Physics Asked by Lagrenge on July 29, 2020
When people talk about topological order and Majorana fermions in a BCS superfluid, do they admit the existence of a massless Goldstone mode? (As I’ve heard that soft Goldstone mode always exists in superfluid)
If this is the case, it would mean that the topological order is unstable. Say in 1D case, we have two degenerate ground states due to unpaired Majorana edge modes $|G_{1,2}rangle$. The nonlocal nature of Majorana guarantees that any local perturbation cannot either split their energy or make a transition between them. However, if the Goldstone mode creation operator is given by $B^dagger_{k}$, an arbitrarily small perturbation $H’=B_k+B^dagger_{k}$ would induce a transition from the ground state to an excited state, destroying the topologically-protected qubit.
No. $B_k$ conserves fermion parity, while the degenerate ground states have opposite fermion parities. So the matrix element of $B_k$ between the two ground states is strictly 0.
Answered by Meng Cheng on July 29, 2020
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