Abstract
Quantum manybody systems whose Hamiltonians are nonstoquastic, i.e., have positive offdiagonal matrix elements in a given basis, are known to pose severe limitations on the efficiency of Quantum Monte Carlo algorithms designed to simulate them, due to the infamous sign problem. We study the computational complexity associated with ‘curing’ nonstoquastic Hamiltonians, i.e., transforming them into signproblemfree ones. We prove that if such transformations are limited to singlequbit Clifford group elements or general singlequbit orthogonal matrices, finding the curing transformation is NPcomplete. We discuss the implications of this result.
Introduction
The “negative sign problem”, or simply the “sign problem”^{1}, is a central unresolved challenge in quantum manybody simulations, preventing physicists, chemists, and material scientists alike from being able to efficiently simulate many of the most profound macroscopic quantum physical phenomena of nature, in areas as diverse as hightemperature superconductivity and material design through neutron stars to lattice quantum chromodynamics. More specifically, the sign problem slows down quantum Monte Carlo (QMC) algorithms^{2,3}, which are in many cases the only practical method available for studying large quantum manybody systems, to the point where they become practically useless. QMC algorithms evaluate thermal averages of physical observables by the (importance) sampling of quantum configuration space via the decomposition of the partition function into a sum of easily computable terms, or weights, which are in turn interpreted as probabilities in a Markovian process. Whenever this decomposition contains negative terms, QMC methods tend to converge exponentially slowly. Most dishearteningly, it is typically the systems with the richest quantum`mechanical behavior that exhibit the most severe sign problem.
In defining the scope under which QMC methods are signproblem free, the concept of “stoquasticity”, first introduced by Bravyi et al.^{4}, has recently become central. The most widely used definition of a local stoquastic Hamiltonian is
Definition 1^{5} A local Hamiltonian, \(H = \mathop {\sum}\nolimits_{a = 1}^M H_a\) is called stoquastic with respect to a basis \({\cal{B}}\), iff all the local terms H_{a} have only nonpositive offdiagonal matrix elements in the basis \({\cal{B}}\).
In the basis \({\cal{B}}\), the partition function decomposition of stoquastic Hamiltonians leads to a sum of strictly nonnegative weights and such Hamiltonians hence do not suffer from the sign problem. For example, in the pathintegral formulation of QMC with respect to a basis \({\cal{B}} = \{ b\}\), the partition function Z at an inverse temperature β is reduced to an Lfold product of sums over complete sets of basis states, {b_{1}}, …, {b_{L}}, which are weighted by the size of the imaginarytime slice Δτ = β/L and the matrix elements of e^{−ΔτH}. Namely, \(Z = \mathop {\prod}\nolimits_{l = 1}^L \mathop {\sum}\nolimits_{b_l} {\langle {b_l \,{\mathrm{e}}^{  \Delta \tau H} b_{l + 1}} \rangle }\), where L is the number of slices and periodic boundary conditions are assumed. For a Hamiltonian H that is stoquastic in the basis \({\cal{B}}\), all the matrix elements of e^{−ΔτH} are nonnegative for any Δτ, leading to nonnegative weights for each time slice. On the other hand, nonstoquastic Hamiltonians, whose local terms have positive offdiagonal entries, induce negative weights and generally lead to the sign problem^{1,6} unless certain symmetries are present.
The concept of stoquasticity is also important from a computational complexitytheory viewpoint. For example, the complexity class StoqMA associated with the problem of deciding whether the groundstate energy of stoquastic local Hamiltonians is above or below certain values, is expected to be strictly contained in the complexity class QMA, that poses the same decision problem for general local Hamiltonians^{4}. In addition, StoqMA appears as an essential part of the complexity classification of twolocal qubit Hamiltonian problem^{7}.
However, stoquasticity does not imply efficient (i.e., polynomialtime) equilibration. For example, finding the groundstate energy of a classical Ising model—which is trivially stoquastic—is already NPhard^{8}. Conversely, nonstoquasticity does not imply inefficiency: there exist numerous cases where an apparent sign problem (i.e., nonstoquasticity) is the result of a naive basis choice that can be transformed away, resulting in efficient equilibration^{7,9,10,11}. Here, we focus on the latter, i.e., whether nonstoquasticity can be “cured”.
To this end, we first propose an alternative definition of stoquasticity that is based on the computational complexity associated with transforming nonstoquastic Hamiltonians into stoquastic ones. Then, we proceed by proving that finding such a transformation for general local Hamiltonians, even if restricted to the singlequbit Clifford group or the singlequbit orthogonal group, is computationally hard. Along the way, we provide several results of independent interest, in particular an algorithm to efficiently group local Hamiltonian terms, and an algorithm to efficiently decide the curing problem using Pauli operators. We conclude by discussing some implications of our results, employing planted solution ideas, and also some potential cryptographic applications.
Results
Computationally stoquastic Hamiltonians
To motivate our alternative definition, we first note that any Hamiltonian can trivially be presented as stoquastic via diagonalization. However, the complexity of finding the diagonalizing basis generally grows exponentially with the size of the system (as noted in ref. ^{6}) and the new basis will generally be highly nonlocal and hence not efficiently representable. We also note that it is straightforward to construct examples where apparent nonstoquasticity may be transformed away. For example, consider the nspin Hamiltonian \(H_{XZ} = {\sum} \tilde J_{ij}X_iX_j  {\sum} J_{ij}Z_iZ_j\) with nonnegative \(J_{ij},\tilde J_{ij}\), where X_{i} and Z_{i} are the Pauli matrices acting on spin i. This Hamiltonian is nonstoquastic, but can easily be converted into a stoquastic form. Denoting the Hadamard gate (which swaps X and Z) by W, consider the transformed Hamiltonian \(H_{ZX} = W^{ \otimes n}H_{XZ}W^{ \otimes n} =  {\sum} J_{ij}X_iX_j + {\sum} \tilde J_{ij}Z_iZ_j\), which is stoquastic. The sign problem of the original Hamiltonian, H_{XZ}, can thus be efficiently cured by a unitdepth circuit of singlequbit rotations. Moreover, thermal averages are invariant under unitary transformations; namely, defining \(\left\langle A \right\rangle _H \equiv \frac{{{\mathrm{Tr}}(e^{  \beta H}A)}}{{{\mathrm{Tr}}(e^{  \beta H})}}\), it is straightforward to check that \(\left\langle A \right\rangle _H = \left\langle {UAU^\dagger } \right\rangle _{UHU^\dagger }\). Therefore, if QMC is run on the transformed, stoquastic Hamiltonian, it is no longer slowed down by the sign problem. Finally, note that Definition 1 implies that a local Hamiltonian, \(H = \mathop {\sum}\nolimits_{a = 1}^M H_a\), is stoquastic if all terms H_{a} are stoquastic. However, there always remains some arbitrariness in the manner in which the total Hamiltonian is decomposed into the various terms. Consider, e.g., H = −2X_{1} + X_{1}Z_{2}. The second term separately is nonstoquastic, whereas the sum is stoquastic. This suggests that the grouping of terms matters (see the Methods section, “Grouping terms without changing the basis”).
The above considerations motivate a reexamination of the concept of stoquasticity from a complexitytheory perspective, which can have important consequences for QMC simulations. (A related approach was discussed in ref. ^{12}.) For example, given a klocal nonstoquastic Hamiltonian \(H = \mathop {\sum}\nolimits_a H_a\) (where each summand is a klocal term, i.e., a tensor product of at most k nonidentity singlequbit Pauli operators), similarly to ref. ^{13}, we may ask whether there exists a constantdepth quantum circuit U such that \(H^{\prime} = UHU^\dagger\) can be written as a k′local stoquastic Hamiltonian \(H^{\prime} = \mathop {\sum}\nolimits_a H^{\prime}_a\) and if so, what the complexity associated with finding it is. It is the answer to the latter question that determines whether the Hamiltonian in question should be considered computationally stoquastic, i.e., whether it is feasible (in a complexity theoretic sense) to find a “curing” transformation U, which would then allow QMC to compute thermal averages with H by replacing it with H′. More formally, we propose the following definition:
Definition 2 A unitary transformation U “cures” a nonstoquastic Hamiltonian H (i.e., removes its sign problem) represented in a given basis if \(H^{\prime} = UHU^\dagger\) is stoquastic, i.e., its offdiagonal elements in the given basis are all nonpositive. A family of local Hamiltonians {H} represented in a given basis is efficiently curable (or, equivalently, computationally stoquastic) if there exists a polynomialtime classical algorithm such that for any member of the family H, the algorithm can find a unitary U and a Hamiltonian H′ with the property that \(H^{\prime} = UHU^\dagger\) is local and stoquastic in the given basis.
As an example, the Hamiltonian H_{XZ} considered above is efficiently curable. General local Hamiltonians are unlikely to be efficiently curable as this would imply the implausible result that QMA=StoqMA^{13}.
Note that given some class of basis transformations, our definition distinguishes between the ability to cure a Hamiltonian efficiently or in principle. For example, deciding whether a Pauli group element \(U = \mathop { \otimes }\nolimits_{i = 1}^n u_i\), where u_{i} belongs to the singlequbit Pauli group \({\cal{P}}_1 = \{ I,X,Y,Z\} \times \{ \pm 1, \pm i\}\), can cure each term {H_{a}} of a klocal Hamiltonian \(H = \mathop {\sum}\nolimits_a H_a\), which can be solved in polynomial time (see the Methods section, “Curing using Pauli operators”). However, the Hamiltonian H = X_{1}Z_{2} cannot be made stoquastic in principle using a Pauli group element, as conjugating it with Pauli operators results in ±X_{1}Z_{2}, both of which are nonstoquastic (see ref. ^{13} for the results on an intrinsic sign problem for local Hamiltonians). Therefore, the fact that general local Hamiltonians are not considered to be computationally stoquastic does not imply that curing is computationally hard for a given class of transformations.
The last example illustrates that, while the curing problem can be efficiently decided for the Pauli group, this group can cure a very limited family of Hamiltonians. This motivates us to consider the curing problem beyond the Pauli group. Our main result is a proof that even for particularly simple local transformations such as the singlequbit Clifford group and realvalued rotations, the problem of deciding whether a family of local Hamiltonians is curable cannot be solved efficiently, in the sense that it is equivalent to solving 3SAT and is hence NPcomplete.
We assume that a klocal Hamiltonian \(H = \mathop {\sum}\nolimits_a H_a\) is described by specifying each of the local terms H_{a}, and the goal is to find a unitary U that cures each of these local terms. In general, a unitary U that cures the total Hamiltonian H may not necessarily cure all H_{a} separately. However, for all of the constructions in this paper, we prove that a unitary U cures H if and only if it cures all H_{a} separately. The decomposition {H_{a}} is merely used to guarantee that verification is efficient and the problem is contained in NP.
Complexity of curing for the singlequbit Clifford group
To study the computational complexity associated with finding a curing transformation U, we shall consider for simplicity singlequbit unitaries \(U = \mathop { \otimes }\nolimits_{i = 1}^n u_i\) and only realvalued Hamiltonian matrices. As we shall show, even subject to these simplifying restrictions, the problem of finding a curing transformation U is computationally hard when \(U\) is not in the Pauli group.
We begin by considering the computational complexity of finding local rotations from a discrete and restricted set of rotations. Specifically, we consider the singlequbit Clifford group \({\cal{C}}_1\) (with group action defined as conjugation by one of its elements), defined as \({\cal{C}}_1 = \{ UUgU^\dagger \in {\cal{P}}_1\;\forall g \in {\cal{P}}_1\}\), i.e., the normalizer of \({\cal{P}}_1\). It is well known that \({\cal{C}}_1\) is generated by W and the phase gate P = diag(1, i)^{14}.
Theorem 1 Let \(U = \mathop { \otimes }\nolimits_{i = 1}^n u_i\), where u_{i}belongs to the singlequbit Clifford group. Deciding whether there exists a curing unitary U for 3local Hamiltonians is NPcomplete.
We prove this theorem by reducing the problem to the canonical NPcomplete problem known as 3SAT (3satisfiability)^{15}, beginning with the following lemma:
Lemma 1 Let u_{i} ∈ {I, W}, where I is the identity operation and W is the Hadamard gate. Deciding whether there exists a curing unitary \(U = \mathop { \otimes }\nolimits_{i = 1}^n u_i\) for 3local Hamiltonians is NPcomplete.
To prove Lemma 1, we first introduce a mapping between 3SAT and 3local Hamiltonians. Our goal is to find an assignment of n binary variables x_{i} ∈ {0, 1} such that the unitary \(W(x) \equiv \mathop { \otimes }\nolimits_{i = 1}^n W_i^{x_i}\) [where x ≡ (x_{1}, …, x_{n})] rotates an input Hamiltonian to a stoquastic Hamiltonian. We use the following 3local Hamiltonian as our building block:
where i, j, and k are three different qubit indices. It is straightforward to check that
is stoquastic (“True”) for any combination of the binary variables (x_{i}, x_{j}, x_{k}) except for (1, 1, 1), which makes Eq. (2) nonstoquastic (“False”).
This is precisely the truth table for the 3SAT clause \((\bar x_i \vee \bar x_j \vee \bar x_k)\), where ∨ denotes the logical disjunction and the bar denotes negation (see Table 1). We can define the other seven possible 3SAT clauses by conjugating \(H_{ijk}^{(111)}\) with Hadamard or identity gates:
The Hamiltonian \(W(x)H_{ijk}^{(\alpha \beta \gamma )}W^\dagger (x)\) is nonstoquastic (corresponds to a clause that evaluates to False) only when (x_{i}, x_{j}, x_{k}) = (α, β, γ), and is stoquastic (True) for any other choice of the variables x. We have thus established a bijection between 3local Hamiltonians \(H_{ijk}^{(\alpha \beta \gamma )}\), with (α, β, γ) ∈ {0, 1}^{3}, and the eight possible 3SAT clauses on three variables (x_{i}, x_{j}, x_{k}) ∈ {0, 1}^{3}. We denote these clauses, which evaluate to False iff (x_{i}, x_{j}, x_{k}) = (α, β, γ), by \(C_{ijk}^{(\alpha \beta \gamma )}\).
The final step of the construction is to add together such “3SATclause Hamiltonians” to form
where C is the set of all M clauses in the given 3SAT instance \(\wedge C_{ijk}^{(\alpha \beta \gamma )}\). Having established a bijection between 3SAT clauses and 3SATclause Hamiltonians, the final step is to show that finding x such that
is stoquastic for every H_{3SAT}, is equivalent to solving the NPcomplete problem of finding satisfying assignments x for the corresponding 3SAT instances. To prove the equivalence, we show (i) that satisfying a 3SAT instance implies that the corresponding H_{3SAT} is cured, and (ii) that if H_{3SAT} is cured, this implies that the corresponding 3SAT instance is satisfied.

(i)
Note that any assignment x that satisfies the given 3SAT instance also satisfies each individual clause. It follows from the bijection we have established that such an assignment cures each corresponding 3SATclause Hamiltonian individually. The stoquasticity of H′ then follows by noting that the tensor product of a stoquastic Hamiltonian with the identity matrix is still stoquastic and the sum of stoquastic Hamiltonians is stoquastic.

(ii)
To simplify the argument, we assume that each clause has exactly three variables. This version of 3SAT, sometimes called EXACT3SAT, remains NPhard^{15}. We prove that an unsatisfied 3SAT instance implies that the corresponding H_{3SAT} is not cured. It suffices to focus on a particular clause \(C_{ijk}^{(\alpha \beta \gamma )}\). The choice of variables that makes this clause False rotates the corresponding 3SATclause Hamiltonian to one that contains a nonstoquastic +X_{i}X_{j}X_{j} term, which generates positive offdiagonal elements in specific locations in the matrix representation of H_{3SAT}. In what follows, we show that no other 3SATclause Hamiltonian in H_{3SAT} contains a ±X_{i}X_{j}X_{k} term and therefore these positive offdiagonal elements cannot be canceled out or made negative regardless of the choice of the other variables in the assignment. To see this, we first note that a 3SATclause Hamiltonian that does not contain x_{i}, x_{j}, and x_{k}, cannot generate a ±X_{i}X_{j}X_{k} term. Second, a choice of assignment for x_{i}, x_{j}, and x_{k} that does not satisfy \(C_{ijk}^{(\alpha \beta \gamma )}\) would satisfy any other 3SAT clause on these three variables. A satisfied 3SAT clause also does not generate an X_{i}X_{j}X_{k} term. Therefore, the rotated Hamiltonian is guaranteed to be nonstoquastic.
This establishes that the problem is NPHard. Checking whether a given U cures all the local terms {H_{a}} is efficient and therefore the problem is NPcomplete.
To complete the proof of Theorem 1, let us consider the modified Hamiltonian
where H_{0} is manifestly stoquastic and c is any number larger than the maximum number of clauses that any variable appears in. As there are M clauses, we simply choose c = O(1)M, which is still a polynomial in the number of variables. (Note that even a restricted variant of 3SAT with each variable restricted to appear at most in a constant number of clauses is still NPComplete^{16} and therefore we can take c to be a constant.) The goal is to find a unitary \(U = \mathop { \otimes }\nolimits_{i = 1}^n u_i\) with \(u_i \in {\cal{C}}_1\) that cures \(\tilde H_{{\mathrm{3SAT}}}\). Note first that any choice of \(U = \otimes _{i = 1}^nW_i^{x_i}\) that cures H_{3SAT} also cures \(\tilde H_{{\mathrm{3SAT}}}\). Second, note that choosing any u_{i} that keeps H_{0} stoquastic is equivalent to choosing one of the elements of \({\cal{C}}_1^\prime \equiv \{ I,X,W,XW\} \subset {\cal{C}}_1\) (e.g., the phase gate, which is an element of \({\cal{C}}_1\), maps X to Y so it is excluded, as is WX, which maps Z to −X). Therefore, by choosing c to be large enough, any choice of \(u_i \in {\cal{C}}_1\backslash {\cal{C}}_1^\prime\) would transform \(\tilde H_{{\mathrm{3SAT}}}\) into a nonstoquastic Hamiltonian. It follows that if \(u_i \in {\cal{C}}_1\) and is to cure \(\tilde H_{{\mathrm{3SAT}}}\) then in fact it must be an element of \({\cal{C}}_1^\prime\).
Next, we note that conjugating a matrix by a tensor product of X or identity operators only shuffles the offdiagonal elements, but never changes their values (for a proof see the Methods section, “Conjugation by a product of X operators”). Therefore, for the purpose of curing a Hamiltonian, applying X is equivalent to applying I and applying XW is equivalent to applying W. With this observation, the set of operators that can cure a Hamiltonian is effectively reduced from {I, W, XW, X} to {I, W}. According to Lemma 1, deciding whether such a curing transformation exists is NPcomplete.
Complexity of curing for the singlequbit orthogonal group
Similarly, we can use Lemma 1 to show that the problem of curing the sign problem remains NPcomplete when the set of allowed rotations is extended to the continuous group of singlequbit orthogonal matrices, i.e., transformations of the form \(Q = \mathop { \otimes }\nolimits_{i = 1}^n q_i\), where \(q_i^Tq_i = I\) ∀i. Namely:
Theorem 2 Deciding whether there exists a curing orthogonal transformation Q for 6local Hamiltonians is NPcomplete.
See the Methods section, “Proof of Theorem 2”, for the proof. In analogy to the proof of Theorem 1, the crucial step is to show that by promoting each Z, X, and W to a twoqubit operator, the continuous set of possible curing transformations reduces to the discrete set considered in Lemma 1.
Implications and applications
An immediate and striking implication of Theorem 1 is that even under the promise that a nonstoquastic Hamiltonian can be cured by onelocal Clifford unitaries (corresponding to trivial basis changes), the problem of actually finding this transformation is unlikely to have a polynomialtime solution.
An interesting implication of Theorem 2 is the possibility of constructing “secretly stoquastic” Hamiltonians. That is, one may generate stoquastic quantum manybody Hamiltonians H_{stoq}, but present these in a “scrambled” nonstoquastic form \(H_{{\mathrm{nonstoq}}} = UH_{{\mathrm{stoq}}}U^\dagger\), where U is a tensor product of singlequbit orthogonal matrices (or in the general case a constant depth quantum circuit). We conjecture that the latter Hamiltonians will be computationally hard to simulate using QMC by parties that have no access to the “descrambling” circuit U. In other words, it is possible to generate efficiently simulable spin models that might be inefficient to simulate unless one has access to the “secret key” to make them stoquastic (see Fig. 1). This observation may potentially have cryptographic applications (see the Methods section, “Encryption based on secretly stoquastic Hamiltonians”).
Our work also has implications for the connection between the sign problem and the NPhardness of a QMC simulation. A prevailing view of this issue associates the origin of the NPhardness of a QMC simulation to the relation between a (“fermionic”) Hamiltonian that suffers from a sign problem and the corresponding (“bosonic”) Hamiltonian obtained by replacing every coupling coefficient by its absolute value. Consider the following example:. \(H_X = \mathop {\sum}\nolimits_{ij} J_{ij}X_iX_j\), with J_{ij} randomly chosen from the set {0, ±J} on a threedimensional lattice, has a sign problem. Deciding whether its groundstate energy is below a given bound is NPcomplete^{8}. Deciding the same for its bosonic and signproblemfree version \(H_{X} = \mathop {\sum}\nolimits_{ij} J_{ij}X_iX_j\) is in BPP (classical polynomial time with bounded error) since this Hamiltonian is that of a simple ferromagnet. The conclusion drawn in ref. ^{6} was that since the bosonic version is easy to simulate, the sign problem is the origin of the NPhardness of a QMC simulation of this model (H_{X}).
The view we advocate here is that a solution to the sign problem is to find an efficiently computable curing transformation that removes it in such a way that the model has the same physics (in general the fermionic and bosonic versions of the same Hamiltonian do not), i.e., conserves thermal averages. In the above example, computing thermal averages via a QMC simulation of H_{X} is the same as for \(H_Z = W^{ \otimes n}H_XW^{ \otimes n} = \mathop {\sum}\nolimits_{ij} J_{ij}Z_iZ_j\), which is stoquastic. Thus, the sign problem of H_{X} is efficiently curable, after which (when it is presented as H_{Z}) deciding its groundstate energy remains NPhard.
Discussion
We have proposed an alternative definition of stoquasticity (or absence of the sign problem) of quantum manybody Hamiltonians that is motivated by computational complexity considerations. We discussed the circumstances under which nonstoquastic Hamiltonians can in fact be made stoquastic by the application of singlequbit rotations and in turn potentially become efficiently simulable by QMC algorithms. We have demonstrated that finding the required rotations is computationally hard when they are restricted to the onequbit Clifford group or onequbit continuous orthogonal matrices.
These results raise multiple questions of interest. It is important to clarify the computational complexity of finding the curing transformation in the case of constantdepth circuits that also allow twobody rotations, whether discrete or continuous. Also, since our NPcompleteness proof involved 3 and 6local Hamiltonians, it is interesting to try to reduce it to 2local building blocks. Another direction into which these results can be extended is to relax the constraints on the offdiagonal elements and require that they are smaller than some small ε > 0. This is relevant when some small positive offdiagonal elements can be ignored in a QMC simulation.
Finally, it is natural to reconsider our results from the perspective of quantum computing. Namely, for nonstoquastic Hamiltonians that are curable, do there exist quantum algorithms that cure the sign problem more efficiently than is possible classically? With the advent of quantum computers, specifically quantum annealers, it may be the case that these can be used as quantum simulators, and as such they will not be plagued by the sign problem. Will such physical implementations of quantum computers offer advantages over classical computing even for problems that are incurably nonstoquastic? We leave these as open questions to be addressed in future studies.
Methods
Proof of Theorem 2
The proof builds on that of Theorem 1, but first we note that the clause Hamiltonians introduced in Eq. (2) can have curing solutions that are orthogonal rotations outside the Clifford group (see Fig. 2). To deal with this richer set of rotations—which is now a continuous group—we promote each Z, X, and W in the clause Hamiltonians to a twoqubit operator: \(Z_i \ \mapsto \ \bar Z_i \equiv Z_{2i  1}Z_{2i}\), \(X_i \ \mapsto \ \bar X_i \equiv X_{2i  1}X_{2i}\), \(W_i^\alpha \ \mapsto \ \overline W_i^\alpha \equiv W_{2i  1}^\alpha \otimes W_{2i}^\alpha\). Thus Eq. (1) becomes
Let \(\bar W(x) \equiv \mathop { \otimes }\nolimits_{i = 1}^n \bar W_i^{x_i}\). It is again straightforward to check that \(\bar W(x)\bar H_{ijk}^{(111)}\bar W^\dagger (x)\) is stoquastic for any combination of the binary variables (x_{i}, x_{j}, x_{k}) except for (1, 1, 1). Likewise, generalizing Eq. (3), we define
\(\bar H_{ijk}^{(\alpha \beta \gamma )}\) is a clause Hamiltonian corresponding to a clause in the 3SAT instance. Similarly, \(\bar W(x)\bar H_{ijk}^{(\alpha \beta \gamma )}\bar W(x)\) is stoquastic for any combination of the binary variables (x_{i}, x_{j}, x_{k}) except for (α, β, γ). Generalizing Eq. (4) we define
where C denotes the corresponding set of clauses in a 3SAT instance, constructed just as in the proof of Lemma 1. This, again, establishes a bijection between 3SAT clauses and “3SATclause Hamiltonians”, now of the form \(\bar H_{{\mathrm{3SAT}}}\).
In lieu of Eq. (4), we consider the 6local Hamiltonian \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\)
where c = O(1). Just as in the proof of Theorem (1), we prove that (i) any satisfying assignment of a 3SAT instance provides a curing Q for the corresponding Hamiltonian \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\), and (ii) any Q that cures \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) provides a satisfying assignment for the corresponding 3SAT instance. Because of the relation between a singlequbit orthogonal matrix and a singlequbit rotation, it suffices to prove the hardness only for pure rotations (see the next subsection for the relation between a singlequbit real–orthogonal matrix and a singlequbit rotation); we let \(R(\theta _i) = \left[ {\begin{array}{*{20}{c}} {{\mathrm{cos}}\theta _i} & {  {\mathrm{sin}}\theta _i} \\ {{\mathrm{sin}}\theta _i} & {{\mathrm{cos}}\theta _i} \end{array}} \right]\) denote a rotation by angle θ_{i}.

(i)
Let P(x) denote the product of 2n singlequbit rotations such that if x_{i} = 0 then qubits 2i − 1 and 2i are unchanged, or if x_{i} = 1 then they are both rotated by \(R\left( {\frac{\pi }{4}} \right)\):
$$P(x) \equiv \otimes _{i = 1}^n\left( {R\left( {\frac{\pi }{4}} \right)^{x_i} \otimes R\left( {\frac{\pi }{4}} \right)^{x_i}} \right),$$(11)where x is a nbit string x = (x_{1}, …, x_{n}). Let us now show that if the 3SAT instance has a satisfying assignment x^{*} then \(P(x^ \ast )\widetilde {\bar H}_{{\mathrm{3SAT}}}P^T(x^ \ast )\) is stoquastic. Note that \(R\left( {\frac{\pi }{4}} \right) = XW\), and as we discussed in the article (under “Complexity of curing for the singlequbit Clifford group”) it is equivalent to W for curing.
To prove the claim, note that x^{*} necessarily satisfies each individual clause of the 3SAT instance, and therefore makes the corresponding clause Hamiltonian stoquastic, i.e., \(P(x^ \ast )\bar H_{ijk}^{(\alpha \beta \gamma )}P^T(x^ \ast )\) is stoquastic \(\forall C_{ijk}^{(\alpha \beta \gamma )}\). Also, \(P(x)H_0^\prime P^T(x)\) is clearly stoquastic for any x, where \(H_0^\prime =  \mathop {\sum}\nolimits_{i = 1}^n 2\bar Z_i + \bar X_i\) [Eq. (10)]. The stoquasticity of \(P(x^ \ast )\widetilde {\bar H}_{{\mathrm{3SAT}}}P^T(x^ \ast )\) then follows immediately.

(ii)
We need to prove that any rotation that cures \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) provides a satisfying assignment for the corresponding 3SAT instance. We do this in two steps:

(a)
Below, under “A useful lemma”, we prove that for any \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\), any curing rotation \(R = \mathop { \otimes }\nolimits_{i = 1}^{2n} R(\theta _i)\) has to satisfy the condition that \((\theta _{2i  1},\theta _{2i}) \in \left\{ {\left( {\frac{\pi }{2},\frac{\pi }{2}} \right),\left( {\frac{\pi }{4},\frac{\pi }{4}} \right),(0,0),\left( {\frac{{  \pi }}{4},\frac{{  \pi }}{4}} \right)} \right\}\) ∀i. This is the crucial step, since it reduces the problem from a continuum of angles to a discrete set. If \((\theta _{2i  1},\theta _{2i}) \in \left\{ {(0,0),\left( {\frac{\pi }{2},\frac{\pi }{2}} \right)} \right\}\), we assign x_{i} = 0, while if \((\theta _{2i  1},\theta _{2i}) \in \left\{ {\left( {  \frac{\pi }{4},  \frac{\pi }{4}} \right),\left( {\frac{\pi }{4},\frac{\pi }{4}} \right)} \right\}\), we assign x_{i} = 1, since rotations with the angles in each pair have the same effect.

(b)
If R cures \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\), x = {x_{i}} satisfies the corresponding 3SAT instance. See Fig. 3 for the discrete set of solutions and the corresponding assignments.
To see this, we first note that if R cures \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) it must cure all the clauses separately: Using step (a), we know that any such solution must be one of the four possible cases. Therefore, if R were to cure \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) but does not cure one of the 3SATclause Hamiltonians, it would result in a \(\bar X_i\bar X_j\bar X_k\) term in the corresponding clause. Since no other 3SATclause Hamiltonian in \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) contains an identical \(\bar X_i\bar X_j\bar X_k\) term, these positive offdiagonal elements cannot be canceled out or made negative, regardless of the choice of the other variables in the assignment. Therefore, if R cures \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\), it also necessarily separately cures all the terms in \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\). By construction, if R cures a term \(\bar H_{ijk}^{(\alpha \beta \gamma )}\), the string x satisfies the corresponding 3SAT clause \(C_{ijk}^{(\alpha \beta \gamma )}\). Thus x satisfies all the clauses in the corresponding 3SAT instance. The decision problem for the existence of R (and hence Q) is therefore NPhard. Given a unitary U and a set of local terms \(\{ H_a\}\), verifying whether U cures all of the terms is clearly efficient and therefore this problem is NPcomplete.
Relation between an orthogonal matrix and a rotation
The condition \(q_iq_i^T = I\) forces each real–orthogonal matrix q_{i} to be either a reflection or a rotation of the form
with a_{i} = +1 (a reflection) or a_{i} = −1 (a rotation). The operators X, Z, and Hadamard, are included in the family with a_{i} = 1; I and iY = XZ are in the family with a_{i} = −1. Note that \(\forall H,\forall \theta _i:{\kern 1pt} q_i(\theta _i)Hq_i^T(\theta _i) = q_i(\theta _i + \pi )Hq_i^T(\theta _i + \pi )\). Therefore, the angles that cure a Hamiltonian are periodic with a period of π. Hence, it suffices to consider the curing solutions only in one period: \(\theta _i \in \left. {\left( {\frac{{  \pi }}{2},\frac{{ + \pi }}{2}} \right.} \right]\).
Next, observe that a reflection by angle θ_{i} can be written as
where
As discussed below (under “Conjugation by a product of X operators”), if \(XR\left( {\frac{\pi }{2}  \theta _i} \right)\) is a curing operator so is \(R\left( {\frac{\pi }{2}  \theta _i} \right)\). Therefore, any curing \(Q = \mathop { \otimes }\nolimits_{i = 1}^n q_i\) provides a curing \(R = \mathop { \otimes }\nolimits_{i = 1}^n R(\theta _i)\). Hence, the NPcompleteness of the decision problem for R implies the NPhardness of the decision problem for Q, which is the statement of Theorem 2.
A useful lemma
Here, we prove that any curing rotation \(R = \mathop { \otimes }\nolimits_{i = 1}^{2n} R(\theta _i)\) for any \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\), as introduced in Eq. (10), must satisfy the condition that \((\theta _{2i  1},\theta _{2i}) \in \left\{ {\left( {\frac{\pi }{2},\frac{\pi }{2}} \right),\left( {\frac{\pi }{4},\frac{\pi }{4}} \right),(0,0),\left( {\frac{{  \pi }}{4},\frac{{  \pi }}{4}} \right)} \right\}\) ∀i. In what follows, we show this for i = 1 (local rotations on the first two qubits), but the proof trivially works for any choice of i.
Our strategy is to expand any locally rotated \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) on the first two qubits and then to find necessary conditions that any curing rotations on these two qubits must satisfy. With this motivation, we introduce the following lemma:
Lemma 2 Let \(\theta _1,\theta _2 \in \left( {  \frac{\pi }{2},\ \frac{\pi }{2}} \right]\). Consider the 2nqubit Hamiltonian
where M_{z}, M_{x}, and M_{I}are Hamiltonians on 2n−2 qubits satisfying the following two conditions:

1.
The absolute value of at least one element of M_{z} is different from the absolute value of the corresponding element in M_{x}.

2.
Both M_{x} and M_{z} have at least one negative element.
Then the only rotation R(θ_{1}) ⊗ R(θ_{2}) that can cure H′ has angles given by the following four points:
To relate this lemma to our construction, note that any locally rotated \(\widetilde {\bar H}_{{\mathrm{3SAT}}}\) is in the form of H′. To be more precise, we will choose \(H{\prime} = R{\prime}\widetilde {\bar H}_{{\mathrm{3SAT}}}R^{\prime T}\), where \(R{\prime} = \mathop { \otimes }\nolimits_{i = 3}^{2n} R(\theta _i)\).
We proceed by first proving the lemma. Then we show that both of the conditions introduced in the lemma are satisfied for our construction.
Proof. It is straightforward to check that
Using this we have
We next find necessary conditions that θ_{1} and θ_{2} must satisfy in order to make H′′ stoquastic.
Let A to E be arbitrary matrices, and let [0] denote the allzero matrix. We note that H′′ can be stoquastic only if the X_{1}Z_{2} ⊗ B term is zero (where we have dropped the tensor product between the first two qubits for notational simplicity). To see this, we first note that the matrix X_{1}Z_{2} ⊗ B has both +B and −B as distinct offdiagonal elements for any nonzero matrix B (a similar observation holds for Z_{1}X_{2} ⊗ C). Second, we note that there are no common offdiagonal elements between X_{1}X_{2} ⊗ A, X_{1}Z_{2} ⊗ B, and Z_{1}X_{2} ⊗ C. Third, there is no common offdiagonal elements between these three matrices and Z_{1}Z_{2} ⊗ D and I_{1}I_{2} ⊗ E. Therefore, these terms cannot make the ±B in X_{1}Z_{2} ⊗ B nonpositive.
Thus, for H′′ to become stoquastic it is necessary to have B = [0]:
Similar reasoning for Z_{1}X_{2} ⊗ C yields
If the absolute value of at least one element of M_{z} is different from the absolute value of the corresponding element in M_{x} (i.e., Condition 1 is satisfied), comparing the corresponding expressions from Eqs. (20) and (21), we can conclude that
Equation (22) gives rise to two possible cases: (i) sin 2θ_{1} = sin 2θ_{2} = 0, or (ii) cos 2θ_{1} = cos 2θ_{2} = 0. For \(\theta _1,\theta _2 \in \left. {\left( {  \frac{\pi }{2},\frac{\pi }{2}} \right.} \right]\). These are the eight possible solutions:
Now, we observe that Condition 2 generates additional constraints on the allowed values of θ_{1} and θ_{2}. To see this, we consider the X_{1}X_{2} ⊗ A term in Eq. (19) in the two possible cases. For case (i), when sin 2θ_{1} = sin 2θ_{2} = 0 and hence cos 2θ_{1} = ±1 and cos 2θ_{2} = ±1, this term becomes X_{1}X_{2} ⊗ cos 2θ_{1}cos 2θ_{2}M_{x}. If M_{x} has any negative element, then the two combinations such that cos 2θ_{1}cos 2θ_{2} = −1 flip the sign of this element and make the term nonstoquastic. That is, if M_{x} has any negative element, the only rotations that can keep H′′ stoquastic satisfy cos 2θ_{1} = cos 2θ_{2} = 1 or cos 2θ_{1} = cos 2θ_{2} = −1. Similarly, for case (ii), when cos 2θ_{1} = cos 2θ_{2} = 0, if M_{z} has any negative elements, the only rotations that can keep H′′ stoquastic satisfy sin 2θ_{1} = sin 2θ_{2} = 1 or sin 2θ_{1} = sin 2θ_{2} = −1.
To summarize, if both conditions hold, the solutions are necessarily one of these four points:
Having proved the lemma, we now proceed with identifying the properties of M_{z} and M_{x} for our construction. As mentioned earlier, we choose \(H{\prime} = R{\prime}\widetilde {\bar H}_{{\mathrm{3SAT}}}R^{\prime T}\), where \(R{\prime} = \mathop { \otimes }\nolimits_{i = 3}^{2n} R(\theta _i)\).
M_{z} can be written as H_{z}−2I, where H_{z} denotes the terms in M_{z} coming from \(\bar H_{{\mathrm{3SAT}}}\). Similarly, M_{x} = H_{x} − I, where H_{x} denotes the terms in M_{x} coming from \(\bar H_{{\mathrm{3SAT}}}\). The term \(\bar Z_1 \otimes H_z\) (recall that \(\bar Z_i \equiv Z_{2i  1}Z_{2i}\) and \(\bar X_i \equiv X_{2i  1}X_{2i}\)) is composed of rotated 3SATclause Hamiltonians that share \(\bar x_1\) in their corresponding 3SAT clauses. Therefore, we have
It is straightforward to check that each of the rotated 3SATclause Hamiltonians has only nonpositive diagonal elements. Namely, using Eq. (17), it is straightforward to check that the max norm, defined as \(\left\Vert A \right\Vert_{max} = max_{ij}[A]_{ij}\), of any rotated Pauli operator is at most 1, and therefore the same is true for any tensor product of rotated Pauli operators. In each 3SATclause Hamiltonian, there are three nonidentity Pauli terms. Therefore, they cannot generate a diagonal element that is larger than 3. There is a −3 term for each clause, guaranteeing that all the diagonal terms remain nonpositive.
As H_{z} is a sum of these matrices with all nonpositive diagonal elements, we conclude that all the diagonal elements of H_{z} are nonpositive. H_{x} is similar to H_{z}, but with a sum over \(C_{1jk}^{(0\beta \gamma )}\). Using similar arguments, we conclude that all the diagonal elements of H_{x} are nonpositive.
Therefore, all the diagonal elements of M_{x} = H_{x} − I and M_{z} = H_{z} − 2I are negative, and we conclude that Condition 2 is satisfied for our construction.
Now, we show that that the first condition also holds. Using the cyclic property of the trace and noting that all the terms in H_{z} except the −3 are traceless, we have Tr(H_{z}) = −3k with \(k \in {\Bbb N}_0\) (\(k = 0\) only if \(H_z = 0\), i.e., when there is no \(\bar x_1\) in any of the 3SAT clauses). Therefore we have \({\mathrm{Tr}}(H_z  2I) =  3k  2^{2n  1}\). Using similar arguments, we conclude that \({\mathrm{Tr}}(H_x) =  3k{\prime}\) with \(k{\prime} \in {\Bbb N}_0\) and \({\mathrm{Tr}}(H_x  I) =  3k{\prime}  2^{2n  2}\) where \(k{\prime} \in {\Bbb N}_0\). (\(k{\prime} = 0\) only if H_{x} = 0, i.e., when there is no x_{1} in any of the 3SAT clauses).
Clearly, the two traces cannot be equal for any value of k and k′. From this, in addition to the alreadyestablished fact that all the diagonal elements of H_{x} − I and H_{z} − 2I are negative, we conclude that at least one diagonal element of H_{x}−I is different from the corresponding element of H_{z} − 2I. Therefore, Condition 1 is also satisfied.
Grouping terms without changing the basis
As discussed here, one ambiguity in the definition of stoquastic Hamiltonians is in the choice of the set {H_{a}}. With this motivation, and ignoring the freedom in choosing a basis, we address the following question.
Problem: We are given the klocal \(H = \mathop {\sum}\nolimits_a {H_a}\), i.e., each H_{a} acts nontrivially on at most k qubits. In the same basis (without any rotation), find a new set \(H_a^\prime\) satisfying \(H = \mathop {\sum}\nolimits_a {H_a^\prime }\), where each H′_{a} is k′local and stoquastic (if such a set exists).
Obviously, if the total Hamiltonian is stoquastic, then considering the total Hamiltonian as one single Hamiltonian is a valid solution with k′ = n. This description of the Hamiltonian requires a 2^{n} × 2^{n} matrix. We would prefer a k′local Hamiltonian, i.e., a set consisting of a polynomial number of terms, each 2^{k′} × 2^{k′}, where k′ is a constant independent of n.
Solution: One simple strategy is to consider any k′local combination of qubits, and to try to find a grouping that makes all of these \(( {\begin{array}{*{20}{c}} n \\ {k\prime } \end{array}} )\) terms stoquastic. To do so, for any k′local combination of qubits, we generate a set of inequalities. First, for a fixed combination of qubits, we add the terms in \(H = \mathop {\sum}\nolimits_a {H_a}\) that act nontrivially only on those k′ qubits, each with an unknown weight that will be determined later. Then we write down the conditions on the weights to ensure that all the offdiagonal elements are nonpositive. This is done for all the \(( {\begin{array}{*{20}{c}} n \\ {k\prime } \end{array}} )\) combinations to get the complete set of linear inequalities. By this procedure, the problem reduces to finding a feasible point for this set of linear inequalities, which can be solved efficiently. (In practice, one can use linear programming optimization tools to check whether such a feasible point exists.) When there is no feasible point for a specific value of k′, we can increase the value of k′ and search again.
Example: Assume we are given H = Z_{1}X_{2}−2X_{2} + X_{2}Z_{3} and the goal is to find a stoquastic description with k′ = 2. We combine the terms acting on qubits 1 and 2 and then the terms acting on qubits 2 and 3 (there is no term on qubits 1 and 3). We construct h_{1,2} = α_{1}Z_{1}X_{2} + α_{2}(−2X_{2}) and h_{2,3} = α_{3}(−2X_{2}) + α_{4}X_{2}Z_{3}. There are two types of constraints: (1) constraints enforcing H = h_{1,2} + h_{2,3}:
and (2) constraints from stoquasticity of each of the two Hamiltonians:
Simplifying these inequalities, we have 0.5 ≤ α_{2}, α_{3} and α_{2} + α_{3} = 1, which clearly has only one feasible point: α_{2} = α_{3} = 0.5. The corresponding terms are H′_{1} = h_{1,2} = Z_{1}X_{2}−X_{2} and H′_{2} = h_{2,3} = −X_{2} + X_{2}Z_{3}. Both of these terms are stoquastic and they satisfy \(H = \mathop {\sum}\nolimits_a {H_a^\prime }\).
Curing using Pauli operators
In the next subsection, we show that conjugating a Hamiltonian by a tensor product of Pauli X operators or identity operators only shuffles the offdiagonal elements without changing their values. Recalling that Y = iXZ, we thus conclude that choosing between Pauli operators to cure a Hamiltonian is equivalent to choosing between I and Z operators. Therefore, given local terms of a klocal Hamiltonian {H_{a}} as input, the goal is to find a string x = (x_{1}, …, x_{n}) such that \(U = \otimes _{i = 1}^nZ^{x_i}\) cures each of the local terms {H_{a}} separately.
Each multiqubit Pauli operator in H_{a} can be decomposed into X components and Z components. We group all the terms in each H_{a} that share the same X component. For example, if H_{a} includes Y_{1}Y_{2}, 3X_{1}X_{2}, and X_{1}X_{2}Z_{3}, we combine them into one single term X_{1}X_{2}(−Z_{1}Z_{2} + 3 + Z_{3}). Conjugating this term with U yields \((  1)^{x_1 + x_2}X_1X_2(  Z_1Z_2 + 3 + Z_3)\). As terms with different X components do not correspond to overlapping offdiagonal elements in H_{a}, the combined Z part fixes a constraint on {x_{i}} based on the positivity or negativity of all its elements (if the combined Z part has both positive and negative elements, we conclude that there is no U that can cure the input H). In this example, \((  1)^{x_1 + x_2}X_1X_2(  Z_1Z_2 + 3 + Z_3)\) becomes stoquastic iff x_{1} + x_{2} ≡ 1 mod 2.
We combine all these linear equations in mod 2 that are generated from terms with different X components, and solve for a satisfying x. This can be done efficiently, e.g., using Gaussian elimination. The absence of a consistent solution implies the absence of a curing Pauli group element.
As the dimension of each of the local terms {H_{a}} is independent of the number of qubits n, and there are at most poly(n) of these terms, the entire procedure takes poly(n) time.
Conjugation by a product of X operators
Here, we show that conjugating a Hamiltonian by a tensor product of X’s or identity operators only shuffles the offdiagonal elements without changing their values.
Lemma 3 Let \(U_X = X^{a_1} \otimes \ldots \otimes X^{a_n}\), where a_{i} ∈ {0, 1}. The set of offdiagonal elements of a general 2^{n} × 2^{n}matrix B, OFF(B) = {[B]_{ij}i, j ∈ {0, 1}^{n}, i ≠ j}, is equal to the set of offdiagonal elements of U_{X}BU_{X}for all possible {a_{i}}.
Proof. Let a = (a_{1},...,a_{n}). Similarly, let i and j represent nbit strings. The elements of the matrix U_{X}BU_{X} are
where \(\odot\) denotes the XOR operation. Clearly for any fixed a, we have \(i \ne j \Leftrightarrow i \odot a \ne j \odot a\) and therefore
A similar argument proves that U_{X} shuffles the diagonal elements: DIA(U_{X}BU_{X}) = DIA(B).
Therefore, conjugating a Hamiltonian by U_{X} does not change the set of inequalities one needs to solve to make a Hamiltonian stoquastic. As a consequence, whenever we want to pick u_{i} as a solution, we can instead choose Xu_{i}.
Onelocal rotations are not enough
Consider, e.g., the three Hamiltonians
The sum of any pair of these Hamiltonians can be cured by singlequbit unitaries (e.g., H_{2} = H_{1,2} + H_{2,3} can be cured by applying U = Z_{2}).
In contrast, the frustrated Hamiltonian H = H_{12} + H_{23} + H_{13} cannot be cured using any combination of singlequbit rotations. To see this, we first note that the partial trace of a stoquastic Hamiltonian is necessarily stoquastic. By partial trace over single qubits of H, we conclude that in order for H to be stoquastic, all three H_{ij}’s must be stoquastic. To find all the solutions that convert each of these Hamiltonians into a stoquastic Hamiltonian, we expand \(R_i(\theta _i) \otimes R_j(\theta _j)H_{ij}R_i^T(\theta _i) \otimes R_j^T(\theta _j)\) and note that it has ±sin 2(θ_{i} − θ_{j}) and cos 2(θ_{i} − θ_{j}) as offdiagonal elements. Demanding that the rotated Hamiltonians are all stoquastic [so that sin 2(θ_{i} − θ_{j}) = 0] forces cos 2(θ_{i} − θ_{j}) = −1 ∀(i, j) ∈ {(1, 2), (2, 3), (3, 1)}. But this set of constraints does not have a feasible point. To see this note that
Therefore, H cannot be made stoquastic using 2 × 2 rotational matrices. Because of the relation between rotation and orthogonal matrices discussed above, we conclude that H cannot be made stoquastic using 2 × 2 orthogonal matrices.
Encryption based on secretly stoquastic Hamiltonians
By generating 3SAT instances with planted solutions (see, e.g., refs. ^{17,18}) and transforming these to nonstoquastic Hamiltonians via the mapping prescribed by Theorems 1 (or 2), one would be able to generate 3local (or 6local) Hamiltonians that are stoquastic, but are computationally hard to transform into a stoquastic form.
This construction may have cryptographic implications. For example, imagine planting a secret nbit message in the (unique by design) ground state of a stoquastic Hamiltonian. Since the solution is planted, Alice automatically knows it. She checks that QMC can find the ground state in a prescribed amount of time \(\tau(n)\), and if this is not the case, she generates a new, random, stoquastic Hamiltonian with the same planted solution and checks again, etc., until this condition is met. Alice and Bob preshare the secret key, i.e., the curing transformation, and after they separate, Alice transmits only the O(n^{2}) coefficients of the nonstoquastic Hamiltonians (transformed via the mapping prescribed by Theorem 1) for every new message she wishes to send to Bob. To discover Alice’s secret message, Bob runs QMC on the cured Hamiltonian. Since Alice verified that QMC can find the ground state in polynomial time, Bob will also find the ground state in polynomial time.
This scheme should be viewed as merely suggestive of a cryptographic protocol, since as it stands it contains several potential loopholes: (i) its security depends on the absence of efficient twoormore qubit curing transformations, as well as the absence of algorithms other than QMC that can efficiently find the ground state of the nonstoquastic Hamiltonians generated by Alice; (ii) the fact that Alice must start from Hamiltonians for which the ground state can be found in polynomial time may make the curing problem easy as well; (iii) this scheme transmits an nbit message using rn^{2} message bits, where r is the number of bits required to specify the n^{2} coefficients of the transmitted nonstoquastic Hamiltonian, so it is less efficient than a onetime pad. Additional research is needed to improve this into a scheme that overcomes these objections.
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Acknowledgements
The research is based upon work (partially) supported by the Office of the Director of National Intelligence (ODNI), Intelligence Advanced Research Projects Activity (IARPA), via the U.S. Army Research Office contract W911NF17C0050. The views and conclusions contained herein are those of the authors and should not be interpreted as necessarily representing the official policies or endorsements, either expressed or implied, of the ODNI, IARPA, or the U.S. Government. The U.S. Government is authorized to reproduce and distribute reprints for Governmental purposes notwithstanding any copyright annotation thereon. We thank Ehsan EmamjomehZadeh, Iman Marvian, Evgeny Mozgunov, Ben Reichardt, and Federico Spedalieri for useful discussions.
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Research Laboratory of Electronics, Massachusetts Institute of Technology, Cambridge, MA, 02139, USA
 Milad Marvian
Department of Electrical and Computer Engineering, University of Southern California, Los Angeles, CA, 90089, USA
 Milad Marvian
 & Daniel A. Lidar
Center for Quantum Information Science & Technology, University of Southern California, Los Angeles, CA, 90089, USA
 Milad Marvian
 , Daniel A. Lidar
 & Itay Hen
Department of Physics and Astronomy, University of Southern California, Los Angeles, CA, 90089, USA
 Daniel A. Lidar
 & Itay Hen
Department of Chemistry, University of Southern California, Los Angeles, CA, 90089, USA
 Daniel A. Lidar
Information Sciences Institute, University of Southern California, Marina del Rey, CA, 90292, USA
 Itay Hen
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Contributions
I.H. conceived of the project. M.M. devised most aspects of the technical proofs. I.H., D.A.L. and M.M. contributed equally to discussions and writing the paper.
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The authors declare no competing interests.
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Correspondence to Milad Marvian.
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