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Protected edge modes without symmetry Michael Levin University of Maryland.

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Presentation on theme: "Protected edge modes without symmetry Michael Levin University of Maryland."— Presentation transcript:

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 –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 = 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 = m 2 – 9 n 2 = 0 Solution: (m n) = (3 -1) U cos( ) = * + 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 = 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

30

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


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