# Does the existence of soft Goldstone mode makes a topological superfluid unstable?

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.

• The name BCS superfluid is a bit confusing : some people call the BCS state a superfluid though it is a superconducting state, but I think you really mean the superfluid state, which is not per se a BCS state. I'm not sure the Goldstone mode would interact with the superfluid condensate as you mention, hence the topological protection : only pairs of particles (or quasi-particles) may enter the condensate, at least below the gap. – FraSchelle Nov 25 '16 at 11:17

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.
• What you have proved is that $B_k$ cannot introduce an error in the code space, which I definitely agree. However, what I'm worried about here is that the perturbation $H'=B_k+B^\dagger_k$ can easily drive a qubit to an excited state, i.e. if we start with $\alpha |0\rangle+\beta |1\rangle$, $H'$ may drive it to something like $|2\rangle$ (completely destroy the quantum information), since $\Delta=E_2-E_0$ is very small. – Lagrenge Dec 10 '16 at 1:57