# Protected edge modes without symmetry Michael Levin University of Maryland.

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Protected edge modes without symmetry Michael Levin University of Maryland

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

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 n R – n L 0 protected edge mode n R = 2 n L = 1

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

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

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

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

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

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

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 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

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 = =

Microscopic analysis: = 8/9 Simplest scattering terms: U 1 m 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 = m 1 - 9n 2, = n 1 + m 2 with: L x t – v 2 ( x ) 2 – v 2 ( x ) 2 + U cos( ) mnmn

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

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

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

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

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

ss

ss

ss

ab 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 WsWs Have: W s |0> = |0>

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

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

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

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

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

Braiding non-degeneracy in bulk t

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

Braiding non-degeneracy at a gapped edge t

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

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

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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