Parent Hamiltonian for the non-Abelian chiral spin liquid

Abstract

We construct a parent Hamiltonian for the family of non-Abelian chiral spin
liquids proposed recently by two of us [PRL 102, 207203 (2009)], which includes
the Abelian chiral spin liquid proposed by Kalmeyer and Laughlin, as the
special case S=1/2. As we use a circular disk geometry with an open boundary,
both the annihilation operators we identify and the Hamiltonians we construct
from these are exact only in the thermodynamic limit.

Full text

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arXiv:1201.5312v2 [cond-mat.str-el] 7 Feb 2012
Parent Hamiltonian for the non-Abelian chiral spin liquid
Martin Greiter,1Darrell F. Schroeter,2and Ronny Thomale3
1Institut f¨ ur Festsk¨ orperphysik, Postfach 3640, KIT, D 76021 Karlsruhe, Germany
2Department of Physics, Reed College, Portland, OR 97202, USA
3Department of Physics, Stanford University, Stanford, CA 94305, USA
We construct a parent Hamiltonian for the family of non-Abelian chiral spin liquids proposed re-
cently by two of us [PRL 102, 207203 (2009)], which includes the Abelian chiral spin liquid proposed
by Kalmeyer and Laughlin, as the special case s =
open boundary, both the annihilation operators we identify and the Hamiltonians we construct from
these, are exact only in the thermodynamic limit.
1
2. As we use a circular disk geometry with an
PACS numbers: 75.10.Jm,75.10.Pq,75.10.Dg
Introduction.—The field of two-dimensional quantum
spin liquids [1–15] is witnessing a renaissance of interest
in present days [16–20]. For one thing, due to advances
in the computer facilities available, evidence for spin liq-
uid states in a range of models is accumulating [21, 22].
At the same time, spin liquids constitute the most intri-
cate, and in general probably least understood, examples
of topological phases [23–27], which themselves estab-
lish another vividly studied branch of condensed matter
physics [28–30]. If a complete description of the elec-
tronic states in the two-dimensional (2D) CuO planes of
high Tcsuperconductors [31] ever emerges, the theory is
likely based on a spin s = 1/2 liquid on a square lattice,
which is stabilized through the kinetic energy of itinerant
holon excitations [1].
Intimately related to the field of topological phases are
the concepts of fractional quantization, and in particu-
lar fractional statistics [32]. This field has experienced
another, seemingly unrelated renaissance of interest in
recent years, due to possible applications of states sup-
porting excitations with non-Abelian statistics [33] to the
rapidly evolving field of quantum computing and cryp-
tography. The paradigm for this class is the Pfaffian
state [34, 35], which has been proposed to describe the
experimentally observed quantized Hall plateau at Lan-
dau level filling fraction ν =
ports quasiparticle excitations which possess Majorana
fermion states at zero energy [36]. Braiding of these half-
vortices yields non-trivial changes in the occupations of
the Majorana fermion states, and hence render the ex-
changes non-commutative or non-Abelian [37, 38]. Since
this “internal” state vector is insensitive to local per-
turbations, it is preeminently suited for applications as
protected qubits in quantum computation [39, 40]. Non-
Abelian anyons are further established in other quantum
Hall states including Read-Rezayi states [41], in the non-
Abelian phase of the Kitaev model [8], the Yao–Kivelson
and Yao–Lee models [10, 18], and in the family of non-
Abelian chiral spin liquid (NACSL) states introduced by
two of us [13]. Very recently, non-Abelian statistics has
been observed numerically in hard-core lattice bosons in
a magnetic field, without reference to explicit wave func-
5
2[35].
The state sup-
tions [42].
In this Letter, we construct a parent Hamiltonian for
the NACSL states [13]. These spin liquids support spinon
excitations with SU(2) level k = 2s statistics for spin s,
i.e., Abelian, Ising, and Fibonacci anyons for s =
and3
2, respectively. The method we employ here is dif-
ferent from the method we used to identify a Hamilto-
nian [43, 44] which singles out the Kalmeyer–Laughlin
chiral spin liquid (CSL) state [2, 45] as its (modulo the
two-fold topological degeneracy) unique ground state for
periodic boundary conditions (PBCs). It is considerably
simpler, applicable to the entire family of spin s NACSL
states, but exact only in the thermodynamic (TD) limit
even if we impose PBCs.
Chiral spin liquid states.—The conceptually simplest
way to construct the non-Abelian chiral spin liquid
(NACSL) state [13] with spin s is to combine 2s identical
copies of Abelian CSL states with spin1
spin on each site onto spin s,
1
2,1,
2, and project the
1
2⊗1
?
2⊗ ... ⊗1
??
2
?
2s
= s ⊕ (2s − 1) · s−1 ⊕ ...
The projection onto the completely symmetric represen-
tation can be carried out conveniently using Schwinger
bosons [7, 46]. For a circular droplet with open boundary
conditions occupying N sites on a triangular or square
lattice, the Abelian CSL state takes the form
|ψKL
0? =
?
?
{z1,...,zM}
ψKL
0(z1,...,zM) S+
z1· ... · S+
zM|↓↓ ... ↓?
=
{z1,...,zM;
w1,...,wM}
ψKL
0(z1,...,zM) a+
z1...a†
zMb+
w1...b†
wM|0?
≡ ΨKL
0[a†,b†]|0?, (1)
where
ψKL
0[z] =
M
?
i<i
(zi− zj)2
M
?
i=1
G(zi)e−1
4|zi|2
(2)

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is a bosonic quantum Hall state in the complex “parti-
cle” coordinates zi≡ xi+ iyisupplemented by a gauge
factor G(zi), M =N
ation operators [7, 46, 47], and the wk’s are those lattice
sites which are not occupied by any of the zi’s. In this
notation, we can write the spin s state obtained by the
projection as
2, a†and b†are Schwinger boson cre-
|ψs
0? =
?
ΨKL
0
?a†,b†??2s
|0?. (3)
The lattice may be anisotropic; we have chosen the lattice
constants such that the area of the unit cell spanned by
the primitive lattice vectors is set to 2π. For a triangular
or square lattice with lattice positions given by ηn,m=
na + mb, where a and b are the primitive lattice vectors
in the complex plane and n and m are integers, the gauge
phases are simply G(ηn,m) = (−1)(n+1)(m+1)[45, 48].
The NACSL state can alternatively be written as
|ψs
0? =
?
{z1,...,zsN}
ψs
0(z1,...,zSN)˜S+
z1· ... ·˜S+
zsN|−s?N,
(4)
where |−s?N≡ ⊗N
which all the spins are maximally polarized in the nega-
tive ˆ z-direction, and˜S+are re-normalized spin flip oper-
ators which satisfy
α=1|s,−s?αis the “vacuum” state in
1
?(2s)!(a†)n(b†)(2s−n)|0? = (˜S+)n|s,−s?.
In a basis in which Szis diagonal, we may write
(5)
˜S+=
1
s − Sz+ 1S+. (6)
Note that (5) implies
S−(˜S+)n|s,−s? = n(˜S+)n−1|s,−s?.
The wave function for the spin s state (3) are then ef-
fectively given by bosonic Read–Rezayi states [41] for
renormalized spin flips,
(7)
ψs
0[z] =
2s
?
m=1
?
mM
?
i<j
i,j=(m−1)M+1
(zi− zj)2
?sN
i=1
?
G(zi)e−1
4|zi|2.
(8)
which we understand to be completely symmetrized over
the “particle” coordinates zi. For s = 1, they take the
form of a Moore–Read state [34, 35]
ψs=1
0
[z] = Pf
?
1
zi− zj
?N
i<j
?
(zi− zj)
sN
?
i=1
G(zi)e−1
4|zi|2.
(9)
For the considerations below, it is convenient to write
the state in the form
|ψs
0? =


?
{z1,...,zM}
ψKL
0(z1,...,zM)˜S+
z1· ... ·˜S+
zM


2s
|0?.
(10)
Since the Abelian KL CSL |ψKL
in the TD limit N → ∞, and is an approximate sin-
glet for finite N, the same holds for the NACSL |ψs
well. This follows from the construction of the Schwinger
boson projection (3), but can also be verified directly us-
ing Perelomov’s identity (see (29) in the supplementary
material) [49]. The Abelian and non-Abelian CSL states
trivially violate parity (P) and and time reversal (T) sym-
metry.
Ground state annihilation operators.—In the TD limit
N → ∞, the NACSL ground states are annihilated by
0? is an exact spin singlet
0? as
Ωs
α=
N
?
β?=α
β=1
1
ηα− ηβ(S−
α)2sS−
β,Ωs
α|ψs
0? = 0 ∀α,(11)
as we will verify now.
Let us consider the action of (S−
in the form (10). Since ψKL
ever two arguments zicoincide, one of the zi’s in each of
the 2s copies in (10) must equal ηα; since ψKL
is symmetric under interchange of the zi’s and we count
each distinct configuration in the sums over {z1,...,zM}
only once, we may take z1= ηα. Regarding the action
of S−
βon (10), we have to distinguish between configura-
tions with n = 0,1,2,...,2s re-normalized spin flips˜S+
at site β. Since the state is symmetric under interchange
of the 2s copies, we may assume that the n spin flips are
present in the first n copies, and account for the restric-
tion through ordering by a combinatorial factor. This
yields
α)2sS−
βon |ψs
0? written
0(z1,...,zM) vanishes when-
0(z1,...,zM)
β
(S−
α)2sS−
β|ψs
0? = (S−
α)2sS−
β
2s
?
n=0
?2s
n
?



?
{z3,...,zM}
ψKL
0(ηα,ηβ,z3,...)˜S+
α˜S+
β˜S+
z3· ... ·˜S+
zM


n
·
?
{z2,...,zM}?=ηβ
ψKL
0(ηα,z2,...)˜S+
α˜S+
z2· ... ·˜S+
zM


2s−n
|0?
2

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= (2s)!2s


?
{z2,...,zM}
ψKL
0(ηα,ηβ,z3,...,zM)˜S+
z3· ... ·˜S+
zM


2s
?
n=1
?2s − 1
n − 1
?
·


?
{z3,...,zM}
ψKL
0(ηα,ηβ,z3,...,zM)˜S+
β˜S+
z3· ... ·˜S+
zM


n−1

?
{z2,...,zM}?=ηβ
ψKL
0(ηα,z2,...,zM)˜S+
z2· ... ·˜S+
zM


2s−n
|0?
= (2s)!2s


?
{z3,...,zM}
ψKL
0(ηα,ηβ,z3,...,zM)˜S+
z3· ... ·˜S+
zM

·


?
{z2,...,zM}
ψKL
0(ηα,z2,...,zM)˜S+
z2· ... ·˜S+
zM


2s−1
|0?,
where we have used (7). This implies
Ωs
α|ψs
0? = (2s)!2s


?
{z3,...,zM}
N
?
?
ψKL
β=1
ψKL
0(ηα,ηβ,z3,...,zM)
ηα− ηβ
??
0(ηα,z2,...,zM)˜S+
?
=0
˜S+
z3· ... ·˜S+
zM


·


?
{z2,...,zM}
z2· ... ·˜S+
zM


2s−1
|0? = 0,
where we have used the Perelomov identity [49], which
states that any infinite lattice sum of e−1
any analytic function of ηβ vanishes.
ing, Perelomov [49] only considered a square lattice. The
identity, however, holds for any 2D lattice with a sin-
gle site per unit cell, as we show in the supplementary
material.)
Parent Hamiltonian.—A Hermitian, positive semi-
definite, and translationally invariant operator which an-
nihilates |ψs
N
?
α?=β,γ
4|ηβ|2G(ηβ) times
(Strictly speak-
0? is given by
Γ ≡
α=1
Ωs
α
†Ωs
α=
?
α,β,γ
ωαβγ(S+
α)2s(S−
α)2sS+
βS−
γ, (12)
where
ωαβγ≡
1
¯ ηα− ¯ ηβ
1
ηα− ηγ. (13)
This operator is not invariant under SU(2) spin rotations,
but rather consists of a scalar, vector, and higher tensor
components up to order 4s+2. Since the NACSL states
|ψs
tensor components must annihilate the state individually.
The scalar component of Γ, which we denote as {Γ}0, pro-
vides us with an SU(2) spin rotationally invariant parent
Hamiltonian.
To obtain the projected operator {Γ}0, we follow the
method described in detail in ref. [50], and summarize
here only the most important steps. With the tensor
content of S+
γgiven by
0? are spin singlets, and are annihilated by Γ, all these
βS−
S+
βS−
γ=2
3SβSγ− i(Sβ× Sγ)z−
1
√6T0
βγ, (14)
where
T0
βγ=
2
√6
?3Sz
βSz
γ− SβSγ
?
(15)
is the m = 0 component of the second order tensor, we
only need to know the scalar, vector and 2nd order tensor
components of (S+
component of Γ. These are given by (see Sec. 5.3.2 of [50])
α)2s(S−
α)2sin order to obtain the scalar
(S+
α)2s(S−
α)2s= a0
?
1 + aSz
α+ bT0
αα+ higher orders
?
(16)
where
a0=(2s)!2
2s + 1, a =
3
s + 1, b =
√6
2
5
(s + 1)(2s + 3). (17)
The scalar component of Γ is hence given by
?Γ?
0= a0
?
α?=β,γ
α,β,γ
ωαβγ
·
?2
3SβSγ−ia
3Sα(Sβ× Sγ) −
b
√6
?T0
ααT0
βγ
?
0
?
(18)
.
With Sβ× Sβ= iSβand (see Sec. 4.5.3 of [50])
0= −4
+ 2?(SαSβ)(SαSγ) + (SαSγ)(SαSβ)?,
5?T0
ααT0
βγ
?
3S2
α(SβSγ) + 2δβγSαSβ
(19)
3

Page 4

we obtain the final parent Hamiltonian
Hs=
?
?
α?=β
ωαββ
?
?
s(s + 1)2+ SαSβ−(SαSβ)2
(s + 1)
?
+
α,β,γ
α?=β?=γ?=α
ωαβγ
(s + 1)SβSγ−
2s + 3
2(s + 1)iSα(Sβ× Sγ)
−(SαSβ)(SαSγ) + (SαSγ)(SαSβ)
2(s + 1)
?
.
(20)
(It is related to (18) via?Γ?
with N lattice sites, and becomes exact in the TD limit
N → ∞, where Hs|ψs
term explicitly breaks P and T. (It would be highly de-
sirable to identify a parent Hamiltonian which is P and T
invariant, such that the ground states violate these sym-
metries spontaneously, but we have so far not succeeded
in finding one.)
0= 2a0/(2s + 3)Hs.) This
Hamiltonian is approximately valid for any finite disk
0? = 0. Note that the Sα(Sβ×Sγ)
The special case s =
Tm
(18) significantly, and yields the parent Hamiltonian
1
2.—Since S+
ααT0
βγ
α
2= 0 for s =
0= 0. This simplifies
1
2,
αα= 0 for all m, and?T0
?
Hs=1
2=
?
ωαβγ
α?=β
ωαββ
?3
4+ SαSβ
?
+
?
α,β,γ
α?=β?=γ?=α
?SβSγ− iSα(Sβ× Sγ)?
(21)
(It is related to (18) via
contrast to the earlier parent Hamiltonian proposed in
ref. [43, 44] (SKTG) for the Abelian KL CSL (2) with
periodic boundary conditions, (21) is not exact for finite
N. It is considerably simpler then the SKTG model, and,
like (20), becomes exact in the TD limit.
?Γ?
0= 2a0/3Hs=1
2.)In
Remarks on periodic boundary conditions.—It is rather
straightforward to formulate the model on a torus. For
simplicity, we choose the lattice constant a real, and b
such that the imaginary part ℑ(b) > 0. We implement
PBCs in both directions by identifying the sites zi, zi+L,
and zi+ Lτ, where L = n1a, Lτ = nτa + mτb, and
ℑ(τ) > 0. n1and mτ are positive integers such that the
number of sites N = n1mτ is even, and nτ is an integer.
We place the lattice sites at positions
ηn,m=
?
n −n1− 1
2
?
a +
?
m −mτ− 1
2
?
b,(22)
with n = 0,1,...,n1−1 and m = 0,1,...,mτ−1. Then
the wave function of the NACSL (8) takes the form
ψs
0[z] =
2s
?
m=1
?
mM
?
i<j
i,j=(m−1)M+1
ϑ 1
2,1
2
?1
L(zi− zj)??τ?2
?
·
2?
ν=1
ϑ 1
2,1
2
?1
L(Zm− Zν,m)??τ?
·
sN
?
i=1
G(zi)e−1
2y2
i,
(23)
where ϑ 1
2,1
2(z|τ) is the odd Jacobi theta function [51], and
Zm≡
mM
?
i=(m−1)M+1
zi,Z1,m= −Z2,m, (24)
are the center-of-mass coordinates and zeros, respec-
tively. The latter can be chosen anywhere within the
principal region bounded by the four points
mτb), and encode the (2s + 1)-fold topological degener-
acy of the NACSL [19]. The gauge factor in (23) is given
by
1
2(±n1a ±
G(ηn,m) = (−1)mτn+me−iπℜ(b)
a
m(mτ−1−m), (25)
where ℜ(b) is the real part of b.
The NACSL (23) is approximately annihilated by
Ωs
α=
N
?
β?=α
β=1
ϑu,v
ϑ 1
2,1
?1
L(ηα− ηβ)??τ?
2
?1
L(ηα− ηβ)??τ?(S−
α)2sS−
β
(26)
for all α, where we can choose any of the three even
Jacobi theta functions in the numerator: (u,v)=(0,0),
(0,1
but only quasiperiodic, due to the shift of the boundary
phases inherent in (26). The statement Ωs
comes exact as N → ∞.
The NACSL (23) is hence the approximate ground
state of (20) (and for s =
2also of (21)) with (13) re-
placed by
2), or (1
2,0). Note that Ωs
α|ψs
0? is not strictly periodic,
α|ψs
0? ≈ 0 be-
1
ωαβγ=
?
ϑu,v
ϑ 1
2,1
?1
L(ηα− ηβ)??τ?
2
?1
L(ηα− ηβ)??τ?
?∗ϑu,v
ϑ 1
?1
L(ηα− ηγ)??τ?
2,1
2
?1
L(ηα− ηγ)??τ?,
(27)
where ∗ denotes complex conjugation. As in the case with
open boundary conditions, the model becomes exact in
the TD limit.
Conclusion.—We have identified a parent Hamiltonian
for the non-Abelian CSL states [13], which becomes ex-
act in the TD limit. This Hamiltonian should allow us
to study the spinon and holon excitations including the
non-Abelian braiding properties within a concise frame-
work. The construction also extends to the Abelian s =1
Kalmeyer–Laughlin CSL [2, 45], where it is likewise exact
2
4

Page 5

only as the number of sites N → ∞, but is considerably
simpler that the SKTG Hamiltonian [43, 44].
Acknowledgments.—MG is supported by the German
Research Foundation under grant FOR 960. RT is sup-
ported by an SITP fellowship at Stanford University.
Note added.—After this work was completed, we be-
came aware of a manuscript by Nielsen, Cirac, and
Sierra [52], in which they derive the s =1
(21) using null operators in the conformal correlators of
the SU(2) level k = 1 Wess–Zumino–Witten model.
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Modern Physics
(Springer,
I.Cirac,and G.Sierra,
5

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Supplementary material
In this supplement, we proof the Perelomov iden-
tity [49] for arbitrary 2D lattices using Fourier transfor-
mation.
The Perelomov identity.—Consider a lattice spanned
by ηn,m= na + mb in the complex plane, with n and m
integer and the area of the unit cell Ω spanned by the
primitive lattice vectors a and b set to 2π,
Ω =??ℑ(a¯b)??= 2π (28)
where ℑ denotes the imaginary part.
(−1)(n+1)(m+1). Then
?
for any polynomial P of ηn,m.
Proof.—It is sufficient to proof the identity for the gen-
erating functional
Let G(ηn,m) =
n,m
P(ηn,m)G(ηn,m)e−1
4|ηn,m|2= 0 (29)
?
n,m
e
1
2ηn,m¯ zG(ηn,m)e−1
4|ηn,m|2= 0. (30)
Since G(ηn,m) takes the value −1 on a lattice with twice
the original lattice constants, we may rewrite this as
?
n,m
e
1
2ηn,m¯ ze−1
4|ηn,m|2− 2
?
n,m
eηn,m¯ ze−|ηn,m|2= 0. (31)
Kalmeyer and Laughlin [45] observed that for the square
lattice, the second sum in (31) can be expressed as a sum
of the Fourier transform of the function we sum over in
the first term. We demonstrate here that their proof can
be extended to arbitrary lattices.
To begin with, we define the Fourier transform in com-
plex coordinates
?
where ℜ denotes the real part and we have used (28).
Since the area of the unit cell of our lattice is taken to be
2π, the reciprocal lattice is given by the original lattice
rotated by
2in the plane without any rescaling of the
lattice constants. In complex coordinates,
˜f(ζ) =d2ηf(η)eiℜ(η¯ζ), (32)
π
ζn′,m′ = i(n′a + m′b), (33)
as this immediately implies
Rn,m· Kn′,m′ = ℜ(ηn,m¯ζn′,m′) =
= ℜ?(na + mb)(−i)(n′¯ a + m′¯b)?
= 2π · integer.
Then
?
= nm′ℑ(a¯b) + mn′ℑ(b¯ a)
n′,m′
˜f(ζn′,m′) = Ω
?
n,m
f(ηn,m). (34)
Eq. (34) follows directly from
?
n′,m′
eiℜ(η¯ζn′,m′)= Ω
?
n,m
δ(2)(ηn,m− η), (35)
which is just the 2D equivalent of the (Dirac comb) iden-
tity
∞
?
n′=−∞
e2πin′x=
∞
?
n=−∞
δ(x − n) (36)
The r.h.s. of (36) is obviously zero if x is not an integer,
and manifestly periodic in x with period 1. To verify the
normalization, observe that since for any N odd,
+N−1
?
2
n′=−N−1
2
e2πin′y/N=
?N for y = N · integer
otherwise.0
This implies
1
N
+N−1
?
2
y=−N−1
2
+N−1
?
2
n′=−N−1
2
e2πin′y/N= 1,
which in the limit N → ∞ is equivalent to
?+N
2
−N
2
dy
N
+N−1
?
2
n′=−N−1
2
e2πin′y/N= 1
Substituting x = y/N yields
?+1
2
−1
2
dx
∞
?
n′=−∞
e2πin′x= 1,
which proves the normalization in (36).
We proceed by evaluation of the Fourier transform of
f(η) = e
1
2η¯ ze−1
4|η|2:
˜f(ζ) =
?
?
d2η e
1
2η¯ ze−1
4|η|2eiℜ(η¯ζ)
=d2η e
1
2η¯ ze−1
4|η|2e
i
2(η¯ζ+¯ ηζ)
= 4πe−|ζ|2+iζ¯ z
(37)
where we have used the integral
?
d2η F(η) e−1
α(|η|2−¯ ηw)
=F(α∂¯ w)
?
?
d2η e−1
α(|η|2−¯ ηw−η ¯ w)
????
¯ w=0
¯ w=0
=F(α∂¯ w)d2η e−1
α(|η−w|2−w ¯ w)
????
=απ F(α∂¯ w)e
1
αw ¯ w= απ F(w)
????
¯ w=0
6

Page 7

with F(η) = e
Substituting (37) into (34) we obtain
1
2η¯ z+i
2η¯ζ, α = 4, and w = 2iζ.
?
n,m
f(ηn,m) = 2
?
n′,m′
e−|ζn′,m′|2+iζn′,m′ ¯ z
(38)
If we now substitute n′= −n, m′= −m, and hence
iζn′,m′ = ηn,m into the r.h.s. of (38), we obtain (31).
This completes the proof.
7