Transcript Fractional topological insulators Michael Levin (Harvard), Ady Stern (Weizmann) Topological vs.

Fractional topological insulators
Michael Levin (Harvard), Ady Stern (Weizmann)
Topological vs. trivial insulators
(Zhang et al, Kane et al)
Non-interacting fermions in a 2D gapped system
(energy bands, Landau levels…)
(Thouless et al, Avron et al)
• An integer index – the Chern number,  xy
• Transition between different Chern numbers requires closing the
energy gap.
• (  xy  0 ) Gapless edge states on the interface with a vacuum.
Refinement:
Impose time reversal symmetry (i.e.,  xy  0),
• Two types of bands – trivial insulators and topological insulators.
• No transition between the two classes unless
either a. the gap closes
or
b. time reversal symmetry is broken
A useful toy model – two copies of the IQHE
Red electrons experience a magnetic field B
Green electrons experience a magnetic field -B
n=odd per each color – a topological insulator
n=even per each color – a trivial insulator
With interactions, gapped systems are more complicated and
have more quantum numbers.
Several states may have the same xy , but differ in other
topological quantum numbers (e.g., charge of quasi-particles,
statistics, thermal Hall conductivity).
A natural question – does time reversal symmetry introduce a
distinction between “trivial” and “topological” classes for
interacting systems?
The modified toy model – two copies of the FQHE
n fractional per each color, artificial type of interaction.
Two quantum numbers characterizing a fractional state:
n– the (spin) Hall conductivity
e* - the smallest charge allowed for an excitation
The question – can the edge states be gapped out without
breaking time reversal symmetry ?
The answer is determined by the parity of n/e*:
Even –Yes
Odd - No
Two ways to analyze this question:
1. A flux insertion argument
2. A microscopic calculation –
1. Write down the edge theory.
2. Look for perturbation(s) that gap(s) these 2n gapless
modes without breaking time reversal symmetry.
(Haldane’s topological stability)
Flux insertion argument – non-interacting case (Fu&Kane)
spin-ups
spin-downs
Turn on a

0
2
in the hole.
• A spin imbalance of ±n (integer number) is created on each
edge.
• Two states that are degenerate in energy and time reversed of
one another. Can they be coupled?
Yes, if the imbalance is even.
No, if it is odd. So, no edge perturbation will lift the degeneracy
and open a gap without breaking time reversal symmetry.
For fractional quantum spin Hall states
spin-ups
spin-downs
• Half a flux quantum leaves the edges coupled.
• We need to insert the flux needed to bring each edge back to
the topological sector from which it started.
• The number of flux quanta needed for that is 1/2e*.
• The imbalance is then n/e*. It is the parity of this number that
determines the protection of the gap.
A microscopic calculation
The unperturbed edge, two copies of the QHE:
Kijti x j  Vij xi x j   n Atini
(Wen…)
The Integer case:
n 1
1 0 
K  

 0  1
n 2
1

0
K 
0

0
  1 
   
 2 
0 0
0

1 0
0
0 1 0 

0 0  1
 1 
 
  2 
 

 3
 
 4
 1
 
.
t 
.
 
 1
The double-FQHE unperturbed edge (N modes each color):
Kijti x j  Vij xi x j   n Atini
 K0
K  
 0
0 

 K0 
 1 


  . 

. 


 2 N 
 1
 
.
t 
.
 
 1
1 0 
 z  
,
 0  1
0 1 
 x  

 1 0
Perturbations:
The perturbation:
a. Charge conserving
b. Possibly strong (“relevance is irrelevant”)
c. So is the relevance to experiments (apologies!)
d. No conservation of momentum, spin
Symmetry with respect to time reversal – a crucial issue
Perturbations:
l


exp i(l )


 
 exp il K

l  2 4  1  5
is an integer vector
 
To conserve charge l  t  0
 
The number of flipped spins Ql   l  t / 2
z
The time reversal operator:
 

l






   xl   Q l
A time reversal symmetric perturbation:


cos (l )

 1
Q l 


cos (xl )

We look for a set of l’s that will gap the edge modes
Start from the double Laughlin state (N=1)
3 0 
K  

 0  3
Perturbation:
 j 
l   
 j
 U ( x) cos3 j ( x)   ( x)   ( x)
j=1: Zeeman field in the
x-direction
j=2: electron-electron
interaction
 U ( x) cos3 j ( x)   ( x)   ( x)
The perturbation gaps the two gapless modes, but
1. For j=1 it explicitly breaks time reversal symmetry
2. For j=2 it spontaneously breaks time reversal symmetry,
giving an expectation value to
expi3 ( x)   ( x)
It is impossible to gap the two modes without breaking symmetry
to time reversal – a topological fractional insulator
For N=1, always n/e*=1
Now for two edge modes in each direction (N=2).
Most generally:
b
 b  su
  


0
b
b  sv



K 0

 b  su
 b  
0

 
b
 b  sv  

b 
 b  su

K  
b  sv 
 b
n
uv
u  v b  uvs
e* 
1
u  v b  uvs
b,s,u,v – integers. u and v have no common factors.
The answer: gapping is possible for even (u+v)
impossible without TRS breaking for odd (u+v)
The method: transform the problem into two decoupled
Luttinger liquids, and then gap them, with two perturbations.
Even (u+v):
The two required vectors are
1
l1  u  1 v  1  u  1 v  1
2
1
l2  u  1 v  1  u  1 v  1
2
Each of the two vectors flips (u+v)/2 spins. Second order splits
the degeneracy. No spontaneous breaking of time reversal
symmetry.
Odd (u+v):
flipping (u+v) spins necessarily break time reversal symmetry
by giving an expectation value to
v
u  v  u
Can a trivial insulator change to a topological insulator
without
but
with
the gap being closed,
breaking of time reversal symmetry?
For the “Integer topological insulator” – yes
For the fractional – more difficult, since the charge e* cannot
change without the gap closing.
The answer:
If 1/e* is odd – yes.
If 1/e* is even – no.
Why?
a relation between
1
e*
and
n
e*
1
e 
q
p
n
q
*
• q flux quanta (quanta of vorticity) in each – topologically trivial
• p units of charge in each
• Statistics – bosonic if
p
The phase of interchange
So, parity of
p
even, fermionic if
 pq
must equal parity of pq
p
odd
e* 
1
q
n
p
q
The phase of interchange
So, parity of
p
 pq
must equal parity of pq
p odd
q odd
q even
p even
only trivial insulators
(By passing – this implies that a n=5/2 state must have quarter
charged excitations in its spectrum)
Summary:
1. Gapless edge modes in “fractional spin Hall insulators” are
protected in the case of odd n/e*, as long as time reversal
symmetry is preserved.
2. These modes may be gapped by spin flipping perturbations
when n/e* is even.