Protected edge modes without symmetry Michael Levin University of Maryland.
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Transcript of Protected edge modes without symmetry Michael Levin University of Maryland.
Protected edge modes without symmetry
Michael LevinUniversity of Maryland
Topological phases
2D quantum many body system with:
Finite energy gap in bulk
Fractional statistics
ei 1
Examples Fractional quantum Hall liquids
Many other examples in principle: Quantum spin systems Cold atomic gases, etc
Edge physics
= 1/3
Some topological phases have robust gapless edge modes:
Edge physics
Fully gapped edge
“Toric code” model
Some phases don’t:
Main question
Which topological phases have protected
gapless edge modes and which do not?
Two types of protected edges
1. Protection based on symmetry(top. insulators, etc.)
2. Protection does not depend on symmetry
Two types of protected edges
1. Protection based on symmetry(top. insulators, etc.)
2. Protection does not depend on symmetry
Edge protection without symmetry
nR – nL 0 protected edge mode
nR = 2nL = 1
Edge protection without symmetry
nR – nL 0 protected edge mode
nR = 2nL = 1
~ KH , “Thermal Hall conductance”
Edge protection without symmetry
What if nR - nL = 0? Can edge ever be
protected?
Edge protection without symmetry
What if nR - nL = 0? Can edge ever be
protected?
Yes!
Examples
= 2/3
nR – nL = 0
Protected edge
Superconductor
Examples
= 8/9
No protected edge
= 2/3
Protected edge
nR – nL = 0nR – nL = 0
SuperconductorSuperconductor
General criterion
An abelian phase with nR = nL can have a gappededge if and only if there exists a subset of quasiparticle types, S = {s1,s2,…} satisfying:
(a) eiss’ = 1 for any s, s’ S “trivial statistics”
(b) If tS, then there exists sS with eist 1 “maximal”
eist
s
t
Examples
= 8/9: Quasiparticle types = {0, e/9, 2e/9,…, 8e/9}
Find S = {0, 3e/9, 6e/9} works
edge can be gapped
2/3: Quasiparticle types = {0, e/3, 2e/3}
Find no subset S works
edge is protected
Microscopic analysis: = 8/9
= 8/9
L = 1/4 x1(t1 – v1 x1) -9/4 x2(t2 – v2 x2)
Electron operators: 1 = ei1
2 = e-9i2
12
Microscopic analysis: = 8/9
L = 1/4 xT (K t- V x)
1 0 v1 0
2 0 –9 0 v2
V =K ==
Microscopic analysis: = 8/9
Simplest scattering terms:
“U 1m 2
n + h.c.” = U Cos(m - 9n)
Will this term gap the edge?
Null vector criterion
Can gap the edge if
(m n) 1 0 = 0 0 -9
Guarantees that we can chg. variables to = m1 - 9n2, = n1 + m2 with:
L x t – v2 (x)2 – v2 (x)2 + U cos()
mn
Null vector criterion
(m n) 1 0 = 0 0 -9
m2 – 9 n2 = 0
Solution: (m n) = (3 -1)
U cos(31 + 92) = “13 2
* + h.c” can gap edge.
mn
Microscopic analysis: = 2/3
= 2/3
L = 1/4 x1(t1 – v1 x1) -3/4 x2(t2 – v2 x2)
Electron operators: 1 = ei1
2 = e-3i2
12
Null vector condition
(m n) 1 0 = 0 0 -3
m2 – 3 n2 = 0
No integer solutions.
(simple) scattering terms cannot gap edge!
mn
General case
Edge can be gapped iff there exist {1,…,N}
satisfying:
iT K j = 0 for all i,j (*)
Can show (*) is equivalent to original criterion
2N
Problems with derivation
1. Only considered simplest kind of backscattering terms
proof that = 2/3 edge is protected is not complete
2. Physical interpretation is unclear
Annihilating particles at a gapped edge
Annihilating particles at a gapped edge
ss
Annihilating particles at a gapped edge
ss
Annihilating particles at a gapped edge
ss
Annihilating particles at a gapped edge
Annihilating particles at a gapped edge
a b
ss
“s, s can be annihilated at the edge”
Annihilating particles at a gapped edge
Define:
S = {s : s can be annihilated at edge}
Constraints from braid statistics
Ws
Have: Ws |0> = |0>
Constraints from braid statistics
Ws Ws’
Have: Ws’ Ws |0> = |0>
Constraints from braid statistics
Ws Ws’
Similarly: Ws Ws’ |0> = |0>
Constraints from braid statistics
On other hand: Ws Ws’ |0> = eiss’ Ws’Ws |0>
Ws Ws’
Constraints from braid statistics
On other hand: Ws Ws’ |0> = eiss’ Ws’Ws |0>
Ws Ws’
Constraints from braid statistics
On other hand: Ws Ws’ |0> = eiss’ Ws’Ws |0>
Ws Ws’
eiss’ = 1 for any s, s’ that can be annihilated at edge
Braiding non-degeneracy in bulk
t
Braiding non-degeneracy in bulk
If t can’t be annihilated (in bulk) thenthere exists s with eist 1
t s
Braiding non-degeneracy at a gapped edge
t
Braiding non-degeneracy at a gapped edge
If t can’t be annihilated at edge then there exists s with eist 1 which CAN be annihilated at edge
t
Braiding non-degeneracy at edge
Have:
(a). eiss’ = 1 for s,s’ S
(b). If t S then there exists s S with eist 1
Proves the criterion
Summary Phases with nL– nR = 0 can have protected
edge
Edge protection originates from braiding statistics
Derived general criterion for when an abelian topological phase has a protected edge mode