Testing product states, quantum Merlin-Arthur games and tensor
optimisation
Aram W. Harrowand Ashley Montanaroy
October 22, 2018
Abstract
We give a test that can distinguish eciently between product states of nquantum systems
and states which are far from product. If applied to a state j iwhose maximum overlap with
a product state is 1  , the test passes with probability 1  (), regardless of nor the local
dimensions of the individual systems. The test uses two copies of j i. We prove correctness of
this test as a special case of a more general result regarding stability of maximum output purity
of the depolarising channel.
A key application of the test is to quantum Merlin-Arthur games with multiple Merlins, where
we obtain several structural results that had been previously conjectured, including the fact that
ecient soundness amplication is possible and that two Merlins can simulate many Merlins:
QMA (k) =QMA (2) fork2. Building on a previous result of Aaronson et al., this implies
that there is an ecient quantum algorithm to verify 3-SAT with constant soundness, given
two unentangled proofs of eO(pn) qubits. We also show how QMA (2) with log-sized proofs is
equivalent to a large number of problems, some related to quantum information (such as testing
separability of mixed states) as well as problems without any apparent connection to quantum
mechanics (such as computing injective tensor norms of 3-index tensors). As a consequence, we
obtain many hardness-of-approximation results, as well as potential algorithmic applications of
methods for approximating QMA (2) acceptance probabilities.
Finally, our test can also be used to construct an ecient test for determining whether a
unitary operator is a tensor product, which is a generalisation of classical linearity testing.
1 Introduction
Entanglement of quantum states presents both an opportunity and a diculty for quantum comput-
ing. To describe a pure state of nqudits (d-dimensional quantum systems) requires a comparable
number of parameters to a classical probability distribution on dnitems. Eective methods are
known for testing properties of probability distributions. However, for quantum states many of
these tools no longer work. For example, due to interference, the probability of a test passing
cannot be simply written as an average over components of the state. Moreover, measuring one
part of a state and conditioning on the measurement outcome may induce entanglement between
other parts of the state that were not previously entangled with each other.
These counter-intuitive properties of entanglement account for many of the outstanding puzzles
in quantum information. In quantum channel coding, the famous additivity violations of [27, 40]
Department of Computer Science & Engineering, University of Washington; aram@cs.washington.edu .
yDepartment of Applied Mathematics and Theoretical Physics, University of Cambridge; am994@cam.ac.uk .
1arXiv:1001.0017v6  [quant-ph]  4 Nov 2012reect how entangled inputs can sometimes have advantages against even uncorrelated noise. For
quantum interactive proofs, the primary diculty is in bounding the ability of provers to cheat using
entangled strategies [48]. Even for QMA (k) (the variant of QMA withkunentangled Merlins [50, 2]),
most important open questions could be resolved by nding a way to control entanglement within
each proof. Here, the recently discovered failure of parallel repetition for entangled provers [49] is
a sort of complexity-theoretic analogue of additivity violations.
The situation is dierent when we consider quantum states that are product across thensystems.
In this case, while individual systems of course behave quantumly, the lack of correlation between
the systems means that classical tools such as Cherno bounds can be used. For example, in channel
coding with product-state inputs, not only does the single-letter Holevo formula give the capacity, so
that there is no additivity problem, but so-called strong converse theorems are known, which prove
that attempting to communicate at a rate above the capacity results in an exponentially decreasing
probability of successfully transmitting a message [61, 73]. Naturally, many of the diculties in
dealing with entangled proofs and quantum parallel repetition would also go away if quantum states
were constrained to be in product form.
1.1 Our results
In this paper, we present a quantum test to determine whether an n-partite statej iis a product
state or far from any product state. We make no assumptions about the local dimensions of j i; in
fact, the local dimension can even be dierent for dierent systems. The test passes with certainty
ifj iis product, and fails with probability ( ) if the overlap between j iand the closest product
state is 1 . An essential feature of our test (or any possible such test, as we will argue in Section
5) is that it requires two copies of j i.
The parameters of our test resemble classical property-testing algorithms [30]. In general, these
algorithms make a small number of queries to some object and accept with high probability if the
object has some property P(completeness ), and with low probability if the object is \far" from
having property P(soundness ). Crucially, the number of queries used and the success probability
should not depend on the size of the object. The main result of this paper is a test for a property of a
quantum state, in contrast to previous work on quantum generalisations of property testing, which
has considered quantum algorithms for testing properties of classical (e.g. [19, 6]) and quantum [59]
oracles (a.k.a. unitary operators, although see Section 7 for an application to this setting). In this
sense, our work is closer to a body of research on determining properties of quantum states directly,
without performing full tomography (e.g. the \pretty good tomography" of Aaronson [1]). The
direct detection of quantities relating to entanglement has received particular attention; see [37]
for an extensive review. However, previous work has generally focused on Bell inequalities and
entanglement witnesses, which are typically designed to distinguish a particular entangled state
from any separable state. By contrast, our product test is generic and will detect entanglement in
any entangled state j i.
The product test is dened in Protocol 1 below, and illustrated schematically in Figure 1. It
uses as a subroutine the swap test for comparing quantum states [18]. This test, which can be
implemented eciently, takes two (possibly mixed) states ,of equal dimension as input. The
test uses an ancilla qubit initialised in state j0iand applies a Hadamard gate to this qubit to
produce the state j+ih+j

. The test proceeds by applying a controlled-SWAP operation
to the latter two registers, controlled by the ancilla qubit, then applies a Hadamard gate on the
ancilla qubit, followed by a computational basis measurement. If the outcome is 0, the output of
the test is \same"; otherwise, the output is \dierent". It is easy to show that this test outputs
2\same" with probability1
2+1
2tr.
Protocol 1 (Product test).
The product test proceeds as follows.
1. Prepare two copies of j i2Cd1

 Cdn; call thesej 1i,j 2i.
2. Perform the swap test on each of the npairs of corresponding subsystems
ofj 1i,j 2i.
3. If all of the tests returned \same", accept. Otherwise, reject.
The product test has appeared before in the literature. It was originally introduced in [58] as
one of a family of tests for generalisations of the concurrence entanglement measure, and has been
implemented experimentally as a means of detecting bipartite entanglement directly [69] (but see
also [66] for caveats). Further, the test was proposed in [59] as a means of determining whether a
unitary operator is product. Our contribution here is to prove the correctness of this test for all n,
as formalised in the following theorem.
Theorem 1. Givenj i2B(Cd1

 Cdn), let
1 = maxfjh j1;:::;nij2:jii2B(Cdi);1ing:
LetPtest(j ih j)be the probability that the product test passes when applied to j i. Then
1 2+2Ptest(j ih j)1 +2+3=2:
For some values of this bound is trivial, but we have the following weaker bound which applies
everywhere:
Ptest(j ih j)1 11
512: (1)
More concisely, Ptest(j ih j) = 1 ().
This result is essentially best possible, in a number of ways. First, we show in Section 5 that
the product test itself is optimal: among all tests for product states that use two copies and have
perfect completeness, the product test has optimal soundness. We also show that there cannot exist
any non-trivial test that uses only one copy of the test state. Second, our analysis of the test cannot
be improved too much, without introducing dependence on nand the local dimensions. When 
is low, we give examples of states j iwhich achieve the upper and lower bounds on Ptest(j ih j),
up to leading order. We also give an example of a bipartite state for which is close to 1, but
Ptest(j ih j)1=2, implying that the constant in our bound cannot be replaced with a function
ofthat goes to 0 as approaches 1. (The bounds on this constant obtained from our proof could
easily be improved somewhat, but we have not attempted to do this.) See Appendix B for all these
examples. Finally, it is unlikely that a similar test could be developed for separability of mixed
states, as the separability problem for mixed states has been shown to be NP-hard [38, 31] (and
indeed we improve on this result, as discussed below).
The proof of Theorem 1 is based on relating the probability of the test passing to the action of
the qudit depolarising channel. In fact, we prove a considerably more general result regarding this
311
22
33
......
nn j 1i
j 2i
Figure 1: Schematic of the product test applied to an n-partite statej i. The swap test (vertical
boxes) is applied to the npairs of corresponding subsystems of two copies of j i(horizontal boxes).
channel. It is known that the maximum output purity of this channel is achieved for product state
inputs [5]; our result, informally, says that any state that is \close" to achieving maximum output
purity must in fact be \close" to a product state. This is a stability result for this channel, which
strengthens the previously known multiplicativity result.
Somewhat more formally, let Dbe thed-dimensional qudit depolarising channel with noise
rate 1 , i.e.
D() = (1 )(tr)I
d+ (2)
fora arbitrary mixed state of one d-dimensional system, and dene the Output Purity of Product
states to be
OPP() = tr(D
n
jihj)2(3)
wherejiis an arbitrary product state. Then our main result, stated more precisely as Theorem
18 in Section 6, is that for small enough  >0, if tr(D
n
j ih j)2(1 ) OPP(), then there is
a product statej1;:::;nisuch thatjh j1;:::;nij21 O().
1.2 Applications and interpretations of the product test
We describe several applications of the product test. The most important of these is that this
test can be used to relate QMA (k) to QMA (2), as we will discuss in Section 3. The complexity
class QMA (k) is dened to be the class of languages that can be decided with bounded error by
a poly-time quantum verier that receives poly-size witnesses from kunentangled provers1[50, 2].
To put QMA (k) inside QMA (2) with constant loss of soundness, we can have two provers simulate k
provers by each submitting kunentangled proofs, whose lack of entanglement can be veried with
our product test. Indeed, this gives an alternate way to understand our test as a method of using
bipartite separability to certify k-partite separability.
Surprisingly, using this result as a building block also allows us to prove amplication for
QMA (k) protocols. It has been conjectured [50, 2] that such protocols can be amplied to expo-
nentially small soundness error. We completely resolve this conjecture, showing that QMA (k) pro-
tocols can be simulated in QMA (2) with exponentially small soundness error, and hence QMA (k) =
QMA (2) fork2. Indeed, we show that this result still holds if the verier's \yes" measurement
operator in a QMA (2) protocol is required to be separable (see Appendix C for the formal denition
of this class of measurements).
As a further corollary, we can improve upon the results of [2, 12] to obtain a protocol in
QMA (2) that veries 3-SAT with constant soundness gap and O(pnpoly log(n)) qubits (where n
1We assume throughout this paper that kis at most polynomial in the input size.
4is the number of clauses). This in turn allows us to prove hardness of approximation results for 19
problems from quantum information theory and elsewhere which turn out to be closely related to
QMA (2). The complete list of equivalent and related problems is given in Section 4.2; while most
had previously been known, we believe that they had not been previously collected in one place.
One example of such a problem is detecting separability, or in other words the weak membership
problem for Sep( d;d), the set of separable quantum states on dddimensions. It was shown in
Ref. [38] that Sep cannot be approximated to precision exp(  d) in time poly( d) unless P=NP. In
Refs. [52, 31], this result was improved to show that approximating Sep to precision 1 =poly(d) is
similarly NP-hard. We show that there is a universal constant  >0 such that, if Kis a convex
set that approximates SEP to within trace distance , then membership in Kcannot be decided in
polynomial time unless 3-SAT 2DTIME (exp(pnlogO(1)(n))). Other such problems for which we
can prove that no polynomial-time algorithm exists, under the same assumption about the hardness
of 3-SAT, are estimating the minimum output entropy of a quantum channel up to a constant, and
estimating the ground-state energy of quantum systems under a mean-eld approximation.
We also prove hardness results for some tensor optimisation problems which are not apparently
related directly to quantum information theory, examples of which include approximating the in-
jective tensor norm of 3-index tensors and estimating the `2!`4norm of a matrix. Our proof
that amplication of QMA (2) protocols is possible implies that one can derive stronger hardness
results for all of these tasks, if one is willing to make stronger assumptions about the hardness of
3-SAT.
Our nal application is that the product test can be used to determine whether a unitary
operator is a tensor product or far from a tensor product in the Hilbert-Schmidt norm, promised
that one of these is the case. This can be seen [59] as one possible generalisation of the well-studied
problem of testing whether a boolean function f0;1gn!f0;1gis linear [13]. This application is
described in Section 7.
These dierent applications of the product test reect the many dierent interpretations of
Ptest(j ih j). It is related in a precise sense to
The purity ofj iafter it is subjected to independent depolarising noise (see Appendix A).
The maximum overlap of j iwith any product state (proved in Appendix B). The logarithm
of this maximum overlap is an important entanglement measure known as the geometric
measure of entanglement (see [71] and references therein).
The overlap ofj i
2with the tensor product of the symmetric subspaces of Cd1
Cd1;:::;Cdn
Cdn(discussed in Section 5).
The average overlap of j iwith a random product state, and a quantum variant of the Gowers
uniformity norm [33] (discussed in Appendix G).
The average purity of j iacross a random partition of [ n] into two subsets (also discussed in
Appendix G).
1.3 Implications for classical computer science
The main result of our paper proposes a quantum solution to a quantum problem. Nevertheless,
there are several implications of our product test that may be of interest to classical computing.
Instead of viewing our results as concerning entangled states of many systems, they may be in-
terpreted in terms of tensors with many indices. These tensors have been studied in the context
5of image processing [4], the planted clique problem [17], constraint satisfaction problems [25] and
many other settings [67].
In this language, our results in Section 4 imply that many central tensor problems, such as
the injective tensor norm (dened in Section 4.2), are hard to approximate even to within constant
factors. On the positive side, our Theorem 11 (together with the equivalences in Section 4.2) implies
that if a heuristic or approximation algorithm existed to optimise over trilinear forms, it could be
extended with little loss of accuracy, to perform optimisations over k-linear forms for general k.
These connections have been further explored in [8], which shows that the `2!`4norm of a matrix
is hard to approximate, and connects this problem to the small-set expansion problem.
1.4 Related work
Our paper addresses a central problem in multipartite entanglement, which is too vast a eld to
reasonably summarise here (one good recent survey is [44]). We therefore concentrate on reviewing
work on quantum Merlin-Arthur games with multiple provers.
The class QMA (k) was rst introduced by Kobayashi, Matsumoto, and Yamakami [50], who
showed that amplication of the soundness gap of QMA (2) protocols implies that QMA (2) =
QMA (k) (a result proven independently in [2]). Both these papers also showed that it is possi-
ble to amplify completeness to exponentially close to 1 (see Lemma 8 for a restatement). Blier
and Tapp showed [12] that graph 3-colourability can be decided using a QMA (2) protocol with
messages of length O(logn) qubits and soundness 1  
(1=n6). This soundness gap was improved
to 
(1=n3+) by Beigi [10], and has recently been improved again to 
(1 =(npolylogn)) by Le Gall,
Nakagawa and Nishimura [51]. By contrast, Aaronson et al. showed [2] that 3-SAT can be solved by
aQMA (k) protocol with constant soundness, at the expense of increasing ktoO(pnpolylog(n)).
Finally, Liu, Christandl and Verstraete have given a problem in QMA (2) which is not obviously in
QMA [53].
Following the conference and arXiv versions of this work, there have been several interesting de-
velopments related to QMA (k). First, it has been shown by Brand~ ao, Christandl and Yard [15, 16]
that QMA (k) protocols, for constant k, are no stronger than QMA protocols if the verier's mea-
surement is restricted to be LOCC (implementable via local operations and classical communica-
tion). One consequence of their work is that, if there existed an ecient LOCC product state test,
QMA (k) = QMA . However, we show in Appendix D that no such test can exist. In the same
work, the authors give a subexponential-time algorithm for optimizing over the set of separable
states [16]; an alternative algorithm for this task has been given by Shi and Wu [64], who also
prove that several special cases of QMA (2) protocols can be simulated in polynomial space.
On the other hand, Chen and Drucker [21] have improved on the result of [2] and given an LOCC
QMA (k) protocol that veries 3-SAT with constant soundness gap for k=O(pnpoly log(n)). (In
fact, their protocol ts in the more restrictive class known as BellQMA (k).) Chiesa and Forbes [22]
recently gave a tight soundness analysis of this protocol, showing that the soundness gap increases
smoothly with k.
McKague has recently used one of our results (that the verier's \yes" measurement operator
may be taken to be separable) to prove that restricting the class QMA (2) to real Hilbert space
does not change its computational power [57]. While it is natural to expect the real case to behave
similarly to the complex case, we note that even for states with real coecients the closest product
state may be complex [23]. Finally, another application of our results was found by [20], who have
presented the only known nontrivial QMA (2)-complete problem: estimating the minimum energy
6of a sparse Hamiltonian over all bipartite product states.
1.5 Organisation
The remainder of this paper is organised as follows. In Section 2, we give an overview of the proofs
of our main results (details are in Appendices A and B). In Section 3, we apply the product test
to prove that QMA (k) =QMA (2) fork2, and we give some complexity-theoretic applications
of this result in Section 4, including an extensive discussion of problems related to QMA (2). In
Section 5 we argue that the product test is essentially optimal, and in Section 6 we state our results
for the depolarising channel. We discuss the use of the product test to test product unitaries in
Section 7, and nish with some open questions in Section 8.
1.6 Notation
For a vector space V, deneB(V) to be the unit vectors in V,L(V) to be the linear operators
fromVtoV, andB(V) to be the density operators on V. More concisely, let B(d) denote the set
ofdddensity matrices. If j iis a vector, let  :=j ih j. Dene the set of separable states on
CdA
CdBto be
Sep(dA;dB) := convf
:ji2B(CdA);ji2B(CdB)g; (4)
where conv( S) denotes the convex closure of a set S. The swap operator on Cd
Cdis denotedF,
and is dened to bePd
i;j=1jiihjj
jjihij.
For1, letkMkdenote the Schatten -norm of a matrix: tr( jMj)1=. For a density matrix
2B(Cd1

 Cdn), andSf1;:::;ng,Sdenotes the density matrix obtained by tracing out
(discarding) the systems not in S. To avoid excessive parenthesization, we write 2
S:= (S)2and
tr2:= tr(2).
2 Overview of the proof of correctness
In this section, we sketch the proof of Theorem 1; a full proof is given in Appendices A and B.
We discuss here only the upper bound on Ptest, since the lower bound follows from continuity
and the fact that product states pass the test with probability 1. First, we make precise the
intuition that the product test is likely to pass precisely when the average subsystem is close to
pure.
Lemma 2. LetPtest(;)denote the probability that the product test passes when applied to two
mixed states ;2B(Cd1

 Cdn). DenePtest() :=Ptest(;). Then
Ptest(;) =1
2nX
S[n]trSS;
and in particular
Ptest() =1
2nX
S[n]tr2
S:
7The proof itself is split into two parts, beginning with the case where is low. We write
j i=p1 j0ni+pjiwithout loss of generality, for some product state j0niand arbitrary state
ji. This allows an explicit expression for tr  2
Sin terms of andjito be obtained. While the
marginals ofjican be complicated (and worse, we need to consider products of expressions of
the form tr Sj0ihj), we can simplify things by only considering the reductions in tr  2
Sthat occur
whenj0iis combined with a state orthogonal to j0i. Thus, we do not need a detailed picture of
ji, but instead will merely split it into a superposition of strings with dierent Hamming weight
(i.e. number of positions orthogonal to j0i). Intuitively, the contribution to ES[n]tr 2
Sof a piece
ofjiwith Hamming weight kshould be exponentially small in k, since each position that diers
from 0 leads to a constant reduction in weight when we project onto the symmetric subspace. In
order to obtain a non-trivial bound from this expression, the nal stage of this part of the proof
is to use the fact that j0niis the closest product state to j ito argue thatjicannot have any
amplitude on basis states of Hamming weight 0 or 1. Ruling out basis states of Hamming weight
0 (i.e.j0ni) is obvious, since otherwise would be smaller. Less obvious is that jicannot have
any amplitude on Hamming weight-1 states, but this too is contradicted by the fact that j0nihas
overlap withjithat is a local maximum among product states, and nonzero amplitude on weight-1
states would mean an innitesimal local rotation could reduce . As a result, we obtain a bound
that is applicable when is small.
Theorem 3. Givenj i2Cd1

 Cdn, let
1 = maxfjh j1;:::;nij2:jii2Cdi;1ing:
Then 1 2+2Ptest(j ih j)1 +2+3=2.
In the case where is high, this result does not yet give a useful upper bound. In the second part
of the proof, we derive a constant bound on Ptest(j ih j) based on considering j ias ak-partite
state, for some k<n .Ptest(j ih j) can be shown to be upper bounded by the probability that the
test for being product across any partition into kparties passes. Informally speaking, if j iis far
from product across the nsubsystems, we show that one can nd a partition such that the distance
from the closest product state (with respect to this partition) falls into the regime where the rst
part of the proof works.
Theorem 4. Givenj i2Cd1

 Cdn, let
1 = maxfjh j1;:::;nij2:jii2Cdi;1ing:
Then, if11=32>0:343,Ptest(j ih j)501=512<0:979.
Between them, Theorems 3 and 4 imply Theorem 1. In fact, we can say precisely that
Ptest(j ih j) = 1 c( )for11
512c( )2.
One feature of our proof that can be generalised is the expectation over S[n]. We eectively
chooseSby ipping a fair coin, but if we use a biased coin then this has an interesting alternate
interpretation in terms of the output purity of the depolarising channel. This yields a similar result,
which is not only that product states maximise the output purity (as was previously known), but
that any state which even approximately maximises the output purity must be approximately
product. See Section 6 for a precise statement. Since the correctness of the product test is a
special case of this more general theorem, we rst prove the result about depolarising channels in
Appendix A and then complete the details necessary for the product test in Appendix B.
This completes the overview of the proof; we now discuss some applications of the product test.
83QMA (2)vs.QMA (k)
In this section, we apply the product test to a problem in quantum complexity theory: whether k
unentangled provers are stronger than 2 unentangled provers. This question can be formalised as
whether the complexity classes QMA (k) and QMA (2) are equal [50, 2]. These classes are dened
as follows.
Denition 1. A language Lis in QMA (k)s;cif there exists a polynomial-time quantum algorithm
Asuch that, for all inputs x2f0;1gn:
1.Completeness: Ifx2L, there exist kwitnessesj 1i;:::;j ki, each a state of poly(n)
qubits, such thatAoutputs \accept" with probability at least con inputjxij 1i:::j ki.
2.Soundness: Ifx =2L, thenAoutputs \accept" with probability at most son inputjxij 1i:::j ki,
for all statesj 1i;:::;j ki.
We use QMA (k)as shorthand for QMA (k)1=3;2=3, and QMA as shorthand for QMA (1). We
always assume 1kpoly(n).
We also dene QMAm(k)s;cto indicate thatj 1i;:::;j kieach involve mqubits, where mmay
be a function of nother than poly(n).
Two of the major open problems related to QMA (k)s;care to determine how the size of the
complexity class depends on kand ons;c. It has been conjectured for some years [50, 2] that in fact
QMA (k) =QMA (2) for 2kpoly(n), and that the soundness and completeness can be amplied
by parallel repetition in a way similar to BPP,BQP,MA,QMA and other complexity classes with
bounded error. In fact, these conjectures are related: 2 kindependent provers can simulate k
independent realisations of a QMA (2) protocol in order to amplify the soundness-completeness gap,
and conversely, [50, 2] proved that QMA (2) amplication implies that QMA (2) = QMA (poly). In
this section, we will fully resolve these conjectures, proving that QMA (2) = QMA (poly) and that
QMA (k) can have its soundness and completeness amplied by a suitable protocol.
The most direct way of putting QMA (k) inside QMA (2) is to ask two provers to each send the
kunentangled proofs that correspond to a QMA (k) protocol. If k= poly(n), then each prover is
still sending only polynomially many qubits. Then the product test can be used to verify that the
states sent were indeed product states and can be used as valid inputs to a QMA (k) protocol. The
specic protocol is described in Protocol 2.
First observe that for YES instances (instances in the language), kMerlins can achieve success
probabilityc, so by sending two copies of this optimal state, two Merlins can achieve completeness
1+c
2cin this modied protocol.
Now consider NO instances. Assume for now that the two Merlins always send the same state.
Then according to Theorem 1, if the Merlins send states that are far from product, they are likely
to fail the product test, whereas basic continuity arguments can show that if they send states that
are nearly product then the success probability will not be much larger than the soundness of the
original protocol. Thus, the soundness does not become too much worse. These ideas (with a
detailed proof in Appendix E) establish
Lemma 5. For anym,k,0s<c1,
QMAm(k)s;cQMAkm(2)s0;c0
wherec0=1+c
2ands0= 1 (1 s)2
100.
9Protocol 2 (QMA (k)toQMA (2)).
TheQMA (2)protocol proceeds as follows.
1. Each of the two Merlins sends j i:=j 1i
:::
j kito Arthur.
2. Arthur performs each of the following tests with probability 1=2.
(a) Arthur runs the product test with the two states as input and accepts
i the test outputs \product."
(b) Arthur randomly chooses one of the states from the two Merlins,
runs the algorithm Aon that state, and outputs the result.
This is already strong enough to achieve amplication up to constant soundness. However,
Protocol 2 has a salutary side eect that will allow us to achieve stronger amplication. To see
this, we will rst introduce a further generalisation of the QMA (k) family. Let Mbe a set of
Hermitian operators Msatisfying 0MI. EachM2Mdenes a binary measurement with M
corresponding to the \accept" outcome and I Mcorresponding to the \reject" outcome. Variants
ofQMA (2) have been considered in which Mis not only restricted to be eciently implementable
on a quantum computer, but also with the further restriction that it belongs to some set M. We will
consider standard classes of measurements such as BELL, LOCC, SEP, ALL, etc., whose denitions
we include in Appendix C.
Formally, dene QMAM
m(k)s;cto be the class QMAm(k)s;cwith Arthur restricted to performing
measurements from Min addition to being restricted to quantum polynomial time. For example,
QMABELL(k) has been introduced under the name BellQMA (k) and proven equal to QMA (for
constantk) by Brand~ ao [14, 2]. Our paper will focus on the case that M= SEP, the class of
measurements such that Mis a separable operator. Observe that M2SEP does not imply that
I Mis separable.
Armed with the denition of QMASEP, we can now see that Protocol 2 produces a protocol that
is not only in QMA (2), but also QMASEP(2). More formally, we can strengthen Lemma 5 to:
Lemma 6. For anym,k,0s<c1,
QMAm(k)s;cQMASEP
km(2)s0;c0
wherec0=1+c
2ands0= 1 (1 s)2
100.
Proof. We again use Protocol 2. By Lemma 5, we know that this protocol has completeness
c0= (1 +c)=2, and has soundness s0= 1 
((1 s)2). It remains only to argue that the \accept"
measurement outcome is a separable operator.
Suppose the rst Merlin sends systems A1;:::;Akand the second Merlin sends systems B1;:::;Bk.
The \accept" outcome of the product test corresponds to the tensor product of projectors onto the
symmetric subspaces of A1B1;A2B2;:::;AkBk. Since the symmetric subspace is spanned by vec-
tors of the form j i
2, it follows that the projector onto each symmetric subspace is separable
across theA:Bcut, and in turn that their tensor product is as well. The other test in the protocol
is to simply apply a measurement either entirely on A1;:::;Akor entirely on B1;:::;Bk, which
is automatically separable. Finally, performing a probabilistic mixture of separable measurements
creates a composite measurement which is also separable.
10The advantage of QMASEP(k) is that it removes the chief diculty with QMA (k) amplication,
which is that conditioning on measurement outcomes can induce entanglement between systems we
have not yet measured. This phenomenon is known as entanglement swapping [45]. However, if we
condition on the outcome of a measurement being M, for someM2SEP, then no entanglement
will be produced in the unmeasured states. As a result, cheating provers cannot gain any advantage
by sending entangled proofs, and we obtain the following lemma.
Lemma 7. For any`1,
QMASEP
m(k)s;cQMASEP
`m(k)s`;c`:
The idea is to simply repeat the original protocol `times in parallel and to accept i each
subprotocol accepts. Since we are considering QMASEPprotocols, obtaining an \accept" outcome
on one proof will not induce any entanglement on the remaining proofs. We give a detailed proof
of Lemma 7 in Appendix E.
From Lemma 6 and Lemma 7, we can almost conclude that strong amplication is possible.
Indeed, when we start with protocols with perfect completeness, we can apply Protocol 2, repeat
p(n) times, and reduce the soundness from stosO(p(n)). For the case of c < 1, we need one
additional argument to keep the completeness from being reduced too much at the same time.
Here we will use a method for completeness amplication proved in both [50, Lemma 5] and [2,
Lemma 6].
Lemma 8 ([50, 2]) .For any`1,
QMAm(k)s;cQMA`m(k)1 c s
3;1 exp( `(c s)2
2):
Our amplication procedure for general c<1 is then to
1. Use Lemma 8 to bring the completeness exponentially close to 1.
2. Use Lemma 6 to convert a general QMA (k) protocol to a QMASEP(2) protocol.
3. Repeat the protocol polynomially many times to make the soundness exponentially small.
This procedure then achieves
Theorem 9. 1. Ifs1 1=poly(n),k= poly(n)andp(n)is an arbitrary polynomial, then
QMA (k)s;1=QMASEP(2)exp( p(n));1.
2. Ifc s1=poly(n),c < 1,k= poly(n)andp(n)is an arbitrary polynomial, then
QMA (k)s;c=QMASEP(2)exp( p(n));1 exp( p(n)).
We prove correctness of Protocol 2 and the rest of Theorem 9 in Appendix E. There are obvious
variants of Theorem 9 to cover the case of limited message size, whose statements we leave implicit.
3.1 QMA (2)and hSep
The complexity of QMAm(2) stems both from the complexity of producing the measurement made
by the verier, and of maximising its acceptance probability over product states. To understand
the complexity of this second step, we dene the support function of the separable states to be
hSep(d;d)(M) := maxftrM:2Sep(d;d)g= maxftrM(
) :ji;ji2B(Cd)g; (5)
11for anyM2L(Cd
Cd). Calculating hSepup to 1=poly(d) accuracy was proven to be NP-hard
by Gurvits [38]. One of our central results will be a weaker hardness result for the problem of
estimating hSeptoconstant additive error; see Theorem 11 below.
Thus, QMAm(2)s;cis the class of languages that can be decided by determining whether
hSep(2m;2m)(M) iscors, whereMis a measurement operator that can be constructed in
polynomial time on a quantum computer. It is instructive to compare the case of QMAm(1),
where the problem can be thought of as computing the largest eigenvalue of a 2m2mmatrix.
There the hardness comes from the fact that the matrix is implicitly specied by a polynomial-size
quantum circuit. By contrast, in the case of QMA (2), there is no known poly( d)-time algorithm
to compute hSep(d;d)(M). As a result, QMA log(1) = BQP [55], but QMA log(2) is not known to
be in BQP (since we do not know how to search over unentangled pairs of log( n)-qubit states
in quantum polynomial time) or NP(since the measurement can depend on a general quantum
poly-time algorithm). The weakest class that we know contains QMA log(2) is NPBQP, by using
theBQP oracle to obtain an explicit description of M. This can be achieved up to error by
running the verier's circuit poly(2m;1=) times and performing tomography. We therefore obtain
that QMAm(2)s;cNTIME (poly(2m;n;1=(c s)))BQP. In particular, QMA (2)NEXP . Unfor-
tunately this cannot be scaled down to place QMA log(2) in NP. This is because the verier in a
QMA log(2) protocol still can perform a poly-time quantum computation. Thus, we only have that
QMA log(2)NPBQP.
Via the connection between QMAm(2) andhSep(2m;2m), all of the results in this section can be
stated in terms of hSep. In particular, we have the following variants of Lemma 7 and Theorem 9.
Lemma 10. LetM2SEP be ad2d2-dimensional separable Hermitian matrix satisfying 0
MI. ThenhSep(dk;dk)(M
k) =hSep(M)k.
Theorem 11. LetMbe ad2d2-dimensional Hermitian matrix satisfying 0MI. Assume
that we are promised that hSep(d;d)(M)is eithercorsfor1c>s> 0. Call these two cases
\Y" and \N." Choose 1c0> s0>0such thatc0= 1 if and only if c= 1. Then there exists a
matrixM0of sizedksuch that
hSep(dk;dk)(M0)(
c0in case Y
s0in case N(6)
Ifc= 1, thenk=O((1 s) 2log(1=s0)), and ifc<1, thenk=O((c s) 3log(1=(1 c0)) log2(1=s0)).
Additionally M02SEP, andM0can be constructed eciently from M, even by a classical log-space
transducer.
4 Complexity-theoretic implications
4.1 Evidence for the hardness of QMA log(2)
A key application of Theorem 9 is to the protocol of Ref. [2] that puts 3-SAT on nclauses inside
the complexity class QMA log(n)(pnpoly log(n))1 
(1);1. Applying Theorem 9 lets us simulate this
using two provers with perfect completeness and arbitrary soundness, so that we obtain
Corollary 12. Let`:N!Nbe polynomially bounded. Then
3-SAT2QMA`(n)pnpoly log(n)(2)2 `(n);1:
12In other words, there is a protocol for 3-SAT instances with nclauses that uses two provers,
`(n)pnpoly log(n)-qubit proofs and has perfect completeness and soundness 2 `(n).
Therefore, making assumptions about the hardness of 3-SAT allows us to prove hardness results
for the complexity class QMA log(2), and stronger assumptions naturally imply stronger hardness
results. We formalise this correspondence as the following corollary.
Corollary 13. The following implications hold.
(i) If 3-SAT on nclauses is not in DTIME (exp(o(n))), then for arbitrary constant >0
QMA log(d)(2) 1
2;1*DTIME (dlog1 d):
(ii) If 3-SAT on nclauses is not in DTIME (exp(o(n))), then
QMA log(d)(2)2 plogd=polylog(log d);1*DTIME (poly(d)):
(iii) If 3-SAT on nclauses is not in DTIME (exp(pnpolylog(n))), then
QMA log(d)(2) 1
2;1*DTIME (poly(d)):
More generally, assume that for some functions `;m :N!N, 3-SAT on nclauses is not
contained in DTIME (m(exp(`(n)pnpolylog(n)))). Then, dening d= 2`(n)pnpolylog(n),
QMA log(d)(2)2 `(n);1*DTIME (m(d)):
Note that the assumptions on the hardness of 3-SAT made in the rst two cases are essentially
equivalent to the (not implausible) Exponential Time Hypothesis of Impagliazzo and Paturi [46],
which states that 3-SAT 62DTIME (exp(`(n))) for any `(n) =o(n).
4.2 QMA log(2)equivalences and reductions
We now discuss a number of problems for which Corollary 13 allows us to prove hardness results.
As described in Section 3.1, QMA log(d)(2) is intimately connected to hSep(d;d). Here we focus solely
on the hardness of estimating hSep(d;d)(M) when 0MIis given explicitly as input. In other
words, we will examine the part of the hardness of QMA log(2) that does notcome from having
access to a poly-time quantum computation.
One denition we will repeatedly use is that of the weak membership problem . IfKis a convex
set, >0 anddis a metric, then WMEM(d)
(K) denotes the following task: given an input x,
determine whether x2Kord(x;K), given the promise that one of these conditions holds.
Hered(x;K) := infy2Kd(x;y). The reason for the is because the complexity of the problem can
depend on the required precision, just as the size of QMA (k)s;cdepends on how close sandcare.
See [35] for more background and equivalent formulations of the weak membership problem for
convex sets. In many cases, dwill be the trace norm distance; in this case, we will simply write
WMEM(K) for the weak membership problem. We also dene Bd(K;) :=fx:d(x;K)g,
and dene the Hausdor distance between (not necessarily convex) sets K;L to bedH(K;L) :=
maxfsupx2Kd(x;L);supx2Ld(x;K)g, or equivalently, inf f0 :XBd(Y;) andYBd(X;)g.
The following equivalences and reductions are a combination of known results ([72, 38, 28, 56,
29, 52, 17, 32, 8], and some unpublished and/or folklore) and consequences of our main theorems.
13Even though many of the reductions are straightforward, we are not aware of any similar list in
the literature, despite many of the quantities being discussed individually.
Equivalent problems
1. GivenMwith 0MI, determine whether
hSep(M) := max
2Sep(d;d)trM= max
ji;ji2B(Cd)trM(
) (7)
iscors. As discussed above, this represents the acceptance probability of a QMA (2)
protocol when the measurement is xed and the provers use an optimal strategy.
This is our reference problem, and we will compare the problems below to this one. However,
we observe that this problem is equivalent (up to a polynomial change of dimension described
below) to versions with dierent choices of candsas long as 0 < s < c < 1 are constants
independent of dimension.
2. Dene ProdSym( d) := convf 
 :j i2B(Cd)g. GivenMwith 0MI, determine
whetherhProdSym(d)(M) isc=4 ors=4.
3. Dene SepSym( d) := convf
:2B(d)g. GivenMwith 0MI, determine whether
hSepSym(d)(M) isc=4 ors=4.
4. The set EW:=EW( d;d) ofentanglement witnesses [65] is the dual cone of Sep, meaning that
EW(d;d) =fW: trW0;82Sep(d;d)g: (8)
GivenM, determine whether min fkM+Wk:W2EW(d;d)giscors.
5. For a quantum channel N, determine whether the superoperator 1 !1 normkNk 1!1is
cors, wherekNk 1!1:= maxkN()k1
kk1.
6. For a quantum channel N, determine whether the superoperator 1 !2 normkNk 1!2is
4c 3 or4s 3, wherekNk 1!2:= maxkN()k2
kk1. (This equivalence is only nontrivial for
some values of c;s.)
7. GivenN, determine whether the minimum output R enyi entropy Smin
1(N) islog(1=s) or
log(1=c). HereSmin
1:= minS1(), whereS1() := logkk1.
8. GivenN, determine whether the minimum output R enyi entropy Smin
2(N) islog(2=ps) or
log(2=pc). HereSmin
2:= minS2(), whereS2() := logkk2.
9. Given a 3-index tensor T2Cd
Cd
Cd, determine whether the injective tensor norm kTkinj
ispcorps. The injective tensor norm is dened here to be
kTkinj= max
x;y;z2B(Cd)jhTjjxi
jyi
jzij (9)
andTshould have the property that for some choice of indices it can be interpreted as a
linear map from Cd!Cd2with operator norm 1.
10. Given a linear map TfromCdtoL(Cd) with operator norm 1, determine whether kTk`2!S1
ispcorps. Here`2is the usual vector 2-norm and we use S1to emphasise that the
output norm is the Schatten 1-norm for operators.
1411. Given a pure state j i2B(Cd
Cd
Cd), dene the geometric measure of entanglement
Egeom(j i) = log max
x;y;z2B(Cd)jh jjxi
jyi
jzij2: (10)
Then determine whether Egeom(j i) islog(d=c) orlog(d=s), forj iABCsatisfying A=
I=d.
12. Given a subspace VCd
Cd, dene the minimum entanglement of Vto be
1(V) := min
j i2B(V)ktr1j ih jk1; (11)
where tr 1means the partial trace over the rst subsystem. Then determine whether 1(V)
iscors).
13. Given a Hermitian K2L(Cd
Cd) with 0KI, dene the mean-eld Hamiltonian
Hn2L((Cd)
n) by
Hn:=1
n(n 1)X
1i6=jnK(i;j); (12)
whereK(i;j)indicates the operator with Kacting on systems i;jand identity matrices
elsewhere. Let min(Hn) denote the smallest eigenvalue of Hn. Then determine whether
limn!1min(Hn) is1 s=2 or1 c=2.
The following problems can be reduced to and from estimating hSep, but unlike the above
problems, the reductions no longer preserve the same completeness and soundness.
Approximately equivalent problems
14. Separability testing: given a state and a promise that it is either separable or a constant
distance away from separable in the trace norm, determine which is the case. In other words,
solve WMEM (Sep(d;d)) for some >0.
15. Weak membership for entanglement witnesses (dened in Eq. (8)), with distance dened in
operator norm; i.e. WMEM1
(EW).
16. Injective tensor norm for k-partite states with k4, geometric measure of entanglement
fork-partite states with k4, mean eld for interactions that are k-local fork3 and
hSep(d;d;d;::: )and weak membership in Sep( d;d;d;::: ) for more systems.
17. Estimating the `2!`4norm of a matrix, dened as kAk`2!`4:= supx6=0kAxk`4=kxk`2, where
kxk`p:= (Pd
i=1jxijp)1=p.
The following problems are at least as easy as hSep, meaning that they can be reduced to
estimating hSep. We will discuss below the specic parameters of the reductions.
Easier problems
18. Deciding 3-SATlog2, which is dened to be the class of 3-SAT instances with log2(d) variables
andO(log2(d)) clauses. By Corollary 12, this reduces to QMA log(2)1=2;1.
19. The planted clique problem is to distinguish a Gn;1=2graph (i.e. an undirected graph with n
vertices in which each edge is present with i.i.d. probability 1 =2) from the union of a Gn;1=2
graph and a random clique of size n. For certain values of , as we discuss below, this
problem is known to reduce to estimating injective tensor norms.
15The following problems are at least as hard as estimating hSep, meaning that hSepcan be reduced
to them, in the special case when c= 1.
Harder problems (when c= 1)
20. Given a channel N, determine whether Smin
(N) = 0 or islog(1=s). The minimum
output R enyi -entropy ofNis dened to be Smin
(N) = minS(N()), whereS() =
1
1 log tr.
21. Determine whether the regularised minimum output R enyi entropy SR;min
 (N) is 0 or
log(1=s). HereSR;min
 (N) = limn!11
nSmin
(N
n).
Before explaining the connections between these problems, we note that Corollary 13 can be
restated in terms of hSep, and thus also in terms of any of the equivalent or harder problems.
Corollary 14. Tasks 1-13 and 20-21 cannot be completed in time poly(d)for any constants 0<
s<c< 1unless 3-SAT2DTIME (exp(pnpoly log(n))).
Explanations
1.ChangingcandsforhSep:This claim follows from Theorem 11. Similarly dicult is the
problem of producing an estimate Xsuch thatjX hSep(M)jfor some>0.
One subtlety is that the c= 1 case is not known to be equivalent to the c<1 case. Soundness,
on the other hand, is always nonzero, since we always have hSep(d;d)(M)trM=d2.
2.Estimating hProdSym :We show this has equivalent diculty to estimating hSep. Initially
assume that we have an algorithm for estimating hProdSym , and given M2L(Cd
Cd), would
like to compute hSep(M). Then dene M0=j01ih01j
M.M0is 4d2-dimensional, and if
0MI, then 0M0I.
To calculate hProdSym , we can without loss of generality let
j i=pp0j0iji+pp1j1iji; (13)
wherep0+p1= 1 andji;ji2B(Cd). Then tr M0( 
 ) =p0p1trM(
). This is
maximised when p0=p1= 1=2. Thus
hProdSym(2 d)(M0) =1
4hSep(d;d)(M):
Conversely, suppose we are given an arbitrary Mand the ability to compute hSep() and
would like to estimate hProdSym (M). First we assume FMF=M. This can be done WLOG
sincehProdSym (M) =hProdSym ((M+FMF)=2). Then dene M0=d;2
sym+M
2. Our desired
equivalence will follow from the following claim:
hSep(M0) =1 +hProdSym (M)
2: (14)
One direction is easy: if hProdSym (M) = trM( 
 ) thenhSep(M0)trM0( 
 ) =
(1 +hProdSym (M))=2. To upper-bound hSep(M0) = max;trM0(
), we dene ;jai;jbi
such that
ji= cos(=2)jai+ sin(=2)jbi
ji= cos(=2)jai sin(=2)jbi:
16To compute tr M0(
), rst we see that tr d;2
sym(
) = (1 + tr )=2 = 1 sin2()=2.
Next, we expand
jiji= cos2(=2)jaai+ sin(=2) cos(=2)(jbai jabi) sin2(=2)jbbi:
When we expand h;jMj;i, the symmetry of Mmeans that terms such as haajM(jbai 
jabi) vanish, and we are left with
cos4(=2)haajMjaai+ sin4(=2)hbbjMjbbi sin2(=2) cos2(=2)(haajMjbbi+hbbjMjaai)
+ 2 sin2(=2) cos2(=2)hbaj habjp
2Mjbai jabip
2:
SincekMk1, and using the denition of hProdSym , we have
trM0(
)1 sin2()
2+(sin4(=2) + cos4(=2))hProdSym (M) + sin2()
2
1 +hProdSym (M)
2:
Maximising over all unit vectors ;, this establishes Eq. (14). We remark that Lemma 5
would also relate hSepandhProdSym but not in this exact fashion.
3.Estimating hSepSym :GivenM, letM0=MA1B1
IA2B2. ThenhProdSym (M0) =hSepSym (M).
For the converse, we use the same construction as hProdSym . Assume that we have an algorithm
for estimating hSepSym , and given M2L(Cd
Cd), would like to compute hSep(M). Again
we deneM0=j01ih01j
M. Letachieve the maximum of tr M0(
), and expand
=j0ih0j
00+j0ih1j
01+j1ih0j
10+j1ih1j
11;for someij2L(Cd). Then
trM0(
) = trM(00
11). Since00;11are proportional to density matrices, and
tr= tr00+ tr11, the rest of the analysis proceeds identically to in the case of hProdSym .
4.Entanglement witnesses: hSep(M) is a convex program whose dual is given by the minimisa-
tion ofkM+WkoverW2EW. See [32] for a discussion of this point.
5.EstimatingkNk 1!1:This connection has been known for some time as folklore and has
appeared before in Ref. [56] (which cites a personal communication from Watrous). Since
the largest value of kN()k1occurs when is pure, nding it corresponds to optimising a
trilinear form over unit vectors [72]; i.e. is equivalent to the injective tensor norm problem
described in task 9. More concretely, dene VN:Cd!Cd
Cdto be the isometric extension
ofN, so that tr EVNVy
N=N(). Then
kNk 1!1= max
;;2B(Cd)j(hj
hj)VNjij2: (15)
This expression equals kTk2
inj(see task 9) forjTi=Pd
i=1jii
VNjii.
6.EstimatingkNk 1!2:DeneM:= (Ny
Ny)(I+F
2). ThenhProdSym (M) = maxf(1 +
tr(N( ))2)=2 :j i 2B(Cd)g= (1 +kNk2
1!2)=2. By Eq. (14), there exists M0with
hSep(M0) = (3 +kNk2
1!2)=4.
7.Estimating Smin
1(N):SinceSmin
1(N) = logkNk 1!1, this is equivalent to task 5.
8.Estimating Smin
2(N):Similarly, this is equivalent to task 6.
179.EstimatingkTkinj:This relates to hSepin a way that is analogous to the relation between the
largest singular value of a matrix Aand the largest eigenvalue of AyA.
kTk2
inj= max
x;y;z2B(Cd)dX
i;j;k=1Ti;j;kxiyjzk2
= max
x;y2B(Cd)dX
i;j;k=1Ti;j;kxiyjjki2
2
= max
x;y2B(Cd)dX
i;j;i0;j0;k=1Ti;j;kT
i0;j0;kxiyjx
i0y
j0
=hSep0
@dX
i;j;i0;j0;k=1Ti;j;kT
i0;j0;kjiihi0j
jjihj0j1
A: (16)
We can think of Tas ad2dmatrix by grouping indices i;jtogether. By doing so, Eq. (16)
becomes simply hSep(TTy) (and by our assumption about the operator norm of the matrix
version ofT, we have that TTyI). To show the equivalence holds in both directions,
observe that any M0 with rankdcan be written as TTyfor somed2dmatrixT. This
rank restriction can be removed either by taking Tto be addd2tensor, or by suitable
padding.
10.EstimatingkTk`2!S1:Observe that
kTk`2!S1= max
x2B(Cd)kT(x)kS1= max
x;y;z2B(Cd)jhyjT(x)jzij:
This last expression is the maximum of a trilinear form over triples of unit vectors, and so is
equivalent to computing an injective tensor norm (see task 9).
11.Estimating Egeom(j i):Treatingj ias a 3-index tensor, it is apparent from the denitions
thatEgeom(j i) = logkj ik2
inj. The condition on  Acorresponds to the requirement thatp
dtimes the resulting tensor should be be an isometry when interpreted as a map from
A!BC. Thus the estimation problems are equivalent. Thep
dfactor also explains why
we need to distinguish the cases Egeomlog(d=c) andlog(d=s). Interestingly, Theorem 1
shows that it is easy to distinguish whether the geometric measurement of entanglement is
orC+for a suciently large constant C.
12.Estimating 1(V):Suppose that dim V=m. DeneTto be an isometry from CmtoV.
Then1(V) =kTk`2!S1, and estimating 1(V) is equivalent to task 10. For simplicity,
one can assume that m=dby padding the appropriate dimensions; this does not aect the
complexity by more than a polynomial factor.
13.Mean-eld Hamiltonians: In Ref. [29], the quantum de Finetti theorem was used to show
that whennd2, then the ground state of His very close to a product state. In the limit,
nding the ground-state energy density of His equivalent to calculating the quantity
max
2B(d)trK(
):
This task is therefore equivalent to task 3.
1814.Separability testing: A classic result in convex optimisation [35] allows one to show that
WMEM(Sep(d;d)) is roughly equivalent to estimating hSep. Unfortunately, known versions
of this result give up 1 =poly(d) factors in the approximation guarantees. This fact has been
used to show the NP-hardness of WMEM 1=poly(Sep) in Refs. [52, 31, 10] and, previously, of
WMEM 1=exp(Sep) by Gurvits [38] (although the connection to QMA log(2) was only observed
by [10]). We conjecture that WMEM (Sep(d;d)) should be NPlog2-hard for some  > 0;
i.e. that Sep( d;d) cannot be approximated to (suciently small) constant accuracy in time
poly(d).
However, we are able to rule out only algorithms that have the further restriction of recog-
nizing a nearly convex set that in turn approximates Sep to constant accuracy. The following
result is an immediate consequence of Corollary 4.3.12 of [35] and Corollary 13.
Proposition 15. Suppose that there exists a constant >0such that for all d, there exists
a convex set Kdwith Hausdor distance toSep(d;d)such that WMEM 1=poly(d)(K)can be
solved in time poly(d). Then 3-SAT2DTIME (exp(pnpoly log(n))).
As a result, one possible alternate title for our paper could have been:
Detecting pure entanglement is easy, so detecting mixed entanglement is hard.
In fact, reductions between WMEM and approximating hSepgo in both directions. We have
to be careful not to assume (as does [47]) that approximation algorithms for hSepoutput an
approximately optimal density matrix. Indeed, some approximations (e.g. [16]) only output
a scalar value approximating hSep. However, we can prove the following reduction.
Proposition 16. Letf(M)be a convex function such that f(0) = 0 andjf(M) hSep(d;d)(M)j
kMk1. Given oracle access to f, we can solve WMEM 2(Sep(d;d))in time poly(d).
Proof. Suppose we are given a density matrix for which we would like to solve WMEM (Sep(d;d)).
The algorithm computes
Z:= maxftrM f(M) : IMIg:
This can be done in polynomial time [35]. If Z, then we declare that 2Sep(d;d), and
ifZ > then we declare that  =2B1(Sep(d;d);).
To analyze the correctness of the algorithm, we prove rather that it is not wrong . In other
words, we need to give the correct answer in the cases: (1) when 62B1(Sep(d;d);2), and (2)
when2Sep(d;d). In case (1), then has trace distance >2from every point in Sep( d;d)
and so there exists a MwithkMk11 for which tr M>h Sep(M) + 2. This implies that
trM>f (M) +, and thatZ > .
On the other hand, in case (2), we have 2Sep(d;d), which implies tr MhSep(d;d)(M)
f(M) +for allM, and thusZ.
15.Weak membership for entanglement witnesses: For a Hermitian matrix M, we havehSep(M)
if and only if M2B1(EW;). HereB1(S;) refers to the points within of a setSin the
Schatten-1norm. This shows that if we can approximate hSepthen we can solve the weak
membership problem for EW. Conversely, if we are given an algorithm for WMEM1
(EW),
then on input Mwe can use binary search to nd approximately the smallest such that
M I2EW. Thiswill be within ofhSep(M).
1916.k-partite tensor norm problems: By adding more systems, we will not make any of the prob-
lems any easier. To reduce from an injective tensor norm on k-tensors to the injective tensor
norm on 3-tensors, we can use Lemma 5. The other reductions claimed in this point are
similar.
When performing these mappings, there is no direct penalty that depends on k. However, the
dimensions of the spaces involved will scale exponentially with k. For example, estimating the
support function of Sep( d1;d2;:::;dk) is harder than estimating hSep(d1;d2), and by Lemma 5
can be reduced to estimating hSep(d;d), whered:= poly(d1d2dk).
17.`2!`4norm: IfA=Pm
i=1jiihijwith eachjii2Cn, then
kAk4
2!4= max
j i2S(Cn)jhij ij2=hProdSym (X
i
2
i) (17)
To show a reduction in the other direction, we need to convert any measurement Minto an
M0with similar hSepsuch thatM0can be written asP
ii
i. Theorem 11 instead allows
us to write Min the formP
ii
i, but the desiredP
ii
iform can be achieved by
takingM0= (p
MA1B1
p
MA2B2)(PA1B1sym
PA2B2sym )(p
MA1B1
p
MA2B2). This construction
is analyzed in [8].
As a result, it is NP-hard to approximate the 2 !4 norm to 1 =poly(d) accuracy, and it is
NPlog2-hard to approximate it to within a constant multiplicative error.
18.3-SAT on log2variables: Corollary 12 explains how 3-SAT instances with O(log2(n)) variables
can be reduced to determining whether hSep(n;n)(M) = 1 or is1=2, for some eciently-
computable matrix M. It is an extremely interesting open question to determine whether the
reverse reduction is also possible.
19.Planted clique problem: In [17], the problem of nding a clique of size 
( n1=rr5log3(n)2)
planted in a Gn;1=2graph was reduced to determining whether kTkinjisror1. Here
1 andTis a tensor in ( Cn)
rwith all1 entries. Thus, we can trivially bound kTTyk1
nr(although we suspect that the norm should typically be O(nr 1)).
For concreteness, let us focus on the case of r= 3. In this case, [17] reduce the problem of
nding a clique of size n1=3+o(1)toQMA log(2)1=n1:5;2=n1:5(orQMA log(2)1=n;2=n, if one assumes
thatEkTTyk1O(nr 1)). Since random graphs typically have no clique of size larger than
2 logn+ 1, the planted clique problem can always be reduced to a Circuit-SAT instance of
size poly(n) withO(log2(n)) input variables. Since
QMA log(2)1=n1:5;2=n3=2QMA log(2)1=2;133-SATlog2;
this implies that the reduction of [17] achieves a reduction that is comparable to the previously
known reduction to Circuit-SAT. It is an interesting open question to determine whether
Circuit-SAT instances with size poly( n) andO(log2(n)) input variables can be placed in
QMA log(2)1=2;1. If this were possible, then it would imply that the reduction of [17] would be
strictly subsumed by the previous reduction of planted clique to Circuit-SAT.
20.Minimum output R enyi entropies: For any0, we have Smin
(N)Smin
1(N) but also
Smin
(N) = 0 iSmin
1(N) = 0. Thus, for any c >0, distinguishing between Smin
= 0 and
Smin
cis at least as hard as distinguishing between Smin
1= 0 andSmin
1c.
2021.Regularised minimum output R enyi entropies: Our hardness result for Smin
immediately gives
us the equivalent hardness result for SR;min
 . The reason is that our proof of amplication for
QMA (2) protocols (see Lemma 7) essentially works by constructing a channel Nfor which
SR;min
1 (N) =Smin
1(N) by design.
4.2.1 Additional remarks
Additivity violations: As a result of the connection between QMA (2) and estimating Smin
1,
the question of whether QMA (2) protocols can be amplied to exponentially small error is di-
rectly related to the question of additivity of the minimum output min-entropy (equivalently,
multiplicativity of the maximum output innity norm). Indeed, additivity violations for Smin
1
(e.g. [42, 41, 36]) translate directly into QMA (2) protocols for which perfect parallel repeti-
tion fails2. Conversely, [56] observed that QMA (2) protocols obey perfect parallel repetition
when the corresponding channel Nis known to have additive Smin
1, for example when Nis
entanglement breaking. Indeed our Lemma 7 is a restatement of this point.
Minimum output entropy: Beigi and Shor previously showed that it is NP-hard to compute the
minimum output entropy up to 1 =poly(d) accuracy [11]. Our result improves their accuracy
requirement, but under a more restrictive complexity assumption. For general channels, we
automatically have SR;min
 (N)Smin
(N); however, the famous failures of the additivity
conjecture imply that sometimes this inequality can be strict, with examples known for 1
[40, 41] and for near 0 [24]. Still, these examples only demonstrate that SR;min
 can deviate
very slightly from Smin
. On the other hand, various lower bounds for SR;min
 are known
[68, 74, 26, 75], and it may be that one of these bounds could be related to Smin
, thereby
proving that SR;min
 cannot be far from Smin
. Our results do not rule out the possibility that
Smin
may be fruitfully related to SR;min
 . However, they do imply that these lower bounds on
SR;min
 (and thereby on Smin
) are unlikely to be eciently computable, or if they are, they
are likely to be extremely loose bounds in general.
Mean-eld approximation: Previous work on the hardness of approximating ground-state
energy of quantum systems generally had dconstant and only ruled out the possibility of
1=poly(n) approximation error. By addressing the case when approximation error is a con-
stant fraction of the overall energy of the system, our result achieves one of the goals of
the conjectured quantum PCP theorem [3]. However, we require dto grow asymptotically,
and we achieve a hardness result much weaker than QMA-hardness. Indeed, due to the
classical PCP theorem combined with the Exponential Time Hypothesis, nding the ground
state of a system of d2log(d) bits (without any symmetry constraint) is likely to require time
exp(d2log(d)), while our results merely imply an exp(
(log2(d))) lower bound. Still, our re-
sult provides a superpolynomial bound on an important class of Hamiltonians that had been
previously considered to be computationally easy to work with.
5 Optimality of the product test
Our product test has perfect completeness in the sense that if j iis exactly a product state then it
will always pass the product test. Soundness could be in principle described by a functional relation
2Note that, taking the standard denition of QMA (2), this is strictly speaking only true if the corresponding
QMA (2) protocol can be implemented in polynomial time.
21between maximum acceptance probability and distance to the nearest product state. However, for
our purposes, we can say that our test has constant soundness in that if j ihas overlap at most
1 with any product state then it will pass the product test with probability at most 1  ().
In fact, if we consider only product-state tests with perfect completeness, then we can show
that our test has optimal soundness: that is, it rejects as often as possible given the constraint
of always accepting product states. More generally, suppose that a product-state test Tis given
j i
kas input. Since the outcome of the test is binary, we can say that Tis an operator on the
nk-qudit Hilbert space with 0 TIand that the test accepts with probability tr T 
k.
LetSbe the set of product states in Cd1

 Cdn, and dene Skto be the span of fji
k:
ji2Sg. For a single system Cd, the span offji
k:ji2Cdgis denoted SymkCd. This is
the symmetric subspace of ( Cd)
k, meaning that it can be equivalently dened to be the set of
vectors in ( Cd)
kthat is invariant under permutation by the symmetric group Sk. This fact allows
the projector onto SymkCd, which we denote sym
d;k, to be implemented eciently [9]. Also, it
implies that Sk= SymkCd1

 SymkCdnand that the projector onto Sk, denoted Sk, is
sym
d1;k

 sym
dn;k.
Now we return to our discussion of product-state tests. If tr T
k= 1 for all 2S, then
TSk. Thus,Twill always accept at least as often as Skwill on any input, or equivalently,
takingT= Skyields the test which rejects as often as possible given the constraint of accepting
every state in Sk.
To understand Sk, note that the projector onto SymkCdis given by1
k!P
2SkPd(), where
Pd() =X
i1;:::;ik2[d]ji1;:::;ikihi(1);:::;i(k)j: (18)
Fork= 1, Sym1Cdsimply equals Cd, and S1is the identity operator on ( Cd)
n. Thus, no
non-trivial product-state test is possible when given one copy of j i.
Whenk= 2, Sym2Cdis the +1 eigenspace of ( I+F)=2, which is the space that passes the
swap test. Thus, the product test (in Protocol 1) performs the projection onto S2and therefore
rejects non-product states as often as possible for a test on j i
2that always accepts when j iis a
product state. These arguments also imply that given j i
k, projecting onto Skyields an optimal
k-copy product-state test of j i. The strength of these tests is strictly increasing with k, but we
leave the problem of analysing them carefully to future work.
Finally, this interpretation of the product test allows us to consider generalisations to testing
membership in other sets S. The general prescription for a test that is given kcopies of a state is
simply to project onto the span of fj i
k:j i2Sg. We will not explore these possibilities further
in this paper, but see [70] for a subsequent paper that considers variations on this theme for the
related problem of testing properties of unitary operators.
6 Stability of the depolarising channel
As discussed in Section 2, the correctness proof of the product test in fact applies to a larger class
of processes acting on two copies of n-partite states. In general, choosing Saccording to a binomial
distribution on [ n] and taking the expectation of tr 2
Sis equivalent to evaluating the output purity
ofnuses of the depolarising channel D(as dened in (2)). A special case of this corresponds to
the probability of the product test passing.
22Lemma 17. We have
tr(D
n
)2=1 2
dnX
S[n]d2
1 2jSj
tr(2
S);
and in particular
tr(D
n
1=p
d+1)2=1
(d+ 1)nX
S[n]tr(2
S);
and for pure product states,
OPP() := tr(D
n
(j 1ih 1j

j nih nj))2=d 1
d2+1
dn
:
We will see that it is possible to prove a more general version of Theorem 1.
Theorem 18. Givenj i2(Cd)
n, let
1 = maxfjh j1;:::;nij2:j1i;:::;jni2Cdg:
Then (recalling the denitions of DandOPP()from equations (2) and (3)),
tr(D
n
j ih j)2OPP() 
1 4(1 )d2(1 2)
(1+(d 1)2)2+ 43=2(1 2)2+d24
(1+(d 1)2)22!
:
In particular,
tr(D
n
1=p
d+1j ih j)2OPP(1=p
d+ 1)
1 +2+3=2
:
The idea of the proof is more or less the same as the outline sketched in Section 2 and the
details can be found in Appendix A.
7 Testing for product unitaries
As well as being useful for testing quantum states, the product test has applications to testing
properties of unitary operators. The results we obtain will be in terms of the normalised Hilbert-
Schmidt inner product, which is dened as hM;Ni:=1
dtrMyNforM;N2M(d), whereM(d)
denotes the set of ddmatrices. Note that, with this normalisation, jhU;Vij 1 for unitary
operatorsU,V.
We consider the problem of testing whether a unitary operator is a tensor product. That is,
we are given access to a unitary Uon the space of nqudits (for simplicity, restricting to the
case where each of the qudits has the same dimension d), and we would like to decide whether
U=U1

Un. This is one possible generalisation of the classical problem of testing linearity of
functionsf:f0;1gn!f0;1g[13]. To see this, observe that fis linear (i.e. f(xy) =f(x)f(y)
for allxandy) if and only if the function g:f0;1gn!f 1gdened byg(x) = ( 1)f(x)is a product
of individual functions gi(x) = ( 1)aixi, forai2f0;1g. Thus the diagonal unitary operator Uon
nqubits dened by Uxx=g(x) is a tensor product if and only if fis linear.
In Protocol 3 we give a test that solves this problem using the product test. The test always
accepts product unitaries, and rejects unitaries that are far from product, as measured by the
23normalised Hilbert-Schmidt inner product. Several papers have proposed property tests with similar
performance for other sets of unitary matrices: e.g. Pauli matrices [59], Cliord gates [54, 70] and
many other sets [70].
The following correspondence (also known as the Choi-Jamio lkowski isomorphism) underlies
our ability to apply the product test to unitaries. Let jibe a maximally entangled state of two
d-dimensional qudits, written as1p
dPd
i=1ji;iiin terms of some basis B= (j1i;:::;jdi). For any
matrixM2M(dn), denejv(M)i:= (M
I)ji
n. Thenhjjhkjv(M)i=hjjMjkip
dn. In particular,
for any matrices M;N2M(dn),hM;Ni=hv(M)jv(N)i= trMyN=dn.
Protocol 3 (Product unitary test).
The product unitary test proceeds as follows.
1. Prepare two copies of the state ji
n, then in both cases apply Uto then
rst halves of each pair of qudits to create two copies of the state jv(U)i2
(Cd2)
n.
2. Return the result of applying the product test to the two copies of jv(U)i,
with respect to the partition into n d2-dimensional subsystems.
Let the probability that this test passes when applied to some unitary UbePtest(U). Then we
have the following theorem, which proves a conjecture from [59].
Theorem 19. GivenU2U(dn), let
1 = maxfjhU;V 1

Vnij2:V1;:::;Vn2U(d)g:
Then, if= 0,Ptest(U) = 1 . If.0:106, thenPtest(U)1 1
4+1
162+1
83=2. If0:106.1,
Ptest(U)501=512. More concisely, Ptest(U) = 1 ().
The proof is given in Appendix F. It is not quite immediate from the previous results; the
key problem is that the closest product state to jv(U)imay not correspond to the closest unitary
operator to U.
Our test is sensitive to the Hilbert-Schmidt distance of a unitary from the set of product
unitaries. One might hope to design a similar test that instead uses a notion of distance based
on the operator norm. However, this is not possible. For example, if we could detect a constant
dierence in the operator norm between an arbitrary unitary Uand the set of product unitaries
then we could nd a single marked item in a set of size dn. By the optimality of Grover's algorithm,
this requires 
( dn=2) queries to U. More generally, any test that uses only a constant number of
black-box queries to Ucan only detect an 
(1) dierence in an 
(1) fraction of the dndimensions
thatUacts upon.
8 Open problems
We conclude with a discussion of open problems related to our work.
1. Our main result can be seen as a \stability" theorem for the output purity of the depolarising
channel (cf. Section 6).. It is an interesting problem to determine whether a similar result
24holds for all output R enyi entropies for the depolarising channel, or even for all channels
where additivity holds.
2. Can Theorem 1 be tightened further, perhaps by improving the constant in the 3=2term?
It would also be interesting to improve the constants in Theorem 1 in the regime of large ,
as at present they are extremely pessimistic. The regime of large is generally somewhat
mysterious: for example, we do not know the minimum value of Ptest, or the largest distance
from any product state that can be achieved by a state of nqudits. This is equivalent to
determining the maximal value of the geometric measure of entanglement [71] which can
be achieved by a pure state of nqudits; see the PhD thesis [7] for a review of recent work
concerning bounds on this quantity.
3. Suppose the goal is to test whether a given state is of the form j'i
nfor some unknown j'i.
Can we substantially improve on the performance of the product test, say with a test whose
acceptance probability decreases exponentially in the number of positions not equal to j'i?
Ideally we would achieve performance comparable to the exponential de Finetti theorem [63],
but without any dependence on dimension. The natural test for this problem is to project
onto the symmetric subspace of all 2 npositions.
4. The relationship between QMA andQMA (2) remains unresolved. Our Theorem 9 proves that
QMASEP(2) = QMA (2), while the result of [15] implies that QMALOCC(2) = QMA . Can
this gap be closed? One possible way to do this would be to improve our results to show
that QMALOCC(2) = QMA (2); but see also Appendix D for a proof that an ecient LOCC
product test does not exist. Alternatively, one might improve the simulation of [15] to apply
to separable measurements instead of only LOCC measurements, but the obvious approaches
to modifying their proof do not appear to work. Finally, if QMA (2) is not shown to be in
QMA , one might hope for any upper bound on its complexity that is better than NEXP .
5. Is there an oracle separation between QMA andQMA (2)? The equalities in the previous point
relativise, so this is equivalent to showing a separation between QMASEP(2) and QMALOCC(2).
A The depolarising channel
LetDbe the qudit depolarising channel as dened in equation (2). We will be interested in
applying the n-fold productD
n
to states of nqudits, and in particular in the purity of the
resulting states. This has the following characterisation.
Lemma 17. We have
tr(D
n
)2=1 2
dnX
S[n]d2
1 2jSj
tr(2
S);
and in particular
tr(D
n
1=p
d+1)2=1
(d+ 1)nX
S[n]tr(2
S);
and for pure product states,
OPP() := tr(D
n
(j 1ih 1j

j nih nj))2=d 1
d2+1
dn
:
25Proof. Consider some Hermitian operator basis for B(Cd) which contains the identity and is or-
thonormal with respect to the normalised Hilbert-Schmidt inner product hA;Bi=1
dtrAyB, and
extend this basis to B((Cd)
n) by tensoring. Expand in terms of the resulting basis as
=X
t2f0;:::;d2 1gn^tt:
where ^t2R,trepresents an element of the tensor product basis corresponding to the string
t2f0;:::;d2 1gn, and the identity is indexed by 0 at each position. Then we have
tr(2
S) =d2n jSj0
@X
t;ti=0;8i2S^2
t1
A;
and hence, for any ,
X
S[n]jSjtr(2
S) =d2nX
S[n](=d)jSj0
@X
t;ti=0;8i2S^2
t1
A=d2nX
t^2
t0
BBB@X
S[n];
ti=0;8i2S(=d)jSj1
CCCA
=d2nX
t^2
t0
@n jtjX
x=0n jtj
x
(=d)x+jtj1
A
=d2nX
t^2
t(=d)jtj(1 +=d)n jtj
= (d(d+))nX
t^2
t(=(+d))jtj
= (d+)ntr(D
np
=(+d))2:
Rearranging completes the proof; the two special cases in the statement of the lemma can be veried
directly.
Using the above lemma, we can see that maximal output purity is obtained only for product
states, since only product states saturate the inequality tr 2
S1 for allS[n]. We will now prove
our main result, which is a \stability" theorem for the depolarising channel: if a state achieves close
to maximal output purity, it must be close to a product state.
Theorem 18. Givenj i2(Cd)
n, let
1 = maxfjh j1;:::;nij2:j1i;:::;jni2Cdg: (19)
Then
tr(D
n
j ih j)2OPP() 
1 4(1 )d2(1 2)
(1 + (d 1)2)2+ 43=2(1 2)2+d24
(1 + (d 1)2)22!
:
In particular,
tr(D
n
1=p
d+1j ih j)2OPP(1=p
d+ 1)
1 +2+3=2
:
26Proof. Without loss of generality assume that one of the states achieving the maximum in Eq. (19)
isj0i
n, which we will abbreviate simply as j0ni, orj0iwhen there is no ambiguity. We thus have
j i=p
1 j0i+pji
for some statejisuch thath0ji= 0, andji=P
x6=0xjxifor somefxg. We write down
explicitly
 :=j ih j= (1 )j0ih0j+p
(1 )(j0ihj+jih0j) +jihj:
By Lemma 17,
tr(D
n
 )2=1 2
dnX
S[n]jSjtr 2
S;
where we set =d2=(1 2) for brevity. Now
X
S[n]jSjtr 2
S=X
S[n]jSj
tr((1 )j0ih0jS+p
(1 )(j0ihjS+jih0jS) +jihjS)2
;
and for any subset S,
tr 2
S= (1 )2trj0ih0j2
S+(1 ) tr(j0ihj+jih0j)2
S+2trjihj2
S
+ 2p(1 )3=2trj0ih0jS(j0ihj+jih0j)S+ 2(1 ) trj0ih0jSjihjS
+ 23=2p
1 trjihjS(j0ihj+jih0j)S:
We now bound the sum over S(weighted by jSj) of each of these terms, in order. Note that we
repeatedly use the notation [ E] for a term which evaluates to 1 if the expression Eis true, and 0
ifEis false.
1. Asj0iis product, clearly
X
S[n]jSjtrj0ih0j2
S=X
S[n]jSj= (1 +)n:
2. We have
tr(j0ihj+jih0j)2
S= trj0ihj2
S+ trjih0j2
S+ 2 trj0ihjSjih0jS:
It is easy to see that the rst two terms must be 0 for all S(as only the o-diagonal entries
of the rst row of the matrix j0ihjcan be non-zero). For the third, we explicitly calculate
j0ihjSjih0jS=X
x6=0jxj2[xi= 0;8i2S]j0ih0j
k;
and hence
X
S[n]jSjtrj0ihjSjih0jS=X
x6=0jxj2X
S[n]jSj[xi= 0;8i2S]
=X
x6=0jxj2nX
k=jxjkn jxj
n k
= (1 +)nX
x6=0jxj2
1 +jxj
:
273. It clearly holds that tr jihj2
S1, so as in part (1),
X
S[n]jSjtrjihj2
S(1 +)n;
and this will be tight if and only if jiis product itself.
4. Using the same argument as in part (2), tr j0ih0jSj0ihjS= trj0ih0jSjih0jS= 0.
5. Write the state =jihjas
=X
x;yx1;:::;ynjx1ihy1j

jxnihynj:
Then, for any S=fi1;:::;ikg,
S=X
x;y[xi=yi;8i2S]x1;:::;ynjxi1ihyi1j

jxikihyikj;
which implies
trj0ih0jSjihjS=X
x[xi= 0;8i2S]jxj2;
and hence, similarly to part (2),
X
S[n]jSjtrj0ih0jSjihjS=X
x6=0jxj2n jxjX
k=0kn jxj
k
= (1 +)nX
x6=0jxj21
1 +jxj
:
6. The last term can be trivially bounded using
jtrjihjS(j0ihj+jih0j)Sj2:
However, it is possible to get a better bound with a bit more work. We expand
X
S[n]jSjtrjihjSj0ihjS=
X
S[n]jSjX
x;y;zx
y
z[zi= 0;i2S][xi=yi;i2S] trjx1ihy1j0ihz1j

jxnihynj0ihznj
=X
S[n]jSjX
x;y;zx
y
z[zi= 0;i2S][xi=yi;i2S][yi= 0;i2S][xi=zi;i2S]
=X
jy^zj=0y_z
y
zX
S[n]jSj[yi= 0;i2S][zi= 0;i2S]
=X
jy^zj=0y_z
y
zjzj(1 +)n jyj jzj:
28This expression can be upper bounded as follows:
X
jy^zj=0y_z
y
zjzj(1 +) (jyj+jzj)
sX
jy^zj=0jyj2jzj2vuutX
jy^zj=02jzj
(1 +)2jy_zjjy_zj2
0
@X
x(1 +) 2jxjjxj20
@X
jy^zj=02jzj[y_z=x]1
A1
A1=2
= X
x1 +2
(1 +)2jxj
jxj2!1=2
: (20)
Combining these terms, we have
X
S[n]jSjtr 2
S(1 +)n((1 )2+ 2(1 )X
x6=0jxj2(1 +) jxj(jxj+ 1) +2+
43=2p
1  X
x1 +2
(1 +)2jxj
jxj2!1=2
):
Note that (1 + ) jxj(jxj+ 1) decreases with jxjfor all >0, as does (1 + 2)jxj(1 +) 2jxj. To
complete the proof, we will show that jihas no weight 1 components (i.e. x= 0 forjxj<2). In
the contribution from Eq. (20), this implies that only the jxj4 terms contribute (since x=y_z
andy^z=;). Therefore,jihaving no weight 1 components would imply that
X
S[n]jSjtr 2
S(1 +)n
1 4
(1 +)2
(1 ) (1 +2)2
(1 +)2
1=2
;
which would imply the theorem. Now, for any ,', we have 1 j(cosh0j+ei'sinh1j)
h0j
n 1j ij2. Pickingsuch that
cos=jh0j ijp
jh0j ij2+jh10n 1j ij2;
and'such thatei'h10n 1j i>0, it is easy to see that
1 jcosh0j i+ei'sinh10n 1j ij2=jh0j ij2+jh10n 1j ij2:
However, we have assumed that 1  =jh0j ij2, so this implies that h10n 1j i= 0. Repeating the
argument for the other n 1 subsystems shows that j iis indeed orthogonal to every state with
Hamming weight at most 1, so jihas no weight 1 components.
B Proof of Theorem 1: correctness of the product test
In this appendix, we prove correctness of the product test (Theorem 1). Let the test be dened as
in Protocol 1. The following lemma from [59] expresses the probability of passing in terms of the
partial traces of the input states; we include a proof for completeness.
29Lemma 2. LetPtest(;)denote the probability that the product test passes when applied to two
mixed states ;2B(Cd1

 Cdn). DenePtest() :=Ptest(;). Then
Ptest(;) =1
2nX
S[n]trSS;
and in particular
Ptest() =1
2nX
S[n]tr2
S:
Ifd1=d2==dn=d, for somed, then
Ptest() =d+ 1
2n
tr(D
n
1=p
d+1)2:
Note that we can in fact assume that d1=d2==dn=dwithout loss of generality by
settingd= max(d1;:::;dn), and embedding each of Cd1;:::;CdnintoCdin the natural way. This
padding operation neither aects the probability of the swap tests passing nor changes the distance
to the closest product state.
Proof. LetFdenote the swap (or ip) operator that exchanges two quantum systems of equal but
arbitrary dimension, with FSdenoting the operator that exchanges only the qudits in the set S.
Then we have
Ptest(;) = tr(
)I+F
2
n
=1
2nX
S[n]tr(
)FS=1
2nX
S[n]trSS:
The second part then follows from Lemma 17.
We now analyse the probability of the product test passing for general n. We rst note that,
in the special case where n= 2, it is possible to analyse the probability of passing quite tightly.
The proof of the following result, which is implicit in previous work of Wei and Goldbart [71], is
essentially immediate from Lemma 2.
Lemma 20. Letj i2Cd1
Cd2, whered1d2, be a bipartite pure state with Schmidt coecientsp1p2p
d1. Then
Ptest(j ih j) =1
2 
1 +X
i2
i!
;
while
1 := max
j1i;j2ijh j1ij2ij2=1:
In particular,
1 +d1
2(d1 1)2Ptest(j ih j)1 +2:
We are nally ready to prove Theorem 1. The proof is split into two parts, which we formalise
as separate theorems. The rst part holds when is small, and depends on the results proven in
Appendix A. The second part holds when is large, and is proved using the rst part.
30Theorem 3. Givenj i2Cd1

 Cdn, let
1 = maxfjh j1;:::;nij2:jii2Cdi;1ing:
Then
1 2+2Ptest(j ih j)1 +2+3=2:
Proof. The lower bound holds by general arguments. It is immediate that, if applied to j1;:::;ni,
the product test succeeds with probability 1. As the test acts on two copies of j i, which has overlap
1 withj1;:::;ni, it must succeed when applied to j iwith probability at least (1  )2. The
upper bound follows from Lemma 2 and Theorem 18. The statement of Theorem 18 only explicitly
covers the case where the dimensions of all the subsystems are the same; however, as noted above,
we can assume this without loss of generality.
This result is close to optimal. At the low end, the state j i=p1 j0ni+pj1nihas
Ptest(j ih j) = 1 2+ 22+o(1). At the high end, for j i=p1 j00i+pj11i,Ptest(j ih j) =
1 +2. We also note that this result does not extend to a test for separability of mixed states;
the maximally mixed state on nqudits is separable but it is easy to verify that Ptest(I=dn) =
((d+ 1)=2d)n, which approaches zero for large n.
Theorem 3 only gives a non-trivial upper bound on the probability of passing when is small
(up to=1
2(3 p
5)0:38). We now show that the product test also works in the case where the
state under consideration is far from any product state. We will need two lemmas.
Lemma 21. Givenj i2Cd1

 Cdn, letPP
test(j ih j)be the probability that the P-product test {
the test for being product across partition P{ passes. Then, for all P,PP
test(j ih j)Ptest(j ih j).
Proof. The subspace corresponding to the usual product test passing is contained within the sub-
space corresponding to the P-product test passing.
Lemma 22. Letj i,jibe pure states such that jh jij2= 1 , and letPsatisfy 0PI;
e.g.Pmight be a projector. Then jh jPj i hjPjijp.
Proof. We can directly calculate1
2kj ih j jihjk1=p. This then gives the claimed upper
bound onjtrP(j ih j jihj)j(see [60, Chapter 9]).
Theorem 4. Givenj i2Cd1

 Cdn, let
1 = maxfjh j1;:::;nij2:jii2Cdi;1ing:
Then, if11=32>0:343,Ptest(j ih j)501=512<0:979.
Proof. For simplicity, in the proof we will use a quadratic upper bound on Ptest(j ih j) that follows
by elementary methods from Theorem 1: Ptest(j ih j)1 3
4+ 22. For a contradiction, assume
thatPtest(j ih j)>p:= 501=512, while11=32.
For any partition Pof [n] into 1knparts, letjPibe the product state (with respect to
P) that maximises jh jij2over all product states ji(with respect to P). If
1 hjh jPij21 `;
where for readability we dene `:= 1=32 andh:= 11=32, then by the quadratic bound given
above theP-product test passes with probability PP
test(j ih j)p, implying by Lemma 21 that
31Ptest(j ih j)p. Therefore, as we are assuming that j iis a counterexample to the present
theorem, there exists a ksuch thatjh jij2>1 `for somejithat is product across kparties,
and yetjh jij2<1 hfor alljithat are product across k+ 1 parties.
So, for this k, letj1ijkibe a state that maximises jh j1;:::;kij2. Thus there is some
0<`such that we can write j ias
j i=p
1 0j1ijki+p
0ji:
Ifk= 1, then trivially j1i=j iand0= 0. Assume without loss of generality that j1iis a state
of two or more qudits. Now we know that
max
j0
1;1i;j0
1;2ijh1j0
1;1ij0
1;2ij2(1 0)<1 h; (21)
orj0
1;1ij0
1;2ij2ijkiwould be a ( k+1)-partite state with overlap at least 1  hwithj i. (Here
we have used the fact that for k>1, by the arguments at the end of Theorem 18, jiis orthogonal
toj0
1;1ij0
1;2ij2ijkifor any choice ofj0
1;1i,j0
1;2i.) Let 1 = maxj0
1;1i;j0
1;2ijh1j0
1;1ij0
1;2ij2.
Then Eq. (21) implies that
1 <1 h
1 0<1 h
1 `=21
31:
Using the exact expression given in Lemma 20, we nd that Ptest(j1ih1j)<751=961 (if 10=31<
1=2, this follows from Ptest(j1ih1j)1++2; if1=2, thenPtest(j1ih1j)3=4 always).
Next we use Lemma 22 to obtain
Ptest(j ih j)Ptest(j1ih1j

jkihkj) +p
0
<P test(j1ih1j) +p
`
<751
961+r
1
32<0:96:
But we previously assumed that Ptest(j ih j)> p > 0:978. We have reached a contradiction, so
the proof is complete.
One might hope that this theorem could be improved to show that, as !1,Ptest(j ih j)
necessarily approaches 0. However, this is not possible. Consider the dd-dimensional bipartite
stateji=1p
dPd
i=1jiii. It is easy to verify using Lemma 20 that Ptest(jihj) = 1=2(1 + 1=d)
while maxj1i;j2ijhj1ij2ij2= 1=d.
Combining Theorems 3 and 4, we obtain Theorem 1 and thus have proven correctness of the
product test. The constants in Theorem 4 have not been optimised as far as possible and could be
improved somewhat.
C Classes of measurement operators
In this appendix, we dene the classes of measurement operators used in our paper and other
relevant literature on QMA (2), such as [15]. Our denitions mostly follow the conventions of
quantum information theory. Each class of measurement operators describes operators on Cd
Cd.
32BELL is the set of Mthat can be expressed as
M=X
(i;j)2Si
j; (22)
whereP
ii=IandP
jj=I, andSis a set of pairs of indices. In other words, the systems
are locally measured, obtaining outcomes iandj, and then the verier accepts if ( i;j)2S.
LOCC 1is the set of Mthat can be realised by measuring the rst system and then choosing
a measurement on the second system conditional on the outcome of the rst measurement.
SuchMcan be written as
M=X
ii
Mi; (23)
whereP
ii=Iand 0MiIfor eachi.
LOCC is the set of Mthat can be realised by alternating partial measurements on the two
systems a nite number of times, choosing each measurement conditioned on the previous
outcomes. An inductive denition is that M is in LOCC if there exist operators fEig;fMig,
withP
iEiIand eachMi2LOCC, such that either M=P
i(pEi
I)Mi(pEi
I) or
M=P
i(I
pEi)Mi(I
pEi). For the base case, it suces to take I2LOCC.
SEP is the set of Msuch that
M=X
ii
i (24)
for some positive semidenite (WLOG rank one) matrices fig;fig. (Note: other works
dene SEP to be the smaller set of Mfor which both MandI Mcan be decomposed as
in Eq. (24), and use the term SEP YESto describe the measurements for which only Mhas
to satisfy Eq. (24).)
SEP-BOTH is the set of Mfor whichM2SEP andI M2SEP.
PPT (positive partial transpose [62, 43]) is the set of Mfor whichM 0, where   is the
partial transpose map dened by ( jiihjj
jkihlj) = (jiihjj
jlihkj). Again note that this
denition does not require I M2PPT.
PPT-BOTH is the set of Mfor whichM2PPT andI M2PPT.
ALL has no restrictions on Mother than 0MI.
We note that SEP-BOTH and PPT-BOTH are natural relaxations of LOCC because they preserve
the property that both MandI Mmust be realisable through local operations and classical
communication. On the other hand, SEP and PPT are more natural when we consider Mby itself
and do not wish to consider additional constraints on I M.
These sets satisfy the following inclusions, all of which are known to be strict
BELLLOCC 1LOCCSEP-BOTHPPT-BOTH
\ \
SEP PPTALL
33D Nonexistence of an LOCC product test
A natural extension of the idea of product state testing is to a distributed setting where two parties,
each of whom receives one copy of an n-partite statej i, must determine whether j iis product
using only local operations and classical communication (LOCC). Indeed, following the completion
of an initial version of this work, it was shown by Brand~ ao, Christandl and Yard that, if there were
an ecient LOCC protocol for product state testing, then QMA (k) =QMA [15, 16].
In this appendix, we show that unfortunately no such LOCC protocol exists. In fact, we rule
out the larger class of PPT-BOTH measurements (dened in Appendix C). Our impossibility result
holds for the easiest version of this task, in which n= 2. For simplicity, here we only consider the
case where the test uses 2 copies of j i; one can show a similar result when the number of copies
is larger but the proof is signicantly more complicated [39].
Formally, we dene a product test as a measurement fM;I Mgthat acts onj i
2=j iA1B1
j iA2B2with outcome Mcorresponding to \product" and I Mcorresponding to \not product."
There is no good canonical way to express the validity of a product test. One rather general way
we might do this is to say that there are functions f() andg() such that ifj ihas overlap 1 
with the closest product state then its probabity Ptestof passing the product test satises
f()Ptestg(): (25)
For example, Theorem 1 shows that our product test satises Eq. (25) with f() = 1 c1and
g() = 1 c2withc1>c2>0.
For our impossibility result, we will use a dierent and simpler success measure. Dene the
completeness cof a product test to be the average probability of accepting a random product state
j i=j Ai
j Bi, and dene the soundness sto be the average probability of accepting a random
bipartite statej i. While strictly speaking a random bipartite state may sometimes be close to
a product state, it has an overwhelmingly high probability of being close to maximally entangled.
Thus, this demand for a product test is nearly as undemanding as possible. Finally, dene the bias
of the test as b=c s.
Theorem 23. Any 2-copy PPT-BOTH product test for bipartite dd-dimensional product states
has bias which is O(1=d).
Proof. Letj ibe a bipartite dd-dimensional state on the system AB. Imagine we have a protocol
which takes as input two copies of j i, writtenj i1j i2, and attempts to determine whether j iis
product across systems A and B. Consider two distributions D0,D1on bipartite ddstates. Let
Mbe a measurement operator which accepts states drawn from D1with probability at least c, and
rejects states from D0with probability at least s. Then
E D1ED0trM( 
  
)c s;
implying
trM(E D1( 
 ) ED0(
))c s:
TakingD1to be the uniform distribution over product states, via a standard calculation we obtain
E D1( 
 ) =E E( A
B)
2=1
d(d+ 1)(I+F)A1A2

1
d(d+ 1)(I+F)B1B2
;
34where as elsewhere Fis the swap operator. Now let D0be the uniform distribution on bipartite
ddstates. In this case we have
ED0(
) =1
d2(d2+ 1)(I+F)12:
Thus
 := E D1( 
 ) ED0(
)
=1
d2(d+ 1)2(IA1A2
FB1B2+FA1A2
IB1B2) 2
d(d2+ 1)(d+ 1)2(I+F12):
Assume that Mis PPT across the 1:2 split. We want to maximise tr M assuming that  I
MIand IM I, where the second is the PPT constraint. Further, as distributions D0
andD1are invariant under product unitaries, we can without loss of generality assume that M
commutes with UA1
UA2
VB1
VB2for all unitaries UandV, which implies that
M=wI+x(FA1A2
IB1B2) +y(IA1A2
FB1B2) +zF12
for somew,x,y,z. By direct calculation
trM =1
d2(d+ 1)2 
2d3w+ (d2+d4)x+ (d2+d4)y+ 2d3z
+O(1=d)
=x+y+O(1=d):
On the other hand,
M =wI+xd(A1A2
IB1B2) +yd(IA1A2
B1B2) +zd2(A1A2
B1B2);
whereji=1p
dPd
i=1jiii. Sojx+yj=O(1=d), and we are done.
E Proof of correctness of the protocol to put QMA (k)inQMA (2)
This section proves several of the claims made in Section 3. First we prove Lemma 5 by showing
the validity of Protocol 2.
Lemma 5 (restatement). For anym,k,0s<c1,
QMAm(k)s;cQMAkm(2)s0;c0
wherec0=1+c
2ands0= 1 (1 s)2
100.
Proof. It is obvious that this protocol achieves completeness (1 + c)=2: if the Merlins follow the
protocol, the product test passes with certainty, so Arthur either accepts with probability 1, or
with the same probability that Aaccepts, which is at least c. Showing soundness is somewhat
more complicated.
Assume that Arthur receives states j1iandj2isuch the maximal overlap of j1i(resp.j2i)
with a product state is 1  1(resp. 1 2), and set=1
2(1+2). Further assume that the product
test would accept j1i
2(resp.j2i
2) with probability 1  1(resp. 1 2).
35Let 1 be the probability that the product test would accept j1i
j2i. We rst show
that this can be upper bounded in terms of the probabilities of accepting j1i
2andj2i
2. The
probability that the product test accepts is
1
2kX
S[k]tr(1)S(2)S1
2kX
S[k]q
tr(1)2
Sq
tr(2)2
S
1
2kX
S[k]tr(1)2
S+ tr(2)2
S
2
=1
2(Ptest(1) +Ptest(2))
= 1 1
2(1+2):
Thus we have the bound from Theorem 1 that 11
512. On the other hand, if Arther chooses to
measureM, then by Lemma 22 his probability of accepting is s+p1+p2
2s+p. Combining
the two tests, we nd that the acceptance probability is
s0max
512
111 + min(1;s+p)
21 (1 s)2
100: (26)
To obtain the last inequality, we observe that the worst case is obtained whenp= 1 s=q
512
11.
As a result of Eq. (26), a k-prover soundness- sprotocol can be simulated by a 2-prover protocol
with soundness s0. Ifkpoly(n), then the messages will still have a polynomial number of
qubits.
Remark. Our choice of protocol in Protocol 2 was carefully designed to yield a separable protocol.
One subtlety in doing so is that measurements in SEP do not compose the same way that separable
operations do. Indeed, if A;B2SEP, then it does notfollow that A1=2BA1=22SEP, even
in the case when Ais a projector. For example, let Aproject onto the symmetric subspace
and letB=j0;1ih0;1j. On the other hand, other choices of protocol could also yield separable
measurements. For example, ( M
I)P(M
I) would work (after using Lemma 8 to amplify
completeness), where Pdenotes the product test, Mis a measurement on Alice's ksystems and I
acts on Bob's ksystems. We are grateful to Fernando Brand~ ao for pointing out this issue to us.
Next, we prove Lemma 7.
Lemma 7 (restatement). For any`1,
QMASEP
m(k)s;cQMASEP
`m(k)s`;c`:
Proof. In the original protocol, Arthur performs a measurement Monkstates, each comprising m
qubits. If the input is a YES instance, then there exists a product state on which Mhas expectation
valuec, whereas if the input is a NO instance, then for all product states, Mhas expectation
values.
For the modied protocol, each of the kprovers submits `mqubits, and Arthur's measurement
isM
`. If the input is a YES instance, then the provers can submit `copies of the optimal input
to the original protocol. This state is still product (across the kprovers) and has probability c`
of being accepted.
36The more interesting case is when the input is a NO instance. In general, the provers can submit
states that are entangled across the `dierent parts of their message. However, there cannot be
any entanglement between the kdierent provers.
For the new protocol, we can imagine Arthur sequentially performing the measurement fM;I 
Mg`times and accepting only if the outcome is Meach time. Since the input is a NO instance, if
this measurement is applied to a state, pure or mixed, that is separable across the kparties, then
outcomeMwill occur with probability s. Therefore the probability that all `measurements have
outcomeMwill bes`as long as conditioning on outcome Mdoes not induce any entanglement
in the unmeasured states. We need not consider the state that remains after outcome I M, since
Arthur rejects immediately when this outcome occurs.
The nal step of the proof is to show that applying the measurement fM;I Mgto the rst
system of a multipartite product state and obtaining outcome Mwill not create any entanglement
across thekprovers. Suppose that the ithMerlin supplies the `m-qubit statej'(i)iPi
1:::Pi
`, where
eachPi
jis anm-qubit system. The original measurement Mis separable, so can be written as
M=X
jajj(1)
jih(1)
jj

j(k)
jih(k)
jj;
foraj0 andj(i)
jiunit vectors on mqubits. This measurement is applied sequentially to P1;:::;k
1,
thenP1;:::;k
2, and so on until P1;:::;k
`. (Here we write P1;:::;k
j as shorthand for P1
jPk
j.) When the
measurement is applied to P1;:::;k
1 and the outcome is M, the residual state is proportional to
trP1;:::;k
1MP1;:::;k
1
j'(1)ih'(1)jP1
1;:::;`

j'(k)ih'(k)jPk
1;:::;`
=X
jajkO
i=1h(1)
jjPi
1
IPi
2;:::;`j'(i)ih'(i)jPi
1;:::;`j(1)
jiPi
1
IPi
2::::;`;
which is separable across the P1
2;:::;`:P2
2;:::;`::Pk
2;:::;`cut. Therefore, by induction, the state
always remain separable as long as outcome Malways occurs.
For the reader's convenience, we briey summarise here the proof of Lemma 8 (originally due
to [50, 2]).
Lemma 8 (restatement). For any`1,
QMAm(k)s;cQMA`m(k)1 c s
3;1 exp( `(c s)2
2):
Proof. The idea is to repeat the basic protocol `times, and accept if there are c+s
2`\accept"
outcomes or reject otherwise. For YES instances, the provers can send `copies of the same proofs,
each of which will be accepted with probability c. Then a Cherno bound yields that the
completeness is1 exp( `(c s)2=2). For NO instances, each of the `copies has probability
sof being accepted. Since the provers may submit entangled states, we can no longer guarantee
that these events are independent, but still Markov's inequality implies that the probability of
c+s
2`accept outcomes is 2s
c+s=1
1+(c s)=21 c s
3. This last step uses the fact that
1=(1 +x=2)1 x=3, for 0x1.
We now complete the proof of Theorem 9.
37Theorem 9 (restatement). 1. Ifs1 1=poly(n),k= poly(n)andp(n)is an arbitrary
polynomial, then QMA (k)s;1=QMASEP(2)exp( p(n));1.
2. Ifc s1=poly(n),c < 1,k= poly(n)andp(n)is an arbitrary polynomial, then
QMA (k)s;c=QMASEP(2)exp( p(n));1 exp( p(n)).
Proof. For the case of perfect completeness, we use Lemma 6 to replace the original k-prover
protocol with a 2-prover protocol that still has perfect completeness, still has 1  1=q(n) soundness
(whereq(n) is still a polynomial of n) and now has a separable measurement performed by the
verier. Now we simply repeat (using Lemma 7) q(n)p(n) times, and obtain soundness (1 
1=q(n))q(n)p(n)e p(n).
When we merely have c s >1=q(n), for some polynomial q(n), then we rst have to apply
Lemma 8 with `=p(n) + log(243 q(n)2p(n)) to replace the soundness with 1  1=3q(n) and the
completeness with 1  exp( p(n))
243q(n)2p(n). Next, we apply Lemma 6 to leave the completeness the same,
reduce the number of provers to 2, guarantee the measurement is in SEP and replace the soundness
with 1 1=243q(n)2. Finally, we repeat 243 q(n)2p(n) times and obtain soundness exp(  p(n)) and
completeness 1 exp( p(n)).
F Proof of correctness of the product unitary test
This appendix is devoted to the proof of Theorem 19. In order to analyse the product unitary test
in Protocol 3, we will need to relate the maximum overlap of an n-qudit unitary with a product
operator to the maximum overlap of that unitary with a product unitary.
Lemma 24. GivenU2U(dn), let
1 = maxfjhU;A 1

Anij2:Ai2M(d);hAi;Aii= 1;1ing:
Then, if1=2, there exist V1;:::;Vn2U(d)such thatjhU;V 1

Vnij2(1 2)2.
Proof. For all 1in, let the polar decomposition of AibejAijCi, wherejAij=q
AiAy
iand
Ci2U(d). SetA=Nn
i=1Ai,C=Nn
i=1Ci. Then
hC;Ai=1
dnnY
i=1trCy
ijAijCi=1
dnnY
i=1trjAij=1
dnmax
V2U(dn)jtrVAjp
1 :
This implies that we can expand
U=p
1 A+D; C =p
1 0A+E
for some0and matrices D;E such thathD;Di=,hE;Ei=0,hA;Di= 0,hA;Ei= 0. So
jhU;Cij=jp
1 p
1 0+hD;Eijjp
1 p
1 0 pp
0j1 2;
for1=2. This implies the lemma.
We are now ready to prove correctness of the product unitary test.
38Theorem 19 (restatement). GivenU2U(dn), let
1 = maxfjhU;V 1

Vnij2:V1;:::;Vn2U(d)g:
Then, if= 0,Ptest(U) = 1 . If.0:106, thenPtest(U)1 1
4+1
162+1
83=2. If0:106.1,
Ptest(U)501=512. More concisely, Ptest(U) = 1 ().
Proof. By the Choi-Jamio lkowski isomorphism, there is a direct correspondence between operators
M2M(d) withjhM;Mij= 1 and normalised quantum states jv(M)i. If we dene
1 0:= maxfjhU;A 1

Anij2:Ai2M(d);hAi;Aii= 1;1ing;
then by Theorem 1, if 0.0:0265,Ptest(U)1 0+02+03=2, and if0&0:0265,Ptest(U)
501=512. If01=2, then the result follows immediately. On the other hand, by Lemma 24,
if01=2, there exist V1;:::;Vn2U(d) such thatjhU;V 1

Vnij2(1 20)21 40.
Thus we have1
40. The theorem follows by combining the bound on and the bound on
Ptest(U).
G Interpretation of the product test as an average over product
states
We have seen (via Lemma 2) that the probability of the product test passing when applied to
some statej i2(Cd)
nis equal to the average purity, across all choices of subsystem S[n],
of trj ih jS. One interpretation of the proof of correctness of the product test is therefore that,
if the average entanglement of j iacross all bipartite partitions of [ n] is low, as measured by the
purity, thenj imust in fact be close to a product state across all subsystems.
In this appendix, we discuss a similar interpretation of our results in terms of an average over
product states, via the following proposition.
Proposition 25. Givenj i2(Cd)
n,
Ptest(j ih j) =d(d+ 1)
2n
Ej1i;:::;jni
jh j1:::nij4
:
Proof. Similarly to before, let the input to the product test be two copies  A, Bof a state
 :=j ih j, and letFdenote the swap operator that exchanges systems A and B. Then
Ej1i;:::;jni
jh j1;:::;nij4
=Ej1i;:::;jni[tr( A
 B)((1

n)A
(1

n)B)]
= tr( A
 B) 
Eji[A
B]
n
= tr( A
 B)I+F
d(d+ 1)
n
=2
d(d+ 1)n
Ptest(j ih j):
We note that, in this interpretation, our main result is reminiscent of the so-called inverse
theorem for the second Gowers uniformity norm [33, 34], which we briey outline. Let f:f0;1gn!
Rbe some function such that1
2nP
xf(x)2= 1, and let the p-norms offon the Fourier side be
39dened ask^fkp=P
x2f0;1gn1
2nP
y2f0;1gn( 1)xyf(y)p1=p
:Then it is straightforward to show
thatk^fk4
1k^fk4
4k^fk2
1, where the quantity in the middle is known as the (fourth power of)
the second Gowers uniformity norm of f. That is,k^fk2
1(representing the largest overlap of fwith
a parity function) is well approximated by k^fk4
4(the average of the squared overlaps with parity
functions). This simple approximation has proven useful in arithmetic combinatorics [33].
Via the correspondence of Proposition 25, Theorem 1 shows that a similar result holds if we
replace the cube f0;1gnwith the space ( Cd)
n: the largest overlap with a product state can be
well approximated by the average squared overlap with product states. Note that if one attempts
to use the classical proof technique for the Gowers uniformity norm to prove this result, one does
not obtain Theorem 1, but a considerably weaker result containing a term exponentially large in n.
Intuitively, this is because the set of overlaps with parity functions for some function f:f0;1gn!R
is essentially arbitrary, whereas the set of overlaps of some state j iwith product states is highly
constrained.
Acknowledgements
We did most of this research while working at the University of Bristol. AM was supported by
the EC-FP6-STREP network QICS and an EPSRC Postdoctoral Research Fellowship. AWH was
supported by the EPSRC grant \QIP-IRC", by NSF grants 0916400, 0829937, 0803478, DARPA
QuEST contract FA9550-09-1-0044 and the IARPA MUSIQC and QCS contracts. We would like
to thank many people for inspiring discussions, including Boaz Barak, Salman Beigi, Fernando
Brand~ ao, Toby Cubitt, Sevag Gharibian, Leonid Gurvits, Julia Kempe, Hirotada Kobayashi, Haru-
michi Nishimura, Thomas Vidick, Andreas Winter and Xiaodi Wu. We would also like to thank
the FOCS and JACM referees for their helpful comments.
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44