\left\{ 1,\ldots,k\right\} $. Once Arthur has an efficient procedure for computing $\widetilde{f}$---that is, a procedure that involves only $\operatorname*{poly}\left( n\right) $ queries to $\mathcal{O}$---he can then verify property (3b) directly. To verify (3c), Arthur runs the algorithm of Ettinger, H\o yer, and Knill~\cite{ehk} for the Hidden Subgroup Problem. Notice that, since $\widetilde{f}:\Gamma\to G$ is a homomorphism, there must be a ``hidden subgroup'' $K\leq\Gamma$---namely the kernel of $\widetilde{f}$---such that $\widetilde{f}$ is constant on cosets of $K$ and distinct on distinct cosets. Furthermore, $\widetilde{f}$ is injective if and only if $K$ is trivial. But deciding whether $K$ is trivial is just an instance of $\operatorname*{HSP}$, and can therefore be solved using $\operatorname*{poly}\left( n\right)$ quantum queries by \expref{Theorem}{ehkthm}. \end{proof} \subsection{Computational complexity\label{COMP}} \expref{Theorem}{hspthm} showed that one can always verify group non-membership using a polynomial-size classical witness, together with polynomially many quantum queries to the group oracle $\mathcal{O}$. Unfortunately, while the \emph{query} complexity is polynomial, the \emph{computational} complexity might be exponential. However, as mentioned in \expref{Section}{RESULTS}, we conjecture that this shortcoming of \expref{Theorem}{hspthm} can be removed, and that $\operatorname*{GNM}$ is in $\mathsf{QCMA}$ for any group oracle $\mathcal{O}$. In our $\mathsf{QCMA}$ protocol, the main computational problem that needs to be solved is not the general $\operatorname*{HSP}$, but rather the Normal Hidden Subgroup Problem ($\operatorname*{NHSP}$)---that is, $\operatorname*{HSP}$\ where the hidden subgroup is normal. This is because the kernel of a homomorphism is always a normal subgroup. Hallgren, Russell, and Ta-Shma~\cite{hrt} showed that $\operatorname*{NHSP}$ is in $\mathsf{BQP}$ for an explicit group $\Gamma$, provided that the quantum Fourier transform over\ $\Gamma$ can be implemented efficiently (and its output can be interpreted). \ Furthermore, Moore, Rockmore, and Russell~\cite{mrr} showed that many classes of finite groups $G$ have an explicit model $\Gamma\cong G$ for which this assumption holds. Even if $\operatorname*{NHSP}$ is in $\mathsf{BQP}$, there are two remaining obstacles to showing that $\operatorname*{GNM}$ is in $\mathsf{QCMA}$. First, we need to be able to verify group non-membership in the explicit model group $\Gamma$, possibly with the help of additional classical information from Merlin. Second, we need an efficient algorithm to compute the function $\widetilde{f}:\Gamma\to G$ for every $\gamma\in\Gamma$, even though $\widetilde{f}$ is explicitly defined only on the generators $\gamma_1,\ldots,\gamma_k$. %%******** new (Sep 10) part of Sec 5.1 begin In the context of computational complexity (as opposed to query complexity), the notion of an ``explicit group'' needs to be better explained. Keeping in mind that this entire section is only one possible path to showing that $\operatorname*{GNM}$ is in $\mathsf{QCMA}$, here is one definition that captures the ideas of previous sections. \begin{definition} A (polylog-time) \emph{explicit sequence} of finite groups is a sequence $\Gamma_n$ such that each term is a group law on the set $\{1,\ldots,|\Gamma_n|\}$. Moreover the multiplication function $m_n(x,y) = xy$ and the inversion function $i_n(x) = x^{-1}$ can both be computed in polynomial time in $\log |\Gamma_n|$. An explicit sequence is \emph{universal} if every finite group is isomorphic to at least one $\Gamma_n$. \end{definition} For example, the symmetric group (sequence) $S_n$ is explicit, because the standard notation for permutations can be compressed to the integers from 1 to $n!$. Likewise the matrix groups $\mathrm{GL}(n,q)$ form an explicit sequence in the joint parameter $(n,a(x))$, where $a(x)$ is a polynomial whose splitting field is $\mathbb{F}_q$. But there is no reason to believe that an explicit model of a group is unique up to polylog-time bijections. On the contrary, if the discrete logarithm problem is hard, then $(\mathbb{F}_q)^\times$ and $\mathbb{Z}/(q-1)$ are inequivalent explicit models for isomorphic groups; and both models appear naturally in the obvious explicit model for matrix groups. It is not known whether there is a universal explicit sequence of finite groups. The current best result for solvable groups is quasipolylogarithmic time \cite{hofling}. Theorem~\ref{iso} implies that the number of isomorphism classes of finite groups does not by itself preclude a universal explicit sequence. %% guarantees that the sheer number of %% isomorphism classes of finite groups is not by itself an obstruction to %% explicit computability. If there is a universal sequence of explicit finite groups with the following additional properties, then following the methods of the previous section, it would show that $\operatorname*{GNM}$ is in $\mathsf{QCMA}$. (We drop the formal subscript $n$.) % (This may be essentially established; see Babai \cite{babai:walk}.) \begin{enumerate} \item[(i)] Each $\Gamma$ has a list of generators $\gamma_1,\ldots,\gamma_k\in\Gamma$ that can be computed in $O(\operatorname*{polylog}|\Gamma|)$ time. Moreover, given an element $\gamma\in\Gamma$, there is a polylog algorithm to express it as a (polylogarithmic length) product of $\gamma_1,\ldots,\gamma_k$. It suffices if this algorithm is polylog time on average for random $\gamma$. A straight-line program rather than a product also suffices. \item[(ii)] $\operatorname*{NHSP}$ over $\Gamma$\ is in $\mathsf{BQP}$. \item[(iii)] $\operatorname*{GNM}$ over $\Gamma$ is in $\mathsf{QCMA}$. \end{enumerate} If the Hallgren-Russell-Ta-Shma algorithm is used for condition (ii), %% (iii), then there should be an algorithm for the quantum Fourier transform over $\Gamma$. In this case the QFT produces randomly-chosen characters of the quotient group $\Gamma/\Lambda$ for some normal subgroup $\Lambda$. The characters must also be listed in some explicit form so that so that $\Lambda$ can be recognized in polylog time, or at least that the triviality of $\Lambda$ can be so recognized. (Here, too, ``explicit'' means that the characters of $\Gamma$ are numbered consecutively and that the relevant algorithms use this numbering.) If $\Gamma$ is the symmetric group $S_n$, or an abelian group expressed as a product of cyclic groups, or if it is a matrix group $\mathrm{GL}(n,q)$, then there is an easy generating set that satisfies (i) %% and (ii) (exercise for the reader). In the abelian case, $\operatorname*{NHSP}$ is in $\mathsf{BQP}$ by the work of Shor \cite{shor}\ and Kitaev \cite{kitaev:meas}; $\operatorname*{GNM}$ is in $\mathsf{P}$ by linear algebra. If $\Gamma = S_n$, then $\operatorname*{NHSP}$ is trivial (since the only normal subgroup is $A_n$) and $\operatorname*{GNM}$\ is in $\mathsf{P}$\ by the work of Sims \cite{sims}. Meanwhile Babai and Szemer\'{e}di \cite{bs:matrix} showed that if every finite simple group has an explicit polylogarithmic presentation, then $\operatorname*{GNM}$ is in $\mathsf{NP}$ for $\mathrm{GL}(n,q)$. Since the point of condition (ii) is to allow Arthur to confirm Merlin's claimed homomorphism from $\Gamma$ to $G$, a polylogarithmic presentation of $\Gamma$ would yield an alternative method that does not rely on the algorithm of Corollary \ref{homtest}. The status of this problem is that the only unknown case among finite simple groups is %%the %%LB ``the'' added Ree groups of type ${}^2G_2(q)$; all other finite simple groups are known to have such a presentation \cite{bgklp,hulpke-seress}. Moreover, it is known that short presentations of a sequence $\{\Gamma_1,\ldots,\Gamma_k\}$ of finite simple groups can be combined to obtain a short presentation for any finite $\Gamma$ composed of $\{\Gamma_1,\ldots,\Gamma_k\}$ \cite{bgklp}. However, there is no known polylog-time algorithm to generate an explicit presentation of each such $\Gamma$, even given a presentation of each simple $\Gamma_j$ as input. In summary, the groups ${}^2G_2(q)$ and the extension problem are the remaining obstructions to a universal, explicit sequence of polylog presentations of finite groups, which would provide a simple alternative to condition (ii). Regardless, all known QFT algorithms employ flags of subgroups, which are structures that can also be used to satisfy condition (ii). Obviously the entire program is far from complete, and each step is open to variations. But we optimistically conjecture that all steps can be completed for arbitrary finite groups. \section{Mimicking random quantum oracles\label{SQO}} We have seen, on the one hand, that there exists a quantum oracle separating $\mathsf{QMA}$ from $\mathsf{QCMA}$; and on the other hand, that separating these classes by a \emph{classical} oracle seems much more difficult. Together, these results raise a general question: how much ``stronger'' are quantum oracles than classical ones? In particular, are there complexity classes $\mathcal{C}$ and $\mathcal{D}$ that can be separated by quantum oracles, but such that separating them by classical oracles is almost as hard as separating them in the unrelativized world? Whatever the answer, we conjecture that $\mathsf{QMA}$ and $\mathsf{QCMA}$ are \emph{not} examples of such classes. The reason is that it seems possible, using only classical oracles, to approximate quantum oracles similar to ones that would separate $\mathsf{QMA}$ from $\mathsf{QCMA}$. To illustrate, let $\sigma$ be the uniform probability measure over $2^{n}\times2^{n}$ unitary diagonal matrices. (In other words, each diagonal entry of $D\in\sigma$ is a random complex number with norm $1$.) Also, let $H^{\otimes n}$ be a tensor product of $n$ Hadamard matrices. Then let $\varsigma_{k}$ be the probability measure over $2^{n}\times2^{n}$ unitary matrices \[ U=D_{k}H^{\otimes n}D_{k-1}H^{\otimes n}\cdots H^{\otimes n}D_{1}H^{\otimes n} \] induced by drawing each $D_{i}$ independently from $\sigma$. In other words, $U\in\varsigma_{k}$ is obtained by first applying a Hadamard gate to each qubit, then a random $2^{n}\times2^{n}$ diagonal matrix, then Hadamard gates again, then another random diagonal matrix, and so on $k$ times. Note that we can efficiently apply such a $U$---at least to polynomially many bits of precision---if given a classical random oracle $A$. To do so, we simply implement the random diagonal matrix $D_{i}$ as \[ \sum_{x\in\left\{ 0,1\right\} ^{n}}\alpha_{x}\left\vert x\right\rangle \to\sum_{x\in\left\{ 0,1\right\} ^{n}}\omega^{A\left( i,x\right) }\alpha_{x}\left\vert x\right\rangle \,, \] where $A\left( i,x\right) $ is a uniformly random $n$-bit integer indexed by $i$ and $x$, and $\omega=e^{2\pi i/2^{n}}$. Now let $\mu$ be the uniform probability measure over $2^{n}\times2^{n}$ unitary matrices. If $k\ll2^{n}$, then $\varsigma_{k}$ is \emph{not} close to $\mu$ in variation distance, since the former has only $\Theta\left( k2^{n}\right) $ degrees of freedom while the latter has $\Theta\left( k4^{n}\right) $.\footnote{Admittedly, it is still conceivable that the finite-precision version of $\varsigma_{k}$ is close in variation distance to the finite-precision version of $\mu$. However, a more sophisticated argument that counts distinguishable unitaries rules out that possibility as well.} On the other hand, we conjecture that a $U$ drawn from $\varsigma_{k}$ will ``look random'' to any polynomial-time algorithm, and that this property can be used to prove a classical oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$. Let us explain what we mean in more detail. Suppose we are given access to an $n$-qubit unitary oracle $U$, and want to decide whether \begin{enumerate} \item[(i)] $U$ was drawn uniformly at random (that is, from $\mu$), or \item[(ii)] $U$ was drawn uniformly at random conditioned on there existing $n/2$-qubit pure states $\left\vert \psi\right\rangle $ and $\left\vert \varphi\right\rangle $ such that $U\left( \left\vert 0\right\rangle ^{\otimes n/2}\left\vert \psi\right\rangle \right) \approx\left\vert 0\right\rangle ^{\otimes n/2}\left\vert \varphi\right\rangle $. \end{enumerate} In case (i), the states $\left\vert \psi\right\rangle $ and $\left\vert \varphi\right\rangle $ will exist only with negligible probability.\footnote{Indeed, the reason we did not ask for $\left(n-1\right)$-qubit states $\left\vert \psi\right\rangle$ and $\left\vert \varphi\right\rangle$ such that $U\left( \left\vert 0\right\rangle \left\vert \psi\right\rangle \right) \approx\left\vert 0\right\rangle \left\vert \varphi\right\rangle$ is that such states will exist (almost) generically. For the choice of $\left\vert \psi\right\rangle$ gives us $2^{n-1}-1$ independent complex variables, whereas the requirement that $U\left( \left\vert 0\right\rangle \left\vert \psi\right\rangle \right)$ have the form $\left\vert 0\right\rangle \left\vert \varphi\right\rangle$ imposes only $2^{n-1}$ constraints. Asking for $\left( n-2\right)$-qubit states $\left\vert \psi\right\rangle$ and $\left\vert \varphi \right\rangle$ such that $U\left( \left\vert 00\right\rangle \left\vert \psi\right\rangle \right) \approx\left\vert 00\right\rangle \left\vert \varphi\right\rangle$ might suffice (since now we have $2^{n-2}-1$ variables versus $3\cdot2^{n-2}$ constraints), but we wish to stay on the safe side.} It follows that the above problem is in $\mathsf{QMA}^{U}$---since if case (ii) holds, then a succinct quantum proof of that fact is just $\left\vert \psi\right\rangle$ itself. We now state three conjectures about this problem, in increasing order of difficulty. \begin{conjecture} \label{notinqcma}The above problem is not in $\mathsf{QCMA}^{U}$. In other words, if case (ii) holds, there is no succinct classical proof of that fact that can be verified with high probability using $\operatorname*{poly}\left( n\right) $ quantum queries to $U$. \end{conjecture} Presumably \expref{Conjecture}{notinqcma} can be proved using ideas similar to those in \expref{Section}{QORACLE}. If so, then the next step is to replace the uniform measure $\mu$ by the ``pseudorandom'' measure $\varsigma_{k}$. \begin{conjecture} \label{secondconj}Suppose that instead of being drawn from $\mu$, the unitary $U$ is drawn from $\varsigma_{k}$ for some $k=\Omega\left( n\right) $. Then the probability that there exist $n/2$-qubit states $\left\vert \psi\right\rangle $ and $\left\vert \varphi\right\rangle $ such that \[ U\left( \left\vert 0\right\rangle ^{\otimes n/2}\left\vert \psi\right\rangle \right) \approx\left\vert 0\right\rangle ^{\otimes n/2}\left\vert \varphi\right\rangle \] is still negligibly small. \end{conjecture} Now suppose we want to decide whether \begin{enumerate} \item[(i')] $U$ was drawn from $\varsigma_{k}$, or \item[(ii')] $U$ was drawn from $\varsigma_{k}$ conditioned on there existing $n/2$-qubit states $\left\vert \psi\right\rangle $ and $\left\vert \varphi\right\rangle $ such that \[ U\left( \left\vert 0\right\rangle ^{\otimes n/2}\left\vert \psi\right\rangle \right) \approx\left\vert 0\right\rangle ^{\otimes n/2}\left\vert \varphi\right\rangle\,. \] \end{enumerate} Also, let $A$ be a classical oracle that encodes the diagonal matrices $D_{1},\ldots,D_{k}$ such that \[ U=D_{k}H^{\otimes n}D_{k-1}H^{\otimes n}\cdots H^{\otimes n}D_{1}H^{\otimes n}\,. \] If \expref{Conjecture}{secondconj} is true, then case (ii') can be verified in $\mathsf{QMA}^{A}$. So to obtain a classical oracle separation between $\mathsf{QMA}$ and $\mathsf{QCMA}$, the one remaining step would be to prove the following. \begin{conjecture} \label{thirdconj}Case (ii') cannot be verified in $\mathsf{QCMA}^{A}$. \end{conjecture} \subsection{From random oracles to random unitaries\label{RORU}} The previous discussion immediately suggests even simpler questions about the ability of classical oracles to mimic quantum ones. In particular, could a $\mathsf{BQP}$ machine use a classical random oracle to prepare a uniformly random $n$-qubit pure state? Also, could it use such an oracle to apply a random $n$-qubit unitary? In this section we answer the first question in the affirmative, and present partial results about the second question. We first need a notion that we call the ``$\varepsilon$-smoothing'' of a probability measure. \begin{definition} \label{esmooth}Let $\sigma$ be a probability measure over $\left\vert \psi\right\rangle \in\mathbb{CP}^{2^{n}-1}$. Then the $\varepsilon $-smoothing of $\sigma$, or $\mathcal{S}_{\varepsilon}\left( \sigma\right) $, is the probability measure obtained by first drawing a state $\left\vert \psi\right\rangle $ from $\sigma$, and then drawing a state $\left\vert \varphi\right\rangle $ uniformly at random subject to $\left\langle \varphi|\psi\right\rangle \geq1-\varepsilon$. \end{definition} Let $\mu$ be the uniform measure over $\mathbb{CP}^{2^{n}-1}$. Also, let $Q$ be a quantum algorithm that queries a classical oracle $A$. Suppose that, given $0^{n}$ as input, $Q^{A}$ outputs the pure state $\left\vert \psi_{A}\right\rangle \in\mathbb{CP}^{2^{n}-1}$. Then we say that $Q$ ``approximates the uniform measure within $\varepsilon$'' if, as we range over uniform random $A\subseteq\left\{ 0,1\right\} ^{n}$, the induced probability measure $\sigma$ over $\left\vert \psi_{A}\right\rangle $ satisfies $\left\Vert \mathcal{S}_{\varepsilon }\left( \sigma\right) -\mu\right\Vert \leq\varepsilon$. \begin{theorem} \label{randstate}For all polynomials $p$, there exists a quantum algorithm $Q$ that runs in polynomial time, and that approximates the uniform measure within $2^{-p\left( n\right) }$. \end{theorem} \begin{proof} [Proof Sketch]The algorithm $Q$ is as follows: first prepare a uniform superposition over $n$-bit strings. Then, using the classical random oracle $A$ as a source of random bits, map this state to \[ \left\vert \Psi\right\rangle =\frac{1}{2^{n/2}}\sum_{x\in\left\{ 0,1\right\} ^{n}}\left\vert x\right\rangle \left( \sqrt{1-\left\vert \alpha _{x}\right\vert ^{2}}\left\vert 0\right\rangle +\alpha_{x}\left\vert 1\right\rangle \right) \, , \] where each $\alpha_{x}$ is essentially a Gaussian random variable. More precisely, let $q\left( n\right) =\left( n+p\left( n\right) \right) ^{2}$. Then each $\alpha_{x}$ is drawn independently from a complex Gaussian distribution with mean $0$ and variance $1/q\left( n\right) $, with the two technicalities that (1) $\alpha_{x}$ is rounded to $q\left( n\right) $ bits of precision, and (2) the cutoff $\left\vert \alpha _{x}\right\vert \leq1$ is imposed. (By a tail bound, with overwhelming probability we will have $\left\vert \alpha_{x}\right\vert \leq1$ for all $x$ anyway.) Next measure the second register of $\left\vert \Psi\right\rangle $ in the standard basis. The outcome $\left\vert 1\right\rangle $ will be observed with probability $\Omega\left( 1/q\left( n\right) \right) $. Furthermore, conditioned on $\left\vert 1\right\rangle $ being observed, one can check that the distribution $\sigma$ over the reduced state of the first register satisfies $\left\Vert \mathcal{S}_{2^{-p\left( n\right) }}\left( \sigma\right) -\mu\right\Vert \leq2^{-p\left( n\right) }$. (We omit the calculation.) Hence it suffices to repeat the algorithm $O\left( q\left( n\right) \right) $ times. \end{proof} \expref{Theorem}{randstate} shows that, by using a classical random oracle $A$, we can efficiently prepare a uniformly random $n$-qubit state $\left\vert \psi_{A}\right\rangle $. But what if we want to use a random oracle to apply a uniformly random $n$-qubit \emph{unitary} $U_{A}$? It is clear that we can do this if we have exponential time: given an oracle $A$, we simply query an exponentially long prefix $A^{\ast}$ of $A$, and then treat $A^{\ast}$ as an explicit description of a quantum circuit for $U_{A}$. But what if we can make only polynomially many quantum queries to $A$? We do not know whether that suffices for applying a random unitary; indeed, we do not even have a conjecture about this. What we \emph{can} show is that a single quantum query to the classical oracle $A$ does not suffice for applying a random unitary. In particular, suppose every entry of an $n$-qubit unitary matrix $U_{A}$ is a degree-$1$ polynomial in the bits of $A$ (as it must be, if $U_{A}$ is the result of a single quantum query). Then $U_{A}$ can assume at most $4^{2^{n}}$ distinct values as we range over the possible $A$'s, as opposed to the $\Omega\bigl(c^{2^{2n}}\bigr)$ that would be needed to approximate every $n$-qubit unitary. To prove this statement, we first need a lemma about matrices satisfying a certain algebraic relation. \begin{lemma} \label{matrixlem}Let $E_{1},\ldots,E_{M}$ be nonzero $N\times N$ matrices over $\mathbb{C}$, and suppose that $E_{i}E_{j}^{\dag}+E_{j}E_{i}^{\dag}=0$ for all $i\neq j$. Then $M\leq2N$. \end{lemma} \begin{proof} Suppose by contradiction that $M>2N$. Let $e_{i}^{\left( k\right) }$ be vector in $\mathbb{C}^{N}$ corresponding to the $k^{th}$ row of $E_{i}$. Then the condition $E_{i}E_{j}^{\dag}+E_{j}E_{i}^{\dag}=0$ implies that \[ e_{i}^{\left( k\right) }\cdot e_{j}^{\left( l\right) }+e_{j}^{\left( k\right) }\cdot e_{i}^{\left( l\right) }=0 \] for all $i\neq j$ and $k,l$, where $\cdot$ denotes the complex inner product. Now for all $i$, let $k\left( i\right) $ be the minimum $k$ such that $e_{i}^{\left( k\right) }\neq0$, and consider the vectors $e_{1}^{\left( k\left( 1\right) \right) },\ldots,e_{M}^{\left( k\left( M\right) \right) }\in\mathbb{C}^{N}$. Certainly these vectors are not all orthogonal---indeed, since $M>2N$, there must exist $i\neq j$ such that $\operatorname{Re}\left( e_{i}^{\left( k\left( i\right) \right) }\cdot e_{j}^{\left( k\left( j\right) \right) }\right) \neq0$. There are now two cases: if $k\left( i\right) =k\left( j\right) $, then \[ e_{i}^{\left( k\left( i\right) \right) }\cdot e_{j}^{\left( k\left( i\right) \right) }+e_{j}^{\left( k\left( i\right) \right) }\cdot e_{i}^{\left( k\left( i\right) \right) }\neq0 \] and we are done. On the other hand, if $k\left( i\right) \neq k\left( j\right) $, then \[ e_{j}^{\left( k\left( i\right) \right) }\cdot e_{i}^{\left( k\left( j\right) \right) }=-e_{i}^{\left( k\left( i\right) \right) }\cdot e_{j}^{\left( k\left( j\right) \right) } \] is nonzero. Hence $e_{j}^{\left( k\left( i\right) \right) }$ and $e_{i}^{\left( k\left( j\right) \right) }$ must themselves be nonzero. But if $k\left( i\right) >k\left( j\right) $, then this contradicts the minimality of $k\left( i\right) $, while if $k\left( i\right)