Download presentation

Presentation is loading. Please wait.

Published byMadison Paley Modified over 3 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

OK

Funded by NSF, Harvard-MIT CUA, AFOSR, DARPA, MURI Takuya Kitagawa Harvard University Mark Rudner Harvard University Erez Berg Harvard University Yutaka.

Funded by NSF, Harvard-MIT CUA, AFOSR, DARPA, MURI Takuya Kitagawa Harvard University Mark Rudner Harvard University Erez Berg Harvard University Yutaka.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on synthesis and degradation of purines and pyrimidines rings Ppt on python programming language Ppt on dubai shopping festival Ppt on van de graaff generator hair Ppt on network security issues Seeley's anatomy and physiology ppt on cells Ppt on building information modeling jobs Ppt on cross docking facility Chapter 1 introduction to human anatomy and physiology ppt on cells Ppt on nanotechnology