Download presentation

Presentation is loading. Please wait.

Published byMadison Paley Modified over 2 years ago

1
Protected edge modes without symmetry Michael Levin University of Maryland

2
Topological phases 2D quantum many body system with: Finite energy gap in bulk Fractional statistics e i 1

3
Examples Fractional quantum Hall liquids Many other examples in principle: Quantum spin systems Cold atomic gases, etc

4
Edge physics = 1/3 Some topological phases have robust gapless edge modes:

5
Edge physics Fully gapped edge Toric code model Some phases dont:

6
Main question Which topological phases have protected gapless edge modes and which do not?

7
Two types of protected edges 1. Protection based on symmetry (top. insulators, etc.) 2. Protection does not depend on symmetry

8
Two types of protected edges 1. Protection based on symmetry (top. insulators, etc.) 2. Protection does not depend on symmetry

9
Edge protection without symmetry n R – n L 0 protected edge mode n R = 2 n L = 1

10
Edge protection without symmetry n R – n L 0 protected edge mode n R = 2 n L = 1 ~ K H, Thermal Hall conductance

11
Edge protection without symmetry What if n R - n L = 0? Can edge ever be protected?

12
Edge protection without symmetry What if n R - n L = 0? Can edge ever be protected? Yes!

13
Examples = 2/3 n R – n L = 0 Protected edge Superconductor

14
Examples = 8/9 No protected edge = 2/3 Protected edge n R – n L = 0 Superconductor

15
General criterion An abelian phase with n R = n L can have a gapped edge if and only if there exists a subset of quasiparticle types, S = {s 1,s 2,…} satisfying: (a) e i ss = 1 for any s, s S trivial statistics (b) If t S, then there exists s S with e i st 1 maximal e i st s t

16
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

17
Microscopic analysis: = 8/9 = 8/9 L = 1/4 x 1 ( t 1 – v 1 x 1 ) -9/4 x 2 ( t 2 – v 2 x 2 ) Electron operators: 1 = e i 1 2 = e -9i 2 1 2

18
Microscopic analysis: = 8/9 L = 1/4 x T (K t - V x ) 1 0 v 1 0 2 0 –9 0 v 2 V =K = =

19
Microscopic analysis: = 8/9 Simplest scattering terms: U 1 m 2 n + h.c. = U Cos(m - 9n ) Will this term gap the edge?

20
Null vector criterion Can gap the edge if (m n) 1 0 = 0 0 -9 Guarantees that we can chg. variables to = m 1 - 9n 2, = n 1 + m 2 with: L x t – v 2 ( x ) 2 – v 2 ( x ) 2 + U cos( ) mnmn

21
Null vector criterion (m n) 1 0 = 0 0 -9 m 2 – 9 n 2 = 0 Solution: (m n) = (3 -1) U cos(3 1 + 9 2 ) = 1 3 2 * + h.c can gap edge. mnmn

22
Microscopic analysis: = 2/3 = 2/3 L = 1/4 x 1 ( t 1 – v 1 x 1 ) -3/4 x 2 ( t 2 – v 2 x 2 ) Electron operators: 1 = e i 1 2 = e -3i 2 1 2

23
Null vector condition (m n) 1 0 = 0 0 -3 m 2 – 3 n 2 = 0 No integer solutions. (simple) scattering terms cannot gap edge! mnmn

24
General case Edge can be gapped iff there exist { 1,…, N } satisfying: i T K j = 0 for all i,j (*) Can show (*) is equivalent to original criterion 2N

25
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

26
Annihilating particles at a gapped edge

27
ss

28
ss

29
ss

31
ab ss s, s can be annihilated at the edge

32
Annihilating particles at a gapped edge Define: S = {s : s can be annihilated at edge}

33
Constraints from braid statistics WsWs Have: W s |0> = |0>

34
Constraints from braid statistics WsWs WsWs Have: W s W s |0> = |0>

35
Constraints from braid statistics WsWs WsWs Similarly: W s W s |0> = |0>

36
Constraints from braid statistics On other hand: W s W s |0> = e i ss W s W s |0> WsWs WsWs

37
Constraints from braid statistics On other hand: W s W s |0> = e i ss W s W s |0> WsWs WsWs

38
Constraints from braid statistics On other hand: W s W s |0> = e i ss W s W s |0> WsWs WsWs e i ss = 1 for any s, s that can be annihilated at edge

39
Braiding non-degeneracy in bulk t

40
If t cant be annihilated (in bulk) then there exists s with e i st 1 t s

41
Braiding non-degeneracy at a gapped edge t

42
If t cant be annihilated at edge then there exists s with e i st 1 which CAN be annihilated at edge t

43
Braiding non-degeneracy at edge Have: (a). e i ss = 1 for s,s S (b). If t S then there exists s S with e i st 1 Proves the criterion

44
Summary Phases with n L – n R = 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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google