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arXiv:1010.2480v1 [quant-ph] 12 Oct 2010

Complete Characterization of the Ground Space Structure of

Two-Body Frustration-Free Hamiltonians for Qubits

Zhengfeng Ji,1,2Zhaohui Wei,3and Bei Zeng4,5

1Perimeter Institute for Theoretical Physics, Waterloo, Ontario, Canada

2State Key Laboratory of Computer Science, Institute of Software, Chinese Academy of Sciences, Beijing, China

3Centre for Quantum Technologies, National University of Singapore, Singapore 117543, Singapore

4Department of Mathematics & Statistics, University of Guelph, Guelph, Ontario, Canada

5Institute for Quantum Computing, University of Waterloo, Waterloo, Ontario, Canada

(Dated: Oct. 12, 2010)

The problem of finding the ground state of a frustration-free Hamiltonian carrying only two-body

interactions between qubits is known to be solvable in polynomial time. It is also shown recently

that, for any such Hamiltonian, there is always a ground state that is a product of single- or two-qubit

states. However, it remains unclear whether the whole ground space is of any succinct structure.

Here, we give a complete characterization of the ground space of any two-body frustration-free

Hamiltonian of qubits. Namely, it is a span of tree tensor network states of the same tree structure.

This characterization allows us to show that the problem of determining the ground state degeneracy

is as hard as, but no harder than, its classical analog.

PACS numbers: 03.67.Lx, 03.67.Mn, 75.10.Jm

Quantum spin models are simplified physical models

for real materials, but are believed to capture some of

their key physical properties, which lie in the heart of

modern condensed matter theory [1]. Ground states of

strongly correlated spin systems is usually highly entan-

gled, even if the system Hamiltonian carries only local

interactions. So in general, finding the ground state of

such a system is intractable with traditional techniques,

such as mean field theory.

In practical spin systems, different local terms in the

Hamiltonian might also compete with each other, a phe-

nomenon called frustration, which makes the system fur-

ther difficult to analyze [2]. However, frustration is not

a necessary factor to cause ground state entanglement.

Frustration-free Hamiltonians can carry lots of interest-

ing physics, ranging from gapped spin chains [3] to topo-

logical orders [4, 5].

During recently years, the active frontier of quantum

information science brings new tools to study quantum

spin systems.In particular, local Hamiltonian prob-

lems are shown to be in general very hard, i.e., QMA-

complete [6].

It is also realized that the study of k-

local frustration-free Hamiltonians for qubits is closely

related to the quantum k-satisfiability problem (Q-k-

SAT) [7], which is the quantum analogy of the classical

k-satisfiability (k-SAT), a problem that is of fundamental

importance and has been extensively studied in theoret-

ical computer science (see, e.g., [8]).

Spin models with two-body interaction are of the most

physical relevance, as two-body interaction, in particular

of nearest neighbor or next nearest neighbors on certain

type of lattices, are the strongest interaction terms in the

real system Hamiltonian. Because two-level systems are

most common in nature, spin-1/2 (qubit) systems are of

particular importance.

It is realized, however, that certain ground states of a

two-body frustration-free (2BFF) Hamiltonian of qubits

could be pretty trivial with almost no entanglement at

all. Algorithmically, the problem of finding the ground

state of a 2BFF Hamiltonian of qubits is known to be

solvable in polynomial time [7]. It is also shown recently

that for any such Hamiltonian, there is always a ground

state that is a product of single- or two-qubit states; and

if there is a genuine entangled ground state, the ground

space must be degenerate [9]. There are also similar ob-

servations of the ground states in random or generic in-

stances [10–13], saying that the entire ground space is of

a trivial structure, which is almost always the fully sym-

metric space, with ground space degeneracy n+1, where

n is the number of qubits [10, 11, 14].

The main purpose of this work is to characterize the

entire ground space in the most general setting. We im-

prove the understanding of the ground space of 2BFF

Hamiltonians of qubits by showing that it is always a

span of tree tensor network states of the same tree struc-

ture. In other words, these states are generated, from

products of single qubit states, by the same series of

isometries (from single qubit to two qubits).

characterization holds for the most general case, it im-

plies that computing the degeneracy of 2BFF Hamilto-

nian (#Q-2-SAT) is in a complexity class called #P [15].

On the other hand, the classical analog #2-SAT of #Q-2-

SAT is #P-hard, therefore #P-complete.This answered

a question raised in [11].

As this

Two-body frustration-free Hamiltonian.— Consider a

system of n qubits labeled by the set V = {1,2,...,n}.

We will be interested in 2BFF Hamiltonians H =?HJ

of the system. The Hamiltonian is called two-body if each

term HJacts non-trivially only on two qubits. The index

J indicates the two qubits on which HJacts. The Hamil-

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tonian H is called frustration-free if its ground state also

minimizes the energy of each term HJ simultaneously.

Without loss of generality, we can assume throughout

the paper that the smallest eigenvalue of each term HJ

is zero by shifting the energy spectrum. In this conven-

tion, the frustration-free Hamiltonian H itself will have

zero ground energy. Specifically, we have

K(H) =

??K(HJ) ⊗ H ¯ J

?,

(1)

where K(H) is the ground space of H and H ¯ Jis the

Hilbert space of the qubits not in J. From this equation,

one easily sees that it is the ground space of each term

HJ, not the structures of excited states, that matters for

the ground space of a frustration-free Hamiltonian H.

In other words, it suffices to consider local terms to be

projections ΠJ for our purpose.

Closely related to the analysis of 2BFF qubit Hamil-

tonians is the quantum 2-SAT problem (Q-2-SAT) first

considered by Bravyi [7]. Naturally generalizing classical

2-SAT, the Q-2-SAT problem asks whether, for a given

set of two-qubit projections {ΠJ} of an n-qubit system,

there is a global state |Ψ? such that ΠJ|Ψ? = 0 for all J.

Apparently, we answer “yes” to the problem if and only if

the Hamiltonian?ΠJ is frustration-free. It was known

that Q-2-SAT is decidable in polynomial time on a clas-

sical computer [7]. The proof of the statement actually

constructs a specific n-qubit state |Ψ? in the ground space

of?ΠJif there is any. Our techniques will be similar to

those used by Bravyi, but we will show a stronger result

that one can not only find one state in the ground space,

but also represent the entire ground space in terms of a

span of special states.

A case study of the rank.— Given a 2BFF Hamiltonian

H =?HJ, what can we say about the ground space

K(H)? First of all, as argued previously, we only need to

consider Hamiltonians of the form H =?ΠJwhere ΠJ’s

are projections onto K(HJ)⊥. We will start our analysis

by considering the rank of the projections ΠJ.

First, if there is a ΠJof rank 3, the only possible state

for the two qubits in J is I − ΠJ of rank 1, and this

reduces to a problem on qubits in V \ J.

If there is a ΠJ of rank 2, the state of qubits in J

is restricted to a two-dimensional subspace. Let |ψ0?a,b

and |ψ1?a,bbe two orthogonal states that span the sub-

space, where a,b are the two qubits in J. One can en-

code qubits a and b by a single qubit d. For this pur-

pose, we define an isometry U in the following form

U : |0?d?→ |ψ0?a,b, |1?d?→ |ψ1?a,b. This procedure pro-

duces a set of constraints on n − 1 qubits. It is easy to

verify that a state |Ψ? is in the ground space of the re-

duced problem if and only if U|Ψ? is in the ground space

of the original problem [7, 9].

When there is no projection of rank larger than 1, we

are dealing with the homogeneous case [7]. It turns out

that the homogeneous case is the hardest and we will dis-

cuss it two separate sections. As we will see, the ground

space of the homogeneous Hamiltonian (more precisely,

the simplified homogeneous Hamiltonian defined later) is

spanned by single-qubit product states. The above case

analysis gives an explicit representation of the ground

space of a general 2BFF qubit Hamiltonian, which is

given by the following

Main Observation — The ground space is always a

span of tree tensor network states of the same tree struc-

ture.

We illustrate this observation in Fig. 1, where the

ground space is viewed as a span of states generated

by the isometries (blue triangles) organized in a forest

form (a collection of trees) acting on product states (in-

put from the left). In the language of tensor network

states [16, 17], one can also represent these states in terms

of tree tensor network after combining the input product

states and the roots of trees in the forest.

Input:

statesthatspanstheground

space of the simplified ho-

mogenous Hamiltonian.

a set of product

FIG. 1: The general structure of the ground space

Homogeneous case with product constraints.— Con-

sider the Hamiltonian H =?ΠJwhere ΠJ’s are rank-1

projections. One can visualize the interactions in H by

a graph G. The graph has n vertices corresponding to

the qubits and two vertices are connected when there is

a non-trivial interaction ΠJacting on them. We will also

distinguish two types of edges in the interaction graph.

Let Π = |φ??φ| be a projection. We will use a solid edge

in the graph when |φ? is entangled and a dashed edge

when |φ? is a product state. Let us first focus on the

homogeneous case with product constraints only.

In this case, the interaction graph consists of dashed

edges. We will show that the ground space is a span

of product of single-qubit states (or, for simplicity, a

product span). It will also be useful to know that the

states we choose are orthogonal up to a local operation

L =?n

acting on the j-th qubit. Note applying L on the 2BFF

Hamiltonian H =?HJ results in HL=?L−1

where LJ =

?

has the same ground state degeneracy as H [7, 9]. The

relation between the ground space of H and HLis

j=1Lj, where Lj is a non-singular local operator

JHJLJ,

j∈JLj. And HL, which is also 2BFF,

L−1K(H) = K(HL).

(2)

Before we actually give the proof, let us first examine

several simple examples. The first example considers a

chain of interactions as in Fig. 2b. Let |αj?⊗|βj? be the

constraint on the j-th edge. We will call it an alternating

chain if |βj−1? and |αj? are linearly independent for all

j. It is easy to see that the solution space is k + 1 for

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an alternating chain of k qubits. The second example

shown in Fig. 2c is called the alternating loop. As its

name suggests, it is a loop where the two constraints

on any vertex are linearly independent. Any alternating

loop has solution space of dimension 2, namely the span

of |00...0? and |11...1? up to the local operation that

maps |αj? and |βj−1? to |0? and |1?. The final example we

consider is called the quasi-alternating loop. It is almost

the same as the alternating loop except that there is one

special vertex on the loop having the same constraint on

the two edges adjacent to it. Figure 2d gives such an

example where the top vertex is special. It is easy to

see that the constraint on the special vertex of a quasi-

alternating loop must be satisfied. In particular, for the

loop in Fig. 2d, the top vertex must be |1? as otherwise

it will be impossible to satisfy all five constraints on the

loop.

(a) An illustration

of a dashed

interaction graph

|β?j

|β?j−1?= |α?j

(b) Alt-chain

01

0

1

0

1

0

1

0

1

(c) Alt-loop

0

1

0

1

0

1

0

1

00

(d)

Quasi-loop

FIG. 2: Dashed interaction graph and three examples

We now start the proof by induction on n, the number

of qubits. For n = 1,2, the observation is trivial. If

there is a vertex a on which the constraints are the same

up to global phases, let the constraints be |0?aand, more

concretely, let the constraints on an edge connects to a be

of the form |0?a|α?bfor some qubit b. We can write any

state in the ground space as |Φ? = |0?a|Φ0? + |1?a|Φ1?.

Obviously, |Φ0? and |Φ1? are both in the ground space of

the constraints not acting on a. Moreover, |Φ0? also needs

to be orthogonal to |α?b’s. By the induction hypothesis,

both |Φ0? and |Φ1? are in a product span. Therefore,

|Φ? is also in a product span. On the other hand, if one

cannot find any vertex whose constraints are the same, we

can find either an alternating loop or a quasi-alternating

loop in the graph. If a quasi-alternating loop is found, we

know the state for the special vertex of the loop and can

use the induction hypothesis on the remaining system.

Otherwise, if an alternating loop is found, we can write

any state in the ground space as

|Φ? = |00···0?|Ψ0? + |11···1?|Ψ1?,

up to local operations on the loop. If a constraint acts on

two qubits on the loop, it can only restricts the loop to

be exactly |00···0? or |11···1?. The analysis is similar

to the first case when a constraint |α?a|β?b acts on one

(3)

qubit a on the loop and another qubit b outside of the

loop. This completes the proof. Notice that the local op-

erations chosen here are determined by the constraints of

alternating loops, and that one will never have two alter-

nating loops giving different local operations for a single

qubit, the orthogonality of the states up to local opera-

tions follows. We note that the orthogonality property

only holds for the product constraints. The symmetric

subspace, for example, is not a span of orthogonal prod-

uct states up to local operations although it is the span

of |00?,|11?,|++? where |+? = (|0? + |1?)/√2.

General homogeneous case.— Given a general homo-

geneous Hamiltonian, the interaction graph will consist

both solid and dashed edges. The main technique is to

simplify the interaction graph in hand without changing

the ground space. Two sliding operations as shown in

Figs. 3a and 3b will be used in the simplification. The

Type-I sliding says that if we have entangled interactions

between 1,2 and 1,3, we can change it to two entan-

gled interactions between 1,2 and 2,3 without affecting

the ground space. The Type-II sliding is of a similar

spirit, but involves both entangled and product interac-

tions. We will only prove the validity of Type-I sliding

as a similar argument holds for the Type-II sliding op-

eration.

=

1

23

1

23

(a) Type-I Sliding

=

1

23

1

23

(b) Type-II Sliding

(c) An example of

simplified interaction

graph

FIG. 3: Simplification of the interaction graph

Let Π12= |φ??φ| and Π13= |ψ??ψ| be the two rank-1

operators acting on qubit 1,2 and 1,3. We will find a

local interaction Π23acting on 2,3 such that Π12+ Π23

has the same ground space as Π12+ Π13. As |φ? and

|ψ? are entangled states, one can find local operations L2

and L3 acting on qubit 2 and 3 respectively such that

|φ? = I1⊗ L2|Y ? and |ψ? = I1⊗ L3|Y ? where |Y ? is

the singlet state (|01? − |10?)/√2. The ground space of

Π12+ Π13is therefore

K(I ⊗ L2|Y ??Y |12I ⊗ L†

= (L†)−1K(|Y ??Y |12+ |Y ??Y |13)

= (L†)−1K(|Y ??Y |12+ |Y ??Y |23)

= K(Π12+ L2⊗ L3|Y ??Y |23L†

where the first equation uses Eq. (2), the second one

is obtained by a direct calculation establishing that

2+ I ⊗ L3|Y ??Y |13I ⊗ L†

3)

2⊗ L†

3),

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K(|Y ??Y |12+|Y ??Y |13) is the symmetric subspace of the

three qubits, and the last step employs Eq. (2) again.

This validates the Type-I sliding operation.

Repeated applications of the two types of sliding oper-

ations can modify an arbitrary graph (a homogeneous

Hamiltonian) with solid and dashed edges to the so-

called simplified interaction graph (simplified homoge-

neous Hamiltonian). The simplified graph has a back-

bone of only dashed edges and several solid-edge tails

attached to the backbone. An example of such a graph

is shown in Fig. 3c. This simplification can be done in

two steps by first changing each connected component of

solid edges into a tail, and then sliding all dashed edges

connected to a tail to one end of the tail. During the pro-

cess of the sliding operations, it may happen that there

is more than one edge between two vertices. If these mul-

tiple edges represent different constraints, one will essen-

tially have a high rank constraint and can deal with it as

before in the case study of rank.

Simplified homogeneous case.— Since sliding opera-

tions do not change the ground space, we only need to

work with simplified interaction graphs. The idea is to

build the entire ground space by extending the ground

space for the dashed backbone.

the case where there is only one tail in the simplified

interaction graph. More specifically, let S be the ground

space of the dashed constraints in the backbone J, and T

be the symmetric subspace confined by the tail of qubit

set K, where J ∩ K has exactly one qubit a, through

which the tail is attached to the backbone. We prove

that R = S ⊗ HK\{a}∩ T ⊗ HJ\{a}is again a product

span. Write S as the direct sum

Let us first consider

(S0⊗ Ha) ⊕

d

?

j=1

Sj⊗ |α⊥

j?a

,

where |αj?a’s are different dashed constraints on vertex

a and d is number of such |αj?a’s. For the basis of S0,

all the constraints in the backbone are already satisfied,

and therefore, the qubit a can be any state. We say that

qubit a is free in this case. For the basis of Sj, qubit a

has to be |α⊥

backbone. In this case, the state can only be extended

to the tail by copying. In summary, the intersection R

contains the space

j? in order to satisfy all the constraints in the

(S0⊗ T) ⊕

d

?

j=1

Sj⊗ |α⊥

j?⊗|T|

.

(4)

We will need to show that this is actually everything in

R.

We first claim that the product basis for Sj’s all to-

gether form a linearly independent set. By orthogonal-

ity (up to local operations), Sj and Sk are orthogonal

if |αj? and |αk? are not. On the other hand, if |αj?

and |αk? are orthogonal, the basis for Sj and Sk are

linearly independent.Otherwise, we will find a state

|ψ? in both Sj and Sk, meaning that |ψ? should be in

S0, a contradiction. Now, for any state |Ψ? in R, we

can write it as |Ψ? =?

early independent product states spanning S. Let |ˆΨj?

be the state on J \{a} when the state on J is |Ψj?. One

can also collect terms according to the state on J \ {a},

that is, |Ψ? =?

viously, |ˆΨk?’s are linearly independent, and we know

?

is indeed in the space of Eq. (4). As the symmetric sub-

space can always be spanned by product states, we have

finished the proof for the case of one tail. For multiple

tails, the proof is essentially the same by an induction on

the number of tails.

Application to the counting of degeneracy.— The re-

sults above actually allow us to prove that counting the

ground state degeneracy of a 2BFF Hamiltonian is in #P.

The class #P contains functions f if there is a polynomial

time algorithm A such that

j|Ψj?|Φj? where |Ψj?’s are lin-

k|ˆΨk??

l|Ψ?a

k,l|Φk,l?. As shown pre-

l|Ψ?a

k,l|Φk,l? is in T for each k. That is, the state |ψ?

f(x) = |{y,A(x,y) accepts.}|,

where y is usually called a proof to the verifier A.

As indicated by the ground space structure in Fig. 1,

the isometries will not change the dimension and we only

need to consider the simplified homogeneous case where

one can actually replace the solid edges of the tails to be

dashed edges forming alternating chains. As long as we

choose the constraint of the tail on the vertex connecting

to the backbone to be different from all other constraints

|αj? of that vertex, the dimension of the solution space

remains unchanged. To understand this, we need to re-

view the extension of the product span with intersection

of symmetric subspaces. If the vertex in the intersection

is free, we will have the whole symmetric subspace on

the tail which is of dimension k + 1 where k the number

of qubits in the tail. This coincides with the dimension

of the alternating chain. If the vertex in the intersection

is not free, we will have a unique extension in the tail,

which again coincides with the case of alternating chain.

It therefore suffices to count the dimension of any

dashed graph. To show that it is in #P, one can choose

the proof to the verifier to be the non-deterministic 0,1

choices in the case of (1) all-the-same-constraint vertex

and (2) alternating loop.

We thank S. Bravyi and X.G. Wen for valuable dis-

cussions. ZJ acknowledges support from NSF of China

(Grant Nos. 60736011 and 60721061); his research at

Perimeter Institute is supported by the Government of

Canada through Industry Canada and by the Province

of Ontario through the Ministry of Research & Innova-

tion. ZW is supported by the grant from the Centre for

Quantum Technology, the WBS grant under contract no.

R-710-000-008-271. BZ is supported by NSERC and CI-

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