Bulk-Edge correspondence and Fractionalization
- Y. Hatsugai
Institute of Physics
- Univ. of Tsukuba
JAPAN
As a topological (spin) insulator with strong interaction
(1,1)
(3,1)
- Dec. 10, 2008 at KITP
iγC(Aψ) =
- C
Aψ
Bulk-Edge correspondence and Fractionalization As a topological - - PowerPoint PPT Presentation
Dec. 10, 2008 at KITP (3,1) (1,1) Bulk-Edge correspondence and Fractionalization As a topological (spin) insulator with strong interaction Y. Hatsugai i C ( A ) = A C Institute of Physics Univ. of Tsukuba JAPAN Plan With
Institute of Physics
JAPAN
(1,1)
(3,1)
iγC(Aψ) =
Aψ
deconfined spinons in 2D & 3D ??
Topological Order
X.G.Wen
Common property of topological ordered states
Common property of topological ordered states
|G =
cJ ⊗ij |Singlet Pairij
|G =
cJ ⊗ij |Singlet Pairij
1 + c† 2)|0
|G =
cJ ⊗ij |Bondij
1 + c† 2)|0
|G =
cJ ⊗ij |Bondij
:Berry Connection :Berry Phase
ieiφi
1|G′ 2ei(φ1−φ2)
No Long Range Order in Spin-Spin Correlation |G =
cJ ⊗ij |Singlet Pairij
|Singlet Pair12 = 1 √ 2(| ↑1↓2 − | ↓1↑2)
Spins disappear as a Singlet pair
No Long Range Order in Spin-Spin Correlation |G =
cJ ⊗ij |Singlet Pairij
|Singlet Pair12 = 1 √ 2(| ↑1↓2 − | ↓1↑2)
Spins disappear as a Singlet pair
Time Reversal Invariant
y) ⊗ (iσ2 y) · · · (iσN y )K
N = (−)N
Mostly N: even Θ2
N = 1
(probability 1/2 in HgTe)
numerically
Calculate the Berry Phases using the Entire Many Spin Wavefunction
Z2 quantization Time Reversal ( Anti-Unitary ) Invariance Require excitation Gap! Only link <ij>
Define a many body hamiltonian by local twist as a parameter U(1)
dxψdx.
|ψ(x) = |ψ′(x)eiΩ(x) Aψ = A′
ψ + idΩ = A′ ψ + idΩ
dx dx
(Abelian)
2π × (integer) if eiΩ is single valued
H(x)
= (x)
JdCJ
J = −AΨ
JCJ = Ψ|Ψ = 1
J |JΘ,
(parameter independent) (Time Reversal, Particle-Hole)
dC∗
JCJ +
C∗
JdCJ =0
Gauge Equivalent(Different Gauge)
θ0 = 0, θN = 2π
∀∆θn → 0
King-Smith & Vanderbilt ’93 (polarization in solids) Luscher ’82 (Lattice Gauge Theory)
n = Ei(xn)|ψi n
Gauge invariant after the discretization
n|ψj n+1
Independent of the choice of the phase
= Im logψ0|ψ1ψ1|ψ · · · = Im log det D1D2 · · · Dn
N−1
non-Abelian non-Abelian ( ) Convenient for Numerics
Introduce interaction between singlets
Strong Coupling Limit of the AF Dimer link is a gapped unique ground state.
Hida Y.H., J. Phys. Soc. Jpn. 75 123601 (2006)
No Symmetry Breaking by the Local Order Parameter “String Order”: Non-Local Order Parameter!
i )2
Describe the Quantum Phase Transition locally c.f. S=1/2, 1D dimers, 2D with Frustrations, Ladders t-J with Spin gap
Y.H., J. Phys. Soc. Jpn. 75 123601 (2006)
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
: dimerization strength
✦ S=1,2 dimerized Heisenberg model
J1 = cos θ, J2 = sin θ
(2,0) (1,1) (0,2) (4,0) (3,1) (2,2) (1,3) (0,4)
S = 2 N = 10 S = 1 N = 14
(4,0) (3,1) (2,2) (1,1)
(2,0) (0,2)
: S=1/2 singlet state : Symmetrization
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
✦ S=2 Heisenberg model with D-term
0.5 1 1.5 2
S = 2 N = 10
:0 magnetization
Red line denotes the non trivial Berry phase
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
|{φi,j} =
ib† j − e−iφij/2b† ia† j
Bij |vac(4)
H({φi,i+1}) =
N
2Bi,i+1
AJP J
i,i+1[φi,i+1],
Twist the link of the generic AKLT model
Contribute to the Entanglement Entropy as of Edge states
T.Hirano, H.Katsura &YH, Phys.Rev.B77 094431’08
(a) (b)
J=1: J=2: = 0:
:
Y.H., J. Phys. Soc. Jpn. 75 123601 (2006), J. Phys. Cond. Matt.19, 145209 (2007)
A B
Direct Product State Entangled State
A B
System = A ⊕ B State =
D
A ⊗ |Ψj B
ρAB = 1 D
D
|Ψj
AΨk A| ⊗| Ψj BΨk B|
Pure State Mixed State
D
AΨj A|
D = 1
Vidal, Latorre, Rico, Kitaev ‘02
Requirement: Finite Energy Gap for the Bulk
Let us assume that the edge states has degrees of freedom DE
A B
(Fermions)
T.Hirano & YH, J. Phys. Soc. Jpn. 76, 1 13601 (2007)
N
extra,
i = S(S + 1)
HS=1
extra =
1 3( Si · Si+1)2 HS=2
extra =
2 9( Si · Si+1)2 + 1 63( Si · Si+1)3 +
L
jb† j+1 − b† ja† j+1)S|vac
Effective Boundary spins
Degrees of Freedom
(Pi + P −1
i
) = S1,i · S2,i + S1,i+1 · S2,i+1 + S1,i · S1,i+1 + S2,i · S2,i+1 + S1,i · S2,i+1 + S2,i · S1,i+1 + 4(S1,i · S2,i)(S1,i+1 · S2,i+1) + 4(S1,i · S1,i+1)(S2,i · S2,i+1) − 4(S1,i · S2,i+1)(S2,i · S1,i+1).
Ferromagnetic Rung Singlet Dominant Collinear Spin K J Dimer LRO Dominant Vector Chirality θ Scalar Chiral LRO
J = 2K
H =
{JrS1,i · S2,i + Jl(S1,i · S1,i+1 + S2,i · S2,i+1) + K(Pi + P −1
i
)}
K = sin θ
θ = 6
Hr =
S1,i · S2,i
Rung singlets
θ = 2.6
Hps =
(S1,i × S2,i) · (S1,i+1 × S2,i+1)
Plaquette singlet (PS)
K : complex conjugate
y) ⊗ (iσ2 y) · · · (iσN y )K
N = 1
K : complex conjugate
y) ⊗ (iσ2 y) · · · (iσN y )K
N = 1