Transcript Document

Reflection Symmetry and Energy-Level
Ordering of Frustrated Ladder Models
The extension of Lieb-Mattis theorem [1962] to a frustrated spin system
Tigran Hakobyan
Yerevan State University & Yerevan Physics Institute
T. Hakobyan, Phys. Rev. B 75, 214421 (2007)
Heisenberg Spin Models
Hamiltonian: H 
J
ij
Si S j
1
J12 2
J 24 4
i, j
J13
i, j :
3
7
8
9
interacting sites
Si  (S  S  S ) :
x
i
J ij :
5
y
i
z
i
spin of i-th site
spin-spin coupling constants
Jij  0:
ferromagnetic bond
Jij  0:
antiferromagnetic bond
6
J 36
Bipartite Lattices
The lattice L is called bipartite if it splits into two disjoint sublattices A and B
such that:
1)
All interactions between the spins of different sublattices are
antiferromagnetic, i. e.
2)
Jij  0 if i  A, j  B or i  B, j  A
All interactions between the spins within the same sublattice are
ferromagnetic, i. e. J  0 if i, j  A or i, j  B
ij
An example of bipartite system:
sublattice A
antiferromagnetic bonds
connect different sites
sublattice B
ferromagnetic bonds
connect similar sites
Classical Ground State: Néel State
Ground state (GS) of the classical Heisenberg model on bipartite lattice is
a Néel state, i. e.
The spins within the same sublattice have the same direction.
The spins of different sublattices are in opposite directions.
Properties of the Néel state:
Néel state minimizes all local interactions in
the classical Hamiltonian.
It is unique up to global rotations.
classical
 S A  SB  .
Its spin is: SNeel  SGS
 SA (SB )  max. spin on
A ( B)
Quantum GS: Lieb-Mattis Theorem
The quantum fluctuations destroy Néel state and the ground state (GS) of
quantum system has more complicated structure.
However, for bipartite spin systems, the quantum GS inherits some
properties of its classical counterpart.
Lieb & Mattis [J. Math. Phys. 3, 749 (1962)] proved that
The quantum GS of a finite-size system is a unique multiplet with total spin
quantum
classical
SGS  S A  SB , i. e. SGS
 SGS
 SNeel.
The lowest-energy
ES
in the sector, where the total spin is equal to S, is a
monotone increasing function of
S for any S  SGS [antiferromagnetic
ordering of energy levels].
All lowest-energy ( ES ) spin-S states form one multiplet for
[nondegeneracy of the lowest levels].
S  SGS
Steps of the Proof
Perron-Frobenius theorem: The lowest eigenvalue of any connected
matrix having negative or vanishing off-diagonal elements is
nondegenerate. Correponding eigenvector is a positive superposition
of all basic states.
After the rotation of all spins on one sublattice on
reads
1

H  J ij  (Si S j  Si S j )  Siz S jz  ,
2

i, j
Unitary
shift
generate negative
off-diagonal elements
 , the Hamiltonian
J ij  0
are diagonal
The matrix of Hamiltonian being restricted to any S z  M subspace is
connected in the standard Ising basis.
Perron-Frobenius theorem is applied to any S  M subspace:
z
Relative
GS

M


i
mi  M
m …m m1 m2 … mN  m …m  0, mi   si ,  si  1, , si
1
N
1
N
Outline of the Proof
[Lieb & Mattis, 1962]
The spin of  M can be found by constructing a trial state being a positive
superposition of (shifted) Ising basic states and having a definite value of the
spin. Then it will overlap with  M. The uniqueness of the relative GS then
implies that both states have the same spin. As a result,

 M  if  M  S Neel  S A  S B 
S   M   SM  

 S Neel if  M  S Neel
The multiplet containing  has the lowest-energy value
M
all states with spin S  S M . It it nondegenerate.
ES among
Antiferromagnetic ordering of energy levels:
ES1  ES2 if S1  S2  S A  SB 
The ground state is a unique multiplet with spin
SGS  SNeel  S A  SB  .
Generalizations
The Lieb-Mattis theorem have been generalized to:
• Ferromagnetic Heisenberg spin chains
B. Nachtergaele and Sh. Starr, Phys. Rev. Lett. 94, 057206 (2005)
• SU(n) symmetric quantum chain with defining representation
T. Hakobyan, Nucl. Phys. B 699, 575 (2004)
• Spin-1/2 ladder model frustrated by diagonal interaction
T. Hakobyan, Phys. Rev. B 75, 214421 (2007)
The topic of this talk
Frustrates Spin Systems
In frustrated spin models, due to competing interactions,
the classical ground state can’t be minimized locally and
usually possesses a large degeneracy.
?
The frustration can be caused by the geometry of the spin
lattice or by the presence of both ferromagnetic and
antiferromagnetic interactions.
Examples of geometrically frustrated systems:
Antiferromagnetic Heisenber spin system on
Triangular lattice,
Kagome lattice,
Square lattice with diagonal interactions.
Frustrated Spin-1/2 Ladder:
Symmetries
J l
Symmetry
axis
S1,l J
l
S 2,l
J l
N 1
N 1
l 1
l 1
N
H   J l (S1l  S1l 1  S2l  S2l 1 )   J (S1l  S2l 1  S1l 1  S2l )   J lS1l  S2l
The total spin

l
l 1
S and reflection parity   1 are good quantum numbers.
So, the Hamiltonian remains invariant on individual sectors with fixed values
of both quantum numbers.
Let
ES 
be the lowest-energy value in corresponding sector.
Frustrated Spin-1/2 Ladder:
Generalized Lieb-Mattis Theorem
J

l
[T. Hakobyan, Phys. Rev. B 75, 214421 (2007)]
S1,N
S1,l J
l
S 2,l
J l  J l 
J

l
S2,N
N = number of rungs
The minimum-energy levels are nondegenerate (except perhaps the one
N 1
with   (1) and S  0 ) and are ordered according to the rule:
 for   (1) N if S1  S2
ES1   ES2   
N 1
for


(

1)
if S1  S2  1

The ground state in entire   (1) N sector is a spin singlet while in
  (1) N 1 sector is a spin triplet. In both cases it is unique.
Rung Spin Operators
N 1
N 1
l 1
l 1
N
H   J l (S1l  S1l 1  S2l  S2l 1 )   J (S1l  S2l 1  S1l 1  S2l )   J lS1l  S2l

l
l 1
The couplings obey:
J l
Sl( s )  S1l  S 2l
J l  J l 
S1,l J
l
Sl( a )  S1l  S 2l
Reflection-symmetric
(antisymmetric) operators
N 1
H   (J S  S
l 1
s
l
(s)
l
(s)
l 1
J S
a
l
J l  J l
where J :
0
2
s
l
S 2,l
(a)
l
1 N  (s) 2
 S )   J l (Sl ) ,
2 l 1
(a)
l 1
J l  J l
J :
0
2
a
l
J l
Symmetry
axis

Construction of Nonpositive Basis:
Rung Spin States
We use the following basis for 4 rung states:
 
1 
0 



2  
 
1)
Rung
singlet
Rung
triplet
1 


,
1 


,
0 
 
1 



2  
 
We use the basis constructed from rung singlet and rung triplet states:
m1  m2 … mN 
ml  1 0 0
The reflection operator R is diagonal in this basis. R  (1)
is the number of rung singlets.
N0
, where N 0
Define unitary operator, which rotates the odd-rung spins around





U  exp i
[( N 1)  2]

l 1
S

(s) z 
2l 1 

z axis on 
Construction of Nonpositive Basis:
Unitary Shift
2)
Apply unitary shift to the Hamiltnian:
generate negative
off-diagonal elements
1 N 1
H  UHU   ( J ls Sl( s ) Sl(s1)   J ls Sl( s )  Sl(s1)   J la Sl( a )  Sl(a1)   J la Sl( a )  Sl(a1)  )
2 l 1
N 1
1 N  (s) 2
s (s) z (s) z
a (a) z (a ) z
 ( J l Sl Sl 1  J l Sl Sl 1 )   J l (Sl ) 
2 l 1
l 1
1
are diagonal
in our basis
All positive off-diagonal elements
become negative after applying a
sign factor to the basic states
Construction of Nonpositive Basis:
Sign Factor
3)
It can be shown that all non-diagonal matrix elements of
nonpositive in the basis
[ N0  2] N00
m1 m2 … mN  (1)
H
become
m1  m2 … mN 
sign factor
N 00
= the number of pairs 0  0
in the sequence m1 m2 … mN
0 is on the left hand side from 0 .
N0
= the number of rung singlets 0 in
m1 m2 … mN .
where
S  M , R   Subspaces and
Relative Ground States
z
z
Due to S and reflection R symmetries, the Hamiltonian is invariant on each
subspace with the definite values of spin projection and reflection operators,
which we call ( M   ) subspace:
S z  M , R   , where M   N   N  1… N and   1
The matrix of the Hamiltonian in the basis m1 m2 … mN being
restricted on any ( M   ) subspace is connected [easy to verify].
Perron-Frobenius theorem can be applied to ( M   ) subspace:
The relative ground state of H in ( M   ) subspace is unique and is a
positive superposition of all basic states:

M 


l N
ml  M
( 1)
0

m …m m1  m2 … mN 
1
N
m …m  0
1
N
Relative ground states
The spin of  M , can be found by constructing a trial state being a positive
superposition of defined basic states and having a definite value of the spin.
 M , . The uniqueness of the relative GS then
Then it will overlap with
implies that both states have the same spin. As a result,
SM 
 1 if M  0 and   (1) N 1

  M  otherwise