Authors Andrew M. Childs, David Gosset, Zak Webb,
License CC-BY-3.0
T HEORY OF C OMPUTING, Volume 11 (20), 2015, pp. 491–603
www.theoryofcomputing.org
The Bose-Hubbard Model is QMA-complete
Andrew M. Childs David Gosset Zak Webb
Received July 31, 2014; Revised June 25, 2015; Published December 31, 2015
Abstract: The Bose-Hubbard model is a system of interacting bosons that live on the
vertices of a graph. The particles can move between adjacent vertices and experience a
repulsive on-site interaction. The Hamiltonian is determined by a choice of graph that
specifies the geometry in which the particles move and interact. We prove that approximating
the ground energy of the Bose-Hubbard model on a graph at fixed particle number is QMA-
complete. In our QMA-hardness proof, we encode the history of an n-qubit computation in
the subspace with at most one particle per site (i. e., hard-core bosons). This feature, along
with the well-known mapping between hard-core bosons and spin systems, lets us prove a
related result for a class of 2-local Hamiltonians defined by graphs that generalizes the XY
model. By avoiding the use of perturbation theory in our analysis, we circumvent the need to
multiply terms in the Hamiltonian by large coefficients.
ACM Classification: F.1.3, F.2.2
AMS Classification: 68Q17, 81P68
Key words and phrases: complexity theory, QMA, quantum computing
1 Introduction
The problem of approximating the ground energy of a given Hamiltonian is a natural quantum analog
of classical constraint satisfaction. Many authors have considered the computational complexity of
such quantum ground state problems. For a variety of classes of Hamiltonians and a suitable notion of
approximation, this task is complete for the complexity class QMA, the quantum version of NP with
two-sided error (see reference [3] for a recent review). These results provide evidence that approximating
the ground energy of such quantum systems is intractable.
The first such example is the Local Hamiltonian problem introduced by Kitaev [21]. A k-local
Hamiltonian acts on a system of n qubits and can be written as a sum of terms, each acting nontrivially
© 2015 Andrew M. Childs, David Gosset, and Zak Webb
c b Licensed under a Creative Commons Attribution License (CC-BY) DOI: 10.4086/toc.2015.v011a020
A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
on at most k qubits. The k-Local Hamiltonian problem is a promise problem related to the task of
approximating the ground energy of a k-local Hamiltonian. Given such a Hamiltonian and two thresholds
a and b, one is asked to determine if the ground energy is below a or above b (promised that one of
these conditions holds). Kitaev’s original work showed that the 5-local Hamiltonian problem is QMA-
complete [21]; subsequent works proved QMA-completeness of the 3-local Hamiltonian problem [20],
the 2-local Hamiltonian problem [19], and the 2-local Hamiltonian problem with interactions between
qubits restricted to a two-dimensional lattice [28].
The complexity of similar computational problems related to other classes of Hamiltonians has also
been considered. These include Hamiltonians in one dimension [1, 14], frustration-free Hamiltonians [4,
13], and stoquastic Hamiltonians (Hamiltonians with no “sign problem”) [5, 6], among others.
The QMA-hardness of ground energy problems for local Hamiltonians acting on qubits has implica-
tions for Hamiltonians acting on indistinguishable particles (bosons or fermions) due to formal mappings
between these systems. By applying such mappings to the Local Hamiltonian problem, one can show that
certain bosonic [30] and fermionic [23] Hamiltonian problems are QMA-hard. A more restrictive class of
QMA-complete fermionic Hamiltonians was considered by Schuch and Verstraete, who showed that the
Hubbard model with a site-dependent magnetic field is QMA-complete [29]. This is a specific model of
interacting electrons (i. e., spin-1/2 fermions) on a two-dimensional lattice, with a magnetic field that
may take different values and point in different directions (in three dimensions) at distinct sites of the
lattice.
Many of the QMA-complete problems considered previously have the property that the form of the
terms in the Hamiltonian is part of the specification of the instance. For example, a 2-local Hamiltonian
is specified by a 2-local Hermitian operator for each pair of qubits. In the Hubbard model considered in
reference [29], there is a similar freedom in the choice of magnetic field at each site. A recent classification
of local Hamiltonian problems [10] likewise applies only to models with adjustable coefficients. In fact,
these results typically require coefficients that grow with the problem size.
In contrast, here we consider a system of interacting bosons with fixed hopping and interaction terms.
Specifically, we consider the Bose-Hubbard model, which has one of the simplest interactions between
particles that conserves total particle number. Although the Bose-Hubbard model is traditionally defined
on a lattice and with negative hopping strength [12], here we consider its extension to a general graph,
with positive hopping strength.
We consider undirected graphs without multiple edges and with at most one self-loop per vertex.
Any such graph G (with vertex set V ) can be specified by its adjacency matrix, a symmetric 0-1 matrix
denoted A(G). The Bose-Hubbard model on G with hopping strength thop and interaction strength Jint has
the Hamiltonian
HG = thop ∑ A(G)i j a†i a j + Jint ∑ nk (nk − 1) (1.1)
i, j∈V k∈V
where a†i creates a boson at vertex i and ni = a†i ai counts the number of bosons at vertex i. The precise
definitions of these operators are given in Section 2.2, where we describe the model in more detail.
Our results apply to the Bose-Hubbard model for any fixed positive hopping strength thop > 0 and any
fixed positive (i. e., repulsive) interaction strength Jint > 0. Unlike in other QMA-completeness results,
in our work the coefficients thop , Jint are not inputs to the problem; rather, each fixed choice defines a
computational problem and we prove QMA-completeness for each of them.
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Observe that the Bose-Hubbard Hamiltonian (1.1) conserves the total number of particles N = ∑k∈V nk .
We focus on the space of N-particle states, which can be identified with the symmetric subspace of
(C|V | )⊗N (as we discuss in more detail in Section 2). The first term in (1.1) allows particles to move
between vertices; the second term is an interaction between particles that assigns an energy penalty
for each vertex occupied by more than one particle. The Bose-Hubbard model is an example of a
multi-particle quantum walk, a generalization of quantum walk to systems with more than one walker.
Recently we showed that the Bose-Hubbard model on a graph can perform efficient universal quantum
computation [8]. Sometimes universality goes hand-in-hand with QMA-completeness, e. g., for local
Hamiltonians, whose dynamics are BQP-complete [11] and whose ground energy problem is QMA-
complete [21]. However, not all classes of Hamiltonians with universal dynamics have QMA-complete
ground energy problems. For example, the dynamics of stoquastic local Hamiltonians are BQP-complete
(as follows from [16] and time reversal), whereas the corresponding ground energy problem is in AM [5]
and hence unlikely to be QMA-hard. Similarly, the ground energy problem for a Bose-Hubbard model
with thop < 0 is also in AM [5], whereas the dynamics of such Hamiltonians are universal [8]. The
ferromagnetic Heisenberg model on a graph provides an even starker contrast: its dynamics are BQP-
complete (as can be inferred from [8] using a correspondence between spins and hard-core bosons) but
its ground energy problem is trivial since the ground space is the symmetric subspace.
1.1 Overview of results
In this paper we define the Bose-Hubbard Hamiltonian problem and characterize its complexity. In this
problem one is given a graph G and a number of particles N and asked to approximate the ground energy
of the Bose-Hubbard Hamiltonian (1.1) in the N-particle sector (in a precise sense described in Section 2).
We prove that this problem is QMA-complete.
To prove QMA-hardness of the Bose-Hubbard Hamiltonian problem, we show that in fact a notable
special case of this problem, called Frustration-Free Bose Hubbard Hamiltonian, is QMA-hard. In this
problem one is asked (roughly) to determine if the ground energy of the Bose-Hubbard Hamiltonian (1.1)
in the N-particle sector is close to N times its single-particle ground energy (i. e., N times the smallest
eigenvalue of the adjacency matrix A(G)). This is always a lower bound on the N-particle energy, and
when it is achieved we say the N-particle ground states are frustration free. A frustration-free state has
the special property that it has minimal energy for both terms in (1.1), and in particular it is annihilated
by the interaction term. Frustration-free states therefore live in the subspace of hard-core bosons, with at
most one boson per vertex.
Furthermore, we prove a reduction from Frustration-Free Bose-Hubbard Hamiltonian to an eigenvalue
problem for a class of 2-local Hamiltonians defined by graphs. The two problems are related by a
well-known mapping between hard-core bosons and spin systems. Specifically, given a graph G (with
vertex set V ) we consider the Hamiltonian
σxi σxj + σyi σyj 1 − σzi
∑ |01ih10| + |10ih01| i j + ∑ |1ih1|i = ∑ + ∑ . (1.2)
A(G)i j =1 A(G) =1 A(G) =1
2 A(G) =1
2
ii ij ii
i6= j i6= j
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Note that this Hamiltonian commutes with the magnetization operator
|V |
1 − σzi
Mz = ∑
i=1 2
and has a sector for each of its eigenvalues Mz ∈ {0, 1, . . . , |V |}. We reduce Frustration-Free Bose-Hubbard
Hamiltonian (with N particles on a graph G) to the problem of approximating the smallest eigenvalue of
(1.2) within the sector with magnetization Mz = N. We call this the XY Hamiltonian problem because of
its connection to the XY model from condensed matter physics. Since this problem is contained in QMA,
our reduction shows it to be QMA-complete.
We also obtain a related result that may be of independent interest. In Section 3 we give a self-
contained proof that computing the smallest eigenvalue of a sparse, efficiently row-computable [2]
symmetric 0-1 matrix (the adjacency matrix of a graph) is QMA-complete. This can alternatively be
viewed as a result about the QMA-completeness of a single-particle quantum walk on a graph with at
most one self-loop per vertex. To prove this, we use a mapping from circuits to graphs that is also used in
our main result. Note that Janzing and Wocjan used a similar construction to design a BQP-complete
problem [16].
1.2 Proof techniques
We prove our main result by direct reduction from quantum circuit satisfiability. We introduce several
new techniques in order to do this using the Bose-Hubbard model on an unweighted graph.
Kitaev’s original proof of QMA-hardness of the Local Hamiltonian problem encodes a QMA veri-
fication circuit using ideas from a computationally universal Hamiltonian proposed by Feynman [11].
This Hamiltonian uses a “clock register” to record the progress of the computation; in an appropriate
basis, the Hamiltonian can be seen as a quantum walk on a path whose vertices represent the steps of the
computation. Other proofs of QMA-hardness have used alternative encodings of the temporal structure
of a verification circuit into a quantum state. In our construction, we encode the history of an n-qubit
verification circuit in the state of n interacting particles on a graph, where each particle encodes a single
qubit.
Our construction uses a class of graphs we define called gate graphs. Gate graphs are built from a basic
subgraph whose single-particle ground states encode the history of a simple single-qubit computation.
By suitably combining copies of this basic unit, we define gadgets with other functionality. (Note that
these gadgets realize some desired behavior exactly; they are not “perturbative gadgets” in the sense
of [19, 17].) In particular, we design gadgets for two-qubit gates such that each ground state of the
two-particle Bose-Hubbard model encodes a two-qubit computation. We now give a high-level description
of how these gadgets work and how we use them to construct a graph for a QMA verification circuit.
For each two-qubit gate U from a fixed universal set, we design a graph GU that can be divided
into four overlapping regions as shown schematically in Figure 1.1(A). (The specific graphs we use
for two-qubit gates each have 4096 vertices and are described using the gate graph formalism.) The
two-particle Bose-Hubbard model on this graph has ground states that encode the two-qubit computation.
To describe them it is helpful to first consider the single-particle ground states, i. e., the ground states
i,U
of the adjacency matrix A(GU ). This matrix has 16 orthonormal single-particle ground states |ρz,a i.
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
1 2
1,2,3,4
overlap
3 4
(A) (B)
Figure 1.1: We design graphs for two-qubit gates with overlapping regions as shown in ( A ). Regions
1 and 2 are associated with the first encoded qubit and regions 3 and 4 are associated with the second
encoded qubit. One could imagine designing a graph for a circuit with two-qubit gates U1 followed by U2
by connecting the corresponding gadgets as in ( B ). In the text we describe a challenge with this approach.
i,U
Each index i ∈ {1, 2, 3, 4} is associated with the corresponding region in the graph, as |ρz,a i is supported
entirely within region i. The index z ∈ {0, 1} corresponds to the computational basis states of a single
encoded qubit. Note that, since A(GU ) is a real matrix, the complex conjugate of any eigenstate is
also an eigenstate with the same eigenvalue. The index a ∈ {0, 1} is associated with this freedom, i. e.,
i,U i,U ∗
|ρz,1 i = |ρz,0 i . The ground space of the two-particle Bose-Hubbard model on GU is spanned by 16
states, indexed by two choices z1 , z2 ∈ {0, 1} of computational basis states for the encoded qubits and two
bits a1 , a2 ∈ {0, 1} associated with complex conjugation. These states can be represented as symmetric
states in the Hilbert space C4096 ⊗ C4096 ; they are
1 1,U 1
(|ρ i|ρ 3,U i + |ρz3,U i|ρz1,U i) + U(a1 )x1 ,x2 ,z1 ,z2 (|ρx2,U i|ρx4,U i + |ρx4,U i|ρx2,U i)
2 z1 ,a1 z2 ,a2 2 ,a2 1 ,a1
2 x1 ,x∑
2 ∈0,1
1 ,a1 2 ,a2 2 ,a2 1 ,a1
where U(0) = U is the two-qubit gate of interest and U(1) = U ∗ is its elementwise complex conjugate.
Observe that each of these states is a superposition of a term where both particles are on the left-hand
side of the graph, encoding a two-qubit input state |z1 i|z2 i, and a term where both particles are on the
right-hand side of the graph, encoding the two-qubit output state U(a1 )|z1 i|z2 i where either U or its
complex conjugate has been applied. In other words, the particles “move together” through the graph
as the gate U(a) is applied. While we might prefer the ground states to only encode the computation
corresponding to U, we must include the possibility of U ∗ because the Hamiltonian is real. The same
issue arises for n-qubit verification circuits. Fortunately, the complex conjugate of a circuit is equally
useful for QMA verification.
It is natural to attempt to construct a graph for an n-qubit verification circuit by combining gadgets
for each of the two-qubit gates. However, there is an obstacle to this approach, as illustrated by the
example of a two-qubit circuit consisting of only two gates U1 and U2 . One could construct a graph for
such a circuit as shown schematically in Figure 1.1(B), where the two-qubit gadgets for U1 and U2 are
connected in some unspecified way in the middle. However, not every ground state of the two-particle
Bose-Hubbard model on such a graph encodes a computation. For example, there could be a ground state
1,U1
where one of the particles is in the single-particle state |ρz,a i localized on the left side of the graph and
2,U2
the other particle is in the state |ρz,a i with support on a disjoint region of the graph on the right-hand
side. To eliminate such spurious ground states, we develop a method to enforce occupancy constraints on
the locations of particles in gate graphs using the Bose-Hubbard interaction. Although this interaction
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
only directly penalizes simultaneous occupation of the same vertex, we show how to simulate terms that
penalize simultaneous occupation of different regions of the graph. We formalize this method by proving
an “Occupancy Constraints Lemma” for gate graphs.
In summary, our construction of the graph for an n-qubit verification circuit proceeds in two steps. We
first construct a graph G by connecting two-qubit gadgets for each of the gates in the circuit. As discussed
above, the ground space of the n-particle Bose-Hubbard model on G includes a subspace of states that
encode computations and a subspace of states that do not. We construct a set of occupancy constraints
that are only satisfied by states in the former subspace. We then apply the Occupancy Constraints Lemma
to obtain a gate graph G where each n-particle ground state encodes a computation. (Note that the
particles are somewhat dilute in this graph: the “filling fraction,” or number of particles n divided by the
size of the graph, approaches 0 as n → ∞.)
Unlike many previous works, we do not use perturbation theory in our analysis. Instead, we use a
“Nullspace Projection Lemma” that characterizes the smallest nonzero eigenvalue of a sum of two positive
semidefinite matrices HA + HB in terms of the smallest nonzero eigenvalue of HA and the smallest nonzero
eigenvalue of HB restricted to the nullspace of HA . This lemma allows us to establish an eigenvalue
promise gap (i. e., to bound the ground energies of yes instances away from those of no instances)
without having to multiply terms in the Hamiltonian by large coefficients, something that is not allowed
in the setting of the Bose-Hubbard model on a graph. Whereas QMA-hardness proofs such as those
of [20, 19, 28, 29] require multiplying terms in the Hamiltonian by unphysical, problem-size dependent
coefficients, our approach avoids this. To the best of our knowledge, our proof would not be much simpler
if we only demanded constant-size coefficients; the further restriction that the model is defined entirely
by a graph is an extra benefit with little additional cost.
Note that the Nullspace Projection Lemma was used implicitly in [25], which claimed to give a
simple proof of the computational universality of adiabatic evolution. That paper encoded a circuit into
the multi-particle ground state of a fermionic Hamiltonian in a way that shares some features with our
approach. Unfortunately, although the encoding from [25] is novel and interesting, the analysis of the
resulting Hamiltonian appears to lack details that are crucial to proving the stated result. (For related
approaches with more complete analysis, see [15, 7].)
1.3 Extensions and open questions
Our result shows that approximating the ground energy of the Bose-Hubbard model on a graph at fixed
particle number is likely intractable. In showing this, we introduce techniques that we expect will be
useful in other contexts. Here we briefly discuss some related questions for future work.
One might consider the complexity of variants of the Bose-Hubbard Hamiltonian problem. For
example, one could consider the problem with negative hopping (i. e., thop < 0), with attractive interactions
(i. e., Jint < 0), or both. For negative hopping, the results of [5] show that the problem is in AM; we do
not know if it is AM-hard. Containment in AM suggests that the problem is easier with thop < 0 than
with thop , Jint > 0. For attractive interactions, the problem is clearly in QMA (the verification procedure
described in Section 4 applies independent of the signs of thop , Jint ), but again we do not know the true
complexity. Figure 1.2 summarizes our knowledge of the complexity of the Bose-Hubbard Hamiltonian
problem with different choices of thop and Jint .
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
thop
QMA-
∈ QMA
complete
Jint
∈ AM ∩ QMA
Figure 1.2: Each choice for the coefficients thop and Jint defines a computational problem. We prove the
problem is QMA-complete for any positive hopping strength thop > 0 and repulsive interaction strength
Jint > 0. For thop < 0 the problem is contained in AM [5]. For any values of Jint and thop the problem is
contained in QMA (see Section 4).
One can define other variants of the Bose-Hubbard Hamiltonian problem by lifting the restriction to
fixed particle number.
One could also consider other classes of graphs. The graphs we consider in this paper are described
by symmetric 0-1 matrices and have at most one self-loop per vertex. We do not know if the model
remains QMA-hard on simple graphs, i. e., without any self-loops. Our reduction from Frustration-Free
Bose-Hubbard Hamiltonian to the ground energy problem for the spin model (1.2) gives a stronger
result for simple graphs, in which case the second term in (1.2) vanishes. Thus, if the Frustration-Free
Bose-Hubbard Hamiltonian problem for simple graphs is QMA-hard, then approximating the ground
energy of the XY model on a (simple) graph at fixed magnetization is QMA-complete.
There are many open questions concerning the complexity of the ground energy problem for other
quantum systems defined by graphs. For example, one could consider fermions or bosons on a graph
with nearest-neighbor interactions. One could also consider quantum spin models such as the XY model
or the antiferromagnetic Heisenberg model defined on graphs. Both of these examples correspond to
Hamiltonians that conserve magnetization, so one could consider the ground energy problem with or
without a restriction to a fixed-magnetization sector. This would complement existing results about the
complexity of computing the lowest-energy configuration of classical spin models defined by graphs (for
example, the antiferromagnetic Ising model on a graph is NP-complete, as it is equivalent to Max Cut).
As emphasized previously, the Hamiltonians we consider are determined entirely by a choice of graph,
with the same type of hopping and interaction terms applied throughout the graph. It might be interesting
to find other QMA-complete problems with similar features, such as a version of Local Hamiltonian with
only one type of local term. Analogous classical constraint satisfaction problems with a fixed type of
constraint are well known (e. g., Exact Cover and Not-All-Equal SAT) and have been widely studied.
Along similar lines, it might be interesting to understand when local Hamiltonian problems remain
QMA-hard with constant-size coefficients. Nagaj and Mozes have shown that the 3-local Hamiltonian
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
problem has this property [27], but whether the same holds for the 2-local Hamiltonian problem remains
open.
1.4 Outline of the paper
The remainder of this paper is organized as follows. In Section 2 we define the class QMA, introduce the
Bose-Hubbard model, and formally state our results. As a warm-up, in Section 3 we prove that computing
the smallest eigenvalue of a (succinctly specified) graph is QMA-complete. In Section 4 we explain why
the ground state problem for the Bose-Hubbard model is contained in QMA. In Section 5 we define the
notion of gate graphs, analyze frustration-free ground states of gate graphs, and introduce the idea of
occupancy constraints. In Section 6 we design and analyze gadgets for implementing gates. In Section 7
we construct the graph that we use to show QMA-hardness of the Bose-Hubbard model. In Section 8 we
perform the spectral analysis needed to prove our main result. In Section 9 we reduce Frustration-Free
Bose-Hubbard Hamiltonian to XY Hamiltonian. In Section 10 we prove the Occupancy Constraints
Lemma. Finally, in Appendix A we provide various necessary technical results, including the Nullspace
Projection Lemma.
Contents
1 Introduction 491
1.1 Overview of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
1.2 Proof techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494
1.3 Extensions and open questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
1.4 Outline of the paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498
2 Definitions and results 500
2.1 Quantum Merlin-Arthur . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
2.2 The Bose-Hubbard model on a graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
2.3 Complexity of the Bose-Hubbard model . . . . . . . . . . . . . . . . . . . . . . . . . . 504
2.4 Complexity of the XY Hamiltonian problem . . . . . . . . . . . . . . . . . . . . . . . . 505
2.5 Complexity of approximating the smallest eigenvalue of a graph . . . . . . . . . . . . . 506
3 Minimum Graph Eigenvalue is QMA-complete 507
3.1 Circuit-to-graph mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508
3.2 Upper bound on the smallest eigenvalue for yes instances . . . . . . . . . . . . . . . . . 511
3.3 Lower bound on the smallest eigenvalue for no instances . . . . . . . . . . . . . . . . . 511
4 Bose-Hubbard Hamiltonian is contained in QMA 513
5 Gate graphs 514
5.1 The graph g0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514
5.2 Gate graphs and gate diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
5.3 Frustration-free states on e1 -gate graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 517
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
5.4 Occupancy constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521
6 Gadgets 522
6.1 The move-together gadget . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523
6.2 Gadgets for two-qubit gates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526
6.3 The boundary gadget . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533
7 Bose-Hubbard Hamiltonian is QMA-hard 535
7.1 The verification circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536
7.2 The gate graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538
7.2.1 Notation for GX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540
7.3 The occupancy constraints graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541
8 Proof of Theorem 7.1 542
8.1 Strategy and outline of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542
8.1.1 Adjacency matrices of the gate graphs . . . . . . . . . . . . . . . . . . . . . . . 543
8.1.2 Building up the Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544
8.2 Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
8.3 Legal configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
8.4 Matrix elements between states with legal configurations . . . . . . . . . . . . . . . . . 551
8.5 From H(G2 , Gocc occ
X , n) to H(G4 , GX , n) . . . . . . . . . . . . . . . . . . . . . . . . . . . 556
8.6 Completeness and soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
8.6.1 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
8.6.2 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563
9 XY Hamiltonian is QMA-complete 564
10 Proof of the Occupancy Constraints Lemma 566
10.1 Definitions and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567
10.2 The gate graph G♦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 570
10.3 The adjacency matrix of the gate graph G4 . . . . . . . . . . . . . . . . . . . . . . . . 572
10.4 The Hamiltonian H(G4 , N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576
10.5 The gate graph G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583
A Technical supporting material 586
A.1 Proof of the Nullspace Projection Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 586
A.2 Proof of Lemma 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
A.3 Computation of matrix elements between states with legal configurations . . . . . . . . 593
A.3.1 Matrix elements of H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593
A.3.2 Matrix elements of H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596
A.3.3 Matrix elements of Hin,i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597
A.3.4 Matrix elements of Hout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598
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2 Definitions and results
In this section we define the complexity class QMA, introduce the Bose-Hubbard model and a related
spin model, and formally state our results.
2.1 Quantum Merlin-Arthur
Quantum Merlin-Arthur, or QMA, is a class of promise problems. Informally, a promise problem is in
QMA if, for any yes instance X, there exists a quantum state |ψwit i (with a number of qubits polynomial
in the input size |X|) that “proves” X is a yes instance. We imagine that all-powerful Merlin prepares the
quantum proof |ψwit i and hands it to polynomially-bounded Arthur, who checks it using his quantum
computer.
Arthur checks Merlin’s proof using a verification circuit CX that is uniformly generated: CX is
computed from X using a deterministic polynomial-time (as a function of |X|) classical algorithm. The
circuit CX has an nin -qubit input register, n − nin ancilla qubits initialized to |0i, and one of the n qubits
designated as an output qubit. The number of qubits n and the number of gates in the circuit are upper
bounded by a polynomial function of |X|. We write UCX for the unitary operation that the verification
circuit implements.
The acceptance probability of CX given some nin -qubit input state |φ i is the probability of measuring
the output qubit to be in the state |1i after the circuit is applied, namely
2
AP(CX , |φ i) = |1ih1|outUCX |φ i|0i⊗n−nin .
Arthur applies the verification circuit to the input state |ψwit i. If he measures the output qubit to be |1i, he
concludes X is a yes instance. QMA is the class of problems for which reliable verification circuits exist,
so that Arthur can trust this conclusion with reasonable probability. In other words, if X is a yes instance
then there exists a state |ψwit i that CX accepts with high probability, whereas if X is a no instance then any
state |φ i has low probability of being accepted. To define the class we must specify which probabilities
are considered “high” and which are “low.” A standard choice uses thresholds of 2/3 and 1/3 (called
completeness and soundness, respectively).
Definition 2.1 (QMA). A promise problem Lyes ∪ Lno ⊂ {0, 1}∗ is contained in QMA if there exists a
uniform polynomial-size circuit family CX with the following two properties. If X ∈ Lyes , there exists an
input state |ψwit i such that
2
AP(CX , |ψwit i) ≥ (completeness).
3
If X ∈ Lno then all input states |φ i satisfy
1
AP(CX , |φ i) ≤ (soundness).
3
It is a nontrivial fact that many other choices for the completeness and soundness thresholds lead to
equivalent definitions of QMA [21, 24]. In particular, we obtain the same class QMA if we change the
completeness threshold 2/3 in the above definition to 1 − (1/2|X| ).
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2.2 The Bose-Hubbard model on a graph
As discussed in Section 1, we consider the Bose-Hubbard model on a graph G, with Hamiltonian
HG = thop ∑ A(G)i j a†i a j + Jint ∑ nk (nk − 1) (2.1)
i, j∈V k∈V
where a†i creates a boson at vertex i and ni = a†i ai counts the number of particles at vertex i. In this
second-quantized formulation of the Bose-Hubbard model, the Hamiltonian HG acts on the Fock space
with orthonormal basis vectors
|l1 , l2 , . . . , l|V | i
where l j ∈ {0, 1, 2, . . .} specifies the number of bosons at vertex j ∈ V . The operator ai is defined by its
action in this basis: p
ai |l1 , l2 , . . . , li , . . . , l|V | i = li |l1 , l2 , . . . , li − 1, . . . , l|V | i . (2.2)
The Hamiltonian (2.1) conserves the total particle number N = n1 + n2 + · · · + n|V | . In the N-particle
sector, the Bose-Hubbard model acts on the finite-dimensional Hilbert space
span{|l1 , . . . , l|V | i : ∑ l j = N} . (2.3)
j
The dimension of this space is
N + |V | − 1
DN = . (2.4)
|V | − 1
Our results apply to the Bose-Hubbard model for any strictly positive hopping and interaction
strengths. Henceforth we set thop = Jint = 1 for convenience, but all our complexity-theoretic results hold
for any fixed thop , Jint > 0.
An equivalent (first-quantized) formulation of the Bose-Hubbard model is as follows. Consider the
Hilbert space
(C|V | )⊗N = span{|i1 , i2 , . . . , iN i : i1 , i2 , . . . , iN ∈ V } (2.5)
where each basis state corresponds to an N-tuple of vertices in the graph. Define the linear operator Sym
that symmetrizes over all N! permutations of the N particles:
1
Sym(|i1 i|i2 i . . . |iN i) = √ ∑ |iπ(1) i|iπ(2) i . . . |iπ(N) i .
N! π∈SN
Note that Sym does not in general preserve the norm (for example, any antisymmetric state is mapped to
zero). Every state in the Hilbert space (2.3) can be identified with a state in
ZN (G) = span{Sym(|i1 , i2 , . . . , iN i) : i1 , i2 , . . . , iN ∈ V }
and vice versa since the two spaces have the same dimension. It is natural to identify states in the two
Hilbert spaces by the following linear mapping, defined by its action on basis states. We identify each
basis state |l1 , . . . , l|V | i with the normalized state
1
p Sym |1i|1i . . . |1i |2i|2i . . . |2i . . . ||V |i||V |i . . . ||V |i . (2.6)
l1 ! l2 ! . . . l|V | ! | {z } | {z } | {z }
l1 l2 l|V |
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The Hamiltonian of the Bose-Hubbard model in the N-particle sector acts as an operator HGN on the
space ZN (G) (see for example [22, §64]):
N
HGN = ∑ A(G)(w) + ∑ n̂k (n̂k − 1) (2.7)
w=1 k∈V
(recall we set thop = Jint = 1) where the number operator is
N
n̂i = ∑ |iihi|(w) . (2.8)
w=1
Here (and throughout the paper) we use the notation
M (w) = 1⊗w−1 ⊗ M ⊗ 1⊗N−w
to indicate that the operator M acts on subsystem w.
While HGN is defined as a |V |N × |V |N matrix in the space (2.5), we consider its restriction
H̄GN = HGN
ZN (G)
to the bosonic N-particle subspace ZN (G) (throughout the paper we write H|W for the restriction of a
Hermitian operator H to a subspace W, which can be written as a dim W × dim W matrix). It is convenient
to add a term proportional to the identity to obtain a positive semidefinite operator. Letting µ(G) denote
the smallest eigenvalue of the adjacency matrix A(G), we consider
H(G, N) = H̄GN − Nµ(G) .
Clearly H(G, N) ≥ 0 since the interaction term is positive semidefinite. Also note that, given the
graph G, the smallest eigenvalue µ(G) of its adjacency matrix can be efficiently approximated using a
classical polynomial-time algorithm, so the complexity of approximating the ground energy of H(G, N)
is equivalent to the complexity of approximating H̄GN . (Note that here the graph is specified explicitly
by its adjacency matrix. In other contexts one might consider a graph specified compactly, e. g., by a
circuit that computes rows of its adjacency matrix. Then the situation is more complex since the input
size can be much smaller than the number of vertices in the graph. Indeed, we prove in Section 3 that
approximating the smallest eigenvalue of such a graph is QMA-complete.)
We write
0 ≤ λN1 (G) ≤ λN2 (G) ≤ . . . ≤ λNDN (G)
for the eigenvalues of H(G, N) and {|λNj (G)i} for the corresponding normalized eigenvectors.
When λN1 (G) = 0, the ground energy of the N-particle Bose-Hubbard model H̄GN is equal to N times
the one-particle energy µ(G). In this case we say that the N-particle Bose-Hubbard model is frustration
free. We also define frustration freeness for N-particle states.
Definition 2.2. If |ψi ∈ ZN (G) satisfies H(G, N)|ψi = 0 then we say |ψi is an N-particle frustration-free
state for G.
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We now present two basic properties of H(G, N). The following lemma states that the ground energy
is non-decreasing as a function of the number of particles N.
1 (G) ≥ λ 1 (G).
Lemma 2.3. For all N ≥ 1, λN+1 N
Proof. Let nbNi be the number operator (2.8) defined in the N-particle space and let nbN+1
i be the corre-
sponding operator in the (N + 1)-particle space. Note that
nbN+1
i = nbNi ⊗ 1 + |iihi|(N+1) ≥ nbNi ⊗ 1 .
Using this and the fact that A(G) ≥ µ(G), we get
HGN+1 − (N + 1) µ(G) ≥ HGN − Nµ(G) ⊗ 1 .
Hence
1
λN+1 (G) = min hψ|HGN+1 − (N + 1) µ(G)|ψi
|ψi∈ZN+1 (G) : hψ|ψi=1
hψ| HGN − Nµ(G) ⊗ 1|ψi
≥ min
|ψi∈ZN (G)⊗C|V | : hψ|ψi=1
= λN1 (G)
(using the fact that ZN+1 (G) ⊂ ZN (G) ⊗ C|V | ).
In this paper we will encounter disconnected graphs G. In the cases of interest, the smallest eigenvalue
of the adjacency matrix for each component is the same. The following lemma shows that the eigenvalues
of H(G, N) on such a graph can be written as sums of eigenvalues for the components. In this lemma
(and throughout the paper), we let [k] = {1, 2, . . . , k}.
Sk
Lemma 2.4. Suppose G = i=1 Gi with µ(G1 ) = µ(G2 ) = · · · = µ(Gk ). The eigenvalues of H(G, N) are
∑ λNyii (Gi )
i∈[k] : Ni 6=0
where N1 , . . . , Nk ∈ {0, 1, 2, . . .} with ∑i Ni = N and yi ∈ [DNi ]. The corresponding eigenvectors are (up to
normalization) !
|λNyii (Gi )i .
O
Sym (2.9)
i∈[k] : Ni 6=0
Proof. Recall that the action of HG − Nµ(G) on the Hilbert space (2.3) is the same as the action of
H(G, N) on the Hilbert space ZN (G). States in these Hilbert spaces are identified via the mapping
described in equation (2.6). It is convenient to prove the lemma by working with the second-quantized
Hamiltonian HG . We then translate our results into the first-quantized picture to obtain the stated claims.
For a graph with k components, equation (2.1) gives
k
HG = ∑ HGi (2.10)
i=1
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where [HGi , HG j ] = 0. Label each vertex of G by (a, b) where b ∈ [k] and a ∈ [|Vb |], where Vb is the vertex
set of the component Gb . An occupation number basis state (2.3) can be written
|l1,1 , . . . , l|V1 |,1 i|l1,2 , . . . , l|V2 |,2 i . . . |l1,k , . . . , l|Vk |,k i. (2.11)
The Hamiltonian HG − Nµ(G) conserves the number of particles Nb in each component b. Within the
sector corresponding to a given set N1 , . . . , Nk with ∑i∈[k] Ni = N, we have
(HG − Nµ(G)) |l1,1 , . . . , l|V1 |,1 i|l1,2 , . . . , l|V2 |,2 i . . . |l1,k , . . . , l|Vk |,k i
= HG1 − N1 µ(G1 )|l1,1 , . . . , l|V1 |,1 i |l1,2 , . . . , l|V2 |,2 i . . . |l1,k , . . . , l|Vk |,k i
+ |l1,1 , . . . , l|V1 |,1 i HG2 − N2 µ(G2 )|l1,2 , . . . , l|V2 |,2 i . . . |l1,k , . . . , l|Vk |,k i + · · ·
+ |l1,1 , . . . , l|V1 |,1 i|l1,2 , . . . , l|V2 |,2 i . . . HGk − Nk µ(Gk )|l1,k , . . . , l|Vk |,k i ,
where we used the fact that µ(Gi ) = µ(G) for i ∈ [k]. From this equation we see that the eigenstates of
HG can be obtained as product states with k factors in the basis (2.11). In each such product state, the
i-th factor is an eigenstate of HGi − Ni µ(Gi ) = HGi − Ni µ(G) in the Ni -particle sector, with eigenvalue
λNjii (Gi ). Rewriting this result in the “first-quantized” language, we obtain the lemma.
2.3 Complexity of the Bose-Hubbard model
Given a K-vertex graph G and a number of particles N, how hard is it to approximate the ground energy
of the N-particle Bose-Hubbard model H̄GN on G? We consider the following decision version of this
computational problem.
Problem 2.5 (Bose-Hubbard Hamiltonian). We are given a K-vertex graph G, a number of particles
N, a real number c, and a precision parameter ε = 1/T . The positive integers N and T are provided in
unary; the graph is specified by its adjacency matrix, which can be any K × K symmetric 0-1 matrix.
We are promised that either the smallest eigenvalue of H̄GN is at most c (yes instance) or is at least c + ε
(no instance) and we are asked to decide which is the case.
Here and elsewhere in this paper, we require inputs such as ε and T to be provided in unary so that the
input size is measured accordingly. The parameter c is provided in a straightforward manner, with enough
precision to resolve ε, i. e., using O(log |c|+log T ) bits. The input size is therefore Θ(K 2 +T +N +log |c|)
bits. We prove that this problem is QMA-complete, providing evidence that approximating the ground
energy of the N-particle Bose-Hubbard model on a graph G is intractable.
Theorem 2.6. Bose-Hubbard Hamiltonian is QMA-complete.
The proof of this theorem has two parts.
The easy part is to show that Bose-Hubbard Hamiltonian is contained in QMA. The basic strategy of
Arthur’s verification protocol is to measure the energy of the Bose-Hubbard Hamiltonian in the state |φ i
given to him by Merlin, using phase estimation and Hamiltonian simulation. Arthur accepts if the energy
is small enough and rejects otherwise. We give a more detailed description of the verification procedure
in Section 4.
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The more involved part is to show that Bose-Hubbard Hamiltonian is QMA-hard. For this we show
that any instance of a QMA problem can be converted (in deterministic polynomial time on a classical
computer) into an equivalent instance of Bose-Hubbard Hamiltonian. In fact, our reduction proves a
slightly stronger result, namely that a notable extremal case of Bose-Hubbard Hamiltonian is already
QMA-hard. We now discuss this special case.
Recall from the previous section that the ground energy of the N-particle Bose-Hubbard model is at
least N times the single-particle ground energy µ(G), i. e., λN1 (G) ≥ 0. We can ask if this inequality is
close to equality, i. e., is the N-particle Bose-Hubbard model close to being frustration free?
Problem 2.7 (Frustration-Free Bose-Hubbard Hamiltonian). We are given a K-vertex graph G, a
number of particles N ≤ K, and a precision parameter ε = 1/T . The integer T ≥ 4K is provided in
unary; the graph is specified by its adjacency matrix, which can be any K × K symmetric 0-1 matrix.
We are promised that either λN1 (G) ≤ ε 3 (yes instance) or λN1 (G) ≥ ε + ε 3 (no instance) and we are
asked to decide which is the case.
For concreteness, we have made some specific choices in defining this problem. Our proof that it is
QMA-hard also applies, for example, to variants of the problem where ε 3 is replaced (in both places it
appears) by ε α for any constant α ∈ {1, 2, 3, . . .}. In Section 2.4, we use the version with α = 3 as stated
above to facilitate a reduction to the XY Hamiltonian problem.
The requirement T ≥ 4K ensures that ε is small so that, for a yes instance, the system is very close to
being frustration free. We choose the specific threshold 4K for concreteness.
The restriction N ≤ K is without loss of generality since the problem is trivial otherwise. To see this,
note that any state with more than K particles is orthogonal to the nullspace of the interaction term since
there are always two or more particles located at one vertex; hence λN1 (G) ≥ 2 whenever N ≥ K + 1.
Frustration-Free Bose-Hubbard Hamiltonian is a special case of Bose-Hubbard Hamiltonian with
c = Nµ(G) + ε 3 . To prove that Bose-Hubbard Hamiltonian is QMA-hard, it therefore suffices to prove
that Frustration-Free Bose-Hubbard Hamiltonian is QMA-hard. The bulk of this paper is concerned with
the proof of this fact.
2.4 Complexity of the XY Hamiltonian problem
We reduce Frustration-Free Bose-Hubbard Hamiltonian to an eigenvalue problem for a class of 2-local
Hamiltonians defined by graphs. The reduction is based on a well-known mapping between hard-core
bosons and spin systems, which we now review.
We define the subspace WN (G) ⊂ ZN (G) of N hard-core bosons on a graph G to consist of the states
where each vertex of G is occupied by either 0 or 1 particle, i. e.,
WN (G) = span{Sym(|i1 , i2 , . . . , iN i) : i1 , . . . , iN ∈ V, i j 6= ik for distinct j, k ∈ [N]} .
A basis for WN (G) is the subset of occupation-number states (2.6) labeled by bit strings l1 . . . l|V | ∈
{0, 1}|V | with Hamming weight ∑ j∈V l j = N. The space WN (G) can thus be identified with the weight-N
subspace
|V |
WtN (G) = span{|z1 , . . . , z|V | i : zi ∈ {0, 1}, ∑ zi = N}
i=1
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of a |V |-qubit Hilbert space. We consider the restriction of HGN to the space WN (G), which can equivalently
be written as a |V |-qubit Hamiltonian OG restricted to the space WtN (G). In particular,
HGN WN (G) = OG WtN (G) (2.12)
where
OG = ∑ |01ih10| + |10ih01| i j + ∑ |1ih1|i
A(G)i j =1 A(G)ii =1
i6= j
σxi σxj + σyi σyj 1 − σzi
= ∑ + ∑ .
A(G) =1
2 A(G) =1
2
ij ii
i6= j
Note that the Hamiltonian OG conserves the total magnetization
|V |
1 − σzi
Mz = ∑
i=1 2
along the z axis.
We define θN (G) to be the smallest eigenvalue of (2.12), i. e., the ground energy of OG in the sector
with magnetization N. We show that approximating this quantity is QMA-complete.
Problem 2.8 (XY Hamiltonian). We are given a K-vertex graph G, an integer N ≤ K, a real number c,
and a precision parameter ε = 1/T . The positive integer T is provided in unary; the graph is specified
by its adjacency matrix, which can be any K × K symmetric 0-1 matrix. We are promised that either
θN (G) ≤ c (yes instance) or else θN (G) ≥ c + ε (no instance) and we are asked to decide which is the
case.
Theorem 2.9. XY Hamiltonian is QMA-complete.
We prove QMA-hardness of XY Hamiltonian by reduction from Frustration-Free Bose-Hubbard
Hamiltonian. The proof of Theorem 2.9 appears in Section 9.
2.5 Complexity of approximating the smallest eigenvalue of a graph
How hard is it to compute the smallest eigenvalue of a symmetric 0-1 matrix (i. e., the adjacency matrix
of a graph)? If the matrix is given explicitly then this can be done in time polynomial in the input
size. Instead, we consider adjacency matrices A specified by a (classical) circuit that takes as input the
index r of a row and outputs the indices { j : Ar j = 1} of the nonzero entries in that row. Note that the
circuit describing a D × D matrix A must have size Ω(log(D)) since it takes as input a row index r ∈ [D].
Efficiently row-computable matrices are those described by circuits of size polynomial in log D [2]. We
consider the following promise problem.
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Problem 2.10 (Minimum Graph Eigenvalue). We are given a D × D symmetric 0-1 matrix A, speci-
fied by a classical deterministic circuit that takes as input a row r ∈ [D] and computes the locations of
the nonzero entries in that row. We are also given a real number a and a precision parameter ε = 1/N,
where N ∈ N is specified in unary. We are promised that either the smallest eigenvalue of A is at most a
(yes instance), or else it is at least a + ε (no instance), and asked to decide which is the case.
The input size of this problem is the size of the circuit that describes A, plus the description size of a,
plus N bits. The interesting set of instances are those for which A is efficiently row-computable, so that
the input size is poly(log D, 1/ε).
Theorem 2.11. Minimum Graph Eigenvalue is QMA-complete.
In the next section we prove this theorem.
3 Minimum Graph Eigenvalue is QMA-complete
The fact that Minimum Graph Eigenvalue is contained in QMA follows from standard techniques
(applying phase estimation to the adjacency matrix) since a Hamiltonian that is a symmetric 0-1 matrix
can be simulated efficiently as a function of the size of the circuit computing its rows [2].
To prove that Minimum Graph Eigenvalue is QMA-hard, we show that an instance of any problem
in QMA can be converted (in deterministic polynomial time on a classical computer) into an equivalent
instance of Minimum Graph Eigenvalue.
Recall that an instance x of a problem in QMA has a verification circuit Cx with n qubits and M
gates, both upper bounded by a polynomial function of |x|. Here we assume that the verification circuit
Cx satisfies a modified version of Definition 2.1 where the completeness parameter 2/3 is replaced by
1 − 1/2|x| (modifying the definition in this way still gives the same class QMA [21, 24]). In Section 3.1
we map Cx to a row-sparse and efficiently row-computable symmetric 0-1 matrix Ax . The mapping we
exhibit has the following two properties (shown in Section 3.2 and Section 3.3, respectively):
√ 1 1
1. If x is a yes instance then the smallest eigenvalue of Ax is at most −1 − 3 2 + .
2M 2|x|
√ 1
2. If x is a no instance then the smallest eigenvalue of Ax is at least −1 − 3 2 + .
120M 4 n
Note that if |x| is larger than some constant, then
1 1 1
|x|
≤ ,
2M 2 240M 4 n
since M and n are both upper bounded by a polynomial in |x|. We assume without loss of generality that
this holds. For any such instance x, we associate an instance of Minimum Graph Eigenvalue defined by
the matrix Ax along with the number
√ 1
a = −1 − 3 2 +
240M 4 n
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and precision parameter
1
ε= .
240M 4 n
Using the two claimed properties of the circuit-to-graph mapping, we see that if x is a yes instance then
the smallest eigenvalue of Ax is at most a, and if it is a no instance then this eigenvalue is at least a + ε.
This shows that Minimum Graph Eigenvalue is QMA-hard.
3.1 Circuit-to-graph mapping
We suppose Cx implements a unitary
UCx = UM . . .U2U1 (3.1)
where each Ui is from the universal gate set
G = {H, HT, (HT )† , (H ⊗ 1) CNOT}
with
1 0 0 0
1 1 1 1 0 0 1 0 0
H=√ , T= π , CNOT = .
2 1 −1 0 ei 4 0 0 0 1
0 0 1 0
The verification circuit Cx has an nin -qubit input register and n − nin ancilla qubits initialized to |0i at the
beginning of the computation. One of these n qubits serves as an output qubit.
It will be convenient to consider
UC†x UCx = W2M . . .W2W1
where (
Ut 1≤t ≤M
Wt = †
U2M+1−t M + 1 ≤ t ≤ 2M.
As in Section 5.1 we start with a version of the Feynman-Kitaev Hamiltonian [11, 21]
√ 2M
Hx = − 2 ∑ Wt† ⊗ |tiht + 1| +Wt ⊗ |t + 1iht| (3.2)
t=1
⊗n
acting on the Hilbert space Hcomp ⊗ Hclock where Hcomp = C2 is an n-qubit computational register
and Hclock = C2M is a 2M-level register with periodic boundary conditions (i. e., we let |2M + 1i = |1i).
Note that
√ 2M
V † HxV = − 2 ∑ (1 ⊗ |tiht + 1| + 1 ⊗ |t + 1iht|) (3.3)
t=1
where
2M 1
V=∑ ∏ W j ⊗ |tiht|
t=1 j=t−1
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and W0 = 1. Since V is unitary, the eigenvalues of Hx are the same as the eigenvalues of (3.3), namely
√
π`
−2 2 cos
M
√
for ` = 0, . . . , 2M − 1. The ground energy of (3.3) is −2 2 and its ground space is spanned by
1 2M
|φ i √ ∑ |ti , |φ i ∈ Λ ,
2M j=1
where Λ is any orthonormal basis for Hcomp . A basis for the ground space of Hx is therefore
1 2M 1 2M
V |φ i √ ∑ |ti = √ ∑ Wt−1Wt−2 . . .W1 |φ i|ti
2M j=1 2M t=1
where |φ i ∈ Λ. The first excited energy of Hx is
√ π
η = −2 2 cos
M
and the gap between ground and first excited energies is lower bounded as
√ √ π2
η +2 2 ≥ 2 2 (3.4)
M
(using the fact that 1 − cos(x) ≤ x2 /2).
Now we modify (3.2) to give a symmetric 0-1 matrix. The trick we use is a variant of one used in
references [16, 18] for similar purposes.
The universal set G is chosen so that each gate has nonzero entries that are integer powers of ω = ei 4 .
π
π
Correspondingly, the nonzero standard basis matrix elements of Hx are also integer powers of ω = ei 4 .
We consider the 8 × 8 shift operator
7
S = ∑ | j + 1 mod 8ih j| (3.5)
j=0
and note that ω is an eigenvalue of S with eigenvector
1 7
|ωi = √ ∑ ω − j | ji . (3.6)
8 j=0
√ √ √ √
We modify Hx as follows. For each operator − 2H, − 2HT , − 2(HT )† , or − 2 (H ⊗ 1) CNOT
appearing in equation (3.2), define another operator that acts on C2 ⊗ C8 or C4 ⊗ C8 (as appropriate) by
replacing nonzero matrix elements with powers of the operator S:
ω k 7→ Sk .
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Matrix elements that are zero are mapped to the 8 × 8 all-zeroes matrix. Write B(W ) for the operators
obtained by making this replacement, e. g.,
√
4
ω ω5
4 5
S S
− 2HT = 4 7→ B(HT ) = 4 .
ω ω S S
Adjoining an 8-level ancilla as a third register and making this replacement in equation (3.2) gives
2M
Hprop = ∑ B(Wt )†13 ⊗ |tiht + 1|2 + B(Wt )13 ⊗ |t + 1iht|2 (3.7)
t=1
which is a symmetric 0-1 matrix (the subscripts indicate which registers the operators act on). Note that
Hprop commutes with S (acting on the 8-level ancilla) and therefore is block diagonal with eight sectors.
In the sector where S has eigenvalue ω, Hprop is identical to the Hamiltonian Hx that we started with (see
equation (3.2)). There is also a sector (where S has eigenvalue ω ∗ ) where the Hamiltonian is the complex
conjugate of Hx . We will add a term to Hprop that introduces an energy penalty for states in any of the
other six sectors, ensuring that none of these states lie in the ground space.
To see what kind of energy penalty is needed, we lower bound the eigenvalues of Hprop . Note that for
each W ∈ G, B(W ) contains at most 2 ones in each row or column. Looking at equation (3.7) and using
this fact, we see that each row and each column of Hprop contains at most four ones (with the remaining
entries all zero). Therefore kHprop k ≤ 4, so every eigenvalue of Hprop is at least −4.
The matrix Ax associated with the circuit Cx acts on the Hilbert space
Hcomp ⊗ Hclock ⊗ Hanc
where Hanc = C8 holds the 8-level ancilla. We define
Ax = Hprop + Hpenalty + Hinput + Houtput (3.8)
where
Hpenalty = 1 ⊗ 1 ⊗ S3 + S4 + S5
(3.9)
is the penalty ensuring that the ancilla register holds either |ωi or |ω ∗ i and the terms
n
Hinput = ∑ |1ih1| j ⊗ |1ih1| ⊗ 1
j=nin +1
Houtput = |0ih0|out ⊗ |M + 1ihM + 1| ⊗ 1
ensure that the ancilla qubits are initialized in the state |0i when t = 1 and that the output qubit is in the
state |1ih1| when the circuit Cx has been applied (i. e., at time t = M + 1). Observe that Ax is a symmetric
0-1 matrix.
Now consider the ground space of the first two terms Hprop + Hpenalty in (3.8). Note that
[Hprop , Hpenalty ] = 0 ,
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so these
√ operators can be simultaneously diagonalized. Furthermore, Hpenalty has smallest eigenvalue
−1 − 2, with eigenspace spanned by |ωi and |ω ∗ i. One can also easily confirm that the first excited
energy of Hpenalty is −1. √
The ground space of Hprop +Hpenalty lives in the sector where Hpenalty has minimal eigenvalue −1− 2.
To see this, note
√ that within this sector Hprop has the same eigenvalues as Hx , and therefore has lowest
eigenvalue −2 2. The minimum eigenvalue e1 of Hprop + Hpenalty in this sector is
√ √ √
e1 = −2 2 + −1 − 2 = −1 − 3 2 = −5.24 . . . , (3.10)
whereas in any other sector Hpenalty has eigenvalue at least −1 and (using the fact that Hprop ≥ −4) the
minimum eigenvalue of Hprop + Hpenalty is at least −5. Thus, an orthonormal basis for the ground space
of Hprop + Hpenalty is furnished by the states
1 2M
√ ∑ Wt−1Wt−2 . . .W1 |φ i|ti|ωi (3.11)
2M t=1
1 2M
√ ∑ (Wt−1Wt−2 . . .W1 )∗ |φ ∗ i|ti|ω ∗ i
2M t=1
(3.12)
where |φ i ranges over the basis Λ for Hcomp and ∗ denotes (elementwise) complex conjugation.
3.2 Upper bound on the smallest eigenvalue for yes instances
Suppose x is a yes instance; then there exists some nin -qubit state |ψin i satisfying AP (Cx , |ψin i) ≥
1 − (1/2|x| ). Let
1 2M
Wt−1Wt−2 . . .W1 |ψin i|0i⊗n−nin |ti|ωi
|witi = √ ∑
2M t=1
and note that this state is in the e1 -energy ground space of Hprop + Hpenalty (since it has the form (3.11)).
One can also directly verify that |witi has zero energy for Hinput . Thus
hwit|Ax |witi = e1 + hwit|Houtput |witi
1
= e1 + hψin |h0|⊗n−nin UC†x |0ih0|outUCx |ψin i|0i⊗n−nin
2M
1
= e1 + (1 − AP(Cx , |ψin i))
2M
1 1
≤ e1 + .
2M 2|x|
3.3 Lower bound on the smallest eigenvalue for no instances
Now suppose x is a no instance. Then the verification circuit Cx has acceptance probability AP (Cx , |ψi) ≤
1/3 for all nin -qubit input states |ψi.
We backtrack slightly to obtain bounds on the eigenvalue gaps of the Hamiltonians Hprop + Hpenalty
and Hprop + Hpenalty + Hinput . We begin by showing that the energy gap of Hprop + Hpenalty is at least an
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inverse polynomial function of M. Subtracting a constant equal to the ground energy times the identity
matrix sets the smallest eigenvalue to zero, and the smallest nonzero eigenvalue satisfies
√ π2
1
γ(Hprop + Hpenalty − e1 · 1) ≥ min 2 2 , −5 − e1 ≥ (3.13)
M 5M 2
since −5 − e1 ≈ 0.24 . . . > 1/5. The first inequality above follows from the fact that every eigenvalue of
Hprop in the range [e1 , −5) is also an eigenvalue of Hx (as discussed above) and the bound (3.4) on the
energy gap of Hx .
Now use the Nullspace Projection Lemma (Lemma 8.1) with
HA = Hprop + Hpenalty − e1 · 1 , HB = Hinput .
Note that HA and HB are positive semidefinite. Let SA be the ground space of HA and consider the
restriction HB |SA . Here it is convenient to use the basis for SA given by (3.11) and (3.12) with |φ i ranging
over the computational basis states of n qubits. In this basis, HB |SA is diagonal with all diagonal entries
equal to 1/(2M) times an integer, so γ(HB |SA ) ≥ 1/(2M). We also have γ(HA ) ≥ 1/(5M 2 ) from equation
(3.13), and clearly kHB k ≤ n. Thus Lemma 8.1 gives
1
1
5M 2 2M 1 1
γ(Hprop + Hpenalty + Hinput − e1 · 1) ≥ 1 1
≥ ≥ . (3.14)
5M 2
+ 2M + n 10M (1 + n) 20M 3 n
3
Now consider adding the final term Houtput . We use Lemma 8.1 again, now setting
HA = Hprop + Hpenalty + Hinput − e1 · 1 , HB = Houtput .
Let SA be the ground space of HA . Note that it is spanned by states of the form (3.11) and (3.12)
where |φ i = |ψi|0i⊗n−nin and |ψi ranges over any orthonormal basis of the nin -qubit input register. The
restriction HB |SA is block diagonal, with one block for states of the form
1 2M
Wt−1Wt−2 . . .W1 |ψi|0i⊗n−nin |ti|ωi
√ ∑ (3.15)
2M t=1
and another block for states of the form
1 2M
(Wt−1Wt−2 . . .W1 )∗ |ψi∗ |0i⊗n−nin |ti|ω ∗ i .
√ ∑ (3.16)
2M t=1
We now show that the minimum eigenvalue of HB |SA is nonzero, and we lower bound it. We consider
the two blocks separately. By linearity, every state in the first block can be written in the form (3.15) for
some state |ψi. Thus the minimum eigenvalue within this block is the minimum expectation of Houtput in
a state (3.15), where the minimum is taken over all nin -qubit states |ψi. This is equal to
1 1
min (1 − AP(Cx , |ψi)) ≥
|ψi 2M 3M
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where we used the fact that AP (Cx , |ψi) ≤ 1/3 for all |ψi. Likewise, every state in the second block can
be written as (3.16) for some state |ψi, and the minimum eigenvalue within this block is
1 1
min (1 − AP(Cx , |ψi)∗ ) ≥
|ψi 2M 3M
(since AP(Cx , |ψi)∗ = AP(Cx , |ψi) ≤ 1/3). Thus we see that HB |SA has an empty nullspace, so its smallest
eigenvalue is equal to its smallest nonzero eigenvalue, namely
1
γ(HB |SA ) ≥ .
3M
Now applying Lemma 8.1 using this bound, the fact that kHB k = 1, and the fact that
1
γ(HA ) ≥
20M 3 n
(from equation (3.14)), we get
1
60M 4 n 1
γ(Ax − e1 · 1) ≥ 1 1
≥ .
20M 3 n
+ 3M + 1 120M 4 n
Since HB |SA has an empty nullspace, Ax − e1 · 1 has an empty nullspace, and this is a lower bound on its
smallest eigenvalue.
4 Bose-Hubbard Hamiltonian is contained in QMA
To prove that Bose-Hubbard Hamiltonian is contained in QMA, we provide a verification algorithm
satisfying the requirements of Definition 2.1. In the definition this algorithm is specified by a circuit
involving only one measurement of the output qubit at the end of the computation. The procedure we
describe below, which contains intermediate measurements in the computational basis, can be converted
into a verification circuit of the desired form by standard techniques.
We are given an instance specified by G, N, c, and ε. We are also given an input state |φ i of nin
qubits, where nin= dlog2 DN e and DN is the dimension of ZN (G) as given in equation (2.4). Note, using
the inequality ab ≤ ab in equation (2.4), that nin = O(K log (N + K)), where K = |V | is the number of
vertices in the graph G. We embed ZN (G) into the space of nin qubits straightforwardly as the subspace
spanned by the first DN standard basis vectors (with lexicographic ordering, say). The first step of the
verification procedure is to measure the projector onto this space ZN (G). If the measurement outcome is
1 then the resulting state |φ 0 i is in ZN (G) and we continue; otherwise we reject.
In the second step of the verification procedure, the goal is to measure H̄GN in the state |φ 0 i. The
Hamiltonian H̄GN is sparse and efficiently row-computable, with norm
H̄GN ≤ HGN ≤ N kA(G)k + ∑ n̂k (n̂k − 1) ≤ NK + N 2 .
k∈V
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We use phase estimation (see for example [9]) to estimate the energy of |φ 0 i, using sparse Hamiltonian
simulation [2] to approximate evolution according to H̄GN . We choose the parameters of the phase
estimation so that, with probability at least 2/3, it produces an approximation E of the energy with error
at most ε/4. This can be done in time poly(N, K, 1/ε). If E ≤ c + ε/2 then we accept; otherwise we
reject.
We now show that this verification procedure satisfies the completeness and soundness requirements
of Definition 2.1. For a yes instance, an eigenvector of H̄GN with eigenvalue e ≤ c is accepted by this
procedure as long as the energy E computed in the phase estimation step has the desired precision. To see
this, note that we measure |E − e| ≤ ε/4, and hence E ≤ c + ε/4, with probability at least 2/3. For a no
instance, write |φ 0 i ∈ ZN (G) for a state obtained after passing the first step. The value E computed by the
subsequent phase estimation step satisfies E ≥ c + (3ε/4) with probability at least 2/3, in which case the
state is rejected. From this we see that the probability of accepting a no instance is at most 1/3.
5 Gate graphs
In Section 5 and Section 6, we develop machinery for proving that Bose-Hubbard Hamiltonian is QMA-
hard (as shown in Section 7). In this section we define a class of graphs (gate graphs) and a diagrammatic
notation for them (gate diagrams). We also discuss the Bose-Hubbard model on these graphs.
Every gate graph is constructed using a specific 128-vertex graph g0 as a building block. This graph is
shown in Figure 5.1 and discussed in Section 5.1. In Section 5.2 we define gate graphs and gate diagrams.
A gate graph is obtained by adding edges and self-loops (in a prescribed way) to a collection of disjoint
copies of g0 .
In Section 5.3 we discuss the ground states of the Bose-Hubbard model on gate graphs. For √ any gate
graph G, the smallest eigenvalue µ(G) of the adjacency matrix A(G) satisfies µ(G) ≥ −1 − 3 2. It is
convenient to define the constant √
e1 = −1 − 3 2 . (5.1)
When µ(G) = e1 we say G is an e1 -gate graph. We focus on the frustration-free states of e1 -gate graphs
(recall from Definition 2.2 that |φ i ∈ ZN (G) is frustration free if and only if H(G, N)|φ i = 0). We show
that all such states live in a convenient subspace (called I(G, N)) of the N-particle Hilbert space. This
subspace has the property that no two (or more) particles ever occupy vertices of the same copy of g0 .
The restriction to this subspace makes it easier to analyze the ground space.
In Section 5.4 we consider a class of subspaces that, like I(G, N), are defined by a set of constraints on
the locations of N particles in an e1 -gate graph G. We state an “Occupancy Constraints Lemma” (proven
in Section 10) that relates a subspace of this form to the ground space of the Bose-Hubbard model on a
graph derived from G.
5.1 The graph g0
Consider the single-qubit circuit C0 with four gates
U1 = U2 = H , U3 = HT , U4 = (HT )† ,
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with
1 1 1 1 0
H=√ , T= π .
2 1 −1 0 ei 4
We use the construction described in Section 3.1 to map this circuit to a graph g0 . The adjacency matrix
of g0 is Hprop + Hpenalty as defined in (3.7) and (3.9). The resulting graph is shown in Figure 5.1.
Each vertex of g0 corresponds to a standard basis vector in the Hilbert space C2 ⊗ C8 ⊗ C8 . We label
the vertices (z,t, j) with z ∈ {0, 1} describing the state of the computational qubit, t ∈ [8] giving the state
of the clock, and j ∈ {0, . . . , 7} describing the state of the ancilla.
√
As shown in Section 3.1, the smallest eigenvalue of A(g0 ) is e1 = −1 − 3 2. According to equations
(3.11) and (3.12), an orthonormal basis for the ground space of A(g0 ) consists of the states
1
|ψz,0 i = √ |zi(|1i + |3i + |5i + |7i) + H|zi(|2i + |8i) + HT |zi(|4i + |6i) |ωi (5.2)
8
|ψz,1 i = |ψz,0 i∗ (5.3)
where z ∈ {0, 1}.
Note that the amplitudes of |ψz,0 i in the above basis contain the result of computing either the identity,
Hadamard, or HT gate acting on the “input” state |zi.
5.2 Gate graphs and gate diagrams
We use three different schematic representations of the graph g0 (defined in Section 5.1), as depicted in
Figure 5.2. We call these figures diagram elements; they are also the simplest examples of gate diagrams,
which we define shortly.
The black and grey circles in a diagram element are called “nodes.” Each node has a label (z,t). The
only difference between the three diagram elements is the labeling of their nodes. In particular, the nodes
in the diagram element U ∈ {1, H, HT } correspond to values of t ∈ [8] where the first register in equation
(5.2) is either |zi or U|zi. For example, the nodes for the H diagram element have labels with t ∈ {1, 3}
(where |ψz,0 i contains the “input” |zi) or t = {2, 8} (where |ψz,0 i contains the “output” H|zi). We draw
the input nodes in black and the output nodes in grey.
The rules for constructing gate diagrams are simple. A gate diagram consists of some number
R ∈ {1, 2, . . .} of diagram elements, with self-loops attached to a subset S of the nodes and edges
connecting a set E of pairs of nodes. A node may have a single edge or a single self-loop attached to
it, but never more than one edge or self-loop and never both an edge and a self-loop. Each node in a
gate diagram has a label (q, z,t) where q ∈ [R] indicates the diagram element it belongs to. An example
is shown in Figure 5.3. Sometimes it is convenient to draw the input nodes on the right-hand side of a
diagram element; e. g., in Figure 6.1 the node closest to the top left corner is labeled (q, z,t) = (3, 0, 2).
To every gate diagram we associate a gate graph G with vertex set
{(q, z,t, j) : q ∈ [R], z ∈ {0, 1}, t ∈ [8], j ∈ {0, . . . , 7}}
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t =1
H H
t =8 t =2
H H
t =7 t =3
(HT )† HT
t =6 t =4
HT (HT )†
t =5
Figure 5.1: The graph g0 .
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(0, 1) (0, 2) (0, 1) (0, 4) (0, 1) (0, 5)
(0, 3) (0, 8) (0, 3) (0, 6) (0, 3) (0, 7)
(1, 1)
H (1, 2) (1, 1)
HT (1, 4) (1, 1)
1 (1, 5)
(1, 3) (1, 8) (1, 3) (1, 6) (1, 3) (1, 7)
(A) (B) (C)
Figure 5.2: Diagram elements from which a gate diagram is constructed. Each diagram element is a
schematic representation of the graph g0 shown in Figure 5.1.
1 2
HT 1
Figure 5.3: A gate diagram with two diagram elements labeled q = 1 (left) and q = 2 (right).
and adjacency matrix
A(G) = 1q ⊗ A(g0 ) + hS + hE (5.4)
hS = ∑ |q, z,tihq, z,t| ⊗ 1 j (5.5)
S
hE = ∑ |q, z,ti + |q0 , z0 ,t 0 i hq, z,t| + hq0 , z0 ,t 0 | ⊗ 1 j .
(5.6)
E
The sums in equations (5.5) and (5.6) run over the set of nodes with self-loops (q, z,t) ∈ S and the set
of pairs of nodes connected by edges {(q, z,t), (q0 , z0 ,t 0 )} ∈ E, respectively. We write 1q and 1 j for the
identity operator on the registers with variables q and j, respectively. We see from the above expression
that each self-loop in the gate diagram corresponds to 8 self-loops in the graph G, and an edge in the gate
diagram corresponds to 8 edges and 16 self-loops in G.
Since a node in a gate graph never has more than one edge or self-loop attached to it, equations (5.5)
and (5.6) are sums of orthogonal Hermitian operators. Therefore
khS k = max |q, z,tihq, z,t| ⊗ 1 j = 1 if S 6= 0/ (5.7)
S
khE k = max |q, z,ti + |q0 , z0 ,t 0 i hq, z,t| + hq0 , z0 ,t 0 | ⊗ 1 j = 2 if E 6= 0/
(5.8)
E
for any gate graph. (Of course, this also shows that khS0 k = 1 and khE0 k = 2 for any nonempty subsets
S0 ⊆ S and E0 ⊆ E.)
5.3 Frustration-free states on e1 -gate graphs
Consider the adjacency matrix A(G) of a gate graph G, and note (from equation (5.4)) that its smallest
eigenvalue µ(G) satisfies
µ(G) ≥ e1
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since hS and hE are positive semidefinite and A(g0 ) has smallest eigenvalue e1 . In the special case where
µ(G) = e1 , we say G is an e1 -gate graph.
Definition 5.1. An e1√
-gate graph is a gate graph G such that the smallest eigenvalue of its adjacency
matrix is e1 = −1 − 3 2.
When G is an e1 -gate graph, a single-particle ground state |Γi of A(G) has minimal energy for each
term in (5.4), i. e., it satisfies
(1 ⊗ A(g0 )) |Γi = e1 |Γi (5.9)
hS |Γi = 0 (5.10)
hE |Γi = 0 . (5.11)
Indeed, to show that a given gate graph G is an e1 -gate graph, it suffices to find a state |Γi satisfying these
conditions. Note that equation (5.9) implies that |Γi can be written as a superposition of the states
q
|ψz,a i = |qi|ψz,a i, z, a ∈ {0, 1}, q ∈ [R] ,
where |ψz,a i is given by equations (5.2) and (5.3). The coefficients in the superposition are then constrained
by equations (5.10) and (5.11).
Example 5.2. As an example, we show the gate graph in Figure 5.3 is an e1 -gate graph. As noted above,
q
equation (5.9) lets us restrict our attention to the space spanned by the eight states |ψz,a i with z, a ∈ {0, 1}
and q ∈ {1, 2}. In this basis, the operators hS and hE only have nonzero matrix elements between states
with the same value of a ∈ {0, 1}. We therefore solve for the e1 energy ground states with a = 0 and
those with a = 1 separately. Consider a ground state of the form
1 1 2 2
τ1 |ψ0,a i + ν1 |ψ1,a i + τ2 |ψ0,a i + ν2 |ψ1,a i
and note that in this case (5.10) implies τ1 = 0. Equation (5.11) gives
!
−τ 1
HT a=0
τ2
−ν1 !
=
∗ −τ1
ν2
(HT ) −ν a = 1.
1
We find two orthogonal e1 -energy states, which are (up to normalization)
π
1 ei 4 2 2
|ψ1,0 i − √ |ψ0,0 i − |ψ1,0 i (5.12)
2
π
1 e−i 4 2 2
|ψ1,1 i − √ |ψ0,1 i − |ψ1,1 i . (5.13)
2
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We interpret each of these states as encoding a qubit that is transformed at each set of input/output nodes
in the gate diagram in Figure 5.3. The encoded qubit begins on the input nodes of the first diagram
element in the state
τ1 0
=
ν1 1
1 i. On the output nodes of diagram element 1, the
because the self-loop penalizes the basis vectors |ψ0,a
encoded qubit is in the state where either HT (if a = 0) or its complex conjugate (if a = 1) has been
applied. The edges in the gate diagram ensure that the encoded qubit on the input nodes of diagram
element 2 is minus the state on the output nodes of diagram element 1.
In this example, each single-particle ground state encodes a single-qubit computation. Later we show
how N-particle frustration-free states on e1 -gate graphs can encode computations on N qubits. Recall
from Definition 2.2 that an N-particle bosonic state |Γi ∈ ZN (G) is said to be frustration free if and only
if H(G, N)|Γi = 0. Note that H(G, N) ≥ 0, so an N-particle frustration-free state is necessarily a ground
state. Putting this together with Lemma 2.3, we see that the existence of an N-particle frustration-free
state implies
λN1 (G) = λN−1
1
(G) = · · · = λ11 (G) = 0 ,
i. e., there are N 0 -particle frustration-free states for all N 0 ≤ N.
We prove that the graph g0 has no two-particle frustration-free states. By Lemma 2.3, it follows that
g0 has no N-particle frustration-free states for N ≥ 2.
Lemma 5.3. λ21 (g0 ) > 0.
Proof. Suppose (for a contradiction) that |Qi ∈ Z2 (g0 ) is a nonzero vector in the nullspace of H(g0 , 2),
so
Hg20 |Qi = A(g0 ) ⊗ 1 + 1 ⊗ A(g0 ) + 2 ∑ |vihv| ⊗ |vihv| |Qi = 2e1 |Qi .
v∈g0
This implies
A(g0 ) ⊗ 1|Qi = 1 ⊗ A(g0 )|Qi = e1 |Qi
since A(g0 ) has smallest eigenvalue e1 and the interaction term is positive semidefinite. We can therefore
write
|Qi = ∑ Qza,xy |ψz,a i|ψx,y i
z,a,x,y∈{0,1}
with Qza,xy = Qxy,za (since |Qi ∈ Z2 (g0 )) and
(|vihv| ⊗ |vihv|) |Qi = 0 (5.14)
for all vertices v = (z,t, j) ∈ g0 . Using this equation with |vi = |0, 1, ji gives
Q00,00 h0, 1, j|ψ0,0 i2 + 2Q01,00 h0, 1, j|ψ0,1 ih0, 1, j|ψ0,0 i + Q01,01 h0, 1, j|ψ0,1 i2
1
Q00,00 i− j + 2Q01,00 + Q01,01 i j
=
64
=0
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for each j ∈ {0, . . . , 7}. The only solution to this set of equations is Q00,00 = Q01,00 = Q01,01 = 0. The
same analysis, now using |vi = |1, 1, ji, gives Q10,10 = Q11,10 = Q11,11 = 0. Finally, using equation (5.14)
with |vi = |0, 2, ji gives
1
h0|H|1ih0|H|0i 2Q10,00 i− j + 2Q10,01 + 2Q11,00 + 2Q11,01 i j
64
1
Q10,00 i− j + Q10,01 + Q11,00 + Q11,01 i j
=
64
=0
for all j ∈ {0, . . . , 7}, which implies that Q10,00 = Q11,01 = 0 and Q11,00 = −Q10,01 . Thus, up to normal-
ization,
|Qi = |ψ1,0 i|ψ0,1 i + |ψ0,1 i|ψ1,0 i − |ψ11 i|ψ00 i − |ψ00 i|ψ11 i .
Now applying equation (5.14) with |vi = |0, 4, ji, we see that the quantity
1 1 iπ π
(2h0|HT |1ih0|(HT )∗ |0i − 2h0|(HT )∗ |1ih0|HT |0i) = e 4 − e−i 4
64 64
must be zero, which is a contradiction. Hence we conclude that the nullspace of H(g0 , 2) is empty.
We now characterize the space of N-particle frustration-free states on an e1 -gate graph G. Define
the subspace I(G, N) ⊂ ZN (G) where each particle is in a ground state of A(g0 ) and no two particles are
located within the same diagram element:
I(G, N) = span{Sym(|ψzq11,a1 i . . . |ψzqNN,aN i) : zi , ai ∈ {0, 1}, qi ∈ [R], qi 6= q j whenever i 6= j}. (5.15)
Lemma 5.4. Let G be an e1 -gate graph. A state |Γi ∈ ZN (G) is frustration free if and only if
(A(G) − e1 )(w) |Γi = 0 for all w ∈ [N] (5.16)
|Γi ∈ I(G, N) . (5.17)
Proof. First suppose that equations (5.16) and (5.17) hold. From (5.17) we see that |Γi has no support on
states where two or more particles are located at the same vertex. Hence
∑ n̂k (n̂k − 1) |Γi = 0 . (5.18)
k∈V
Putting together equations (5.16) and (5.18), we get
H(G, N)|Γi = HGN − Ne1 |Γi = 0 ,
so |Γi is frustration free.
To complete the proof, we show that if |Γi is frustration free, then conditions (5.16) and (5.17) hold.
By definition, a frustration-free state |Γi satisfies
!
N
H(G, N)|Γi = ∑ (A(G) − e1 )(w) + ∑ n̂k (n̂k − 1) |Γi = 0 . (5.19)
w=1 k∈V
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Since both terms in the large parentheses are positive semidefinite, they must both annihilate |Γi; similarly,
each term in the first summation must be zero. Hence equation (5.16) holds. Let Grem be the graph
obtained from G by removing all of the edges and self-loops in the gate diagram of G. In other words,
R
A(Grem ) = ∑ |qihq| ⊗ A(g0 ) = 1 ⊗ A(g0 ) .
q=1
Noting that
H(G, N) ≥ H(Grem , N) ≥ 0 ,
we see that equation (5.19) also implies
H(Grem , N)|Γi = 0 . (5.20)
Since each of the R components of Grem is an identical copy of g0 , the eigenvalues and eigenvectors of
H(Grem , N) are characterized by Lemma 2.4 (along with knowledge of the eigenvalues and eigenvectors
of g0 ). By Lemma 5.3 and Lemma 2.3, no component has a two- (or more) particle frustration-free state.
Combining these two facts, we see that in an N-particle frustration-free state, every component of Grem
must contain either 0 or 1 particles, and the nullspace of H(Grem , N) is the space I(G, N). From equation
(5.20) we get |Γi ∈ I(G, N).
Note that if I(G, N) is empty then Lemma 5.4 says that G has no N-particle frustration-free states.
For example, this holds for any e1 -gate graph G whose gate diagram has R < N diagram elements.
A useful consequence of Lemma 5.4 is the fact that every k-particle reduced density matrix of an
N-particle frustration-free state |Γi on an e1 -gate graph G (with k ≤ N) has all of its support on k-particle
frustration-free states. To see this, note that for any partition of the N registers into subsets A (of size k)
and B (of size N − k), we have
I(G, N) ⊆ I(G, k)A ⊗ ZN−k (G)B .
Thus, if condition (5.17) holds, then all k-particle reduced density matrices of |Γi are contained in I(G, k).
Furthermore, (5.16) is a statement about the single-particle reduced density matrices, so it also holds
for each k-particle reduced density matrix. From this we see that each reduced density matrix of |Γi is
frustration free.
5.4 Occupancy constraints
Lemma 5.4 says in particular that a frustration-free state on an e1 -gate graph has no support on states
where multiple particles occupy the same diagram element. Indeed, the lemma allows us to restrict our
attention to the subspace I(G, N) when solving for N-particle frustration-free states. Here we consider
restrictions to other subspaces corresponding to more general constraints on the locations of particles in a
gate graph.
For any e1 -gate graph G with R diagram elements and any simple R-vertex graph Gocc with edge set
E(Gocc ), we define a subspace
I(G, Gocc , N) = span{Sym(|ψzq11,a1 i . . . |ψzqNN,aN i) : zi , ai ∈ {0, 1}, qi 6= q j for i 6= j, {qi , q j } ∈
/ E(Gocc )} .
(5.21)
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In this subspace, no two particles occupy the same diagram element and no two particles occupy diagram
elements connected by an edge in Gocc . Note that
I(G, Gocc , N) ⊆ I(G, N) ⊆ ZN (G) .
We say that Gocc specifies a set of occupancy constraints on G. If I(G, Gocc , N) is not empty, we
define the restriction
H(G, Gocc , N) = H(G, N) I(G,Gocc ,N)
and write λN1 (G, Gocc ) for its smallest eigenvalue.
We now explain how the subspace I(G, Gocc , N) relates to the Bose-Hubbard model on a graph
derived from G and Gocc . Specifically, the following lemma states that we can construct another graph
G (depending on G and Gocc ) so that the ground state of the Bose-Hubbard model on G effectively
“simulates” the restriction to I(G, Gocc , N). The ground energy λN1 (G, Gocc ) of H(G, Gocc , N) is zero if
and only if the ground energy λN1 (G ) of H(G , N) is zero. The lemma also shows that the two energies
are related even if they are nonzero.
Lemma 5.5 (Occupancy Constraints Lemma). Let G be an e1 -gate graph specified as a gate diagram
with R ≥ 2 diagram elements. Let N ∈ [R], let Gocc specify a set of occupancy constraints on G, and
suppose the subspace I(G, Gocc , N) is nonempty. Then there exists an efficiently computable e1 -gate
graph G with at most 7R2 diagram elements such that
a
1. If λN1 (G, Gocc ) ≤ a then λN1 (G ) ≤ .
R
b
2. If λN1 (G, Gocc ) ≥ b with b ∈ [0, 1], then λN1 (G ) ≥ .
(13R)9
Note that the lemma stipulates N ≤ R without loss of generality, since I(G, Gocc , N) is empty otherwise.
The proof of this lemma appears in Section 10. In the proof we show how to construct the gate graph G
from G and Gocc .
6 Gadgets
In Example 5.2 we saw how a single-particle ground state can encode a single-qubit computation. In this
section we see how a two-particle frustration-free state on a suitably designed e1 -gate graph can encode a
two-qubit computation. We design specific e1 -gate graphs (called gadgets) that we use in Section 7 to
prove that Bose-Hubbard Hamiltonian is QMA-hard. For each gate graph we discuss, we show that the
smallest eigenvalue of its adjacency matrix is e1 and we solve for all of the frustration-free states.
We first design a gate graph where, in any two-particle frustration-free state, the locations of the parti-
cles are synchronized. This “move-together” gadget is presented in Section 6.1. In Section 6.2, we design
gadgets for two-qubit gates using four move-together gadgets, one for each two-qubit computational
basis state. Finally, in Section 6.3 we describe a small modification of a two-qubit gate gadget called the
“boundary gadget.”
The circuit-to-gate graph mapping described in Section 7 uses a two-qubit gate gadget for each gate
in the circuit, together with boundary gadgets in parts of the graph corresponding to the beginning and
end of the computation.
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1
3 α 5
H β H H
α
4 6 = β
γ W δ
2
H δ
H H
γ
Figure 6.1: The gate diagram for the move-together gadget.
6.1 The move-together gadget
The gate diagram for the move-together gadget is shown in Figure 6.1. Using equation (5.4), we write the
adjacency matrix of the corresponding gate graph GW as
6
A(GW ) = ∑ |qihq| ⊗ A(g0 ) + hE (6.1)
q=1
where hE is given by (5.6) and E is the set of edges in the gate diagram (in this case hS = 0 as there are no
self-loops).
We begin√by solving for the single-particle ground states, i. e., the eigenvectors of (6.1) with eigenvalue
e1 = −1 − 3 2. As in Example 5.2, we can solve for the states with a = 0 and a = 1 separately, since
j i
hψx,1 |hE |ψz,0 i=0
for all i, j ∈ {1, . . . , 6} and x, z ∈ {0, 1}. We write a single-particle ground state as
6
i i
∑ τi |ψ0,a i + νi |ψ1,a i
i=1
and solve for the coefficients τi and νi using equation (5.11) (in this case equation (5.10) is automatically
satisfied since hS = 0). Enforcing (5.11) gives eight equations, one for each edge in the gate diagram:
1
τ3 = −τ1 √ (τ1 + ν1 ) = −τ6
2
1
τ4 = −ν1 √ (τ1 − ν1 ) = −τ5
2
1
ν3 = −τ2 √ (τ2 + ν2 ) = −ν5
2
1
ν4 = −ν2 √ (τ2 − ν2 ) = −ν6 .
2
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There are four linearly independent solutions to this set of equations, given by
1 1
Solution 1: τ1 = 1 τ3 = −1 τ5 = − √ τ6 = − √ all other coefficients 0
2 2
1 1
Solution 2: ν1 = 1 τ4 = −1 τ5 = √ τ6 = − √ all other coefficients 0
2 2
1 1
Solution 3: ν2 = 1 ν4 = −1 ν5 = − √ ν6 = √ all other coefficients 0
2 2
1 1
Solution 4: τ2 = 1 ν3 = −1 ν5 = − √ ν6 = − √ all other coefficients 0.
2 2
For each of these solutions, and for each a ∈ {0, 1}, we find a single-particle state with energy e1 . This
result is summarized in the following lemma.
Lemma 6.1. GW is an e1 -gate graph. A basis for the eigenspace of A(GW ) with eigenvalue e1 is
1 1 1 3 1 5 1 6
|χ1,a i = √ |ψ0,a i − √ |ψ0,a i − √ |ψ0,a i − √ |ψ0,a i (6.2)
3 3 6 6
1 1 1 4 1 5 1 6
|χ2,a i = √ |ψ1,a i − √ |ψ0,a i + √ |ψ0,a i − √ |ψ0,a i (6.3)
3 3 6 6
1 2 1 4 1 5 1 6
|χ3,a i = √ |ψ1,a i − √ |ψ1,a i − √ |ψ1,a i + √ |ψ1,a i (6.4)
3 3 6 6
1 2 1 3 1 5 1 6
|χ4,a i = √ |ψ0,a i − √ |ψ1,a i − √ |ψ1,a i − √ |ψ1,a i (6.5)
3 3 6 6
where a ∈ {0, 1}.
In Figure 6.1 we have used a shorthand α, β , γ, δ to identify four nodes of the move-together gadget;
these are the nodes with labels (q, z,t) = (1, 0, 1), (1, 1, 1), (2, 1, 1), (2, 0, 1), respectively. We view α and
γ as “input” nodes and β and δ as “output” nodes for the move-together gadget. It is natural to associate
each single-particle state |χi,a i with one of these four nodes. Each node corresponds to 8 vertices of the
underlying graph, e. g.,
Sα = {(1, 0, 1, j) : j ∈ {0, . . . , 7}} .
Looking at equation (6.2) (and referring back to equation (5.2)) we see that |χ1,a i has support on vertices
in Sα but not on vertices in Sβ , Sγ , or Sδ . Looking at the picture on the right-hand side of the equality
sign in Figure 6.1, we think of |χ1,a i as localized at the node α, with no support on the other three nodes.
The states |χ2,a i, |χ3,a i, |χ4,a i are similarly localized at nodes β , γ, δ . We view |χ1,a i and |χ3,a i as input
states and |χ2,a i and |χ4,a i as output states.
Now we turn our attention to the two-particle frustration-free states of the move-together gadget, i. e.,
the states |Φi ∈ Z2 (GW ) in the nullspace of H(GW , 2). Using Lemma 5.4 we can write
|Φi = ∑ C(I,a),(J,b) |χI,a i|χJ,b i (6.6)
a,b∈{0,1}, I,J∈[4]
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where the coefficients are symmetric, i. e.,
C(I,a),(J,b) = C(J,b),(I,a) , (6.7)
and where
q q
hψz,a |hψx,b |Φi = 0 (6.8)
for all z, a, x, b ∈ {0, 1} and q ∈ [6].
The move-together gadget is designed so that each solution |Φi to these equations is a superposition
of a term where both particles are in input states and a term where both particles are in output states. The
particles move from input nodes to output nodes together. We now solve equations (6.6)–(6.8) and prove
the following.
Lemma 6.2. A basis for the nullspace of H(GW , 2) is
1 1
|Φa,b i = Sym √ |χ1,a i|χ3,b i + √ |χ2,a i|χ4,b i , a, b ∈ {0, 1} . (6.9)
2 2
There are no N-particle frustration-free states on GW for N ≥ 3, i. e.,
λN1 (GW ) > 0 for N ≥ 3 .
Proof. The states |Φa,b i manifestly satisfy equations (6.6) and (6.7), and one can directly verify that they
also satisfy (6.8) (the nontrivial cases to check are q = 5 and q = 6).
To complete the proof that (6.9) is a basis for the nullspace of H(GW , 2), we verify that any state
satisfying these conditions must be a linear combination of these four states. Applying equation (6.8)
gives
1 1 1 1 1 1
hψ0,a |hψ0,b |Φi = C(1,a),(1,b) = 0 hψ1,a |hψ1,b |Φi = C(2,a),(2,b) = 0
3 3
2 2 1 2 2 1
hψ1,a |hψ1,b |Φi = C(3,a),(3,b) = 0 hψ0,a |hψ0,b |Φi = C(4,a),(4,b) = 0
3 3
1 1 1 2 2 1
hψ0,a |hψ1,b |Φi = C(1,a),(2,b) = 0 hψ0,a |hψ1,b |Φi = C(4,a),(3,b) = 0
3 3
3 3 1 4 4 1
hψ0,a |hψ1,b |Φi = C(1,a),(4,b) = 0 hψ0,a |hψ1,b |Φi = C(2,a),(3,b) = 0
3 3
for all a, b ∈ {0, 1}. Using the fact that all of these coefficients are zero, and using equation (6.7), we get
|Φi = ∑ C(1,a),(3,b) (|χ1,a i|χ3,b i + |χ3,b i|χ1,a i) +C(2,a),(4,b) (|χ2,a i|χ4,b i + |χ4,b i|χ2,a i) .
a,b∈{0,1}
Finally, applying equation (6.8) again gives
6 6 1 1
hψ0,a |hψ1,b |Φi = C(2,a),(4,b) − C(1,a),(3,b) = 0 .
6 6
Hence
|Φi = ∑ C(1,a),(3,b) (|χ1,a i|χ3,b i + |χ3,b i|χ1,a i + |χ2,a i|χ4,b i + |χ4,b i|χ2,a i) ,
a,b∈{0,1}
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which is a superposition of the states |Φa,b i.
Finally, we prove that there are no frustration-free ground states of the Bose-Hubbard model on
GW with more than two particles. By Lemma 2.3, it suffices to prove that there are no frustration-free
three-particle states.
Suppose (for a contradiction) that |Γi ∈ Z3 (GW ) is a normalized three-particle frustration-free state.
Write
|Γi = ∑ D(i,a),( j,b),(k,c) |χi,a i|χ j,b i|χk,c i .
Note that each reduced density matrix of |Γi on two of the three subsystems must have all of its support
on two-particle frustration-free states (see the remark following Lemma 5.4), i. e., on the states |Φa,b i.
Using this fact for the subsystem consisting of the first two particles, we see in particular that
(i, j) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ D(i,a),( j,b),(k,c) = 0 (6.10)
(since |Φa1 ,a2 i only has support on vectors |χi,a i|χ j,b i with i, j ∈ {(1, 3), (3, 1), (2, 4), (4, 2)}).
Using this fact for subsystems consisting of particles 2, 3 and 1, 3, respectively, gives
( j, k) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ D(i,a),( j,b),(k,c) = 0 (6.11)
(i, k) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ D(i,a),( j,b),(k,c) = 0 . (6.12)
Putting together equations (6.10), (6.11), and (6.12), we see that |Γi = 0. This is a contradiction, so no
three-particle frustration-free states exist.
Next we show how the move-together gadget can be used to build gadgets that implement two-qubit
gates.
6.2 Gadgets for two-qubit gates
In this section we define a gate graph for each of the two-qubit unitaries
{CNOT12 , CNOT21 , CNOT12 (H ⊗ 1) , CNOT12 (HT ⊗ 1)}.
Here CNOT12 is the standard controlled-not gate with the second qubit as a target, whereas CNOT21 has
the first qubit as target.
We define the gate graphs by exhibiting their gate diagrams. For the three cases
U = CNOT12 (Ũ ⊗ 1)
with Ũ ∈ {1, H, HT }, we associate U with the gate diagram shown in Figure 6.2(A). In the figure we also
indicate a shorthand used to represent this gate diagram. As one might expect, for the case U = CNOT21 ,
we use the same gate diagram as for U = CNOT12 ; however, we use the slightly different shorthand
shown in Figure 6.2(B).
Roughly speaking, the two-qubit gate gadgets work as follows. In Figure 6.2(A) there are four move-
together gadgets, one for each two-qubit basis state |00i, |01i, |10i, |11i. These enforce the constraint
that two particles must move through the graph together. The connections between the four diagram
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5 6
1 1
1 00 2
α W ε
β Ũ 01 1 ζ α
Ũ
ε
W β ζ
3 10 4 =
γ W η γ η
δ 1 11 1 θ
δ θ
W
7 8
1 1
(A)
γ η
δ θ
α ε
β ζ
(B)
Figure 6.2: ( A ) Gadget for the two-qubit unitary U = (Ũ ⊗ 1)CNOT12 with Ũ ∈ {1, H, HT }. ( B ) For
the U = CNOT21 gate (first qubit is the target), we use the same gate graph as in ( A ) with Ũ = 1; we
represent it schematically as shown.
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elements labeled 1, 2, 3, 4 and the move-together gadgets ensure that certain frustration-free two-particle
states encode two-qubit computations, while the connections between diagram elements 1, 2, 3, 4 and
5, 6, 7, 8 ensure that there are no additional frustration-free two-particle states (i. e., states that do not
encode computations).
To describe the frustration-free states of the gate graph depicted in Figure 6.2(A), first recall the
definition of the states |χ1,a i, |χ2,a i, |χ3,a i, |χ4,a i from equations (6.2)–(6.5). For each of the move-together
gadgets xy ∈ {00, 01, 10, 11} in Figure 6.2(A), write
xy
|χL,a i
for the state |χL,a i with support (only) on the gadget labeled xy. Write
(
U if a = 0
U(a) = ∗
U if a = 1
and similarly for Ũ (we use this notation throughout the paper to indicate a unitary or its elementwise
complex conjugate).
The following lemma shows that GU is an e1 -gate graph and characterizes its frustration-free states.
Lemma 6.3. Let U = CNOT12 (Ũ ⊗ 1) where Ũ ∈ {1, H, HT }. The corresponding gate graph GU is
defined by its gate diagram shown in Figure 6.2(A). The adjacency matrix A(GU ) has ground energy e1 ;
a basis for the corresponding eigenspace is
r
1,U 1 1 1 5+z 3 1 yx
|ρz,a i = √ |ψz,a i − √ |ψz,a i − ∑ Ũ(a)yz |χ1,a i (6.13)
8 8 8 x,y=0
r
2,U 1 2 1 6−z 3 1 zx
|ρz,a i = √ |ψz,a i − √ |ψz,a i − ∑ |χ2,a i (6.14)
8 8 8 x=0
r
3,U 1 3 1 7 3 1 xz
|ρz,a i = √ |ψz,a i − √ |ψz,a i − ∑ |χ3,a i (6.15)
8 8 8 x=0
r
4,U 1 4 1 8 3 1 x(z⊕x)
|ρz,a i = √ |ψz,a i − √ |ψz,a i − ∑ |χ4,a i (6.16)
8 8 8 x=0
where z, a ∈ {0, 1}. A basis for the nullspace of H(GU , 2) is
Sym(|TzU1 ,a,z2 ,b i), z1 , z2 , a, b ∈ {0, 1} (6.17)
where
1
1 3,U 1
|TzU1 ,a,z2 ,b i = √ |ρz1,U
,a i|ρz2 ,b i + √ ∑ U(a)x1 x2 ,z1 z2 |ρx2,U
1 ,a
i|ρx4,U
2 ,b
i (6.18)
2 1 2 x1 ,x2 =0
for z1 , z2 , a, b ∈ {0, 1}. There are no N-particle frustration-free states on GU for N ≥ 3, i. e.,
λN1 (GU ) > 0 for N ≥ 3 .
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We view the nodes labeled α, β , γ, δ in Figure 6.2(A) as “input” nodes and those labeled ε, ζ , η, θ as
i,U
“output nodes.” Each of the states |ρx,y i is associated with one of the nodes, depending on the values of
1,U 1,U
i ∈ {1, 2, 3, 4} and x ∈ {0, 1}. For example, the states |ρ0,0 i and |ρ0,1 i are associated with input node α
since they both have nonzero amplitude on vertices of the gate graph that are associated with α (and zero
amplitude on vertices associated with other labeled nodes).
The two-particle state Sym(|TzU1 ,a,z2 ,b i) is a superposition of a term
1 1,U 3,U
Sym √ |ρz ,a i|ρz2 ,b i
2 1
with both particles located on vertices corresponding to input nodes and a term
!
1
Sym √ ∑ U(a)x1 x2 ,z1 z2 |ρx2,U
1 ,a
i|ρx4,U
2 ,b
i
2 x1 ,x2 ∈{0,1}
with both particles on vertices corresponding to output nodes. The two-qubit gate U(a) is applied as the
particles move from input nodes to output nodes.
Proof of Lemma 6.3. Recall that the gate graph GU is specified by its gate diagram, shown in Fig-
ure 6.2(A). The adjacency matrix of the gate graph GU is of the form in equation (5.4). There are 6
diagram elements for each of the move-together gadgets, so there are 32 diagram elements in total. We
will need to refer to those diagram elements labeled q ∈ [8] in Figure 6.2(A) (i. e., those not contained in
the move-together gadgets).
Write
A(GU ) = A(GU0 ) + hE0
where GU0 is the gate graph obtained from GU by removing all 24 edges shown in Figure 6.2(A) (GU0 does
include the edges within each of the move-together gadgets). Here hE0 is given by equation (5.6) with E0
the set of 24 edges shown in Figure 6.2(A).
One basis for the e1 -energy ground space of A(GU0 ) is given by the 64 states
q
|ψz,a i, q ∈ [8], z, a ∈ {0, 1}
xy
|χL,a i, x, y, a ∈ {0, 1}, L ∈ [4] .
However, it is convenient to work with the following slightly different basis for this space:
q
|ψz,a i, q ∈ [8], z, a ∈ {0, 1}
xy
∑ Ũ(a)xz |χ1,a i, y, z, a ∈ {0, 1}
x∈{0,1}
xy
|χL,a i, x, y, a ∈ {0, 1}, L ∈ {2, 3, 4} .
Here some of the states are in a superposition corresponding to the output of the single-qubit unitary Ũ.
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We are interested in the intersection of the ground space of A(GU0 ) with the nullspace of hE0 , so we
compute the matrix elements of hE0 in the above basis. The resulting 64 × 64 matrix is block diagonal
with sixteen 4 × 4 blocks. Each block is identical, with entries
3 1 1 1 √ √
8 8 8 3 8 3
1 1
0 0
8 8
. (6.19)
√1 0 1
0
8 3 24
1
√ 1
8 3
0 0 24
The four states involved in each block are given by (in order from left to right as in the matrix above):
1 5+z x0 x1
|ψz,a i, |ψz,a i, ∑ Ũ(a)xz |χ1,a i, ∑ Ũ(a)xz |χ1,a i
x∈{0,1} x∈{0,1}
2 6−z z0 z1
|ψz,a i, |ψz,a i, |χ2,a i, |χ2,a i
3 7 0z 1z
|ψz,a i, |ψz,a i, |χ3,a i, |χ3,a i
4 8 0z 1(z⊕1)
|ψz,a i, |ψz,a i, |χ4,a i, |χ4,a i.
The unique zero eigenvector of the matrix (6.19) is
1
1 −1
√ .
√
8 −√3
− 3
1,U 2,U 3,U 4,U
Constructing this vector within each of the 16 blocks, we get the states {|ρz,a i, |ρz,a i, |ρz,a i, |ρz,a i}.
Now consider the two-particle sector. Using Lemma 5.4 we can write any two-particle frustration-free
state as
I,U J,U
|Θi = ∑ ∑ B(z,a,I),(x,b,J) |ρz,a i|ρx,b i (6.20)
z,a,x,b∈{0,1} I,J∈[4]
where
B(z,a,I),(x,b,J) = B(x,b,J),(z,a,I) (6.21)
and
q q
hψx,a |hψz,b |Θi = 0 (6.22)
for all x, z, a, b ∈ {0, 1} and q ∈ [32]. To enforce equation (6.22) we consider the diagram elements
q ∈ [8] (as labeled in Figure 6.2(A)) separately from the other 24 diagram elements (those inside the
move-together gadgets).
Using equation (6.22) with q ∈ {1, 2, 3, 4, 7, 8} and x, z, a, b ∈ {0, 1} gives
B(x,a,I),(z,b,I) = 0 I ∈ [4], x, z, a, b ∈ {0, 1} . (6.23)
Using q = 5, x = 0, and z = 1 in equation (6.22) gives
5 5 1
hψ0,a |hψ1,b |Θi = B(0,a,1),(1,b,2) = 0 , (6.24)
8
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for a, b ∈ {0, 1}, while q = 6, x = 0, and z = 1 gives
6 6 1
hψ0,a |hψ1,b |Θi = B(0,a,2),(1,b,1) = 0 . (6.25)
8
Applying equation (6.22) with q = 5 or q = 6 and other choices for x and z does not lead to any additional
independent constraints on the state |Θi.
Now consider the constraint (6.22) for diagram elements inside the move-together gadgets in Fig-
ure 6.2(A). Let Πxy be the projector onto two-particle states where both particles are located at vertices
contained within the move-together gadget labeled xy ∈ {00, 01, 10, 11}. Using the results of Lemma 6.2,
we see that for diagram elements inside the move-together gadgets, (6.22) is satisfied if and only if
xy xy xy xy
Πxy |Θi ∈ span Sym |χ1,a i|χ3,b i + |χ2,a i|χ4,b i , a, b ∈ {0, 1} .
Since we already know
xy
i|χ xy
Πxy |Θi ∈ span Sym |χi,a j,b i , i, j ∈ [4], a, b ∈ {0, 1}
we get
xy xy
hχK,a |hχK,b |Θi = 0 K ∈ [4] (6.26)
xy xy
hχK,a |hχL,b |Θi = 0 (K, L) ∈ {(1, 2), (2, 3), (3, 4), (1, 4)} (6.27)
xy xy xy xy
hχ1,a |hχ3,b | − hχ2,a |hχ4,b | |Θi = 0 (6.28)
for all a, b ∈ {0, 1}. Note that (6.26) is automatically satisfied whenever (6.23) holds.
Applying equation (6.27) with (K, L) = (1, 2) and a, b, x, y ∈ {0, 1}, we get
xy xy 3 3
hχ1,a |hχ2,b |Θi = ∑ Ũ(a)xz B(z,a,1),(x,b,2) = Ũ(a)xx B(x,a,1),(x,b,2) = 0 . (6.29)
8 z∈{0,1} 8
In the second equality we used the fact that B(z,a,1),(x,b,2) is zero whenever z 6= x (from equations (6.21),
(6.24), and (6.25)). Since Ũ ∈ {1, H, HT } we have Ũ(a)xx 6= 0, and it follows that
B(x,a,1),(x,b,2) = 0 (6.30)
for all x, a, b ∈ {0, 1}.
Applying equation (6.27) with (K, L) = (1, 4) gives
xy xy 3
hχ1,a |hχ4,b |Θi = ∑ Ũ(a)xz B(z,a,1),(x⊕y,b,4) = 0 , x, y, a, b ∈ {0, 1}. (6.31)
8 z∈{0,1}
By taking appropriate combinations of these equations, we have
x(y⊕x) x(y⊕x)
∑ Ũ(a)†wx hχ1,a |hχ4,b |Θi = B(w,a,1),(y,b,4) = 0 , w, y, a, b ∈ {0, 1}. (6.32)
x∈{0,1}
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Applying equation (6.27) with (K, L) = (2, 3) and (K, L) = (3, 4) gives
xy xy 3
hχ2,a |hχ3,b |Θi = B(x,a,2),(y,b,3) = 0 (6.33)
8
xy xy 3
hχ3,a |hχ4,b |Θi = B(x,a,3),(x⊕y,b,4) = 0 (6.34)
8
for all x, y, a, b ∈ {0, 1}.
Now putting together equations (6.23), (6.24), (6.25), (6.30), (6.32), (6.33), and (6.34) (and using the
symmetrization (6.21)), we get
B(x,a,I),(z,b,J) = 0 for all x, z, a, b ∈ {0, 1}, where I = J or {I, J} ∈ {1, 2}, {1, 4}, {2, 3}, {3, 4} ,
so |Θi can be written
1,U 3,U 3,U 1,U 2,U 4,U 4,U 2,U
∑ B(z,c,1),(w,d,3) |ρz,c i|ρw,d i + |ρw,d i|ρz,c i + B(z,c,2),(w,d,4) |ρz,c i|ρw,d i + |ρw,d i|ρz,c i .
z,c,w,d∈{0,1}
(6.35)
Now
xy xy 1,U 3,U 3
hχ1,a |hχ3,b |ρz,c i|ρw,d i = δa,c δb,d Ũ(a)xz δy,w
8
xy xy 2,U 4,U 3
hχ2,a |hχ4,b |ρz,c i|ρw,d i = δa,c δb,d δx,z δy,w⊕x ,
8
so enforcing equation (6.28) gives
∑ Ũ(a)xz B(z,a,1),(y,b,3) = B(x,a,2),(x⊕y,b,4)
z∈{0,1}
for each x, y, a, b ∈ {0, 1}. In other words
B(z,c,2),(w,d,4) = ∑ Ũ(c)zx B(x,c,1),(z⊕w,d,3) = ∑ U(c)zw,xy B(x,c,1),(y,d,3)
x∈{0,1} x,y∈{0,1}
where we used U(a) = CNOT12 (Ũ(a) ⊗ 1). Plugging this into (6.35) gives
1,U 3,U 3,U 1,U
|Θi = ∑ B(z,c,1),(w,d,3) |ρz,c i|ρw,d i + |ρw,d i|ρz,c i
z,c,w,d∈{0,1}
2,U 4,U 4,U 2,U
+ ∑ U(c)zw,xy B(x,c,1),(y,d,3) |ρz,c i|ρw,d i + |ρw,d i|ρz,c i
x,y∈{0,1}
1,U 3,U 3,U 1,U
= ∑ B(z,c,1),(w,d,3) |ρz,c i|ρw,d i + |ρw,d i|ρz,c i
z,c,w,d∈{0,1}
2,U 4,U 4,U 2,U
+ ∑ U(c)xy,zw |ρx,c i|ρy,d i + |ρy,d i|ρx,c i
x,y∈{0,1}
= ∑ 2B(z,c,1),(w,d,3) Sym(|Tz,c,w,d i) .
z,c,w,d∈{0,1}
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This is the general solution to equations (6.20)–(6.22), so the space of two-particle frustration-free states
for GU is spanned by the 16 orthonormal states (6.17).
Finally, we show that there are no three-particle frustration-free states. By Lemma 2.3, this implies
that there are no frustration-free states for more than two particles. Suppose (to reach a contradiction)
that |Γi is a normalized three-particle frustration-free state. Write
q r s
|Γi = ∑ E(x,a,q),(y,b,r),(z,c,s) |ρx,a i|ρy,b i|ρz,c i
and note that each reduced density matrix of |Γi on two of the three subsystems must have all of its
support on two-particle frustration-free states (see the remark following Lemma 5.4). Using this fact for
each two-particle subsystem we get
(q, r) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ E(x,a,q),(y,b,r),(z,c,s) = 0
(q, s) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ E(x,a,q),(y,b,r),(z,c,s) = 0
(r, s) ∈
/ {(1, 3), (3, 1), (2, 4), (4, 2)} =⇒ E(x,a,q),(y,b,r),(z,c,s) = 0
which together imply that |Γi = 0 (a contradiction). Hence no three-particle frustration-free state
exists.
6.3 The boundary gadget
The boundary gadget is shown in Figure 6.3. This gate diagram is obtained from Figure 6.2(A) (with
Ũ = 1) by adding self-loops. Although the boundary gadget will be used for a different purpose than
the other gadgets, for technical reasons it is convenient to construct the boundary gadget similarly. The
adjacency matrix is
A(Gbnd ) = A(GCNOT12 ) + hS
where
1
hS = ∑ (|1, z, 1ih1, z, 1| ⊗ 1 j + |2, z, 5ih2, z, 5| ⊗ 1 j + |3, z, 1ih3, z, 1| ⊗ 1 j ) .
z=0
i,U
The single-particle ground states (with energy e1 ) are superpositions of the states |ρz,a i from Lemma 6.3
that are in the nullspace of hS . Note that
j,U i,U 1 1
hρx,b |hS |ρz,a i = δa,b δx,z (δi,1 δ j,1 + δi,2 δ j,2 + δi,3 δ j,3 ) ·
8 8
(one factor of 1/8 comes from the normalization in equations (6.13)–(6.16) and the other factor comes
from the normalization in equation (5.2)), so the only single-particle ground states are
bnd 4,U
|ρz,a i = |ρz,a i
with z, a ∈ {0, 1}. Thus there are no two- (or more) particle frustration-free states, because no superposi-
tion of the states (6.17) lies in the subspace
4,U 4,U
span{Sym(|ρz,a i|ρx,b i) : z, a, x, b ∈ {0, 1}}
of states with single-particle reduced density matrices in the ground space of A(Gbnd ). We summarize
these results as follows.
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5 6
1 1
1 00 2
W
1 01 1
W α
3 10 4 = Bnd
γ
W α β
δ
1 11 1 β
W
7 8
γ
1 1 δ
Figure 6.3: The gate diagram for the boundary gadget is obtained from Figure 6.2(A) by setting Ũ = 1
and adding 6 self-loops.
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Lemma 6.4. The smallest eigenvalue of A(Gbnd ) is e1 , with corresponding eigenvectors
r
bnd 1 4 1 8 3 x(z⊕x)
|ρz,a i = √ |ψz,a i − √ |ψz,a i − ∑ |χ4,a i (6.36)
8 8 8 x=0,1
for z, a ∈ {0, 1}. There are no frustration-free states with two or more particles, i. e., λN1 (Gbnd ) > 0 for
N ≥ 2.
7 Bose-Hubbard Hamiltonian is QMA-hard
Since Frustration-Free Bose-Hubbard Hamiltonian is a special case of Bose-Hubbard Hamiltonian, to
prove the latter is QMA-hard it suffices to prove the former is QMA-hard. To achieve this, we efficiently
map any sufficiently large instance of any problem in QMA to an equivalent instance of Frustration-Free
Bose-Hubbard Hamiltonian.
For any instance X of a problem in QMA there is a verification circuit CX . We require the verification
circuit to be of a certain form described in Section 7.1. Since the class of circuits we consider is
universal, this choice is without loss of generality. We also assume without loss of generality [21, 24]
that CX satisfies a stronger version of Definition 2.1 where the completeness threshold 2/3 is replaced by
1 − (1/2|X| ), i. e.,
1
• If X ∈ Lyes then there exists a state |ψwit i with AP(CX , |ψwit i) ≥ 1 − .
2|X|
1
• If X ∈ Lno then AP(CX , |φ i) ≤ for all |φ i.
3
In this section we exhibit an efficiently computable mapping from the n-qubit, M-gate verification circuit
CX to an e1 -gate graph GX with R = 32(M +2n−2) diagram elements and an occupancy constraints graph
Gocc occ 1 occ
X on R vertices. We consider the Hamiltonian H(GX , GX , n) and its smallest eigenvalue λn (GX , GX ).
In Section 8 we prove the following.
Theorem 7.1. If there exists a state |ψwit i with AP(CX , |ψwit i) ≥ 1 − (1/2|X| ), then
1
λn1 (GX , Gocc
X )≤ (7.1)
2|X|
On the other hand, if AP(CX , |φ i) ≤ 1/3 for all |φ i, then
K
λn1 (GX , Gocc
X )≥ (7.2)
n4 M 4
where K ∈ (0, 1] is an absolute constant.
The number K appearing in this theorem can in principle be computed (see Section 8) but we will not
need to know its value.
Note that if X is a yes instance then (7.1) is satisfied and if X is a no instance then (7.2) is satisfied. By
applying the Occupancy Constraints Lemma (Lemma 5.5), with gate graph GX and occupancy constraints
graph Gocc
X , and using the fact that R = 32 (M + 2n − 2) ≤ 64(M + n), we get an efficiently computable
e1 -gate graph G 2 2
X with at most 7 · 64 (n + M) diagram elements such that the following holds.
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Theorem 7.2. If X is a yes instance, then
1
λn1 (G
X) ≤ .
2|X|
On the other hand, if X is a no instance, then
K
λn1 (G
X) ≥
n4 M 4 8329 (M + n)9
where K ∈ (0, 1] is the absolute constant from Theorem 7.1.
This theorem is sufficient to prove that Frustration-Free Bose-Hubbard Hamiltonian is QMA-hard. To
see this, first let Q = d1/Ke and define the precision parameter
1 1
ε= .
2Q n4 M 4 8329 (M + n)9
Note that 1/ε is at least four times the number of vertices in the graph (i. e., at least 4 · 128 · 7 · 642 (n + M)2 ,
since each diagram element corresponds to 128 vertices); this condition is required in the definition of
Frustration-Free Bose-Hubbard Hamiltonian. We now show that solving the instance of Frustration-Free
Bose-Hubbard Hamiltonian with graph G X , number of particles n, and precision parameter ε is sufficient
to solve the original instance X (for |X| at least some constant).
If X is a no instance then the theorem states that λn1 (G 3
X ) ≥ 2ε ≥ ε + ε , so the corresponding instance
of Frustration-Free Bose-Hubbard Hamiltonian is also a no instance. Furthermore, since M and n are
upper bounded by a polynomial function of the instance size |X|, there exists a constant instance size
above which
1
≤ ε3 ,
2|X|
so when X is a sufficiently large yes instance, the theorem says λn1 (G 3
X ) ≤ ε , i. e., the corresponding
instance of Frustration-Free Bose-Hubbard Hamiltonian is also a yes instance. This proves that Frustration-
Free Bose-Hubbard Hamiltonian is QMA-hard, which implies that Bose-Hubbard Hamiltonian is QMA-
hard.
The remainder of this section describes the ingredients used to prove Theorem 7.1. In Section 7.1 we
describe the verification circuit CX , in Section 7.2 we define the gate graph GX , and in Section 7.3 we
define the occupancy constraints graph Gocc X . We prove Theorem 7.1 in Section 8.
7.1 The verification circuit
We take the verification circuit CX to be from the following universal circuit family. Write n for the
number of qubits and M for the number of gates in the circuit
UCX = UMUM−1 . . .U1 .
The qubits labeled 1, . . . , nin hold the input state. The remaining n − nin ancilla qubits are each initialized
to |0i. We require each qubit to be involved in at least one gate. The qubit labeled 2 is the output qubit
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that contains the result of the computation after the circuit is applied. The qubit labeled 1 mediates
two-qubit gates: each gate U j is a two-qubit gate acting nontrivially on the qubit labeled 1 and another
qubit s( j) ∈ {2, . . . , n}. Each unitary U j is chosen from the set
{CNOT1s( j) , CNOTs( j)1 , CNOT1s( j) (H ⊗ 1) , CNOT1s( j) (HT ⊗ 1)} , (7.3)
but no two consecutive gates (U j and U j+1 ) act between the same two qubits, i. e.,
s( j) 6= s( j + 1) , j ∈ [M − 1] . (7.4)
As discussed in Section 2, the acceptance probability for the circuit acting on input state |ψin i ∈ (C2 )⊗nin
is the probability that a final measurement of the output qubit in the computational basis gives the value 1:
2
AP (CX , |ψin i) = |1ih1|(2)UCX |ψin i|0i⊗n−nin .
We now establish that circuits of this form are universal. We show that any quantum circuit (with
n ≥ 4 qubits) expressed using the universal gate set
{CNOT, H, HT } (7.5)
can be efficiently rewritten in the prescribed form without increasing the number of qubits and with at
most a constant factor increase in the number of gates.
First we map a circuit from the gate set (7.5) to the gate set (7.3) (without necessarily satisfying
condition (7.4)). A SWAP gate between qubits 1 and k can be performed using the identity
SWAP1k = CNOT1k CNOTk1 CNOT1k .
To perform a CNOTik gate between two qubits i, k (neither of which is qubit 1), we swap qubits 1 and i,
apply CNOT1k , and then swap back. Similarly, we can apply a single-qubit gate U ∈ {H, HT } to some
qubit k 6= 1 using the sequence of gates
SWAP1k CNOT12 (CNOT12U ⊗ 1) SWAP1k .
Applying these replacement rules, we obtain a circuit over the gate set (7.3). However, the resulting
circuit will not in general satisfy equation (7.4). To enforce this condition, we insert a sequence of four
gates equal to the identity, namely
1 = CNOT1a CNOT1b CNOT1a CNOT1b ,
between any two consecutive gates U j and U j+1 with s( j) = s( j + 1), where a 6= b 6= s( j). For example,
CNOT15 CNOT51 −→ CNOT15 CNOT12 CNOT13 CNOT12 CNOT13 CNOT51 .
Thus we map a circuit over the gate set (7.5) into the prescribed form. Note that n ≥ 4 is needed to ensure
that any quantum circuit can be efficiently rewritten so that s( j) 6= s( j + 1). (There do exist circuits of the
desired of the form with n = 3, such as in the example shown in Figure 7.1.)
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7.2 The gate graph
For any n-qubit, M-gate verification circuit CX of the form described above, we associate a gate graph
GX . The gate diagram for GX is built using the gadgets described in Section 6; specifically, we use M
two-qubit gadgets and 2(n − 1) boundary gadgets. Since each two-qubit gadget and each boundary gadget
contains 32 diagram elements, the total number of diagram elements in GX is R = 32(M + 2n − 2).
We now present the construction of the gate diagram for GX . We also describe some gate graphs
obtained as intermediate steps that are used in our analysis in Section 8. The reader may find this
description easier to follow by looking ahead to Figure 7.1, which illustrates the construction for a specific
3-qubit circuit.
1. Draw a grid with columns labeled j = 0, 1, . . . , M + 1 and rows labeled i = 1, . . . , n (this grid is
only used to help describe the diagram).
2. Place gadgets in the grid to mimic the quantum circuit. For each j = 1, . . . , M, place a gadget
for the two-qubit gate U j between rows 1 and s( j) in the j-th column. Place boundary gadgets in
rows i = 2, . . . , n of column 0 and in the same rows of column M + 1. Write G1 for the gate graph
associated with the resulting diagram.
3. Connect the nodes within each row. First add edges connecting the nodes in rows i = 2, . . . , n;
call the resulting gate graph G2 . Then add edges connecting the nodes in row 1; call the resulting
gate graph G3 .
4. Add self-loops to the boundary gadgets. In this step we add self-loops to enforce initialization
of ancillas (at the beginning) and the proper output of the circuit (at the end). For each row
k = nin + 1, . . . , n, add a self-loop to node δ (as shown in Figure 6.3) of the corresponding boundary
gadget in column r = 0, giving the gate diagram for G4 . Finally, add a self-loop to node α of the
boundary gadget (as in Figure 6.3) in row 2 and column M + 1, giving the gate diagram for GX
(Recall from the previous section that qubit 2 is always taken to be the output qubit).
Figure 7.1 illustrates the step-by-step construction of GX using a simple 3-qubit circuit with four
gates
CNOT12 (CNOT13 HT ⊗ 1) CNOT21 CNOT13 .
In this example, two of the qubits are input qubits (so nin = 2), while the third qubit is an ancilla initialized
to |0i. Following the convention described in Section 7.1, we take qubit 2 to be the output qubit. (In
this example the circuit is not meant to compute anything interesting; its only purpose is to illustrate our
method of constructing a gate graph.)
We made some choices in designing this circuit-to-gate graph mapping that may seem arbitrary (e. g.,
we chose to place boundary gadgets in each row except the first). We have tried to achieve a balance
between simplicity of description and ease of analysis, but we expect that other choices could be made to
work.
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j=0 j=1 j=2 j=3 j=4 j=5
i=1 1 HT 1
i=2
Bnd Bnd
i=3
Bnd Bnd
(A)
j=0 j=1 j=2 j=3 j=4 j=5
i=1 1 HT 1
i=2
Bnd Bnd
i=3
Bnd Bnd
(B)
j=0 j=1 j=2 j=3 j=4 j=5
i=1 1 HT 1
i=2
Bnd Bnd
i=3
Bnd Bnd
(C)
Figure 7.1: Step-by-step construction of the gate diagram for GX for the three-qubit example circuit
described in the text. ( A ) The gate diagram for G1 . ( B ) Add edges in all rows except the first to obtain the
gate diagram for G2 . ( C ) Add edges in the first row to obtain the gate diagram for G3 . See the following
figure for the remainder of the construction.
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
j=0 j=1 j=2 j=3 j=4 j=5
i=1 1 HT 1
i=2
Bnd Bnd
i=3
Bnd Bnd
(D)
Figure 7.1: Continuation of step-by-step construction. ( D ) Add self-loops to the boundary gadgets to
obtain the gate diagram for GX (the diagram for G4 in this case differs from ( D ) by removing the self-loop
in column 5; this diagram is not shown).
7.2.1 Notation for GX
We now introduce some notation that allows us to easily refer to a subset L of the diagram elements in
the gate diagram for GX .
Recall from Section 6 that each two-qubit gate gadget and each boundary gadget is composed of
32 diagram elements. This can be seen by looking at Figure 6.2(A) and Figure 6.3 and noting (from
Figure 6.1) that each move-together gadget comprises 6 diagram elements.
For each of the two-qubit gate gadgets in the gate diagram for GX , we focus our attention on the
four diagram elements labeled 1–4 in Figure 6.2(A). In total there are 4M such diagram elements in the
gate diagram for GX : in each column j ∈ {1, . . . , M} there are two in row 1 and two in row s( j). When
U j ∈ {CNOT1s( j) , CNOT1s( j) (H ⊗ 1) , CNOT1s( j) (HT ⊗ 1)} the diagram elements labeled 1, 2 are in row
1 and those labeled 3, 4 are in row s( j); when U j = CNOTs( j)1 those labeled 1, 2 are in row s( j) and those
labeled 3, 4 are in row 1. We denote these diagram elements by triples (i, j, d). Here i and j indicate
(respectively) the row and column of the grid in which the diagram element is found, and d indicates
whether it is the leftmost (d = 0) or rightmost (d = 1) diagram element in this row and column. We define
Lgates = {(i, j, d) : i ∈ {1, s( j)}, j ∈ [M], d ∈ {0, 1}} (7.6)
to be the set of all such diagram elements.
For example, in Figure 7.1 the first gate is
U1 = CNOT13 ,
so the gadget from Figure 6.2(A) (with Ũ = 1) appears between rows 1 and 3 in the first column. The
diagram elements labeled 1, 2, 3, 4 from Figure 6.2(A) are denoted by (1, 1, 0), (1, 1, 1), (3, 1, 0), (3, 1, 1),
respectively. The second gate in Figure 7.1 is U2 = CNOT21 , so the gadget from Figure 6.2(B) (with
Ũ = 1) appears between rows 2 and 1; in this case the diagram elements labeled 1, 2, 3, 4 in Figure 6.2(A)
are denoted by (2, 2, 0), (2, 2, 1), (1, 2, 0), (1, 2, 1), respectively.
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We also define notation for the boundary gadgets in GX . For each boundary gadget. we focus on
a single diagram element, labeled 4 in Figure 6.3. For the left hand-side and right-hand side boundary
gadgets, respectively, we denote these diagram elements as
Lin = {(i, 0, 1) : i ∈ {2, . . . , n}} (7.7)
Lout = {(i, M + 1, 0) : i ∈ {2, . . . , n}} . (7.8)
Definition 7.3. Let L be the set of diagram elements
L = Lin ∪ Lgates ∪ Lout
where Lin , Lgates , and Lout are given by equations (7.7), (7.6), and (7.8), respectively.
Finally, it is convenient to define a function F that describes horizontal movement within the rows of
the gate diagram for GX . The function F takes as input a two-qubit gate j ∈ [M], a qubit i ∈ {2, . . . , n},
and a single bit and outputs a diagram element from the set L. If the bit is 0 then F outputs the diagram
element in row i that appears in a column 0 ≤ k < j with k maximal (i. e., the closest diagram element in
row i to the left of column j):
(
(i, k, 1) where 1 ≤ k < j is the largest k such that s(k) = i, if it exists
F(i, j, 0) = (7.9)
(i, 0, 1) otherwise.
On the other hand, if the bit is 1, then F outputs the diagram element in row i that appears in a column
j < k ≤ M + 1 with k minimal (i. e., the closest diagram element in row i to the right of column j).
(
(i, k, 0) where j < k ≤ M is the smallest k such that s(k) = j, if it exists
F(i, j, 1) = (7.10)
(i, M + 1, 0) otherwise.
7.3 The occupancy constraints graph
In this section we define an occupancy constraints graph Gocc X . Along with GX and the number of
particles n, this determines a subspace I(GX , Gocc
X , n) ⊂ Z n (GX ) through equation (5.21). We will see in
Section 8 how low-energy states of the Bose-Hubbard model that live entirely within this subspace encode
computations corresponding to the quantum circuit CX . This fact is used in the proof of Theorem 7.1,
which shows that the smallest eigenvalue λn1 (GX , GoccX ) of
H(GX , Gocc
X , n) = H(GX , n) I(GX ,Gocc ,n)X
is related to the maximum acceptance probability of the circuit.
We encode quantum data in the locations of n particles in the graph GX as follows. Each particle
encodes one qubit and is located in one row of the graph GX . Since all two-qubit gates in CX involve
the first qubit, the location of the particle in the first row determines how far along the computation has
proceeded. We design the occupancy constraints graph to ensure that low-energy states of H(GX , Gocc X , n)
have exactly one particle in each row (since there are n particles and n rows), and so that the particles
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in rows 2, . . . , n are not too far behind or ahead of the particle in the first row. To avoid confusion, we
emphasize that not all states in the subspace I(GX , Gocc X , n) have the desired properties—for example,
there are states in this subspace with more than one particle in a given row. We see in the next section that
states with low energy for H(GX , n) that also satisfy the occupancy constraints (i. e., low-energy states of
H(GX , Gocc
X , n)) have the desired properties.
We now define Gocc X , which is a simple graph with a vertex for each diagram element in GX . Each
edge in Gocc
X places a constraint on the locations of particles in GX . The graph Gocc X only has edges
between diagram elements in the set L from Definition 7.3; we define the edge set E(Gocc X ) by specifying
pairs of diagram elements L1 , L2 ∈ L. We also indicate (in bold) the reason for choosing the constraints.
1. No two particles in the same row. For each i ∈ [n] we add constraints between diagram elements
(i, j, c) ∈ L and (i, k, d) ∈ L in row i but in different columns, i. e.,
{(i, j, c) , (i, k, d)} ∈ E(Gocc
X ) whenever j 6= k . (7.11)
2. Synchronization with the particle in the first row. For each j ∈ [M] we add constraints between
row 1 and row s( j):
{(1, j, c), (s( j), k, d)} ∈ E(Gocc
X ) whenever k 6= j and (s( j), k, d) 6= F(s( j), j, c) .
For each j ∈ [M] we also add constraints between row 1 and rows i ∈ [n] \ {1, s( j)}:
{(1, j, c), (i, k, d)} ∈ E(Gocc
X ) whenever (i, k, d) ∈
/ {F(i, j, 0), F(i, j, 1)} .
8 Proof of Theorem 7.1
We write γ(H) for the smallest nonzero eigenvalue of a positive semidefinite matrix H.
8.1 Strategy and outline of the proof
Theorem 7.1 bounds the smallest eigenvalue λn1 (GX , Gocc occ
X ) of H(GX , GX , n). To prove the theorem, we
investigate a sequence of Hamiltonians starting with H(G1 , n) and H(G1 , GoccX , n) and then work our way
up to the Hamiltonian H(GX , GoccX , n) by adding positive semidefinite terms.
For each Hamiltonian we consider, we solve for the nullspace and the smallest nonzero eigenvalue.
To go from one Hamiltonian to the next, we use the following “Nullspace Projection Lemma,” which was
used (implicitly) in reference [25]. The lemma bounds the smallest nonzero eigenvalue γ(HA + HB ) of a
sum of positive semidefinite Hamiltonians HA and HB using knowledge of the smallest nonzero eigenvalue
γ(HA ) of HA and the smallest nonzero eigenvalue γ(HB |S ) of the restriction of HB to the nullspace S of
HA .
Lemma 8.1 (Nullspace Projection Lemma [25]). Let HA , HB ≥ 0. Suppose the nullspace S of HA is
non-empty and
γ(HB |S ) ≥ c > 0 and γ(HA ) ≥ d > 0 .
Then
cd
γ(HA + HB ) ≥ . (8.1)
c + d + kHB k
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We prove the lemma in Section A.1. When we apply this lemma, we are usually interested in an
asymptotic limit where c, d kHB k and the right-hand side of (8.1) is Ω(cd/kHB k).
Our proof strategy, using repeated applications of the Nullspace Projection Lemma, is analogous
to that of reference [19], where the so-called Projection Lemma was used similarly. Our technique has
the advantage of not requiring the terms we add to our Hamiltonian to have “unphysical” problem-size
dependent coefficients (it also has this advantage over the method of perturbative gadgets [19, 17]). This
allows us to prove results about the “physically realistic” Bose-Hubbard Hamiltonian. A similar technique
based on Kitaev’s Geometric Lemma was used in reference [13] (however, that method is slightly more
computation intensive, requiring a lower bound on γ(HB ) as well as bounds on γ(HA ) and γ(HB |S )).
8.1.1 Adjacency matrices of the gate graphs
We begin by discussing the graphs
G1 , G2 , G3 , G4 , GX
(as defined in Section 7; see Figure 7.1) in more detail and deriving some properties of their adjacency
matrices.
The graph G1 has a component for each of the two-qubit gates j ∈ [M], for each of the boundary
gadgets i = 2, . . . , n in column 0, and for each of the boundary gadgets i = 2, . . . , n in column M + 1. In
other words ! ! !
n
[ M
[ n
[
G1 = Gbnd ∪ GU j ∪ Gbnd . (8.2)
i=2 j=1 i=2
| {z } | {z } | {z }
left boundary two-qubit gates right boundary
We use our knowledge of the adjacency matrices of the components Gbnd and GU j to understand the
ground space of A(G1 ). Recall (from Section 6) that the smallest eigenvalue of A(GU j ) is
√
e1 = −1 − 3 2
(with degeneracy 16) which is also the smallest eigenvalue of A(Gbnd ) (with degeneracy 4). For each
diagram element L ∈ L and pair of bits z, a ∈ {0, 1} there is an eigenstate |ρz,a
L i of A(G ) with this
1
minimal eigenvalue e1 . In total we get sixteen eigenstates
(1, j,0) (1, j,1) (s( j), j,0) (s( j), j,1)
|ρz,a i, |ρz,a i, |ρz,a i, |ρz,a i, z, a ∈ {0, 1} ,
for each two-qubit gate j ∈ [M], four eigenstates
(i,0,1)
|ρz,a i, z, a ∈ {0, 1} ,
for each boundary gadget i ∈ {2, . . . , n} in column 0, and four eigenstates
(i,M+1,0)
|ρz,a i, z, a ∈ {0, 1} ,
for each boundary gadget i ∈ {2, . . . , n} in column M + 1. The set
|ρz,a i : z, a ∈ {0, 1}, L ∈ L
L
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is an orthonormal basis for the ground space of A(G1 ).
We write the adjacency matrices of G2 , G3 , G4 , and GX as
n
A(G2 ) = A(G1 ) + h1 A(G4 ) = A(G3 ) + ∑ hin,i
i=nin +1
A(G3 ) = A(G2 ) + h2 A(GX ) = A(G4 ) + hout .
From step 3 of the construction of the gate diagram in Section 7.2, we see that h1 and h2 are both sums of
terms of the form
|q, z,ti + |q0 , z,t 0 i hq, z,t| + hq0 , z,t 0 | ⊗ 1 j ,
where h1 contains a term for each edge in rows 2, . . . , n and h2 contains a term for each of the 2(M − 1)
edges in the first row. The operators
hin,i = |(i, 0, 1), 1, 7ih(i, 0, 1), 1, 7| ⊗ 1 hout = |(2, M + 1, 0), 0, 5ih(2, M + 1, 0), 0, 5| ⊗ 1 (8.3)
correspond to the self-loops added in the gate diagram in step 4 of Section 7.2.
We prove that G1 , G2 , G3 , G4 , and GX are e1 -gate graphs.
Lemma 8.2. The smallest eigenvalues of G1 , G2 , G3 , G4 and GX are
µ(G1 ) = µ(G2 ) = µ(G3 ) = µ(G4 ) = µ(GX ) = e1 .
Proof. We showed in the above discussion that µ(G1 ) = e1 . The adjacency matrices of G2 , G3 , G4 ,
and GX are obtained from that of G1 by adding positive semidefinite terms (h1 , h2 , hin,i , and hout are all
positive semidefinite). It therefore suffices to exhibit an eigenstate |ρi of A(G1 ) with
h1 |ρi = h2 |ρi = hin,i |ρi = hout |ρi = 0
(for each i ∈ {nin + 1, . . . , n}). There are many states |ρi satisfying these conditions; one example is
(1,1,0)
|ρi = |ρ0,0 i
which is supported on vertices where h1 , h2 , hin,i , and hout have no support.
8.1.2 Building up the Hamiltonian
We now outline the sequence of Hamiltonians considered in the following sections and describe the
relationships between them. As a first step, in Section 8.2 we exhibit a basis Bn for the nullspace of
H(G1 , n) and we prove that its smallest nonzero eigenvalue is lower bounded by a positive constant. We
then discuss the restriction
H(G1 , Gocc
X , n) = H(G1 , n) I(G ,Gocc ,n) (8.4)
1 X
in Section 8.3, where we prove that a subset Blegal ⊂ Bn is a basis for the nullspace of (8.4), and that its
smallest nonzero eigenvalue is also lower bounded by a positive constant.
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For the remainder of the proof we use the Nullspace Projection Lemma (Lemma 8.1) four times,
using the decompositions
H(G2 , Gocc occ
X , n) = H(G1 , GX , n) + H1 I(G ,Gocc ,n) (8.5)
2 X
H(G3 , Gocc occ
X , n) = H(G2 , GX , n) + H2 I(G ,Gocc ,n) (8.6)
3 X
n
H(G4 , Gocc occ
X , n) = H(G3 , GX , n) + ∑ Hin,i I(G ,G ,n) occ (8.7)
4 X
i=nin +1
H(GX , Gocc occ
X , n) = H(G4 , GX , n) + Hout I(GX ,Gocc ,n) (8.8)
X
where
n n n n
(w) (w) (w) (w)
H1 = ∑ h1 , Hin,i = ∑ hin,i , H2 = ∑ h2 , Hout = ∑ hout
w=1 w=1 w=1 w=1
are all positive semidefinite, with h1 , h2 , hin,i , hout as defined in Section 8.1.1. Note that in writing
equations (8.5), (8.6), (8.7), and (8.8), we have used the fact (from Lemma 8.2) that the adjacency
matrices of the graphs we consider all have the same smallest eigenvalue e1 . Also note that
I (Gi , Gocc
X , n) = I (GX , GX , n)
occ
for i ∈ [4] since the gate diagrams for each of the graphs G1 , G2 , G3 , G4 and GX have the same set of
diagram elements.
Let Sk be the nullspace of H(Gk , Gocc
X , n) for k = 1, 2, 3, 4. Since these positive semidefinite Hamilto-
nians are related by adding positive semidefinite terms, their nullspaces satisfy
S4 ⊆ S3 ⊆ S2 ⊆ S1 ⊆ I (GX , Gocc
X , n) .
We solve for S1 = span(Blegal ) in Section 8.3 and we characterize the spaces S2 , S3 , and S4 in Section 8.5
in the course of applying our strategy.
For example, to use the Nullspace Projection Lemma to lower bound the smallest nonzero eigenvalue
of H(G2 , Gocc
X , n), we consider the restriction
H1 I(G ,Gocc ,n) = H1 S . (8.9)
2 X S1 1
We also solve for S2 , which is equal to the nullspace of (8.9). To obtain the corresponding lower bounds
on the smallest nonzero eigenvalues of H(Gk , Gocc occ
X , n) for k = 2, 3, 4 and H(GX , GX , n), we consider
restrictions
n
H2 S , ∑ Hin,i S , and Hout S .
2 3 4
i=nin +1
Analyzing these restrictions involves extensive computation of matrix elements. To simplify and organize
these computations, we first compute the restrictions of each of these operators to the space S1 . We present
the results of this computation in Section 8.4; details of the calculation can be found in Section A.3. In
Section 8.5 we proceed with the remaining computations and apply the Nullspace Projection Lemma
three times using equations (8.5), (8.6), and (8.7). Finally, in Section 8.6 we apply the lemma again using
equation (8.8) and we prove Theorem 7.1.
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8.2 Configurations
In this section we use Lemma 2.4 to solve for the nullspace of H(G1 , n), i. e., the n-particle frustration-free
states on G1 . Lemma 2.4 describes how frustration-free states for G1 are built out of frustration-free states
for its components.
To see how this works, consider the example from Figure 7.1(A). In this example, with n = 3, we
construct a basis for the nullspace of H(G1 , 3) by considering two types of eigenstates. First, there are
frustration-free states
Sym(|ρzL11,a1 i|ρzL22,a2 i|ρzL33,a3 i) (8.10)
where Lk = (ik , jk , dk ) ∈ L belong to different components of G1 . That is to say, jw 6= jt unless jw = jt ∈
{0, 5}, in which case iw 6= it (in this case the particles are located either at the left or right boundary, in
different rows of G1 ). There are also frustration-free states where two of the three particles are located in
the same two-qubit gadget J ∈ [M] and one of the particles is located in a diagram element L1 from a
different component of the graph. These states have the form
Sym(|TzJ1 ,a1 ,z2 ,a2 i|ρzL31,a3 i) (8.11)
where
1 (1,J,0) (s(J),J,0) 1 (1,J,1) (s(J),J,1)
|TzJ1 ,a1 ,z2 ,a2 i = √ |ρz1 ,a1 i|ρz2 ,a2 i+ √ ∑ UJ (a1 )x1 x2 ,z1 z2 |ρx1 ,a1 i|ρx2 ,a2 i (8.12)
2 2 x1 ,x2 ∈{0,1}
and L1 = (i, j, k) ∈ L satisfies j 6= J (using Lemma 6.3 and also the fact that the controlled-not gate is
real for the case UJ = CNOTs(J)1 ).
Each of the states (8.10) and (8.11) is specified by 6 “data” bits z1 , z2 , z3 , a1 , a2 , a3 ∈ {0, 1} and a
“configuration” indicating where the particles are located in the graph. The configuration is specified
either by three diagram elements L1 , L2 , L3 ∈ L from different components of G1 or by a two-qubit gate
J ∈ [M] along with a diagram element L1 ∈ L from a different component of the graph.
We now define the notion of a configuration for general n. Informally, we can think of an n-particle
configuration as a way of placing n particles in the graph G1 subject to the following restrictions. We first
place each of the n particles in a component of the graph, with the restriction that no boundary gadget
may contain more than one particle and no two-qubit gadget may contain more than two particles. For
each particle on its own in a component (i. e., in a component with no other particles), we assign one of
the diagram elements L ∈ L associated to that component. We therefore specify a configuration by a set
of two-qubit gadgets J1 , . . . , JY that contain two particles, along with a set of diagram elements Lk ∈ L
that give the locations of the remaining n − 2Y particles. We choose to order the Js and the Ls so that
each configuration is specified by a unique tuple (J1 , . . . , JY , L1 , . . . , Ln−2Y ). For concreteness, we use
the lexicographic order on diagram elements in the set L: LA = (iA , jA , dA ) and LB = (iB , jB , dB ) satisfy
LA < LB iff either iA < iB , or iA = iB and jA < jB , or (iA , jA ) = (iB , jB ) and dA < dB .
Definition 8.3 (Configuration). An n-particle configuration on the gate graph G1 is a tuple
(J1 , . . . , JY , L1 , . . . , Ln−2Y )
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j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1
×
×
i=2
× i=2
i=3
× i=3
×
(A) (B)
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1
×
i=2
× i=2
× ×
i=3
× i=3
(C) (D)
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1 ×
i=2
× × i=2
×
i=3 i=3
×
(E) (F)
Figure 8.1: Diagrammatic depictions of configurations for the example where G1 is the gate graph
from Figure 7.1(A). The figures show the locations of each of the three particles in the gate graph.
×
The symbol × indicates a single-particle state and the symbol × indicates a two-particle state. ( A )
((1, 1, 1), (2, 2, 0), (3, 3, 0)). ( B ) (2, (3, 1, 1)). ( C ) ((1, 1, 1), (2, 0, 1), (3, 3, 0). ( D ) (3, (2, 0, 1)). ( E )
((1, 3, 0), (2, 2, 1), (2, 4, 0)). ( F ) ((1, 1, 1), (2, 2, 0), (3, 5, 0)). The examples in ( A ), ( B ), and ( C ) show
legal configurations whereas the examples in ( D ), ( E ), and ( F ) are illegal, as defined in Section 8.3 .
with Y ∈ {0, . . . , bn/2c}, ordered integers
1 ≤ J1 < J2 < · · · < JY ≤ M ,
and lexicographically ordered diagram elements
L1 < L2 < · · · < Ln−2Y , Lk = (ik , jk , dk ) ∈ L .
We further require that each Lk is from a different component of G1 , i. e.,
jw = jt =⇒ jw ∈ {0, M + 1} and iw 6= it ,
and we require that ju 6= Jv for all u ∈ [n − 2Y ] and v ∈ [Y ].
In Figure 8.1 we give some examples of configurations (for the example from Figure 7.1(A) with
n = 3) and we introduce a diagrammatic notation for them.
For any configuration and n-bit strings ~z and ~
a, there is a state in the nullspace of H(G1 , n), given by
Sym(|TzJ11,a1 ,z2 ,a2 i . . . |TzJ2YY −1 ,a2Y −1 ,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i) . (8.13)
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The ordering in the definition of a configuration ensures that each distinct choice of configuration and
n-bit strings ~z, ~
a gives a different state.
Definition 8.4. Let Bn be the set of all states of the form (8.13), where (J1 , . . . , JY , L1 , . . . , Ln−2Y ) is a
configuration and ~z, ~
a ∈ {0, 1}n .
Lemma 8.5. The set Bn is an orthonormal basis for the nullspace of H(G1 , n). Furthermore,
γ(H(G1 , n)) ≥ K0 (8.14)
where K0 ∈ (0, 1] is an absolute constant.
Proof. Each component of G1 is either a two-qubit gadget or a boundary gadget (see equation (8.2)).
The single-particle states of A(G1 ) with energy e1 are the states |ρz,a L i for L ∈ L and z, a ∈ {0, 1}, as
discussed in Section 8.1.1. Each of these states has support on only one component of G1 . In addition,
G1 has a two-particle frustration-free state for each two-qubit gadget J ∈ [M] and bits z, a, x, b, namely
J
Sym(|Tz,a,x,b i). Furthermore, no component of G1 has any three- (or more) particle frustration-free states.
Using these facts and applying Lemma 2.4, we see that Bn spans the nullspace of H(G1 , n).
Lemma 2.4 also expresses each eigenvalue of H(G1 , n) as a sum of eigenvalues for its components.
We use this fact to obtain the desired lower bound on the smallest nonzero eigenvalue. Our analysis
proceeds on a case-by-case basis, depending on the occupation numbers for each component of G1 (the
values N1 , . . . , Nk in Lemma 2.4).
First consider any set of occupation numbers where some two-qubit gate gadget J ∈ [M] contains 3
or more particles. By Lemma 2.3 and Lemma 2.4, any such eigenvalue is at least λ31 (GUJ ), which is a
positive constant by Lemma 6.3. Next consider a case where some boundary gadget contains more than
one particle. The corresponding eigenvalues are similarly lower bounded by λ21 (Gbnd ), which is also a
positive constant by Lemma 6.4. Finally, consider a set of occupation numbers where each two-qubit
gadget contains at most two particles and each boundary gadget contains at most one particle. The
smallest eigenvalue with such a set of occupation numbers is zero. The smallest nonzero eigenvalue is
either
γ(H(GUJ , 1)), γ(H(GUJ , 2)) for some J ∈ [M], or γ(H(Gbnd , 1)) .
However, these quantities are at least some positive constant since H(GUJ , 1), H(GUJ , 2), and H(Gbnd , 1)
are nonzero constant-sized positive semidefinite matrices.
Now combining the lower bounds discussed above and using the fact that, for each J ∈ [M], the
two-qubit gate UJ is chosen from a fixed finite gate set (given in (7.3)), we see that γ(H(G1 , n)) is lower
bounded by the positive constant
min{λ31 (GU ), λ21 (Gbnd ), γ(H(GU , 1)), γ(H(GU , 2)), γ(H(Gbnd , 1)) : U is from the gate set (7.3)} .
(8.15)
The condition K0 ≤ 1 can be ensured by setting K0 to be the minimum of 1 and (8.15).
Note that the constant K0 can in principle be computed using (8.15): each quantity on the right-hand
side can be evaluated by diagonalizing a specific finite-dimensional matrix.
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8.3 Legal configurations
In this section we define a subset of the n-particle configurations that we call legal configurations, and
we prove that the subset of the basis vectors in Bn that have legal configurations spans the nullspace of
H(G1 , Gocc
X , n).
We begin by specifying the set of legal configurations. Every legal configuration
(J1 , . . . , JY , L1 , . . . , Ln−2Y )
has Y ∈ {0, 1}. The legal configurations with Y = 0 are
((1, j, d1 ), F(2, j, d2 ), F(3, j, d3 ), . . . , F(n, j, dn )) (8.16)
where j ∈ [M] and where d~ = (d1 , . . . , dn ) satisfies di ∈ {0, 1} and d1 = ds( j) . (Recall that the function F,
defined in equations (7.9) and (7.10), describes horizontal movement of particles.) The legal configurations
with Y = 1 are
( j, F(2, j, d2 ), . . . , F(s( j) − 1, j, ds( j)−1 ), F(s( j) + 1, j, ds( j)+1 ), . . . , F(n, j, dn )) (8.17)
where j ∈ {1, . . . , M} and di ∈ {0, 1} for i ∈ [n] \ {1, s( j)}. Although the values d1 and ds( j) are not used
in equation (8.17), we choose to set them to
d1 = ds( j) = 2
for any legal configuration with Y = 1. In this way we identify the set of legal configurations with the set
of pairs j, d~ with j ∈ [M] and
d~ = (d1 , d2 , d3 , . . . , dn )
satisfying
d1 = ds( j) ∈ {0, 1, 2} and di ∈ {0, 1} for i ∈
/ {1, s( j)} .
The legal configuration is given by equation (8.16) if d1 = ds( j) ∈ {0, 1} and equation (8.17) if d1 =
ds( j) = 2.
The examples in Figures 8.1(A), ( B ), and ( C ) show legal configurations whereas the examples
in Figures 8.1(D), ( E ), and ( F ) are illegal. The legal examples correspond to j = 1, d~ = (1, 1, 1);
j = 2, d~ = (2, 2, 0); and j = 1, d~ = (1, 0, 1), respectively. We now explain why the other examples
are illegal. Looking at (8.17), we see that the configuration (3, (2, 0, 1)) depicted in Figure 8.1(D)
is illegal since F(2, 3, 0) = (2, 2, 1) 6= (2, 0, 1) and F(2, 3, 1) = (2, 4, 0) 6= (2, 0, 1). The configuration
in Figure 8.1(E) is illegal since there are two particles in the same row. Looking at equation (8.16),
we see that the configuration ((1, 1, 1), (2, 2, 0), (3, 5, 0)) in Figure 8.1(F) is illegal since (3, 5, 0) ∈ /
{F(3, 1, 0), F(3, 1, 1)} = {(3, 0, 1), (3, 3, 0)}.
We now identify the subset of basis vectors Blegal ⊂ Bn that have legal configurations. We write each
such basis vector as
n
(1, j,d1 ) O F(i, j,di )
Sym |ρz ,a
1 1 i |ρ z ,a
i i i d1 = ds( j) ∈ {0, 1}
i=2
~ ~z, ~
| j, d, ai = n (8.18)
j
O F(i, j,di )
Sym |Tz1 ,a1 ,zs( j) ,as( j) i
|ρzi ,ai i d1 = ds( j) = 2
i=2
i6=s( j)
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where j, d~ specifies the legal configuration and ~z,~
a ∈ {0, 1}n . (Note that the bits in ~z and ~
a are ordered
slightly differently than in equation (8.13); here the labeling reflects the indices of the encoded qubits.)
Definition 8.6. Let
~ ~z, ~
Blegal = | j, d, / {1, s( j)}, ~z, ~
a ∈ {0, 1}n
ai : j ∈ [M], d1 = ds( j) ∈ {0, 1, 2} and di ∈ {0, 1} for i ∈
and Billegal = Bn \ Billegal .
The basis Bn = Blegal ∪ Billegal is convenient when considering the restriction to the subspace
I(G1 , Gocc
X , n). Letting Π0 be the projector onto I(G1 , GX , n), the following lemma (proven in Sec-
occ
tion A.2) shows that the restriction
Π0 span(Bn ) (8.19)
is diagonal in the basis Bn . The lemma also bounds the diagonal entries.
Lemma 8.7. Let Π0 be the projector onto I(G1 , Gocc ~ z, ~
ai ∈ Blegal , we have
X , n). For any | j, d,~
~ ~z, ~
Π0 | j, d, ~ ~z, ~
ai = | j, d, ai. (8.20)
Furthermore, for any two distinct basis vectors |φ i, |ψi ∈ Billegal , we have
255
hφ |Π0 |φ i ≤ (8.21)
256
hφ |Π0 |ψi = 0 . (8.22)
We use this lemma to characterize the nullspace of H(G1 , Gocc
X , n) and bound its smallest nonzero
eigenvalue.
Lemma 8.8. The nullspace S1 of H(G1 , Gocc X , n) is spanned by the orthonormal basis Blegal . Its smallest
nonzero eigenvalue is
K0
γ(H(G1 , Gocc
X , n)) ≥ (8.23)
256
where K0 ∈ (0, 1] is the absolute constant from Lemma 8.5.
Proof. Recall from Section 7.3 that
H(G1 , Gocc
X , n) = H(G1 , n)|I(G1 ,Gocc
X ,n)
.
Its nullspace is the space of states |κi satisfying
Π0 |κi = |κi and H(G1 , n)|κi = 0
(recall that Π0 is the projector onto I(G1 , Gocc
X , n), the states satisfying the occupancy constraints). Since
Bn is a basis for the nullspace of H(G1 , n), to solve for the nullspace of H(G1 , Gocc
X , n) we consider the
restriction (8.19) and solve for the eigenspace with eigenvalue 1. This calculation is simple because (8.19)
is diagonal in the basis Bn , according to Lemma 8.7. We see immediately from the lemma that Blegal
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
spans the nullspace of H(G1 , Gocc
X , n); we now show that Lemma 8.7 also implies the lower bound (8.23).
Note that
γ(H(G1 , Gocc
X , n)) = γ(Π0 H(G1 , n)Π0 ) .
Let Πlegal and Πillegal project onto the spaces spanned by Blegal and Billegal respectively, so Πlegal + Πillegal
projects onto the nullspace of H(G1 , n). The operator inequality
H(G1 , n) ≥ γ(H(G1 , n)) · 1 − Πlegal − Πillegal
implies
Π0 H(G1 , n)Π0 ≥ γ(H(G1 , n)) · Π0 (1 − Πlegal − Πillegal )Π0 .
Since the operators on both sides of this inequality are positive semidefinite and have the same nullspace,
their smallest nonzero eigenvalues are bounded as
γ(Π0 H(G1 , n)Π0 ) ≥ γ(H(G1 , n)) · γ(Π0 (1 − Πlegal − Πillegal )Π0 ) .
Hence
X , n)) = γ(Π0 H(G1 , n)Π0 ) ≥ K0 · γ(Π0 (1 − Πlegal − Πillegal )Π0 )
γ(H(G1 , Gocc (8.24)
where we used Lemma 8.5. From equations (8.21) and (8.22) we see that
255
Π0 |gi = |gi and Πillegal | f i = | f i =⇒ h f |gihg| f i ≤ . (8.25)
256
The nullspace of
Π0 1 − Πlegal − Πillegal Π0 (8.26)
is spanned by
Blegal ∪ {|τi : Π0 |τi = 0} .
To see this, note that (8.26) commutes with Π0 , and the space of +1 eigenvectors of Π0 that are annihilated
by (8.26) is spanned by Blegal (by Lemma 8.7). Any eigenvector |g1 i corresponding to the smallest
nonzero eigenvalue of this operator therefore satisfies Π0 |g1 i = |g1 i and Πlegal |g1 i = 0, so
1
γ(Π0 (1 − Πlegal − Πillegal )Π0 ) = 1 − hg1 |Πillegal |g1 i ≥
256
using equation (8.25). Plugging this into equation (8.24) gives the lower bound (8.23).
8.4 Matrix elements between states with legal configurations
We now consider
H1 |S1 , H2 |S1 , Hin,i |S1 , Hout |S1 (8.27)
where these operators are defined in Section 8.1.2 and
S1 = span(Blegal )
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
is the nullspace of H(G1 , Gocc
X , n).
We specify the operators (8.27) by their matrix elements in an orthonormal basis for S1 . Although the
basis Blegal was convenient in Section 8.3, here we use a different basis in which the matrix elements of
H1 and H2 are simpler. We define
~ In(~z), ~ ~~
h~x|Ū j,d1 (a1 )|~zi | j, d, x, ~
| j, d, ai = ∑ ai (8.28)
~
x∈{0,1}n
where
(
U j−1 (a1 )U j−2 (a1 ) . . .U1 (a1 ) if d1 ∈ {0, 2}
Ū j,d1 (a1 ) = (8.29)
U j (a1 )U j−1 (a1 ) . . .U1 (a1 ) if d1 = 1.
In each of these states the quantum data (represented by the ~x register on the right-hand side) encodes
the computation in which the unitary Ū j,d1 (a1 ) is applied to the initial n-qubit state |~zi (the notation In(~z)
indicates that ~z is the input). The vector ~
a is only relevant insofar as its first bit a1 determines whether
or not each two-qubit unitary is complex conjugated; the other bits of ~ a go along for the ride. Letting
~z, ~
a ∈ {0, 1}n , j ∈ [M], and
d~ = (d1 , . . . , dn ) with d1 = ds( j) ∈ {0, 1, 2} and di ∈ {0, 1}, i ∈
/ {1, s( j)} ,
we see that the states (8.28) form an orthonormal basis for S1 . In Section A.3 we compute the matrix
elements of the operators (8.27) in this basis, which are reproduced below.
Roughly speaking, the nonzero off-diagonal matrix elements of the operator H1 in the basis (8.28)
~ In(~z), ~
occur between states | j, d, ai where the legal configurations j, d~ and j,~c are
ai and | j,~c, In(~z), ~
related by horizontal motion of a particle in one of the rows i ∈ {2, . . . , n}.
Matrix elements of H1
c = d~
n−1
64 ~
n
1
di 6= ci for some i ∈ [n] \ {1, s( j)}
64 ∏ δdr ,cr
r=1
hk,~c, In(~x),~b|H1 | j, d,
~ In(~z), ~ r6=i
ai = δk, j δ~a,~b δ~z,~x · n
1√
(c1 , d1 ) ∈ {(2, 0), (0, 2), (1, 2), (2, 1)}
64 2 ∏ dr ,cr
δ
r=2
r6=s( j)
0 otherwise.
(8.30)
From this expression we see that H1 |S1 is block diagonal in the basis (8.28), with a block for each
~z, ~
a ∈ {0, 1}n and j ∈ [M]. Moreover, the submatrix for each block is the same. In Figure 8.2 we illustrate
some of the off-diagonal matrix elements of H1 |S1 for the example from Figure 7.1.
Next, we present the matrix elements of H2 .
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 i=1
× ×
i=2
×
−→ i=2
×
i=3
× i=3
×
(A) h j,~c, In(~z), ~ ~ In(~z), ~
a|H1 | j, d, 1
ai = 64 ; here j = 2 and d~ = (2, 2, 0) → ~c = (2, 2, 1).
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1
−→ ×
i=2
× i=2
× ×
i=3
× i=3
(B) h j,~c, In(~z), ~ ~ In(~z), ~
a|H1 | j, d, ai = 1√
; here j = 3 and d~ = (0, 0, 0) → ~c = (2, 0, 2).
64 2
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1
−→ ×
i=2
× i=2 × ×
i=3
× i=3
(C) h j,~c, In(~z), ~ ~ In(~z), ~
a|H1 | j, d, ai = 1√
; here j = 1 and d~ = (1, 1, 1) → ~c = (2, 1, 2).
64 2
Figure 8.2: Examples of matrix elements of H1 in the basis (8.28) of S1 . The relevant matrix elements (as
indicated above) are computed using ( A ) the second case and ( B ), ( C ) the third case in equation (8.30).
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
Matrix elements of H2
hk,~c, In(~x),~b|H2 | j, d,
~ In(~z), ~
ai = fdiag(d,
~ j)·δ j,k δ ~ δ~z,~x δ ~ (8.31)
~
a,b ~
c,d
~ j) · δk, j−1 + foff-diag (d,
+ foff-diag (~c, d, ~~
c, k) · δk−1, j δ~a,~b δ~z,~x
where
0
d1 = 0 and j = 1, or d1 = 1 and j = M
~ j) =
fdiag (d, 1
d1 = 2 and j ∈ {1, M} (8.32)
128
1
64 otherwise
and
1√
64 2
(c1 , cs( j) , d1 , ds( j−1) ) ∈ {(2, 0, 0, 0), (1, 1, 2, 1)}
n 1
~ j) = (c1 , cs( j) , d1 , ds( j−1) ) = (1, 0, 0, 1)
foff-diag (~c, d, ∏δdr ,cr · 641 (8.33)
r=2 128 (c1 , cs( j) , d1 , ds( j−1) ) = (2, 1, 2, 0)
r∈{s(
/ j−1),s( j)}
0 otherwise.
This shows that H2 |S1 is block diagonal in the basis (8.28), with a block for each ~z, ~
a ∈ {0, 1}n . Also
note that (in contrast with H1 ) H2 connects states with different values of j. In Figure 8.3 we illustrate
some of the off-diagonal matrix elements of H2 |S1 , for the example from Figure 7.1.
Finally, we present the matrix elements of Hin,i (for i ∈ {nin + 1, . . . , n}) and Hout :
Matrix elements of Hin,i For each ancilla qubit i ∈ {nin + 1, . . . , n}, define
jmin,i = min { j ∈ [M] : s( j) = i}
to be the index of the first gate in the circuit that involves this qubit (recall from Section 7 that we
consider circuits where each ancilla qubit is involved in at least one gate). The operator Hin,i is diagonal
in the basis (8.28), with entries
(
1
~ In(~z), ~ ~ In(~z), ~ j ≤ jmin,i , zi = 1, and di = 0
h j, d, a|Hin,i | j, d, ai = 64 (8.34)
0 otherwise.
Matrix elements of Hout Let jmax = max{ j ∈ [M] : s( j) = 2} be the index of the last gate U jmax in the
circuit that acts between qubits 1 and 2 (the output qubit). Then
(
h~x|UC†X (a1 )|0ih0|2UCX (a1 )|~zi 64
1
j ≥ jmax and d2 = 1
hk,~c, In(~x),~b|Hout | j, d,
~ In(~z), ~
ai = δ j,k δ~c,d~δ~a,~b
0 otherwise.
(8.35)
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1
−→ ×
i=2
× i=2
× ×
i=3
× i=3
(A) h j − 1,~c, In(~z), ~ ~ In(~z), ~
a|H2 | j, d, ai = 1√
; here j = 4 and d~ = (0, 0, 0) → ~c = (2, 0, 2)
64 2
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 i=1 ×
× −→
i=2 × × i=2
×
i=3 i=3
×
(B) h j − 1,~c, In(~z), ~ ~ In(~z), ~
a|H2 | j, d, ai = 1√ ; here j = 3 and d~ = (2, 1, 2) → ~c = (1, 1, 1)
64 2
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 × i=1 ×
−→
i=2
× i=2
×
i=3
× i=3
×
(C) h j − 1,~c, In(~z), ~ ~ In(~z), ~
a|H2 | j, d, 1
ai = 64 ; here j = 3 and d~ = (0, 1, 0) → ~c = (1, 1, 0)
j=0 j=1 j=2 j=3 j=4 j=5 j=0 j=1 j=2 j=3 j=4 j=5
i=1 i=1
× ×
×
−→ ×
i=2 i=2
×
i=3
× i=3
(D) h j − 1,~c, In(~z), ~ ~ In(~z), ~
a|H2 | j, d, 1
ai = 128 ; here j = 2 and d~ = (2, 2, 0) → ~c = (2, 1, 2)
Figure 8.3: Examples of matrix elements of H2 in the basis (8.28) of S1 . The relevant matrix elements (as
indicated above) are computed using the ( A ), ( B ) first, ( C ) second, and ( D ) third cases in equation (8.33).
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
8.5 From H(G2 , Gocc occ
X , n) to H(G4 , GX , n)
Define the (n − 2)-dimensional hypercubes
Dkj = (d1 , . . . , dn ) : d1 = ds( j) = k, di ∈ {0, 1} for i ∈ [n] \ {1, s( j)}
for j ∈ {1, . . . , M} and k ∈ {0, 1, 2}, and the superpositions
1 n
~ In(~z), ~
|Cubek ( j,~z, ~
a)i = √ ∑ (−1)∑i=1 di | j, d, ai
2 n−2
~
d∈D
j
k
for k ∈ {0, 1, 2}, j ∈ [M], and ~z, ~
a ∈ {0, 1}n . For each j ∈ [M] and ~z, ~
a ∈ {0, 1}n , let
1 1 1
a)i = |Cube0 ( j,~z, ~
|C( j,~z, ~ a)i − √ |Cube2 ( j,~z, ~
a)i + |Cube1 ( j,~z, ~ a)i. (8.36)
2 2 2
We prove
Lemma 8.9. The Hamiltonian H(G2 , Gocc
X , n) has nullspace S2 spanned by the states
|C( j,~z, ~
a)i
for j ∈ [M] and ~z, ~
a ∈ {0, 1}n . Its smallest nonzero eigenvalue is
K0
γ(H(G2 , Gocc
X , n)) ≥
35000n
where K0 ∈ (0, 1] is the absolute constant from Lemma 8.5.
Proof. Recall from the previous section that H1 |S1 is block diagonal in the basis (8.28), with a block for
a ∈ {0, 1}n . That is to say, hk,~c, In(~x),~b|H1 | j, d,
each j ∈ [M] and ~z, ~ ~ In(~z), ~ a = ~b, k = j,
ai is zero unless ~
and ~z = ~x. Equation (8.30) gives the nonzero matrix elements within a given block, which we use to
compute the frustration-free ground states of H1 |S1 .
Looking at equation (8.30), we see that the matrix for each block can be written as a sum of n
commuting matrices: (n−1)/ 64 times the identity matrix (case 1 in equation (8.30)), n − 2 terms that each
/ {1, s( j)} of d~ (case 2), and a term that changes the value of the “special” components
flip a single bit i ∈
d1 = ds( j) ∈ {0, 1, 2} (case 3). Thus
h j,~c, In(~z), ~ ~ In(~z), ~
a|H1 | j, d, ai =
* " # +
1 ~ In(~z), ~
j,~c, In(~z), ~
a (n − 1) + ∑ Hflip,i + Hspecial, j j, d, a
64 i∈[n]\{1,s( j)}
where
h j,~c, In(~z), ~ ~ In(~z), ~
a|Hflip,i | j, d, ai = δci ,di ⊕1 ∏ δc ,d r r
r∈[n]\{i}
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
and
√1
(c1 , d1 ) ∈ {(2, 0), (0, 2), (1, 2), (2, 1)}
h j,~c, In(~z), ~ ~ In(~z), ~
a|Hspecial, j | j, d, ai = 2 and dr = cr for r ∈ [n] \ {1, s( j)}
0 otherwise.
Note that these n matrices are mutually commuting, each eigenvalue of Hflip,i is ±1, and each eigenvalue
of Hspecial, j is equal to one of the eigenvalues of the matrix
0 0 1
1
√ 0 0 1 ,
2 1 1 0
which are {−1, 0, 1}. Thus we see that the eigenvalues of H1 |S1 within a block for some j ∈ [M] and
~z, ~
a ∈ {0, 1}n are
1
n − 1 + ∑ yi + w (8.37)
64 i∈{1,s(
/ j)}
where yi ∈ ±1 for each i ∈ [n] \ {1, s( j)} and w ∈ {−1, 0, 1}. In particular, the smallest eigenvalue within
the block is zero (corresponding to yi = w = −1). The corresponding eigenspace is spanned by the
simultaneous −1 eigenvectors of each Hflip,i for i ∈ [n] \ {1, s( j)} and Hspecial, j . The space of simulta-
neous −1 eigenvectors of Hflip,i for i ∈ [n] \ {1, s( j)} within the block is spanned by {|Cube0 ( j,~z, ~ a)i,
|Cube1 ( j,~z, ~
a)i, |Cube2 ( j,~z, ~
a)i}. The state |C( j,~z, ~
a)i is the unique superposition of these states that is a
−1 eigenvector of Hspecial, j . Hence, for each block we obtain a unique state |C( j,~z, ~ a)i in the space S2 .
Ranging over all blocks j ∈ [M] and ~z, ~ n
a ∈ {0, 1} , we get the basis described in the lemma.
The smallest nonzero eigenvalue within each block is 1/64 (corresponding to yi = −1 and w = 0 in
equation (8.37)), so
1
γ(H1 |S1 ) = . (8.38)
64
To get the stated lower bound, we use Lemma 8.1 with H(G2 , Gocc
X , n) = HA + HB where
HA = H(G1 , Gocc
X , n) , HB = H1 |I(G2 ,Gocc
X ,n)
(as in equation (8.5)), along with the bounds
K0 1
γ(HA ) ≥ , γ(HB |S1 ) = γ(H1 |S1 ) = , kHB k ≤ kH1 k ≤ n kh1 k = 2n ,
256 64
from Lemma 8.8, equations (8.9) and (8.38), and the fact that kh1 k = 2 from (5.8). This gives
K0 K0
γ(H(G2 , Gocc
X , n)) ≥ ≥
64K0 + 256 + 2n · 64 · 256 35000n
where we used the facts that K0 ≤ 1 and n ≥ 1.
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
For each ~z, ~
a ∈ {0, 1}n define the uniform superposition
1 M
|H(~z, ~
a)i = √ ∑ |C ( j,~z, ~
a)i
M j=1
that encodes (somewhat elaborately) the history of the computation that consists of applying either UCX
a determines whether the circuit CX or its complex conjugate is
or UC∗X to the state |~zi. The first bit of ~
applied.
Lemma 8.10. The Hamiltonian H(G3 , Gocc
X , n) has nullspace S3 spanned by the states
|H(~z, ~
a)i
for ~z, ~
a ∈ {0, 1}n . Its smallest nonzero eigenvalue is
K0
γ(H(G3 , Gocc
X , n)) ≥
107 n2 M 2
where K0 ∈ (0, 1] is the absolute constant from Lemma 8.5.
Proof. Recall that
H(G3 , Gocc occ
X , n) = H(G2 , GX , n) + H2 |I(G3 ,Gocc
X ,n)
with both terms on the right-hand side positive semidefinite. To solve for the nullspace of H(G3 , Gocc
X , n),
it suffices to restrict our attention to the space
S2 = span{|C( j,~z, ~
a)i : j ∈ [M], ~z, ~
a ∈ {0, 1}n } (8.39)
of states in the nullspace of H(G2 , Gocc
X , n). We begin by computing the matrix elements of H2 in the
basis for S2 given above. We use equations (8.31) and (8.36) to compute the diagonal matrix elements:
1 1
hC ( j,~z, ~
a) |H2 |C ( j,~z, ~
a)i = hCube0 ( j,~z, ~
a)|H2 |Cube0 ( j,~z, ~
a)i + hCube1 ( j,~z, ~
a)i|H2 |Cube1 ( j,~z, ~
a)i
4 4
1
+ hCube2 ( j,~z, ~ a)|H2 |Cube2 ( j,~z, ~
a)i (8.40)
2
1 1
0 + 256 + 256
j=1
1 1 1
= 256 + 256 + 128 j ∈ {2, . . . , M − 1} (8.41)
1 1
256 + 0 + 256 j=M
(
1
j ∈ {1, M}
= 128
1
(8.42)
64 j ∈ {2, . . . , M − 1}.
In the second line we used equation (8.32). Looking at equation (8.31), we see that the only nonzero
off-diagonal matrix elements of H2 in this basis are of the form
hC( j − 1,~z, ~
a)|H2 |C( j,~z, ~
a)i or hC( j,~z, ~
a)|H2 |C( j − 1,~z, ~
a)i = hC( j − 1,~z, ~ a)i∗
a)|H2 |C( j,~z, ~
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
for j ∈ {2, . . . , M}, ~z, ~
a ∈ {0, 1}n . To compute these matrix elements we first use equation (8.33) to
evaluate
hCubew ( j − 1,~z, ~
a)|H2 |Cubev ( j,~z, ~
a)i
for v, w ∈ {0, 1, 2} and j ∈ {2, . . . , M}. For example, using the second case of equation (8.33), we get
hCube1 ( j − 1,~z, ~
a)|H2 |Cube0 ( j,~z, ~
a)i
1 ~ In(~z), ~
= n−2 ∑ ∑ (−1)∑i∈[n] (ci +di ) h j − 1,~c, In(~z), ~
a|H2 | j, d, ai
2 ~ j j−1
d∈D0 ~c∈D1
1 1 1
= ∑ (−1) · =− .
2n−2 ~ j 64 128
d∈D0 :ds( j−1) =1
To go from the first to the second line we used the fact that, for each d~ ∈ D0j with ds( j−1) = 1, there is one
c ∈ D1j−1 for which h j − 1,~c, In(~z), ~
~ ~ In(~z), ~
a|H2 | j, d, ai = 1/64 (with all other such matrix elements equal
~ so
to zero). This ~c satisfies c1 = cs( j−1) = 1 and cs( j) = 0, with all other bits equal to those of d,
n
(−1)∑i=1 (ci +di ) = (−1)c1 +cs( j) +cs( j−1) +d1 +ds( j) +ds( j−1) = −1
for each nonzero term in the sum.
We perform a similar calculation using cases 1, 3, and 4 in equation (8.33) to obtain
1
− 128 (w, v) = (1, 0)
1√
(w, v) ∈ {(2, 0), (1, 2)}
hCubew ( j − 1,~z, ~
a)|H2 |Cubev ( j,~z, ~
a)i = 128 2
1
− 256 (w, v) = (2, 2)
0 otherwise.
Hence
hC ( j − 1,~z, ~
a) |H2 |C ( j,~z, ~
a)i
1 1
= hCube1 ( j − 1,~z, ~ a)|H2 |Cube0 ( j,~z, ~a)i − √ hCube2 ( j − 1,~z, ~
a)|H2 |Cube0 ( j,~z, ~
a)i
4 2 2
1 1
+ hCube2 ( j − 1,~z, ~ a)|H2 |Cube2 ( j,~z, ~
a)i − √ hCube1 ( j − 1,~z, ~
a)|H2 |Cube2 ( j,~z, ~
a)i
2 2 2
1
=− .
128
Combining this with equation (8.42), we see that H2 |S2 is block diagonal in the basis (8.39), with a block
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
for each pair of n-bit strings ~z, ~
a ∈ {0, 1}n . Each of the 22n blocks is equal to the M × M matrix
1 −1 0 0 ··· 0
−1 2 −1 0 · · · 0
.. ..
1 0 −1 2 −1 . .
.. .. .
128 0
0 −1 . . 0
. .. .. ..
.. . . . 2 −1
0 0 ··· 0 −1 1
This matrix is just 1/128 times the Laplacian of a path of length M, whose spectrum is well known.
In particular, it has a unique eigenvector with eigenvalue zero (the all-ones vector) and its eigenvalue
gap is 2(1 − cos(π/M)) ≥ 4/M 2 . Thus for each of the 22n blocks there is an eigenvector of H2 |S2 with
eigenvalue 0, equal to the uniform superposition |H(~z, ~
a)i over the M states in the block. Furthermore,
the smallest nonzero eigenvalue within each block is at least 1/(32M 2 ). Hence
1
γ(H2 |S2 ) ≥ . (8.43)
32M 2
To get the stated lower bound on γ(H(G3 , Gocc
X , n)), we apply Lemma 8.1 with
HA = H(G2 , Gocc
X , n) , HB = H2 |I(G3 ,Gocc
X ,n)
,
and
K0 1
γ(HA ) ≥ , γ(HB |S2 ) = γ(H2 |S2 ) ≥ , kHB k ≤ kH2 k ≤ nkh2 k = 2n , (8.44)
35000n 32M 2
from Lemma 8.9, equation (8.43), and the fact that kh2 k = 2 from (5.8). This gives
K0
γ(H(G3 , Gocc
X , n)) ≥
32M K0 + 35000n + 2n(35000n)(32M 2 )
2
K0 K0
≥ 2 2
≥ 7 2 2.
M n (32 + 35000 + 70000 · 32) 10 M n
Lemma 8.11. The nullspace S4 of H(G4 , Gocc
X , n) is spanned by the states
|H (~z, ~
a)i where ~z = z1 z2 . . . znin 00 . . . 0}
| {z (8.45)
n−nin
for ~
a ∈ {0, 1}n and z1 , . . . , znin ∈ {0, 1}. Its smallest nonzero eigenvalue satisfies
K0
γ(H(G4 , Gocc
X , n)) ≥
1010 M 3 n3
where K0 ∈ (0, 1] is the absolute constant from Lemma 8.5.
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Proof. Using equation (8.34), we find
1
hC(k,~x,~b)|Hin,i |C( j,~z, ~
a)i = δk, j δ~x,~z δ~a,~b hCube0 ( j,~z, ~
a)|Hin,i |Cube0 ( j,~z, ~
a)i
4
1
+ hCube1 ( j,~z, ~ a)|Hin,i |Cube1 ( j,~z, ~
a)i
4
1
+ hCube2 ( j,~z, ~ a)|Hin,i |Cube2 ( j,~z, ~
a)i
2
1 1 1 1 1 1
1 4 · 2 + 4 · 2 + 2 · 2
j < jmin,i
= δk, j δ~x,~z δ~a,~b 1
4 +0+0 j = jmin,i
δz ,1
64 i
0+0+0 j > jmin,i
for each i ∈ {nin + 1, . . . , n}. Hence
n n
1 1 jmin,i − 1 1
hH(~x,~b)| ∑ Hin,i |H(~z, ~
a)i = δ~x,~z δ~a,~b ∑ + δzi ,1 .
i=nin +1 M i=nin +1 64 2 4
Therefore
n
∑ Hin,i S 3
i=nin +1
is diagonal in the basis {|H(~z, ~
a)i : ~z, ~
a ∈ {0, 1}n }. The zero eigenvectors are given by equation (8.45),
and the smallest nonzero eigenvalue is
!
n
1
γ ∑ Hin,i S3 ≥ 256M (8.46)
i=nin +1
since jmin,i ≥ 1. To get the stated lower bound we now apply Lemma 8.1 with
n
HA = H(G3 , Gocc
X , n) , HB = ∑ Hin,i I(G ,G ,n) occ
4 X
i=nin +1
and
K0 1 n
γ(HA ) ≥ , γ(HB |S3 ) ≥ , kHB k ≤ n ∑ hin,i = n ,
107 M 2 n2 256M i=nin +1
where we used Lemma 8.10, equation (8.46), and the fact that ∑ni=nin +1 hin,i = 1 (from equation (5.7)).
This gives
K0
γ (H(G4 , Gocc
X , n)) ≥
256MK0 + 10 n M + n(256M)(107 n2 M 2 )
7 2 2
K0 K0
≥ ≥ .
(M 3 n3 )(256 + 107 + 256 · 107 ) 1010 M 3 n3
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8.6 Completeness and soundness
We now finish the proof of Theorem 7.1. Using equation (8.35) we get
1
hC(k,~x,~b)|Hout |C( j,~z, ~
a)i = δk, j δ~a,~b hCube0 ( j,~x, ~
a)|Hout | Cube0 ( j,~z, ~
a)i
4
1
+ hCube1 ( j,~x, ~ a)|Hout | Cube1 ( j,~z, ~
a)i
4
1
+ hCube2 ( j,~x, ~ a)|Hout | Cube2 ( j,~z, ~
a)i
2
1 1 1 1 1 1
δk, j δ~a,~b 4 · 2 + 4 · 2 + 2 · 2
j > jmax
= h~x|UC†X (a1 ) (|0ih0|2 )UCX (a1 )|~zi 0 + 14 + 0 j = jmax
64
0+0+0 j < jmax
and
1
hH(~x,~b)|Hout |H(~z, ~
a)i = δ~a,~b (M − jmax + 12 )h~x|UC† (a1 ) (|0ih0|2 )UC (a1 )|~zi. (8.47)
128M
For any nin -qubit state |φ i, define
|H(φ
b ,~ h~z|φ i|0i⊗n−nin |H(~z, ~
a)i = ∑ a)i. (8.48)
~z∈{0,1}n
Note (from Lemma 8.11) that |H(φ
b ,~a)i is in the nullspace of H(G4 , Gocc
X , n).
8.6.1 Completeness
Suppose there exists an nin -qubit state |ψwit i such that AP(CX , |ψwit i) ≥ 1 − (1/2|X| ), i. e.,
1
hψwit |h0|n−nin UC†X |0ih0|2UCX |ψwit i|0in−nin ≤ . (8.49)
2|X|
Then (letting ~0 denote the all-zeros vector)
b wit ,~0)|H(GX , Gocc , n)|H(ψ
hH(ψ b wit ,~0)i
X
b wit ,~0)|H(G4 , Gocc , n) + Hout |H(ψ
= hH(ψ b wit ,~0)i
X
b wit ,~0)|Hout |H(ψ
= hH(ψ b wit ,~0)i
1
= (M − jmax + 21 )hψwit |h0|n−nin UC†X |0ih0|2UCX |ψwit i|0in−nin
128M
1
≤ |X|
2
(using equations (8.47) and (8.48) to go from the second to the third line, and equation (8.49) in the last
line). Hence λn1 (GX , Gocc |X|
X ) ≤ 1/2 , establishing equation (7.1).
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8.6.2 Soundness
Now suppose AP(CX , |φ i) ≤ 31 for all normalized nin -qubit states |φ i, i. e.,
2
hφ |h0|n−nin UC†X |0ih0|2UCX |φ i|0in−nin ≥ for all normalized |φ i ∈ (C2 )⊗nin . (8.50)
3
Complex-conjugating this equation gives
2
hφ |∗ h0|n−nin UC†∗X |0ih0|2UC∗X |φ i∗ |0in−nin ≥ for all normalized |φ i ∈ (C2 )⊗nin ,
3
or equivalently (replacing |φ i with its complex conjugate),
2
hφ |h0|n−nin UC†∗X |0ih0|2UC∗X |φ i|0in−nin ≥ for all normalized |φ i ∈ (C2 )⊗nin . (8.51)
3
Recall that S4 is the nullspace of H(G4 , Gocc
X , n) and consider the restriction
Hout S . (8.52)
4
We now show that the smallest eigenvalue of (8.52) is strictly positive. This implies that the nullspace
of H(GX , Gocc
X , n) is empty, which can be seen from (8.8) since both terms in this equation are positive
semidefinite and S4 is the nullspace of the first term. We then use Nullspace Projection Lemma to lower
bound the smallest eigenvalue λn1 (GX , Gocc occ
X ) of H(GX , GX , n).
By Lemma 8.11, the states
|H(
b ~z, ~
a)i = |H(z1 z2 . . . znin 00 . . . 0}, ~
| {z a)i , ~
a ∈ {0, 1}n , ~z ∈ {0, 1}nin ,
n−nin
are a basis for S4 and in this basis Hout is block diagonal with a block for each ~
a ∈ {0, 1}n , as can be seen
using equation (8.47):
b ~x,~b)|Hout |H( 1
hH( b ~z, ~
a)i = δ~a,~b (M − jmax + 12 )h~x|h0|n−nin UC†X (a1 )|0ih0|2UCX (a1 )|~zi|0in−nin .
128M
From equation (8.48) we can see that any normalized state that lives in the block corresponding to some
a ∈ {0, 1}n can be written as |H(φ
fixed ~ b ,~ a)i for some normalized nin -qubit state |φ i. The smallest
eigenvalue of Hout |S4 is therefore
b ,~
hH(θ b ,~
α )|Hout |H(θ α )i
for some normalized nin -qubit state |θ i and some ~
α ∈ {0, 1}n . Now
b ,~ b ,~ 1
hH(θ α )|Hout |H(θ α )i = (M − jmax + 12 )hθ |h0|n−nin UC†X (α1 )|0ih0|2UCX (α1 )|θ i|0in−nin
128M
1 2
≥ · (8.53)
256M 3
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using equation (8.50) if α1 = 0 and equation (8.51) if α1 = 1. Since (8.53) is a lower bound on the
smallest eigenvalue within each block, the nullspace of Hout |S4 is empty and
1
γ(Hout |S4 ) ≥ . (8.54)
384M
As noted above, the fact that this matrix has strictly positive eigenvalues implies the same property of
H(GX , Gocc
X , n), i. e.,
λn1 (GX , Gocc occ
X ) = γ(H(GX , GX , n)) .
To lower bound this quantity we apply Lemma 8.1 with
HA = H(G4 , Gocc
X , n) , HB = Hout I(GX ,Gocc ,n) ,
X
and we use the bound (8.54) as well as
K0
γ(HA ) ≥
10 n3 M 3
10
(from Lemma 8.11) and kHB k ≤ kHout k ≤ n khout k = n (using equation (8.3)). Applying the lemma gives
λn1 (GX , Gocc occ
X ) = γ(H(GX , GX , n))
K0
≥
1010 n3 M 3 + 384MK 10 3 3
0 + n(10 n M )(384M)
K0
≥
n M (10 + 384 + 1010 · 384)
4 4 10
K0
≥ .
1013 n4 M 4
Now choosing K (the constant in the statement of Theorem 7.1) to be equal to K0 /1013 (recall K0 ∈ (0, 1]
is the absolute constant from Lemma 8.5) proves equation (7.2). This completes the proof of Theorem 7.1.
9 XY Hamiltonian is QMA-complete
In this section we prove Theorem 2.9, showing that XY Hamiltonian is QMA-complete.
Proof. An instance of XY Hamiltonian can be verified by the standard QMA verification protocol for
the Local Hamiltonian problem [21] with one slight modification: before running the protocol Arthur
measures the magnetization of the witness and rejects unless it is equal to N. Thus the problem is
contained in QMA.
To prove QMA-Hardness, we show that the solution (yes or no) of an instance of Frustration-Free
Bose-Hubbard Hamiltonian with input G, N, ε is equal to the solution of the instance of XY Hamiltonian
with the same graph G and integer N, with precision parameter ε/4 and c = Nµ(G) + ε/4.
We separately consider yes instances and no instances of Frustration-Free Bose-Hubbard Hamiltonian
and show that the corresponding instance of XY Hamiltonian has the same solution in both cases.
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
Case 1: no instances
First consider a no instance of Frustration-Free Bose-Hubbard Hamiltonian, for which λN1 (G) ≥ ε + ε 3 .
We have
λN1 (G) = min hφ |HGN − Nµ(G)|φ i (9.1)
|φ i∈ZN (G)
hφ |φ i=1
≤ min hφ |HGN − Nµ(G)|φ i (9.2)
|φ i∈WN (G)
hφ |φ i=1
= min hφ |OG − Nµ(G)|φ i (9.3)
|φ i∈WtN (G)
hφ |φ i=1
= θN (G) − Nµ(G) (9.4)
where in the inequality we used the fact that WN (G) ⊂ ZN (G). Hence
ε
θN (G) ≥ Nµ(G) + λN1 (G) ≥ Nµ(G) + ε + ε 3 ≥ Nµ(G) + ,
2
so the corresponding instance of XY Hamiltonian is a no instance.
Case 2: yes instances
Now consider a yes instance of Frustration-Free Bose-Hubbard Hamiltonian, so 0 ≤ λN1 (G) ≤ ε 3 .
We consider the case λN1 (G) = 0 separately from the case where it is strictly positive. If λN1 (G) = 0
then any state |ψi in the ground space of HGN satisfies
N
hφ | ∑ (A(G) − µ(G))(w) + ∑ nbk (b
nk − 1)|φ i = 0 .
w=1 k∈V
Since both terms are positive semidefinite, the state |φ i has zero energy for each of them. In particular, it
has zero energy for the second term, or equivalently, |φ i ∈ WN (G). Therefore
λN1 (G) = min hφ |HGN − Nµ(G)|φ i = min hφ |OG − Nµ(G)|φ i = θN (G) − Nµ(G) ,
|φ i∈WN (G) |φ i∈WtN (G)
hφ |φ i=1 hφ |φ i=1
so θN (G) = Nµ(G), and the corresponding instance of XY Hamiltonian is a yes instance.
Finally, suppose 0 < λN1 (G) ≤ ε 3 . Then λN1 (G) is also the smallest nonzero eigenvalue of H(G, N),
which we denote by γ(H(G, N)). (Here and throughout this paper we write γ(M) for the smallest nonzero
eigenvalue of a positive semidefinite matrix M.) Note that λN1 (G) > 0 also implies (by the inequalities
(9.1)–(9.4)) that θN (G) − Nµ(G) > 0, so
θN (G) − Nµ(G) = γ (OG − Nµ(G)) WtN (G) .
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To upper bound θN (G) we use the Nullspace Projection Lemma (Lemma 8.1). We apply this lemma
using the decomposition H(G, N) = HA + HB where
N
nk − 1) ZN (G) ,
HA = ∑ nbk (b HB = ∑ (A(G) − µ(G))(w) ZN (G) .
k∈V w=1
Note that HA and HB are both positive semidefinite, and that the nullspace S of HA is equal the space
WN (G) of hard-core bosons. To apply the lemma we compute bounds on γ(HA ), kHB k, and γ(HB |S ).
We use the bounds γ(HA ) = 2 (since the operators {b
nk : k ∈ V } commute and have nonnegative integer
eigenvalues),
kHB k ≤ NkA(G) − µ(G)k ≤ N(kA(G)k + µ(G)) ≤ 2NkA(G)k ≤ 2KN ≤ 2K 2
(where we used the fact that kA(G)k is at most the maximum degree of G, which is at most the number
of vertices K), and
!
N
γ(HB |S ) = γ ∑ (A(G) − µ(G))(w) W (G) N
w=1
= γ (OG − Nµ(G)) WtN (G)
= θN (G) − Nµ(G) .
Now applying the lemma, we get
2(θN (G) − Nµ(G))
λN1 (G) = γ(H(G, N)) ≥ .
2 + (θN (G) − Nµ(G)) + 2K 2
Rearranging this inequality gives
2(K 2 + 1)
θN (G) − Nµ(G) ≤ λN1 (G) ≤ 4K 2 λN1 (G) ≤ 4K 2 ε 3
2 − λN1 (G)
where in going from the second to the third inequality we used the fact that 1 ≤ K 2 in the numerator
and λN1 (G) ≤ ε 3 < 1 in the denominator. Now using the fact (from the definition of Frustration-Free
Bose-Hubbard Hamiltonian) that ε ≤ 1/(4K), we get
ε
θN (G) ≤ Nµ(G) + ,
4
i. e., the corresponding instance of XY Hamiltonian is a yes instance.
10 Proof of the Occupancy Constraints Lemma
In this section we prove the Occupancy Constraints Lemma.
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Lemma 10.1 (Occupancy Constraints Lemma). Let G be an e1 -gate graph specified as a gate diagram
with R ≥ 2 diagram elements. Let N ∈ [R], let Gocc specify a set of occupancy constraints on G, and
suppose the subspace I(G, Gocc , N) is nonempty. Then there exists an efficiently computable e1 -gate
graph G with at most 7R2 diagram elements such that
a
1. If λN1 (G, Gocc ) ≤ a then λN1 (G ) ≤ .
R
b
2. If λN1 (G, Gocc ) ≥ b with b ∈ [0, 1], then λN1 (G ) ≥ .
(13R)9
10.1 Definitions and notation
In this section we establish notation and we describe how the gate graph G is constructed from G and
Gocc . We also define two related gate graphs G4 and G♦ that we use in our analysis.
Let us first fix notation for the gate graph G and the occupancy constraints graph Gocc . Write the
adjacency matrix of G as (see equation (5.4))
R
A(G) = ∑ |qihq| ⊗ A(g0 ) + hEG + hSG
q=1
where hEG and hSG are determined (through equations (5.6) and (5.5)) by the sets EG and SG of edges
and self-loops in the gate diagram for G, and where g0 is the 128-vertex graph from Figure 5.1. Recall
that the occupancy constraints graph Gocc is a simple graph with vertices labeled q ∈ [R], one for each
diagram element in G. We write
occ [R]
E(G ) ⊆ = {{x, y} : x, y ∈ [R], x 6= y}
2
for the edge set of Gocc .
Definition of G
To ensure that the ground space has the appropriate form, the construction of G is slightly different
depending on whether R is even or odd. The following description handles both cases.
1. Replace each diagram element q ∈ [R] in the gate diagram for G as shown in Figure 10.1, with
diagram elements labeled qin , qout and d(q, s) where q, s ∈ [R] and q 6= s if R is even. Each node
(q, z,t) in the gate diagram for G is mapped to a new node new(q, z,t) as shown by the black and
grey arrows, i. e.,
(
(qin , z,t) if (q, z,t) is an input node
new(q, z,t) = (10.1)
(qout , z,t) if (q, z,t) is an output node.
Edges and self-loops in the gate diagram for G are replaced by edges and self-loops between the
corresponding nodes in the modified diagram.
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
d(q, q) is omitted if R is even
q qin d(q, 1) d(q, 2) d(q, R) qout
−→ 1 1 1 .. .. .. 1
Ũ Ũ
Figure 10.1: The first step in constructing the gate diagram of G from that of G is to replace each
diagram element as shown. The four input nodes (black arrow) and four output nodes (grey arrow) on the
left-hand side are identified with nodes on the right-hand side as shown.
to d(q, s) ei j (q, s) to d(q, s) ei j (q, s) to e00 (q, s) d(q, s) to e01 (q, s)
d(q, s) d(q, qb)
1 1 1 1 1
to d(s, q)
to e10 (q, s) to e11 (q, s)
(A) {q, s} ∈ E(Gocc ), q < s / E(Gocc )
(B) {q, s} ∈ (C)
Figure 10.2: Edges and self-loops added in step 4 of the construction of the gate diagram of G . When
{q, s} ∈ E(Gocc ) with q < s, we add two outgoing edges to ei j (q, s) as shown in ( A ). Note that if q > s
and {q, s} ∈ E(Gocc ) then ei j (q, s) = e ji (s, q). When {q, s} ∈
/ E(Gocc ) we add a self-loop and a single
outgoing edge from ei j (q, s) as shown in ( B ). Each diagram element d(q, s) has eight outgoing edges
(four of which are added in step 4), as shown in ( C ).
2. For each edge {q1 , q2 } ∈ E(Gocc ) in the occupancy constraints graph we add four diagram elements
of the type shown in Figure 5.2(C) (i. e., diagram elements corresponding to the identity). We refer
to these diagram elements by labels ei j (q1 , q2 ) with i, j ∈ {0, 1}. For these diagram elements the
labeling function is symmetric, i. e., ei j (q1 , q2 ) = e ji (q2 , q1 ) whenever {q1 , q2 } ∈ E(Gocc ).
3. For each non-edge {q1 , q2 } ∈ / E(Gocc ) with q1 , q2 ∈ [R] and q1 6= q2 we add 8 diagram elements of
the type shown in Figure 5.2(C). We refer to these diagram elements as ei j (q1 , q2 ) and ei j (q2 , q1 )
with i, j ∈ {0, 1}; when {q1 , q2 } ∈/ E(Gocc ) the labeling function is not symmetric, i. e., ei j (q1 , q2 ) 6=
e ji (q2 , q1 ). If R is odd we also add 4R diagram elements labeled ei j (q, q) with i, j ∈ {0, 1} and
q ∈ [R].
4. Finally, we add edges and self-loops to the gate diagram as shown in Figure 10.2. This gives the
gate diagram for G .
The set of diagram elements in the gate graph for G is indexed by
L = Qin ∪ D ∪ Eedges ∪ Enon-edges ∪ Qout (10.2)
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where
Qin = {qin : q ∈ [R]} (10.3)
D = {d(q, s) : q, s ∈ [R] and q 6= s if R is even} (10.4)
Eedges = ei j (q, s) : i, j ∈ {0, 1}, {q, s} ∈ E(Gocc ) and q < s
/ E(Gocc ) and q 6= s if R is even
Enon-edges = ei j (q, s) : i, j ∈ {0, 1}, {q, s} ∈
Qout = {qout : q ∈ [R]} . (10.5)
The total number of diagram elements in G is
|L | = |Qin | + |D| + |Eedges | + |Enon-edges | + |Qout |
(
R + R2 + 4|E(Gocc )| + 4 R2 − 2|E(Gocc )| + R
R odd
=
R + R (R − 1) + 4|E(Gocc )| + 4 (R(R − 1) − 2|E(Gocc )|) + R R even
(
5R2 + 2R − 4|E(Gocc )| R odd
=
5R2 − 3R − 4|E(Gocc )| R even.
In both cases this is upper bounded by 7R2 as claimed in the statement of the lemma. Write
A(G ) = ∑ |lihl| ⊗ A(g0 ) + hS + hE (10.6)
l∈L
where S and E are the sets of self-loops and edges in the gate diagram for G .
We now focus on the input nodes of diagram elements in Qin and the output nodes of the diagram
elements in Qout . These are the nodes indicated by the black and grey arrows in Figure 10.1. Write
E0 ⊂ E and S0 ⊂ S for the sets of edges and self-loops that are incident on these nodes in the gate
diagram for G . Note that the sets E0 and S0 are in one-to-one correspondence with (respectively) the
sets EG and SG of edges and self-loops in the gate diagram for G. The other edges and self-loops in G
do not depend on the sets of edges and self-loops in G. Writing
S4 = S \ S0 , E4 = E \ E0 ,
we have
hS = hS0 + hS4 , hE = hE0 + hE4 . (10.7)
Definition of G4
The gate diagram for G4 is obtained from that of G by removing all edges and self-loops attached to
the input nodes of the diagram elements in Qin and the output nodes of the diagram elements in Qout . Its
adjacency matrix is
A(G4 ) = ∑ |lihl| ⊗ A(g0 ) + hS4 + hE4 . (10.8)
l∈L
Note that G4 = G whenever the gate diagram for G contains no edges or self-loops.
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1
H
1
2
HT
2
3
1
3
(A) (B)
Figure 10.3: An example ( A ) Gate diagram for a gate graph G and ( B ) Occupancy constraints graph Gocc .
In the text we describe how these two ingredients are mapped to a gate graph G ; the gate diagram for
G is shown in Figure 10.4.
Definition of G♦
We also define a gate graph G♦ with gate diagram obtained from that of G4 by removing all edges (but
leaving the self-loops). Note that G♦ has a component for each diagram element l ∈ L . The components
corresponding to diagram elements without a self-loop (those with l ∈ L \ Enon-edges ) have adjacency
matrix A(g0 ); those with a self-loop (l ∈ Enon-edges ) have adjacency matrix A(g0 ) + |1, 1ih1, 1| ⊗ 1, so
A(G♦ ) = ∑ |lihl| ⊗ A(g0 ) + hS4 (10.9)
l∈L
= ∑ |lihl| ⊗ A(g0 ) + ∑ |lihl| ⊗ (A(g0 ) + |1, 1ih1, 1| ⊗ 1) . (10.10)
l∈L \E non-edges
l∈Enon-edges
Example
We provide an example of this construction in Figure 10.3 (which shows a gate graph and an occupancy
constraints graph) and Figure 10.4 (which describes the derived gate graphs G , G4 , and G♦ ).
10.2 The gate graph G♦
We now solve for the e1 -energy ground states of the adjacency matrix A(G♦ ). Write g1 for the graph with
adjacency matrix
A(g1 ) = A(g0 ) + |1, 1ih1, 1| ⊗ 1
(i. e., g0 with 8 self-loops added), so (recalling equation (10.10)) each component of G♦ is either g0 or g1 .
Recall from Section 5.1 that A(g0 ) has four orthonormal e1 -energy ground states |ψz,a i with z, a ∈ {0, 1}.
It is also not hard to verify that the e1 -energy ground space of A(g1 ) is spanned by two of these states
|ψ0,a i for a ∈ {0, 1}. Now letting |ψz,a l i = |li|ψ i, we choose a basis W for the e -energy ground space
z,a 1
of A(G♦ ) where each basis vector is supported on one of the components:
W = |ψz,a i : z, a ∈ {0, 1}, l ∈ L \ Enon-edges ∪ |ψ0,a
l l
i : a ∈ {0, 1}, l ∈ Enon-edges . (10.11)
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e00 (1, 1) e01 (1, 1) e00 (1, 3) e01 (1, 3)
1 1 1 1
1in d(1, 1) d(1, 2) d(1, 3) 1out
1 1 1 1 H
1 1 1 1
e01 (1, 2) e00 (1, 2) e10 (1, 2) e11 (1, 2)
e10 (1, 1) e11 (1, 1) e10 (1, 3) e11 (1, 3)
1 1 1 1
e00 (2, 2) e01 (2, 2) e00 (2, 3) e01 (2, 3)
1 1 1 1
2in d(2, 1) d(2, 2) d(2, 3) 2out
1 1 1 1 HT
1 1 1 1
e10 (2, 2) e11 (2, 2) e10 (2, 3) e11 (2, 3)
e00 (3, 1) e01 (3, 1) e00 (3, 2) e01 (3, 2) e00 (3, 3) e01 (3, 3)
1 1 1 1 1 1
3in d(3, 1) d(3, 2) d(3, 3) 3out
1 1 1 1 1
1 1 1 1 1 1
e10 (3, 1) e11 (3, 1) e10 (3, 2) e11 (3, 2) e10 (3, 3) e11 (3, 3)
Figure 10.4: The gate diagram for G4 (only solid lines) and G (including dotted lines) derived from the
example gate graph G and occupancy constraints graph Gocc from Figure 10.3. The gate diagram for G♦
is obtained from that of G4 by removing all edges (but leaving the self-loops).
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The eigenvalue gap of A(G♦ ) is equal to that of either A(g0 ) or A(g1 ). Since g0 and g1 are specific 128-
vertex graphs we can calculate their eigenvalue gaps using a computer; we get γ(A(g0 ) − e1 ) = 0.7785 . . .
and γ(A(g1 ) − e1 ) = 0.0832 . . .. Hence
1
γ(A(G♦ ) − e1 ) ≥ 0.0832 . . . > . (10.12)
13
The ground space of A(G♦ ) has dimension
(
4 5R2 + 2R − 4|E(Gocc )| − 2 4R2 − 8|E(Gocc )|
R odd
|W| = 4 L − 2 Enon-edges = 2 occ
occ
4 5R − 3R − 4|E(G )| − 2 (4R(R − 1) − 8|E(G )|) R even
(
12R2 + 8R R odd
= (10.13)
12R2 − 4R R even.
We now consider the N-particle Hamiltonian H(G♦ , N) and characterize its nullspace.
Lemma 10.2. The nullspace of H(G♦ , N) is
I♦ = span{Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqNN,aN i) : |ψzqii,ai i ∈ W and qi 6= q j for all distinct i, j ∈ [N]}
where W is given in equation (10.11). The smallest nonzero eigenvalue satisfies γ(H(G♦ , N)) > 1/300.
Proof. For the first part of the proof we use the fact that the basis vectors |ψz,a l i ∈ W span the e -
1
eigenspace of the component G♦ l of G ♦ corresponding to the diagram element l ∈ L , i. e., the nullspace
of H(G♦ l , 1). Furthermore, no component of G♦ supports a two-particle frustration-free state, i. e.,
λ2 (g0 ) > 0 and λ2 (g1 ) > 0 (by Lemma 5.3). Now applying Lemma 2.4 we see that I♦ is the nullspace
1 1
of H(G♦ , N). We also see that the smallest nonzero eigenvalue γ(H(G♦ , N)) is either λ21 (g0 ), λ21 (g1 ),
γ(H(g0 , 1)), or γ(H(g1 , 1)). These constants can be calculated numerically using a computer; they are
λ21 (g0 ) = 0.0035 . . . , λ21 (g1 ) = 0.0185 . . . , γ(H(g0 , 1)) = 0.7785 . . . ,
and γ(H(g1 , 1)) = 0.0832 . . . .
Hence
1
γ(H(G♦ , N)) ≥ min{λ21 (g0 ), λ21 (g1 ), γ(H(g0 , 1)), γ(H(g1 , 1))} > .
300
10.3 The adjacency matrix of the gate graph G4
We begin by solving for the e1 -energy ground space of the adjacency matrix A(G4 ). From equations
(10.8) and (10.9) we have
A(G4 ) = A(G♦ ) + hE4 . (10.14)
Recall the e1 -energy ground space of A(G♦ ) is spanned by W from equation (10.11). Since hE4 ≥ 0 it
follows that A(G4 ) ≥ e1 . To solve for the e1 -energy ground space of A(G4 ) we construct superpositions
of vectors from W that are in the nullspace of hE4 . To this end we consider the restriction
hE4 span(W) . (10.15)
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We now show that it is block diagonal in the basis W and we compute its matrix elements.
First recall from equation (5.6) that
|l, z,ti + |l 0 , z0 ,t 0 i hl, z,t| + hl 0 , z0 ,t 0 | ⊗ 1 .
hE4 = ∑ (10.16)
{(l,z,t),(l 0 ,z0 ,t 0 )}∈E4
The edges {(l, z,t), (l 0 , z0 ,t 0 )} ∈ E4 can be read off from Figure 10.1 and Figure 10.2, respectively
(referring back to Figure 5.2 for our convention regarding the labeling of nodes on a diagram element).
The edges from Figure 10.1 are
(qin , z,t), (d(q, 1), z,t 0 ) , (d(q, 2), z,t), (d(q, 3), z,t 0 ) , . . . , (d(q, R), z,t), (qout , z,t 0 )
(10.17)
with q ∈ [R] and (z,t,t 0 ) = (0, 7, 3) or (1, 5, 1), and where d(q, q) does not appear if R is even (i. e.,
d(q, q − 1) is followed by d(q, q + 1)). The edges from Figure 10.2 are
{(d(q, s), 0, 1) , (e00 (q, s), α(q, s), 1)}, {(d(q, s), 1, 3) , (e10 (q, s), α(q, s), 1)} , (10.18)
{(d(q, s), 0, 5) , (e01 (q, s), α(q, s), 1)}, {(d(q, s), 1, 7) , (e11 (q, s), α(q, s), 1)}
with q, s ∈ [R] and q 6= s if R is even, and where
(
1 q > s and {q, s} ∈ E(Gocc )
α(q, s) =
0 otherwise.
The set E4 consists of all edges (10.17) and (10.18).
We claim that (10.15) is block diagonal with a block W(z,a,q) ⊆ W of size
(
3R + 2 R odd
W(z,a,q) =
3R − 1 R even
for each for each triple (z, a, q) with z, a ∈ {0, 1} and q ∈ [R]. Using equation (10.13) we confirm that
|W| = 4R W(z,a,q) , so this accounts for all basis vectors in W. The subset of basis vectors for a given
block is
d(q,s)
W(z,a,q) = |ψz,a
q qout
in
i, |ψz,a i ∪ |ψz,a i : s ∈ [R], s 6= q if R even
ezx (q,s)
∪ |ψα(q,s),a i : x ∈ {0, 1}, s ∈ [R], s 6= q if R even . (10.19)
Using equation (10.16) and the description of E4 from (10.17) and (10.18), one can check by direct
inspection that (10.15) only has nonzero matrix elements between basis vectors in W from the same
block. We also compute the matrix elements between vectors from the same block. For example, if R is
odd or if R is even and q 6= 1, there are edges {(qin , 0, 7), (d(q, 1), 0, 3)} , {(qin , 1, 5), (d(q, 1), 1, 1)} ∈ E4 .
l i = |li|ψ i where |ψ i is given by (5.2) and (5.3), we compute the relevant
Using the fact that |ψz,a z,a z,a
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matrix elements:
qin d(q,1)
hψz,a |hE4 |ψz,a i
!
qin 0 0 0
0 0 0
d(q,1)
= hψz,a | ∑ |qin , z ,ti + |d(q, 1), z ,t i hqin , z ,t| + hd(q, 1), z ,t | ⊗ 1 |ψz,a i
(z0 ,t,t 0 )∈{(0,7,3),(1,5,1)}
1
hψz,a | |z0 ,tihz0 ,t 0 | ⊗ 1 |ψz,a i = .
= ∑
(z0 ,t,t 0 )∈{(0,7,3),(1,5,1)}
8
Continuing in this manner, we compute the principal submatrix of (10.15) corresponding to the set
W(z,a,q) . This matrix is shown in Figure 10.5(A). In the figure each vertex is associated with a state in
the block and the weight on a given edge is the matrix element between the two states associated with
vertices joined by that edge. The diagonal matrix elements are described by the weights on the self-loops.
The matrix described by Figure 10.5(A) is the same for each block.
q
For each triple (z, a, q) with z, a ∈ {0, 1} and q ∈ [R], define |φz,a i to be
√ 1 qin j d(q, j) ez0 (q, j) ez1 (q, j) qout
3R+2
|ψz,a i + ∑ j∈[R] (−1) |ψ z,a i − |ψ α(q, j),a i − |ψ α(q, j),a i + |ψ z,a i if R odd,
qin d(q, j) e (q, j) e (q, j) qout
√ 1 i + ∑ j<q − ∑ j>q (−1) j |ψz,a i − |ψα(q,
z0 z1
3R−1
|ψz,a j),a i − |ψα(q, j),a i + |ψz,a i if R even.
(10.20)
Next we show that these states span the ground space of A(G4 ). The choice to omit d(q, q) for R even
qin qout
ensures that |ψz,a i and |ψz,a i have the same sign in these ground states.
Lemma 10.3. An orthonormal basis for the e1 -energy ground space of A(G4 ) is given by the states
q
|φz,a i : z, a ∈ {0, 1}, q ∈ [R]
defined by equation (10.20). The eigenvalue gap is bounded as
1
γ(A(G4 ) − e1 ) > . (10.21)
(30R)2
Proof. The e1 -energy ground space of A(G4 ) is equal to the nullspace of (10.15). Since this operator is
block diagonal in the basis W, we can solve for the eigenvectors in the nullspace of each block. Thus, to
prove the first part of the lemma, we analyze the |W(z,a,q) | × |W(z,a,q) | matrix described by Figure 10.5(A)
and show that (10.20) is the unique vector in its nullspace. We first rewrite it in a slightly different basis
obtained by multiplying some of the basis vectors by a phase of −1. Specifically, we use the basis
n o
qin d(q,1) ez0 (q,1) ez1 (q,1) d(q,2) ez0 (q,2) ez1 (q,2) qout
|ψz,a i, −|ψz,a i, |ψα(q,1),a i, |ψα(q,1),a i, |ψz,a i, −|ψα(q,2),a i, −|ψα(q,2),a i, . . . , |ψz,a i
where the state associated with each vertex on one side of a bipartition of the graph is multiplied by −1;
these are the phases appearing in equation (10.20). Changing to this basis replaces the weight 1/8 on
each edge in Figure 10.5(A) by −1/8 and does not change the weights on the self-loops. The resulting
matrix is (1/8)L0 , where L0 is the Laplacian matrix of the graph shown in Figure 10.5(B). Now we use
the fact that the Laplacian of any connected graph has smallest eigenvalue zero, with a unique eigenvector
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e (q,1)
z0 z0 e (q,2) e (q,R)
z0
|ψα(q,1),a i |ψα(q,2),a i |ψα(q,R),a i
1/8 1/8 1/8
1/8 1/8 1/8
1/2 1/2 1/2
1/8 1/8
1/8 1/8 1/8
qin
|ψz,a i ... qout
|ψz,a i
d(q,1) d(q,2) d(q,R)
|ψz,a i |ψz,a i |ψz,a i
1/8 1/8 1/8
1/8 1/8 1/8
ez1 (q,1) ez1 (q,2) ez1 (q,R)
|ψα(q,1),a i |ψα(q,2),a i |ψα(q,R),a i
(A) The matrix h4 E |span(W) is block diagonal in the basis W, with a block W(z,a,q) for each
z, a ∈ {0, 1} and q ∈ {1, . . . , R}. The states involved in a given block and the matrix elements
between them are depicted.
R (for R odd) or R − 1 (for R even)
...
(B) After multiplying some of the basis vectors by −1, the matrix
depicted in ( A ) is transformed into 1/8 times the Laplacian matrix
of this graph.
Figure 10.5
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equal to the all-ones vector. Hence for each block we get an eigenvector in the nullspace of (10.15)) given
by (10.20). Ranging over all z, a ∈ {0, 1} and q ∈ [R] gives the claimed basis for the e1 -energy ground
space of A(G4 ).
To prove the lower bound, we use the Nullspace Projection Lemma (Lemma 8.1) with
HA = A(G♦ ) − e1 , HB = hE4 ,
and where S = span(W) is the nullspace of HA as shown in Section 10.2. Since it is block diagonal in the
basis W, the smallest nonzero eigenvalue of (10.15) is equal to the smallest nonzero eigenvalue of one of
its blocks. The matrix for each block is 18 L0 . Thus we can lower bound the smallest nonzero eigenvalue
of HB |S using standard bounds on the smallest nonzero eigenvalue of the Laplacian L of a graph G. In
particular, Theorem 4.2 of reference [26] shows that
4 4
γ(L) ≥ ≥
|V (G)| diam(G) |V (G)|2
(where diam(G) is the diameter of G). Since the size of the graph in Figure 10.5(B) is either 3R − 1 or
3R + 2, we have
1 1 4 1
γ(HB |S ) = γ(L0 ) ≥ 2
≥
8 8 (3R + 2) 32R2
since R ≥ 2. Using this bound and the fact that γ(HA ) > 1/13 (from equation (10.12)) and kHB k = 2
(from equation (5.8)) and plugging into Lemma 8.1 gives
1
· 12 1 1
γ(A(G4 ) − e1 ) ≥ 1 13 32R
1
≥ 2
> .
13 + 32R2 + 2
(32 + 13 + 832)R (30R)2
10.4 The Hamiltonian H(G4 , N)
We now consider the N-particle Hamiltonian H(G4 , N) and solve for its nullspace. We use the following
fact about the subsets W(z,a,q) ⊂ W defined in equation (10.19).
Definition 10.4. We say W(z1 ,a1 ,q1 ) and W(z2 ,a2 ,q2 ) overlap on a diagram element if there exists l ∈ L
such that
|ψxl 1 ,b1 i ∈ W(z1 ,a1 ,q1 ) and |ψxl 2 ,b2 i ∈ W(z2 ,a2 ,q2 )
for some x1 , x2 , b1 , b2 ∈ {0, 1}.
Fact 10.5 (Key property of W(z,a,q) ). Sets W(z1 ,a1 ,q1 ) and W(z2 ,a2 ,q2 ) overlap on a diagram element if and
only if q1 = q2 or {q1 , q2 } ∈ E(Gocc ).
This fact can be confirmed by direct inspection of the sets W(z,a,q) . If q1 = q2 = q the diagram
element l on which they overlap can be chosen to be l = qin ; if q1 6= q2 and {q1 , q2 } ∈ E(Gocc ) then
l = ez1 z2 (q1 , q2 ) = ez2 z1 (q2 , q1 ).
Conversely, if {q1 , q2 } ∈ / E(Gocc ) with q1 6= q2 , then there is no overlap.
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We show that the nullspace I4 of H(G4 , N) is
I4 = span{Sym(|φzq11,a1 i|φzq22,a2 i . . . |φzqNN,aN i) : zi , ai ∈ {0, 1}, qi ∈ [R], qi 6= q j , and {qi , q j } ∈
/ E(Gocc )}.
(10.22)
Note that I4 ⊂ ZN (G4 ) is very similar to I(G, Gocc , N) ⊂ ZN (G) (from equation (5.21)) but with each
q q
single-particle state |ψz,a i ∈ ZN (G) replaced by |φz,a i ∈ ZN (G4 ).
Lemma 10.6. The nullspace of H(G4 , N) is I4 as defined in equation (10.22). Its smallest nonzero
eigenvalue is
1
γ(H(G4 , N)) > . (10.23)
(17R)7
In addition to Fact 10.5, we use the following simple fact in the proof of the lemma.
√
Fact 10.7. Let |pi = c|α0 i + 1 − c2 |α1 i with hαi |α j i = δi j and c ∈ [0, 1]. Then
|pihp| = c2 |α0 ihα0 | + M
where kMk ≤ 1 − (3/4)c4 .
To prove this fact, one can calculate
1 1p
kMk = (1 − c2 ) + 1 + 2c2 − 3c4
2 2
√
and use the inequality 1 + x ≤ 1 + x/2 for x ≥ −1.
Proof of Lemma 10.6. Using equation (10.14) and the fact that the smallest eigenvalues of A(G♦ ) and
A(G4 ) are the same (equal to e1 , from Section 10.2 and Lemma 10.3), we have
N
(w)
H(G4 , N) = H(G♦ , N) + ∑ hE4 . (10.24)
w=1 ZN (G4 )
Recall from Lemma 10.2 that the nullspace of H(G♦ , N) is I♦ . We consider
N
(w)
∑ hE 4 . (10.25)
w=1 I♦
We show that its nullspace is equal to I4 (establishing the first part of the lemma), and we lower bound
its smallest nonzero eigenvalue. Specifically, we prove
!
N
(w) 1
γ ∑ hE4 > . (10.26)
w=1 I♦ (9R)6
Now we prove equation (10.23) using this bound. We apply the Nullspace Projection Lemma
(Lemma 8.1) with HA and HB given by the first and second terms in equation (10.24); in this case
the nullspace of HA is S = I♦ (from Lemma 10.2). Now applying Lemma 8.1 and using the bounds
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γ(HA ) > 1/300 (from Lemma 10.2), kHB k ≤ N khE4 k = 2N ≤ 2R (from equation (5.8) and the fact that
N ≤ R), and the bound (10.26) on γ(HB |S ), we find
1
300(9R)6 1 1 1
γ(H(G4 , N)) ≥ 1 1
≥ 6 + 300 + 600 · 96 7
> 7
.
300 + (9R)6 + 2R 9 R (17R)
To complete the proof we must establish that the nullspace of (10.25) is I4 and prove the lower bound
(10.26). To analyze (10.25) we use the fact (established in Section 10.3) that (10.15) is block diagonal
with a block W(z,a,q) ⊂ W for each triple (z, a, q) with z, a ∈ {0, 1} and q ∈ [R]. The operator (10.25)
inherits a block structure from this fact. For any basis vector
Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqNN,aN i) ∈ I♦ , (10.27)
we define a set of occupation numbers
N = N(x,b,r) : x, b ∈ {0, 1}, r ∈ [R]
where
q
N(x,b,r) = |{ j : |ψz jj,a j i ∈ W(x,b,r) }| .
Now observe that (10.25) conserves the set of occupation numbers and is therefore block diagonal with a
block for each possible set N.
For a given block corresponding to a set of occupation numbers N, we write I♦ (N) ⊂ I♦ for the
subspace spanned by basis vectors (10.27) in the block. We classify the blocks into three categories
depending on N.
Classification of the blocks of (10.25) according to N Consider the following two conditions on a set
N = {N(x,b,r) : x, b ∈ {0, 1}, r ∈ [R]} of occupation numbers:
(a) N(x,b,r) ∈ {0, 1} for all x, b ∈ {0, 1} and r ∈ [R]. If this holds, write (yi , ci , si ) for the nonzero
occupation numbers (with some arbitrary ordering), i. e., N(yi ,ci ,si ) = 1 for i ∈ [N].
(b) The sets W(yi ,ci ,si ) and W(y j ,c j ,s j ) do not overlap on a diagram element for all distinct i, j ∈ [N].
We say a block is of type 1 if N satisfies (a) and (b). We say it is of type 2 if N does not satisfy (a). We
say it is of type 3 if N satisfies (a) but does not satisfy (b).
Note that every block is either of type 1, 2, or 3. We consider each type separately. Specifically, we
show that each block of type 1 contains one state in the nullspace of (10.25) and, ranging over all blocks
of this type, we obtain a basis for I4 . We also show that the smallest nonzero eigenvalue within a block
of type 1 is at least 1/(32R2 ). Finally, we show that blocks of type 2 and 3 do not contain any states in
the nullspace of (10.25) and that the smallest eigenvalue within any block of type 2 or 3 is greater than
1/(9R)6 . Hence, the nullspace of (10.25) is I4 and its smallest nonzero eigenvalue is lower bounded as
in equation (10.26).
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Type 1 Note (from Definition 10.4) that (b) implies q 6= r whenever
q
|ψx,b i ∈ W(yi ,ci ,si ) and |ψz,a
r
i ∈ W(y j ,c j ,s j )
for distinct i, j ∈ [N]. Hence
q
I♦ (N) = span{Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqNN,aN i) : qi 6= q j and |ψz jj,a j i ∈ W(y j ,c j ,s j ) }
q
= span{Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqNN,aN i) : |ψz jj,a j i ∈ W(y j ,c j ,s j ) } .
From this we see that
(
N
(3R + 2)N R odd
dim(I♦ (N)) = ∏ W(y j ,c j ,s j ) =
j=1 (3R − 1)N R even.
We now solve for all the eigenstates of (10.25) within the block.
It is convenient to write an orthonormal basis of eigenvectors of the |W(z,a,q) | × |W(z,a,q) | matrix
described by Figure 10.5(A) as
q
|φz,a (u)i , u ∈ [|W(z,a,q) |] , (10.28)
and their ordered eigenvalues as
θ1 ≤ θ2 ≤ . . . ≤ θ|W(z,a,q) | .
q q
From the proof of Lemma 10.3, the eigenvector with smallest eigenvalue θ1 = 0 is |φz,a i = |φz,a (1)i and
2
θ2 ≥ 1/(32R ). For any u1 , u2 , . . . , uN ∈ [|W(z,a,q) |], the state
Sym(|φys11,c1 (u1 )i|φys22,c2 (u2 )i . . . |φysNN,cN (uN )i)
is an eigenvector of (10.25) with eigenvalue ∑Nj=1 θ j . Furthermore, states corresponding to different
choices of u1 , . . . , uN are orthogonal, and ranging over all dim(I♦ (N)) choices we get every eigenvector
in the block. The smallest eigenvalue within the block is Nθ1 = 0 and there is a unique vector in the
nullspace, given by
Sym(|φys11,c1 i|φys22,c2 i . . . |φysNN,cN i) (10.29)
q q
(recall |φz,a i = |φz,a (1)i). The smallest nonzero eigenvalue of (10.25) within the block is
1
(N − 1)θ1 + θ2 = θ2 ≥
.
32R2
Finally, we show that the collection of states (10.29) obtained from all blocks of type 1 spans the
space I4 . Each block of type 1 corresponds to a set of occupation numbers
N(y1 ,c1 ,s1 ) = N(y2 ,c2 ,s2 ) = · · · = N(yN ,cN ,sN ) = 1 (with all other occupation numbers zero)
and gives a unique vector (10.29) in the nullspace of H(G4 , N). The sets W(yi ,ci ,si ) and W(y j ,c j ,s j ) do not
overlap on a diagram element for all distinct i, j ∈ [N]. Using Fact 10.5 we see this is equivalent to si 6= s j
/ E(Gocc ) for distinct i, j ∈ [N]. Hence the set of states (10.29) obtained from of all blocks of
and {si , s j } ∈
type 1 is
/ E(Gocc )
Sym(|φys11,c1 i|φys22,c2 i . . . |φysNN,cN i) : yi , ci ∈ {0, 1}, si ∈ [R], si 6= s j , {si , s j } ∈
which spans I4 .
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Type 2 If N is of type 2 then there exist x, b ∈ {0, 1} and r ∈ [R] such that N(x,b,r) ≥ 2. We show there
are no eigenvectors in the nullspace of (10.25) within a block of this type and we lower bound the smallest
eigenvalue within the block. Specifically, we show
N
(w) 1
min hκ| ∑ hE4 |κi > . (10.30)
|κi∈I♦ (N) w=1 (9R)6
First note that all |κi ∈ I♦ satisfy (A(G♦ ) − e1 )(w) |κi = 0 for each w ∈ [N], which can be seen using
the definition of I♦ and the fact that W spans the nullspace of A(G♦ ) − e1 . Using this fact and equation
(10.14), we get
N N (w)
(w)
min hκ| ∑ hE4 |κi = min hκ| ∑ A(G4 ) − e1 |κi . (10.31)
|κi∈I♦ (N) w=1 |κi∈I♦ (N) w=1
Now we use the operator inequality
!
N
(w) N
(w)
∑ A(G4 ) − e1 ≥ γ ∑ A(G4 ) − e1 · 1 − Π4
w=1 w=1
1
= γ(A(G4 ) − e1 ) · 1 − Π4 >1 − Π4
, (10.32)
(30R)2
(w)
where Π4 is the projector onto the nullspace of ∑Nw=1 A(G4 ) − e1 , and where in the last step we
used Lemma 10.3. Plugging equation (10.32) into equation (10.31) gives
N
(w) 1 4
min hκ| ∑ hE4 |κi > 1 − max hκ|Π |κi . (10.33)
|κi∈I♦ (N) w=1 (30R)2 |κi∈I♦ (N)
In the following we show that hκ|Π4 |κi = hκ|Π4 4
N |κi for all |κi ∈ I♦ (N), where ΠN is a Hermitian
operator with
3 1 4
4 3
1 − ΠN ≥ = . (10.34)
4 4R 1024R4
Plugging this into (10.33) gives
N
(w) 3 1
min hκ| ∑ hE4 |κi > > .
|κi∈I♦ (N) w=1 (30R)2 · 1024R4 (9R)6
To complete the proof, we exhibit the operator Π4
N and show that its norm is bounded as (10.34).
Using Lemma 10.3 we can write Π4 explicitly as
Π4 = ∑ P(~z,~a,~q) (10.35)
(~z,~
a,~
q)∈Q
where
P(~z,~a,~q) = |φzq11,a1 ihφzq11,a1 | ⊗ |φzq22,a2 ihφzq22,a2 | ⊗ · · · ⊗ |φzqNN,aN ihφzqNN,aN |
Q = {(z1 , . . . zN , a1 , . . . , aN , q1 , . . . , qN ) : zi , ai ∈ {0, 1} and qi ∈ [R]} .
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For each (~z, ~ q) ∈ Q we also define a space
a, ~
S(~z,~a,~q) = span(W(z1 ,a1 ,q1 ) ) ⊗ span(W(z2 ,a2 ,q2 ) ) ⊗ · · · ⊗ span(W(zN ,aN ,qN ) ) .
Note that P(~z,~a,~q) has all of its support in S(~z,~a,~q) , and that
S(~z,~a,~q) ⊥ S(~z0 ,~a0 ,~q0 ) for distinct (~z, ~ q), (~z0 , ~
a, ~ a0 , ~
q0 ) ∈ Q . (10.36)
Therefore P(~z,~a,~q) P(~z0 ,~a0 ,~q0 ) = 0 for distinct (~z, ~ q), (~z0 , ~
a, ~ q0 ) ∈ Q. (Below we use similar reasoning to
a0 , ~
obtain a less obvious result.) Note that P(~z,~a,~q) is orthogonal to I♦ (N) unless
j : (z j , a j , q j ) = (w, u, v) = N(w,u,v) for all w, u ∈ {0, 1}, v ∈ [R] . (10.37)
We restrict our attention to the projectors that are not orthogonal to I♦ (N). Letting Q(N) ⊂ Q be the set
of (~z, ~
a, ~
q) satisfying equation (10.37), we have
hκ| ∑ P(~z,~a,~q) |κi = hκ| ∑ P(~z,~a,~q) |κi for all |κi ∈ I♦ (N) . (10.38)
(~z,~
a,~
q)∈Q (~z,~
a,~
q)∈Q(N)
Since N(x,b,r) ≥ 2, note that in each term P(~z,~a,~q) with (~z, ~ q) ∈ Q(N), the operator
a, ~
r r r r
|φx,b ihφx,b | ⊗ |φx,b ihφx,b |
appears between two of the N registers (tensored with rank-1 projectors on the other N − 2 registers).
r i as a sum of states from W
Using equation (10.20) we may expand |φx,b (x,b,r) . This gives
rin rin 1
r
|φx,b r
i|φx,b i = c0 |ψx,b i|ψx,b i + 1 − c20 2 |Φrx,b i
rin rin
where c0 is either 1/(3R + 2) (if R is odd) or 1/(3R − 1) (if R is even), and where |ψx,b i|ψx,b i is orthogonal
r r r r r r
to |Φx,b i. Note that each of the states |φx,b i|φx,b i, |ψx,b i|ψx,b i, and |Φx,b i lie in the space
in in
span(W(x,b,r) ) ⊗ span(W(x,b,r) ) . (10.39)
Now applying Fact 10.7 gives
rin rin rin rin
r
|φx,b r
ihφx,b r
| ⊗ |φx,b r
ihφx,b | = c20 |ψx,b ihψx,b | ⊗ |ψx,b ihψx,b r
| + Mx,b (10.40)
r is a Hermitian operator with all of its support on the space (10.39) and
where Mx,b
4
3 3 1 3 1
r
Mx,b ≤ 1 − c40 ≤ 1 − ≤ 1− (10.41)
4 4 3R + 2 4 (4R)4
since R ≥ 2. For each (~z, ~ q) ∈ Q(N) we define PM
a, ~ (~z,~ q) to be the operator obtained from P(~z,~
a,~ q) by
a,~
replacing
r r r r r
|φx,b ihφx,b | ⊗ |φx,b ihφx,b | 7→ Mx,b
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on two of the registers (if N(x,b,r) > 2 there is more than one way to do this; we fix one choice for each
(~z, ~ q) ∈ Q(N)). Note that PM
a, ~ (~z,~ q) has all of its support in the space S(~z,~
a,~ q) . Using (10.36) gives
a,~
PM
(~z,~ q) P(~z0 ,~
a,~
M
q0 ) = 0 for distinct (~
a0 ,~ z, ~ q), (~z0 , ~
a, ~ a0 , ~
q0 ) ∈ Q(N) .
Using equation (10.40) and the fact that
rin rin (w1 ) rin rin (w2 )
hκ| |ψx,b ihψx,b | |ψx,b ihψx,b | |κi = 0 for all |κi ∈ I♦ (N) and distinct w1 , w2 ∈ [N]
(which can be seen from the definition of I♦ ), we have
hκ|P(~z,~a,~q) |κi = hκ|PM
(~z,~ q) |κi for all |κi ∈ I♦ (N) .
a,~
Hence, letting
Π4
N = ∑ PM
(~z,~ q) ,
a,~ (10.42)
(~z,~
a,~
q)∈Q(N)
we have hκ|Π4 |κi = hκ|Π4 N |κi for all |κi ∈ I♦ (N). To obtain the desired bound (10.34) on the norm of
4
ΠN , we use the fact that the norm of a sum of pairwise orthogonal Hermitian operators is upper bounded
by the maximum norm of an operator in the sum, so
3 1
Π4
N = ∑ PM
(~z,~ q) =
a,~ max PM
(~z,~
a,~
r
q) = Mx,b ≤ 1 − . (10.43)
(~z,~
a,~
q)∈Q(N) (~z,~
a,~
q)∈Q(N) 4 (4R)4
Type 3 If N is of type 3 then N(x,b,r) ∈ {0, 1} for all x, b ∈ {0, 1} and r ∈ [R], and
N(y,c,s) = N(t,d,u) = 1
for some (y, c, s) 6= (t, d, u) with either u = s or {u, s} ∈ E(Gocc ) (using property (b) and Fact 10.5). We
show there are no eigenvectors in the nullspace of (10.25) within a block of this type and we lower bound
the smallest eigenvalue within the block. We establish the same bound (10.30) as for blocks of Type 2.
The proof is very similar to that given above for blocks of Type 2. In fact, the first part of proof is
identical, from equation (10.31) up to and including equation (10.38). That is to say, as in the previous
case we have
hκ| ∑ P(~z,~a,~q) |κi = hκ| ∑ P(~z,~a,~q) |κi for all |κi ∈ I♦ (N) . (10.44)
(~z,~
a,~
q)∈Q (~z,~
a,~
q)∈Q(N)
In this case, since N(y,c,s) = N(t,d,u) = 1, in each term P(~z,~a,~q) with (~z, ~ q) ∈ Q(N), the operator
a, ~
s s u u
|φy,c ihφy,c | ⊗ |φt,d ihφt,d |
appears between two of the N registers (tensored with rank 1 projectors on the other N − 2 registers).
√
s i and |φ u i as superpositions (with amplitudes ±1/ 3R + 2
Using equation (10.20) we may expand |φy,c
√ t,d
if R is odd or ±1/ 3R − 1 if R is even) of the basis states from W(y,c,s) and W(t,d,u) respectively. Since
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
W(y,c,s) and W(t,d,u) overlap on some diagram element, there exists l ∈ L such that |ψxl 1 ,b1 i ∈ W(y,c,s)
and |ψxl 2 ,b2 i ∈ W(t,d,u) for some x1 , x2 , b1 , b2 ∈ {0, 1}. Hence
1
s
|φy,c u
i|φt,d i = c0 ±|ψxl 1 ,b1 i|ψxl 2 ,b2 i + 1 − c20 2 |Φs,u
y,c,t,d i
where c0 is either 1/(3R + 2) (if R is odd) or 1/(3R − 1) (if R is even). Now applying Fact 10.7 we get
s,u
s
|φy,c s
ihφy,c u
| ⊗ |φt,d u
ihφt,d | = c20 |ψxl 1 ,b1 ihψxl 1 ,b1 | ⊗ |ψxl 2 ,b2 ihψxl 2 ,b2 | + My,c,t,d (10.45)
where 4
s,u 3 1
kMy,c,t,d k ≤ 1− .
4 4R
For each (~z, ~ q) ∈ Q(N) we define PM
a, ~ (~z,~ q) to be the operator obtained from P(~z,~
a,~ q) by replacing
a,~
s s u u s,u
|φy,c ihφy,c | ⊗ |φt,d ihφt,d | 7→ My,c,t,d
on two of the registers and we let Π4 4
N be given by (10.42). Then, as in the previous case, hκ|Π |κi =
hκ|Π4 N |κi for all |κi ∈ I♦ (N) and using the same reasoning as before, we get the bound (10.34) on
4
kΠN k. Using these two facts we get the same bound on the smallest eigenvalue within a block of type 3
as the bound we obtained for blocks of type 2:
N
(w) 1 4
1
min hκ| ∑ hE4 |κi > 2
1 − ΠN > .
|κi∈I♦ (N) w=1 (30R) (9R)6
10.5 The gate graph G
We now consider the gate graph G and prove Lemma 5.5. We first show that G is an e1 -gate graph.
From equations (10.6), (10.7), and (10.8) we have
A(G ) = A(G4 ) + hE0 + hS0 . (10.46)
q
Lemma 10.3 characterizes the e1 -energy ground space of G4 and gives an orthonormal basis {|φz,a i : z, a ∈
{0, 1}, q ∈ [R]} for it. To solve for the e1 -energy ground space of A(G ), we solve for superpositions of
q
the states {|φz,a i} in the nullspace of hE0 + hS0 .
Recall the definition of the sets E0 and S0 . From Section 10.1, each node (q, z,t) in the gate diagram
for G is associated with a node new(q, z,t) in the gate diagram for G as described by (10.1). This
mapping is depicted in Figure 10.1 by the black and grey arrows. Applying this mapping to each pair of
nodes in the edge set EG and each node in the self-loop set SG of the gate diagram for G, we get the sets
E0 and S0 . Hence, using equations (5.5) and (5.6),
hS0 = ∑ |new(q, z,t)ihnew(q, z,t)| ⊗ 1 (10.47)
(q,z,t)∈SG
|new(q, z,t)i + |new(q0 , z0 ,t 0 )i hnew(q, z,t)| + hnew(q0 , z0 ,t 0 )| ⊗ 1 . (10.48)
hE 0 = ∑
{(q,z,t),(q0 ,z0 ,t 0 )}∈EG
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Using equation (10.20), we see that for all nodes (q, z,t) in the gate diagram for G and for all
j ∈ {0, . . . , 7}, x, b ∈ {0, 1}, and r ∈ [R],
(
rin
r √ hqin , z,t, j|ψx,b i if (q, z,t) is an input node
hnew(q, z,t), j|φx,b i = c0 rout
hqout , z,t, j|ψx,b i if (q, z,t) is an output node
√
= c0 δr,q hz,t, j|ψx,b i (10.49)
where c0 is 1/(3R + 2) if R is odd or 1/(3R − 1) if R is even, and where |ψx,b i is defined by equations
(5.2) and (5.3). The matrix element on the left-hand side of this equation is evaluated in the Hilbert
space Z1 (G ) where each basis vector corresponds to a vertex of the graph G ; these vertices are labeled
(l, z,t, j) with l ∈ L , z ∈ {0, 1}, t ∈ [8], and j ∈ {0, . . . , 7}. However, from (10.49) we see that
r √ r
hnew(q, z,t), j|φx,b i = c0 hq, z,t, j|ψx,b i (10.50)
| {z } | {z }
in Z1 (G ) in Z1 (G)
where the right-hand side is evaluated in the Hilbert space Z1 (G).
Putting together equations (10.47), (10.48), and (10.50) gives
(
1
q r q r 3R+2 R odd
hφz,a |hE0 + hS0 |φx,b i = hψz,a |hEG + hSG |ψx,b i· 1
(10.51)
3R−1 R even
for all z, a, x, b ∈ {0, 1} and q, r ∈ [R]. On the left-hand side of this equation, the Hilbert space is Z1 (G );
on the right-hand side it is Z1 (G).
We use equation (10.51) to relate the e1 -energy ground states of A(G) to those of A(G ). Since G is
an e1 -gate graph, there is a state
q
|Γi = ∑ αz,a,q |ψz,a i ∈ Z1 (G)
z,a,q
that satisfies A(G)|Γi = e1 |Γi and hence hEG |Γi = hSG |Γi = 0. Letting
|Γ0 i = ∑ αz,a,q |φz,a
q
i ∈ Z1 (G )
z,a,q
and using equation (10.51), we see that hΓ0 |hE0 + hS0 |Γ0 i = 0 and therefore hΓ0 |A(G )|Γ0 i = e1 . Hence
G is an e1 -gate graph. Moreover, the linear mapping from Z1 (G) to Z1 (G ) defined by
q q
|ψz,a i 7→ |φz,a i (10.52)
maps each e1 -energy eigenstate of A(G) to an e1 -energy eigenstate of A(G ).
Now consider the N-particle Hamiltonian H(G , N). Using equation (10.46) and the fact that both
A(G ) and A(G4 ) have smallest eigenvalue e1 , we have
N
H(G , N) = H(G4 , N) + ∑ (hE0 + hS0 )(w) .
w=1 ZN (G )
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Recall from Lemma 10.3 that the nullspace of the first term is I4 . The N-fold tensor product of the
mapping (10.52) acts on basis vectors of I(G, Gocc , N) as
Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqNN,aN i) 7→ Sym(|φzq11,a1 i|φzq22,a2 i . . . |φzqNN,aN i) , (10.53)
/ E(Gocc ). Clearly this defines an invertible linear map between
where zi , ai ∈ {0, 1}, qi 6= q j , and {qi , q j } ∈
the two spaces I(G, Gocc , N) and I4 . Let |Θi ∈ I(G, Gocc , N) and write |Θ0 i ∈ I4 for its image under the
map (10.53). Then
(
N N 1
0 0 0 (w) 0 (w) R odd
hΘ |H(G , N)|Θ i = hΘ | ∑ (hE0 + hS0 ) |Θ i = hΘ| ∑ (hEG + hSG ) |Θi· 3R+2 1
(10.54)
w=1 w=1 3R−1 R even
where in the first equality we used the fact that |Θ0 i is in the nullspace I4 of H(G4 , N) and in the second
q r i = hψ q |ψ r i. We now complete the proof
equality we used equation (10.51) and the fact that hφz,a |φx,b z,a x,b
of Lemma 5.5 using equation (10.54).
Case 1: λN (G, Gocc ) ≤ a.
In this case there exists a state |Θi ∈ I(G, Gocc , N) satisfying
N
hΘ| ∑ (hEG + hSG )(w) |Θi ≤ a .
w=1
From equation (10.54) we see that the state |Θ0 i ∈ I4 satisfies
a a
hΘ0 |H(G , N)|Θ0 i ≤ ≤ .
3R − 1 R
Case 2: λN (G, Gocc ) ≥ b.
In this case
N
λN (G, Gocc ) = min hΘ|H(G, Gocc , N)|Θi = min hΘ| ∑ (hEG + hSG )(w) |Θi ≥ b .
|Θi∈I(G,Gocc ,N) |Θi∈I(G,Gocc ,N) w=1
Now applying equation (10.54) gives
N
1
min hΘ0 |H(G , N)|Θ0 i = min hΘ0 | ∑ (hE0 + hS0 )(w) |Θ0 i ≥ b. (10.55)
0
|Θ i∈I4 0 |Θ i∈I4 w=1 3R + 2
This establishes that the nullspace of H(G , N) is empty, i. e., λN1 (G ) > 0, so λN1 (G ) = γ(H(G , N)).
We lower bound λN1 (G ) using the Nullspace Projection Lemma (Lemma 8.1) with
N
HA = H(G4 , N) , HB = ∑ (hE0 + hS0 )(w) ,
w=1 ZN (G )
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A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
and where the nullspace of HA is S = I4 . We apply Lemma 8.1 and use the bounds γ(HA ) > 1/(17R)7
(from Lemma 10.6), γ(HB |S ) ≥ b/(3R + 2) (from equation (10.55)), and kHB k ≤ N khE0 + hS0 k ≤ 3N ≤
3R (using equations (5.8) and (5.7) and the fact that N ≤ R) to find
λN1 (G ) = γ(H(G , N))
b
≥ 1 b
(3R + 2)(17R)7 ( (17R)7 + 3R+2 + 3R)
b 1
≥ ·
R9 3 + 2 + b · (17)7 + 3 · (3 + 2) (17)7
b
>
(13R)9
where in the denominator we used the fact that b ≤ 1.
Acknowledgments
This work was supported in part by NSERC; the Ontario Ministry of Research and Innovation; the Ontario
Ministry of Training, Colleges, and Universities; and the US ARO.
A Technical supporting material
A.1 Proof of the Nullspace Projection Lemma
The following proof fills in the details of the argument given in reference [25]. Recall that we write γ(M)
for the smallest nonzero eigenvalue of a positive semidefinite matrix M.
Lemma A.1 (Nullspace Projection Lemma [25]). Let HA , HB ≥ 0. Suppose the nullspace S of HA is
non-empty and
γ(HB |S ) ≥ c > 0 and γ(HA ) ≥ d > 0 .
Then
cd
γ(HA + HB ) ≥ . (8.1)
c + d + kHB k
Proof. Let |ψi be a normalized state satisfying
hψ|HA + HB |ψi = γ(HA + HB ) .
Let ΠS be the projector onto the nullspace of HA . First suppose that ΠS |ψi = 0, in which case
hψ|HA + HB |ψi ≥ hψ|HA |ψi ≥ γ(HA )
and the result follows. On the other hand, if ΠS |ψi 6= 0 then we can write
|ψi = α|ai + β |a⊥ i
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
with |α|2 + |β |2 = 1, α 6= 0, and two normalized states |ai and |a⊥ i such that |ai ∈ S and |a⊥ i ∈ S⊥ . (If
β = 0 then we may choose |a⊥ i to be an arbitrary state in S⊥ but in the following we fix one specific
choice for concreteness.) Note that any state |φ i in the nullspace of HA + HB satisfies HA |φ i = 0 and
hence hφ |a⊥ i = 0. Since hφ |ψi = 0 and α 6= 0 we also see that hφ |ai = 0. Hence any state
| f (q, r)i = q|ai + r|a⊥ i
is orthogonal to the nullspace of HA + HB , and
γ(HA + HB ) = min h f (q, r)|HA + HB | f (q, r)i .
|q|2 +|r|2 =1
The operator HA + HB acts on the two-dimensional space spanned by |ai and |a⊥ i as the matrix
w v∗
v y+z
where w = ha|HB |ai, v = ha⊥ |HB |ai, y = ha⊥ |HA |a⊥ i, and z = ha⊥ |HB |a⊥ i. The smallest eigenvalue of
this matrix is q
w + y + z − (w + y + z)2 + 4 (|v|2 − wy − wz)
γ(HA + HB ) = .
2
Since HB is positive semidefinite, its principal minors are nonnegative, and in particular wz − |v|2 ≥ 0.
Using this inequality in the above gives
s !
w+y+z 4wy wy 1
γ(HA + HB ) ≥ 1− 1− 2
≥ = 1 1 z
(A.1)
2 (w + y + z) w+y+z w + y + wy
√
where in the second step we used the fact that 1 − x ≤ 1 − x/2 for x ∈ [0, 1]. Since |ai is orthogonal to
the nullspace of HA + HB , we have
w ≥ γ(HB |S ) ≥ c .
We also have y ≥ γ(HA ) ≥ d and z ≤ kHB k. Since the right-hand side of (A.1) increases monotonically
with w and y and decreases monotonically with z, we find
cd
γ(HA + HB ) ≥
c + d + kHB k
as claimed.
A.2 Proof of Lemma 8.7
Here we prove Lemma 8.7. We begin with the following lemma relating the occupancy constraints
graph Gocc
X to the illegal configurations. The proof of this lemma uses the definitions of a configuration
(Definition 8.3), the sets of legal and illegal configurations (from Section 8.3), and the occupancy
constraints graph Gocc
X (from Section 7.3).
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Lemma A.2. For any illegal configuration
(J1 , . . . , JY , L1 , . . . , Ln−2Y ) (A.2)
there exist diagram elements {Q1 , Q2 } ∈ E(Gocc
X ) satisfying at least one of the following conditions:
(i) Q1 = (1, Jk , 0) and Q2 = (1, Jl , 0) for some k, l ∈ [Y ].
(ii) Y ∈ {0, 1}, Q1 = Ls , and Q2 = Lt for some s,t ∈ [n − 2Y ].
(iii) Y = 1 and Q1 = (i, J1 , d) and Q2 = Lt for some i ∈ {1, s(J1 )}, t ∈ [n − 2] and d ∈ {0, 1}.
Proof. We prove the contrapositive: we suppose the configuration (A.2) violates each of the conditions
(i), (ii), and (iii) for all {Q1 , Q2 } ∈ E(Gocc
X ) and show it is a legal configuration.
From part (1) of the definition of the occupancy constraints graph in Section 7.3, we see that
{(1, j, 0), (1, k, 0)} ∈ E(Gocc
X )
for all j, k ∈ [M] with j 6= k. If the configuration (A.2) has Y ≥ 2, then we may choose Q1 = (1, J1 , 0) and
Q2 = (1, J2 , 0) so that {Q1 , Q2 } ∈ E(GoccX ) and condition (i) is satisfied (note J1 6= J2 follows from the
definition of a configuration). Since by assumption, (A.2) violates condition (i) for all {Q1 , Q2 } ∈ E(Gocc
X ),
this implies that Y ∈ {0, 1}. We consider the cases Y = 0 and Y = 1 separately.
First suppose Y = 0, so (A.2) is equal to
(L1 , . . . , Ln ) .
/ E(Gocc
Since (ii) is violated, {Ls , Lt } ∈ occ
X ) for all s,t ∈ [n]. Using part (1) of the definition of GX and the
definition of a configuration, this implies that each diagram element is in a different row, i. e., Li = (i, ji , ci )
for each i ∈ [n]. From part (2) of the definition of Gocc / E(Gocc
X , we see in particular that {L1 , Lt } ∈ X ) for
each t ∈ {2, . . . , n} implies
Ls( j1 ) = (s( j1 ), j1 , c1 ) and Li = F(i, j1 , di )
for i ∈ [n] \ {1, s( j1 )} and bits d2 , . . . , dn ∈ {0, 1}, i. e., the configuration is legal.
Now suppose Y = 1, so (A.2) is
(J1 , L1 , . . . , Ln−2 ) .
Since (ii) is violated, each Li for i ∈ [n − 2] is from a different row. Since (iii) is violated, none of these
diagram elements are in rows 1 or s(J1 ). Now applying part (2) of the definition of Gocc X , we see that the
configuration is legal:
(J1 , L1 , . . . , Ln−2 )
= (J1 , F(2, J1 , d2 ), . . . , F(s(J1 ) − 1, J1 , ds(J1 )−1 ), F(s(J1 ) + 1, J1 , ds(J1 )+1 ), . . . , F(n, J1 , dn ))
where di ∈ {0, 1} for i ∈ [n] \ {1, s(J1 )}.
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
Lemma A.3. Let Π0 be the projector onto I(G1 , Gocc ~ z, ~
ai ∈ Blegal , we have
X , n). For any | j, d,~
~ ~z, ~
Π0 | j, d, ~ ~z, ~
ai = | j, d, ai. (8.20)
Furthermore, for any two distinct basis vectors |φ i, |ψi ∈ Billegal , we have
255
hφ |Π0 |φ i ≤ (8.21)
256
hφ |Π0 |ψi = 0 . (8.22)
Proof. We begin with equation (8.20). Recall from (5.21) that
I(G1 , Gocc q1 q2 qn
/ E(Gocc
X , n) = span{Sym(|ψz1 ,a1 i|ψz2 ,a2 i . . . |ψzn ,an i) : zi , ai ∈ {0, 1}, qi 6= q j , and {qi , q j } ∈ X )}
which can alternatively be characterized as the subspace of
I(G1 , n) = span{Sym(|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqnn,an i) : zi , ai ∈ {0, 1}, qi 6= q j }
consisting of zero eigenvectors of each of the operators
q q
|ψs,t ihψs,t r
| ⊗ |ψu,v r
ihψu,v | ⊗ 1⊗n−2 , {q, r} ∈ E(Gocc
X ), s,t, u, v ∈ {0, 1} . (A.3)
Now using equation (8.18) and the fact that
L̃ L 1 1 L̃ L
hψx,b |ρz,a i = √ δL̃,L δx,z δa,b = √ hρx,b |ρz,a i (A.4)
8 8
for all x, z, a, b ∈ {0, 1} and L̃, L ∈ L (from equations (6.13)–(6.16), and (6.36)), we get
~ ~z, ~ q q r r ~ ~z, ~
| ⊗ 1⊗n−2 | j, d,
h j, d, a| |ψs,t ihψs,t | ⊗ |ψu,v ihψu,v ai
1 ~ ~z, ~ q q ~ ~z, ~
r r
| ⊗ 1⊗n−2 | j, d,
= h j, d, a| |ρs,t ihρs,t | ⊗ |ρu,v ihρu,v ai
64
= 0 if {q, r} ∈ E(Gocc X ).
In the last line we used equations (8.12) and (8.18) and the definition of the occupancy constraints graph
~ z, ~
ai ∈ I(G1 , n) is a zero eigenvector of each of the
X from Section 7.3. Hence each legal state | j, d,~
Gocc
~ ~z, ~
operators (A.3), so | j, d, ai ∈ I(G1 , GX , n). This gives equation (8.20).
occ
Now we prove equation (8.21). For each illegal configuration we associate two diagram elements Q1
and Q2 with (Q1 , Q2 ) ∈ E(Gocc X ) as in Lemma A.2 (if there is more than one such pair we fix a specific
choice). Likewise for each basis vector |φ i ∈ Billegal we associate the two diagram elements Q1 and Q2
corresponding to its (illegal) configuration. Let Pφ be the projector onto the space
span{Sym(|ψzQ1 ,a
1
1
i|ψzQ2 ,a
2
2
i|ψzq33,a3 i . . . |ψzqnn,an i) : zi , ai ∈ {0, 1}, qi ∈
/ {Q1 , Q2 }}
where (exactly) one particle is located at Q1 and (exactly) one particle is located at Q2 . We show that
1
hφ |Pφ |φ i ≥ . (A.5)
256
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Note that Π0 Pφ = 0 since Π0 projects onto a subspace for which no two (or more) particles are simultane-
ously located at Q1 and Q2 . Therefore
hφ |Π0 |φ i + hφ |Pφ |φ i ≤ 1 ,
and applying (A.5) gives (8.20). Equation (A.5) can be shown by considering cases (i), (ii), and (iii) from
Lemma A.2. It is convenient to define
Q1 Q1 Q2 Q2
ΠQ1 = ∑ |ψx,y ihψx,y |, ΠQ2 = ∑ |ψx,y ihψx,y |.
x,y∈{0,1} x,y∈{0,1}
In case (i) we have Q1 = (1, Jk , 0) and Q2 = (1, Jl , 0) for some k, l ∈ [Y ]. Here we consider the case
k = 1, l = 2 without loss of generality. Then
Pφ |φ i = Sym Pφ (|TzJ11,a1 ,z2 ,a2 i|TzJ32,a3 ,z4 ,a4 i . . . |TzJ2YY −1 ,a2Y −1 ,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i)
= Sym (ΠQ1 ⊗ 1)|TzJ11,a1 ,z2 ,a2 i(ΠQ2 ⊗ 1)|TzJ32,a3 ,z4 ,a4 i . . . |TzJ2YY −1 ,a2Y −1 ,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i
(A.6)
for some configuration and some ~z, ~
a, where in the first line we used the fact that Pφ commutes with any
permutation of the n registers and in the second line we used the fact that
(w)
ΠQ1 |TzJ11,a1 ,z2 ,a2 i|TzJ32,a3 ,z4 ,a4 i . . . |TzJ2YY −1 ,a2Y −1 ,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i = 0 unless w = 1
(w)
ΠQ2 |TzJ11,a1 ,z2 ,a2 i|TzJ32,a3 ,z4 ,a4 i . . . |TzJ2YY −1 ,a2Y −1 ,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i = 0 unless w = 3.
We find
hφ |Pφ |φ i = hTzJ11,a1 ,z2 ,a2 | (ΠQ1 ⊗ 1) |TzJ11,a1 ,z2 ,a2 i · hTzJ32,a3 ,z4 ,a4 | (ΠQ2 ⊗ 1) |TzJ32,a3 ,z4 ,a4 i
2
1 (1,J1 ,0) (s(J1 ),J1 ,0) (1,J1 ,0) (s(J1 ),J1 ,0)
= hρz ,a |hρz2 ,a2 |ΠQ1 ⊗ 1|ρz1 ,a1 i|ρz2 ,a2 i
2 1 1
2
1 1
= = , (A.7)
16 256
where in the second line we used the fact that both terms in the product are equal and in the third line we
used equation (A.4).
In case (ii) we have Y ∈ {0, 1} and Q1 = Ls , Q2 = Lt for some s,t ∈ [n − 2Y ]. By a similar argument
as in (A.6),
1 1 1
hφ |Pφ |φ i = hρzLssas |ΠQ1 |ρzLss,as i · hρzLt tat |ΠQ2 |ρzLt t,at i = · = . (A.8)
8 8 64
In case (iii) we have Y = 1, Q1 = (i, J1 , d), and Q2 = Lt for some i ∈ {1, s(J1 )}, t ∈ [n − 2], and
d ∈ {0, 1}. If i = 1 then, again by a similar reasoning as in (A.6),
hφ |Pφ |φ i = hTzJ11,a1 ,z2 ,a2 |ΠQ1 ⊗ 1|TzJ11,a1 ,z2 ,a2 i · hρzL2Yt +t a2Y +t |ΠQ2 |ρzL2Yt +t a2Y +t i (A.9)
1 1 1
= · = . (A.10)
16 8 128
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If i = s(J1 ) then ΠQ1 ⊗ 1 should be replaced with 1 ⊗ ΠQ1 in (A.9) but the lower bound in (A.10) is the
same.
From equations (A.7), (A.8), and (A.10), we see that equation (A.5) holds in cases (i), (ii), and (iii),
respectively, thereby establishing (8.21).
Finally, we prove equation (8.22), showing that Π0 |span(Bn ) is diagonal in the basis Bn . Let
|φ i = Sym(|TzJ11,a1 ,z2 ,a2 i . . . |TzJ2YY −1 ,a2Y −1,z2Y ,a2Y i|ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i) (A.11)
˜ ˜ L̃
|ψi = Sym(|TxJ11,b1 ,x2 ,b2 i . . . |TxJ2K−1
K L̃1 n−2K
,b2K −1,x2K ,b2K i|ρx2K+1 ,b2K+1 i . . . |ρxn ,bn i) (A.12)
be distinct vectors from Bn (note it is possible that K = 0 or Y = 0 or both). Expand each of the |T i states
using equation (8.12), which we can also write as
J 1 1 (1,J,c) (s(J),J,c)
|Tz,a,y,b i = √ ∑ UJ (a)c |ρz,a i|ρy,b i
2 c=0
where we use the shorthand
(1,J,1) (s(J),J,1) (1,J,1) (s(J),J,1)
UJ (a)|ρz,a i|ρy,b i= ∑ UJ (a)x1 x2 ,zy |ρx1 ,a i|ρx2 ,b i. (A.13)
x1 ,x2 ∈{0,1}
For the state |φ i, this gives the expansion
1 Y
(J ,...,J ,L ,...,L ),(c ,...,c )
|φ i = √ ∑ Sym(|O~z,~a1 Y 1 n−2Y 1 Y i)
2 c1 ,...,cY ∈{0,1}
where
!
Y
(J ,...,J ,L ,...,L ),(c ,...,c ) O
ci (1,J ,c ) (s(J ),J ,c )
|O~z,~a1 Y 1 n−2Y 1 Y i = UJi (a2i−1 ) |ρz2i−1i ,ai2i−1 i|ρz2i ,ai2i i i i |ρzL2Y1 +1 ,a2Y +1 i . . . |ρzLnn−2Y
,an i .
i=1
(A.14)
Define the projector
PL1 = L
∑ ∑ |ψz,a L
ihψz,a |
z,a∈{0,1} L∈L
which has support only on diagram elements contained in L, and let PL0 = 1 − PL1 . Note that for each
L ∈ L and z, a ∈ {0, 1}, we can write
L
|ρz,a i = PL1 |ρz,a
L
i + PL0 |ρz,a
L
i (A.15)
where (from equation (A.4))
1
PL1 |ρz,a
L L
i = √ |ψz,a i.
8
L i are orthonormal, and similarly for the states |ψ L i, we get
Since the states |ρz,a z,a
(
1
L̃ L δz,x δa,b δL,L̃ α = 1
hρx,b |PLα |ρz,a i = 87 (A.16)
8 δz,x δa,b δL,L̃ α = 0.
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Inserting n copies of the identity PL1 + PL0 = 1 gives
1 Y
(J ,...,J ,L ,...,L ),(c ,...,c )
|φ i = √ ∑ ∑ Sym(PLα1 ⊗ · · · ⊗ PLαn |O~z,~a1 Y 1 n−2Y 1 Y i) . (A.17)
2 c1 ,...,cY ∈{0,1} α1 ,...,αn ∈{0,1}
Likewise for |ψi we get
1 K
β β (J˜ ,...,J˜ ,L̃ ,...,L̃ ),(e ,...,e )
|ψi = √ ∑ ∑ Sym(PL1 ⊗ · · · ⊗ PLn |O ~1 K 1 n−2K 1 K i) . (A.18)
2 e1 ,...,eK ∈{0,1} β1 ,...,βn ∈{0,1} ~
x,b
Using equations (A.4), (A.14), and (A.13), we see that the states
(J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY ) (J˜ ,...,J˜K ,L̃1 ,...,L̃n−2K ),(e1 ,...,eK )
|O~z,~a1 i and |O ~1 i
~
x,b
are orthogonal for any choice of bit strings c1 , . . . , cY and e1 , . . . , eK , since |φ i 6= |ψi implies that
a) 6= ((J˜1 , . . . , J˜K , L̃1 , . . . , L̃n−2K ),~x,~b) .
((J1 , . . . , JY , L1 , . . . , Ln−2Y ),~z, ~
Using equation (A.16), we have
(J˜ ,...,J˜K ,L̃1 ,...,L̃n−2K ),(e1 ,...,eK ) (J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY )
hO ~1 |PLα1 ⊗ · · · ⊗ PLαn |O~z,~a1 i
~
x,b
∑ni=1 αi n−∑ni=1 αi
1 7 (J˜ ,...,J˜ ,L̃ ,...,L̃ ),(e ,...,e ) (J ,...,J ,L ,...,L ),(c ,...,c )
= hO ~1 K 1 n−2K 1 K |O~z,~a1 Y 1 n−2Y 1 Y i ,
8 8 ~
x,b
so the states
(J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY )
PLα1 ⊗ · · · ⊗ PLαn |O~z,~a1 i (A.19)
and
β β (J˜ ,...,J˜K ,L̃1 ,...,L̃n−2K ),(e1 ,...,eK )
PL1 ⊗ · · · ⊗ PLn |O ~1 i (A.20)
~
x,b
are orthogonal for each choice of bit strings α1 , . . . , αn , β1 , . . . , βn , c1 , . . . , cY , and e1 , . . . , eK . Furthermore,
observe that
(J˜ ,...,J˜K ,L̃1 ,...,L̃n−2K ),(e1 ,...,eK ) β β (J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY )
Sym(hO ~1 |PL1 ⊗· · ·⊗PLn ) Sym(PLα1 ⊗· · ·⊗PLαn |O~z,~a1 i)
~
x,b
(J˜ ,...,J˜K ,L̃1 ,...,L̃n−2K ),(e1 ,...,eK ) β β (J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY )
= (hO ~1 |PL1 ⊗ · · · ⊗ PLn )(PLα1 ⊗ · · · ⊗ PLαn |O~z,~a1 i) .
~
x,b
Thus each symmetrized state in the sum (A.17) is orthogonal to each symmetrized state in the sum (A.18).
To complete the proof, we show that
Π0 |φ i
is a superposition of a subset of the states in the sum (A.17) and hence is orthogonal to |ψi. To see this,
first note that |φ i ∈ I(G1 , n) by Lemma 5.4 since it is in the nullspace of H(G1 , n) (by Lemma 8.5) and
G1 is an e1 -gate graph (by Lemma 8.2). Now comparing I(G1 , n) (defined in (5.15)) and I(G1 , Gocc X , n)
(defined in (5.21)), we see that
0 |Γi for all |Γi ∈ I(G1 , n)
Π0 |Γi = Πocc
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
where Πocc
0 projects onto the space
/ E(Gocc
span{|ψzq11,a1 i|ψzq22,a2 i . . . |ψzqnn,an i : zi , ai ∈ {0, 1}, qi ∈ [R], {qi , q j } ∈ X )} . (A.21)
In particular, Π0 |φ i = Πocc
0 |φ i. We claim that this quantity is a superposition of a subset of the states in
the sum (A.17).
The diagram elements q1 , . . . , qn appearing in (A.21) range over the set of all R diagram elements in
the gate graph G1 ; however, recall that E(Gocc X ) only contains edges between diagram elements in the
subset L of these diagram elements. Since the PLα either project onto this set of diagram elements or onto
the complement, each state
(J ,...,JY ,L1 ,...,Ln−2Y ),(c1 ,...,cY )
Sym(PLα1 ⊗ · · · ⊗ PLαn |O~z,~a1 i)
is an eigenvector of Πocc
0 . Hence Π0 |φ i is a superposition of the terms in (A.17) that are +1 eigenvectors
of Πocc
0 (as the 0 eigenvectors are annihilated). It follows that hψ|Π0 |φ i = 0 since we established above
that each such term is orthogonal to |ψi.
A.3 Computation of matrix elements between states with legal configurations
In this section we compute the matrix elements of
H1 S , H2 S , Hin,i S , Hout S (A.22)
1 1 1 1
in the basis of S1 consisting of the states
~ In(~z), ~ ~~
h~x|Ū j,d1 (a1 )|~zi | j, d, x, ~
| j, d, ai = ∑ ai (A.23)
~
x∈{0,1}n
for ~z, ~
a ∈ {0, 1}, j ∈ [M], and
d~ = (d1 , . . . , dn ) with d1 = ds( j) ∈ {0, 1, 2} and di ∈ {0, 1} i ∈
/ {1, s( j)} ,
where (
U j−1 (a1 )U j−2 (a1 ) . . .U1 (a1 ) if d1 ∈ {0, 2}
Ū j,d1 (a1 ) = (A.24)
U j (a1 )U j−1 (a1 ) . . .U1 (a1 ) if d1 = 1.
Specifically, we prove the results stated in the four boxes in Section 8.4.
A.3.1 Matrix elements of H1
We begin by computing the matrix elements of
n
(w)
H1 = ∑ h1
w=1
in the basis Blegal ; then we use them to compute the matrix elements of H1 in the basis (A.23).
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Note that since H1 is symmetric under permutations of the n registers,
n
(1, j,d1 ) N F(i, j,di )
Sym H1 |ρz 1 ,a 1 i |ρ zi ,ai i d1 = ds( j) ∈ {0, 1}
i=2
~ ~z, ~
!
H1 | j, d, ai = n
F(i, j,d ) (A.25)
Sym H1 |Tz1j ,a1 ,zs( j) ,as( j) i |ρzi ,ai i i
N
d1 = ds( j) = 2
i=2
i6=s( j)
and recall that
1 (1, j,0) (s( j), j,0) 1 (1, j,1) (s( j), j,1)
|Tz1j ,a1 ,zs( j) ,as( j) i = √ |ρz1 ,a1 i|ρzs( j) ,as( j) i + √ ∑ U j (a1 )x1 x2 ,z1 zs( j) |ρx1 ,a1 i|ρx2 ,as( j) i . (A.26)
2 2 x1 ,x2 ∈{0,1}
To compute hk,~c,~x,~b|H1 | j, d,
~ ~z, ~
ai, we first evaluate the matrix elements of h1 between single-particle
states of the form
(1, j,d) (s( j), j,d) F(i, j,d)
|ρz,a i, |ρz,a i, |ρz,a i
(for j ∈ [M], i ∈ {2, . . . , n}, and z, a, d ∈ {0, 1}) that appear in equation (A.25). To evaluate these matrix
elements, we use the fact that h1 is of the form (5.6), where E is the set of edges in rows 2, . . . , n that are
added to the gate diagram in step 3 of Section 7.2.
We have
F(i, j,0) F(i, j,0) 1 F(i, j,0) F(i, j,0) 1
hρx,b |h1 |ρz,a i = hψx,b |h1 |ψz,a i = δx,z δa,b (A.27)
8 64
F(i, j,1) F(i, j,1) 1
hρx,b |h1 |ρz,a i = δx,z δa,b (A.28)
64
for all i ∈ {2, . . . , n}, j ∈ [M], and x, z, a, b ∈ {0, 1}. Similarly,
F(i, j,0) F(i, j,1) F(i, j,1) F(i, j,0) 1
hρx,b |h1 |ρz,a i = hρx,b |h1 |ρz,a i= δx,z δa,b (A.29)
64
for all i ∈ [n] \ {1, s( j)}, j ∈ [M], and z, x, a, b ∈ {0, 1}. Furthermore,
(1, j,d)
h1 |ρz,a i=0 (A.30)
for all j ∈ [M] and z, a, d ∈ {0, 1}, and
F(s( j), j,c) 1
(s( j), j,d)
hρx,b |h1 |ρz,a i=
δx,z δa,b δc,d (A.31)
64
(s( j), j,c) (s( j), j,d) 1
hρx,b |h1 |ρz,a i = δx,z δa,b δc,d (A.32)
64
for all j ∈ [M] and z, x, a, b, c, d ∈ {0, 1}.
Using equations (A.27), (A.28), (A.30), and (A.32), we compute the diagonal matrix elements of H1 :
n F(i, j,di ) F(i, j,d )
∑i=2 hρzi ,ai
|h1 |ρzi ,ai i i d1 ∈ {0, 1}
~
h j, d,~z, ~ ~
a|H1 | j, d,~z, ~
n
ai = hTz ,a ,z ,a |1 ⊗ h1 |Tz ,a ,z ,a i + ∑ hρz ,a i |h1 |ρz ,a i i d1 = 2
j j F(i, j,d ) F(i, j,d )
1 1 s( j) s( j)
1 1 s( j) s( j)
i=2
i i i i
i6=s( j)
n−1
= (A.33)
64
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
where in the last line we used equation (A.26) and the fact that U j (a1 ) is unitary. We use equations (A.29)
and (A.31) to compute the nonzero off-diagonal matrix elements of H1 between states in Blegal . We get
c = d~
1
64 δ~x,~z (n − 1) ~
n
1 δ ∏δ
∃i ∈ [n] \ {1, s( j)}, ci 6= di
64 ~x,~z cr ,dr
r=1
r6=i
n
1
√ δ~x,~z ∏ δc ,d (c1 , d1 ) ∈ {(2, 0), (0, 2)}
r r
64 2
r=2
hk,~c,~x,~b|H1 | j, d,
~ ~z, ~ r6=s( j)
ai = δ j,k δ~a,~b · n
√ U j (a1 )∗
1
∏ δcr ,dr δxr ,zr (c1 , d1 ) = (2, 1)
64 2 z1 zs( j) ,x1 xs( j)
r=2
r6=s( j)
n
1
√ U j (a1 )x x ,z z
64 2 1 s( j) 1 s( j) ∏ δcr ,dr δxr ,zr (c1 , d1 ) = (1, 2)
r=2
r6=s( j)
0 otherwise.
(A.34)
For the second case we used equation (A.29), for the third case we used equation (A.31) to get
(1, j,0) F(s( j), j,0) 1
hTxj1 ,b1 ,xs( j) ,bs( j) |1 ⊗ h1 |ρz1 ,a1 i|ρzs( j) as( j) i = δx1 ,z1 δb1 ,a1 δxs( j) ,zs( j) δbs( j) ,as( j) √ ,
64 2
and for the fourth and fifth cases we used equation (A.31) to get
(1, j,1) F(s( j), j,1) 1
hTxj1 ,b1 ,xs( j) ,bs( j) |1 ⊗ h1 |ρz1 ,a1 i|ρzs( j) as( j) i = δb1 ,a1 δbs( j) ,as( j) √ U j (a1 )∗z1 zs( j) ,x1 xs( j) .
64 2
In the remaining case, (c1 , d1 ) ∈ {(1, 0), (0, 1)} and the matrix element is 0.
We now compute the matrix elements of H1 in the basis (A.23). We have
h j,~c, In(~x), ~ ~ In(~z), ~
a|H1 | j, d, ai = ∑ h j,~c,~x0 , ~ ~ ~z0 , ~
a|H1 | j, d, aih~x|Ū j,d1 (a1 )† |~x0 ih~z0 |Ū j,d1 (a1 )|~zi.
~
x0 ,~z0 ∈{0,1}n
(A.35)
Using this with (A.34) gives
c = d~
n−1
64 ~
n
1
∃i ∈ [n] \ {1, s( j)}, ci 6= di
64 ∏ δcr ,dr
r=1
hk,~c, In(~x),~b|H1 | j, d,
~ In(~z), ~ r6=i
ai = δ j,k δ~a,~b δ~x,~z · n
1
√ ∏ δc ,d
64 2 r=2 r r (c1 , d1 ) ∈ {(2, 0), (0, 2), (2, 1), (1, 2)}
r6=s( j)
0 otherwise
as claimed in equation (8.30). Note that in the basis Blegal , H1 has nonzero matrix elements between
states with different values of ~z; the basis (A.23) is convenient because H1 only connects basis states with
the same value of ~z.
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A.3.2 Matrix elements of H2
Recall that
n
(w)
H2 = ∑ h2
w=1
and note, just as in (A.25), that because H2 is permutation invariant,
n
(1, j,d1 ) N F(i, j,di )
Sym H2 |ρz 1 ,a 1 i |ρzi ,ai i d1 = ds( j) ∈ {0, 1}
i=2
~ ~z, ~
!
H2 | j, d, ai = n
F(i, j,d ) (A.36)
Sym H2 |Tz1j ,a1 ,zs( j) ,as( j) i |ρzi ,ai i i
N
d1 = ds( j) = 2.
i=2
i6=s( j)
Also recall that h2 is of the form (5.6), where E is the set of edges in row 1 that are added in step 3 of
Section 7.2.
To compute hk,~c,~x,~b|H2 | j, d, ~ ~z, ~
ai we first evaluate the matrix elements of h2 between the relevant
single-particle states |ρz,a i with L ∈ L and z, a ∈ {0, 1}. The only such matrix elements that are nonzero
L
are
(
(1, j,0) (1, j,0) 0 j=1
hρx,b |h2 |ρz,a i = 1
64 δz,x δa,b j ∈ {2, . . . , M}
( (A.37)
1
(1, j,1) (1, j,1) δz,x δa,b j ∈ {1, . . . , M − 1}
hρx,b |h2 |ρz,a i = 64
0 j=M
for z, a, x, b ∈ {0, 1} and
(1, j−1,1) (1, j,0) (1, j,0) (1, j−1,1) 1
hρx,b |h2 |ρz,a i = hρz,a |h2 |ρx,b i= δz,x δa,b (A.38)
64
for j ∈ {2, . . . , M} and z, x, a, b ∈ {0, 1}.
Using equations (A.36) and (A.37), we compute the diagonal matrix elements of H2 in the basis
Blegal :
0
d1 = 0 and j = 1, or d1 = 1 and j = M
~
h j, d,~z, ~ ~
a|H2 | j, d,~z, ~ 1
ai = 128 d1 = 2 and j ∈ {1, M} (A.39)
1
64 otherwise.
Using equations (A.36) and (A.38), we compute the nonzero off-diagonal matrix elements, which are all
of the form
~ ~z, ~ ~ ~z, ~ ~ ~z, ~
∗
h j − 1,~c,~x, ~
a|H2 | j, d, ai or h j, d, a|H2 | j − 1,~c,~x, ~
ai = h j − 1,~c,~x, ~
a|H2 | j, d, ai
for j ∈ {2, . . . , M}, ~x,~z, ~
a ∈ {0, 1}n , and
d~ = (d1 , . . . , dn ) with d1 = ds( j) ∈ {0, 1, 2} and di ∈ {0, 1} for all i ∈
/ {1, s( j)} .
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
We get
h j − 1,~c,~x, ~ ~ ~z, ~
a|H2 | j, d, ai
1
64 δ~x,~z (c1 , cs( j) , d1 , ds( j−1) ) = (1, 0, 0, 1)
n
1√
U j−1 (a1 )∗z1 zs( j−1) ,x1 xs( j−1) ∏ δxr ,zr (c1 , cs( j) , d1 , ds( j−1) ) = (2, 0, 0, 0)
64 2
n
r=2
r6=s( j−1)
= ∏ δdr ,cr 1 n
r=2 128 U j−1 (a1 )∗z1 zs( j−1) ,x1 xs( j−1) ∏ δxr ,zr (c1 , cs( j) , d1 , ds( j−1) ) = (2, 1, 2, 0)
r∈{s(
/ j),s( j−1)}
r=2
r6=s( j−1)
1√ δ
(c1 , cs( j) , d1 , ds( j−1) ) = (1, 1, 2, 1).
64 2 ~x,~z
(A.40)
Now we compute the diagonal matrix elements of H2 in the basis (A.23) using equations (A.35) and
(A.39):
0
d1 = 0 and j = 1, or d1 = 1 and j = M
~
h j, d, In(~z), ~ ~
a|H2 | j, d, In(~z), ~ 1
ai = 128 d1 = 2 and j ∈ {1, M} (A.41)
1
64 otherwise.
The nonzero off-diagonal matrix elements are (using equations (A.35) and (A.40))
h j − 1,~c, In(~x),~b|H2 | j, d,
~ In(~z), ~ ~ In(~z), ~
ai = h j, d, a|H2 | j − 1,~c, In(~x),~bi
1
64 (c1 , cs( j) , d1 , ds( j−1) ) = (1, 0, 0, 1)
n 1√ (c1 , c , d1 , d
s( j−1) ) = (2, 0, 0, 0)
s( j)
= δ~x,~z δ~a,~b ∏ δdr ,cr · 641 2 (c , c , d , d (A.42)
r=2 128 1 s( j) 1 s( j−1) ) = (2, 1, 2, 0)
r∈{s(
/ j),s( j−1)}
1√ (c , c , d , d
) = (1, 1, 2, 1).
64 2 1 s( j) 1 s( j−1)
Combining equations (A.41) and (A.42) gives the result claimed in equations (8.31), (8.32), and (8.33).
A.3.3 Matrix elements of Hin,i
We now consider
n
(w)
Hin,i = ∑ hin,i
w=1
where i is from the set of indices of the ancilla qubits, i. e., i ∈ {nin + 1, . . . , n}. Using equation (8.3) we
get
L2 L1 1 L2 L1 1
hρx,b |hin,i |ρz,a i = hψx,b |hin,i |ψz,a i = δx,1 δz,1 δa,b δL1 ,(i,0,1) δL2 ,(i,0,1)
8 64
for L1 , L2 ∈ L and a, b, x, z ∈ {0, 1}. Thus Hin,i is diagonal in the basis Blegal with entries
(
1
~ ~z, ~ ~ ~z, ~ di = 0, zi = 1, and F(i, j, 0) = (i, 0, 1)
h j, d, a|Hin,i | j, d, ai = 64 (A.43)
0 otherwise.
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Note that F(i, j, 0) = (i, 0, 1) if and only if none of the gates U1 ,U2 , . . . ,U j−1 acts on the i-th qubit, i. e.,
j ≤ jmin,i
where
jmin,i = min{ j ∈ [M] : s( j) = i} .
Now using this fact and equations (A.35) and (A.43), we get the following expression for the nonzero
matrix elements of Hin,i in the basis (A.23):
~ In(~
h j, d, x), ~ ~ In(~z), ~
a|Hin,i | j, d, ai
= ∑ †
h~x|Ū j,d ~ ih~
(a1 )|w ~w
y|Ū j,d1 (a1 )|~zih j, d, ~ ,~ ~~
a|Hin,i | j, d, y, ~
ai
1
~ ,~
w y∈{0,1}n
†
∑ h~x|Ū j,d1
(a1 )|~yih~y|Ū j,d1 (a1 )|~zi 64
1
δyi ,1 j ≤ jmin,i and di = 0
= ~
y∈{0,1}n
0 otherwise
(
†
h~x|Ū j,d (a1 )|1ih1|iŪ j,d1 (a1 )|~zi 64
1
j ≤ jmin,i and di = 0
= 1
0 otherwise
† † ( j < jmin,i and di = 0 and d1 ∈ {0, 2})
h~x|U1 (a1 ) . . .U j−1 (a1 )|1ih1|iU j−1 (a1 ) . . .U1 (a1 )|~zi 64
1
or ( j = jmin,i and di = 0)
= † †
h~x|U1 (a1 ) . . .U j (a1 )|1ih1|iU j (a1 ) . . .U1 (a1 )|~zi 64
1
j < jmin,i and di = 0 and d1 = 1
0 otherwise.
In the last line we use the fact that d1 = di when j = jmin,i (since s( jmin,i ) = i). Since [UJ (a1 ), |1ih1|i ] = 0
for J < jmin,i , we have
(
~ In(~ ~ In(~z), ~
1
h~x| (|1ih1|i ) |~zi j ≤ jmin,i and di = 0
h j, d, x), ~
a|Hin,i | j, d, ai = 64 (A.44)
0 otherwise
(
1
δ~x,~z δxi ,1 j ≤ jmin,i and di = 0
= 64 (A.45)
0 otherwise
(with all other matrix elements equal to zero), which confirms the result stated in equation (8.34).
A.3.4 Matrix elements of Hout
Finally, consider
n
(w)
Hout = ∑ hout
w=1
where (from equation (8.3))
L2 L1 1
hρx,b |hout |ρz,a i= δa,b δx,0 δz,0 δL1 ,(2,M+1,0) δL2 ,(2,M+1,0)
64
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T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
for L1 , L2 ∈ L and z, a, x, b ∈ {0, 1}. From this we see that Hout is diagonal in the basis Blegal , with entries
(
1
~ ~z, ~ ~ ~z, ~ 64 d2 = 1, F(2, j, 1) = (2, M + 1, 0) and z2 = 0
h j, d, a|Hout | j, d, ai =
0 otherwise.
Note that F(2, j, 1) = (2, M + 1, 0) if and only if j ≥ jmax , where
jmax = max{ j ∈ [M] : s( j) = 2} .
Using this fact we compute the nonzero matrix elements of Hout in the basis (A.23):
~ In(~
h j, d, x), ~ ~ In(~z), ~
a|Hout | j, d, ai
= ∑ †
h~x|Ū j,d ~ ih~
(a1 )|w ~w
y|Ū j,d1 (a1 )|~zih j, d, ~ ,~ ~~
a|Hout | j, d, y, ~
ai
1
~ ,~
w y∈{0,1}n
(
†
h~x|Ū j,d (a1 )|0ih0|2Ū j,d1 (a1 )|~zi 64
1
j ≥ jmax and d2 = 1
= 1
0 otherwise
† †
h~x|U1 (a1 ) . . .U j (a1 )|0ih0|2U j (a1 ) . . .U1 (a1 )|~zi 64
1
j ≥ jmax and d1 = d2 = 1
† †
= h~x|U1 (a1 ) . . .U j−1 (a1 )|0ih0|2U j−1 (a1 ) . . .U1 (a1 )|~zi 64 j > jmax and d2 = 1 and d1 ∈ {0, 2}
1
0 otherwise
(
† †
h~x|U1 (a1 ) . . .UM (a1 )|0ih0|2UM (a1 ) . . .U1 (a1 )|~zi 64 1
j ≥ jmax and d2 = 1
=
0 otherwise
(
h~x|UC†X (a1 )|0ih0|2UCX (a1 )|~zi 64
1
j ≥ jmax and d2 = 1
=
0 otherwise.
In going from the second to the third equality we use the fact that j = jmax implies d1 = d2 (since
s( jmax ) = 2). In the next-to-last line we use the fact that [UJ (a1 ), |0ih0|2 ] = 0 for J > jmax . This confirms
the result stated in equation (8.35).
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T HEORY OF C OMPUTING, Volume 11 (20), 2015, pp. 491–603 601
A NDREW M. C HILDS , DAVID G OSSET, AND Z AK W EBB
AUTHORS
Andrew M. Childs
Department of Combinatorics and Optimization
and
Institute for Quantum Computing
University of Waterloo
amchilds umd edu
https://cs.umd.edu/~amchilds/
David Gosset
Department of Combinatorics and Optimization
and
Institute for Quantum Computing
University of Waterloo
dngosset gmail com
http://www.davidgosset.com
Zak Webb
Department of Physics & Astronomy
and
Institute for Quantum Computing
University of Waterloo
zakwwebb gmail com
http://zakwebb.me
ABOUT THE AUTHORS
A NDREW C HILDS received his Ph. D. in Physics from MIT in 2004, advised by Edward
Farhi. His research focuses on quantum algorithms, including quantum walks on graphs,
quantum simulation, quantum algorithms for algebraic problems, and quantum query
complexity. From 2007 to 2014 he worked at the University of Waterloo, Department
of Combinatorics and Optimization and the Institute for Quantum Computing. He is
now at the University of Maryland, where he co-directs the Joint Center for Quantum
Information and Computer Science.
T HEORY OF C OMPUTING, Volume 11 (20), 2015, pp. 491–603 602
T HE B OSE -H UBBARD M ODEL IS QMA- COMPLETE
DAVID G OSSET completed his Ph. D. in 2011, under the supervision of Edward Farhi at
MIT. His research interests include quantum algorithms, quantum walks, Hamiltonian
complexity, quantum circuit synthesis, and other areas in the theory of quantum compu-
tation. He is currently a postdoctoral fellow at the Institute for Quantum Information and
Matter and the Burke Institute for Theoretical Physics at Caltech.
Z AK W EBB is a graduate student in physics at the Institute for Quantum Computing at
the University of Waterloo, advised by Andrew Childs. His research interests include
quantum walks on graphs and quantum Hamiltonian complexity.
T HEORY OF C OMPUTING, Volume 11 (20), 2015, pp. 491–603 603