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I am interested in the relation between the following three phases of matter (in 2D):

  • chiral $p$-wave superconductor (spineless $p_x + i p_y$ pairing)
  • $\nu=5/2$ fractional quantum Hall state
  • A-phase of ${}^3$He

I have read that all of these are topological ordered having Ising anyons as elementary excitations. All of them are at mean-field level described by a BCS theory.

However, I notice the following important difference:

  • the condensate in the superconductor is charged whereas ${}^3$He is a superfluid (so only the former has a Higgs-mechanism)
  • in $\nu=5/2$ composite fermions condense

So my question is if these three states are really equivalent (which is not clear to me). In particular, I am interested in the following:

  • do all show a degeneracy of the ground state when put on the torus? (for the superconductor this is not clear as typically, we need a vortex to bind a zero energy state)
  • is there a superflow of pairs of composite fermions in $\nu=5/2$ and what this corresponds to physically?

I am looking forward to your insights. As I have browed the internet already for some time without any good explanation, I would be also grateful if you could simply post some reference where the differences are discussed.

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  • $\begingroup$ In 5/2, pairing happens between composite fermions which is a charge attached to a vortex. A change of magnetic field here corresponds to application of a field in a supercond. Resistance to this ie incompressibility then corresponds to Meissner effect, and quantization of quasihole excitations correspond to the flux quantization. Since vortices are expelled in Meissner phase, and charges are attached to vortices, any charge current is expelled to the edge, so I think, the edge current of the 5/2 system corresponds to a supercurrent. $\endgroup$ – symanzik138 Jan 28 '16 at 19:03
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Chiral $p$-wave superconductor and He A phase can be considered being equivalent phases of matter, for the following reason: The fact that the superconductor is charged does not make too much difference in this regard, because it only affects the electromagnetic response (one has Meissner effect, the other does not), but electromagnetic field is an external probe, and thus is not a part of the dynamics of the system itself. It is important to clarify what "topological order" means in this context. The superconductor/superfluid do not have topological entanglement entropy in the ground state; they are essentially non-interacting fermions. They do not have Ising anyons as true low-energy excitations of the Hamiltonian, but as vortices which can be induced externally. If the electromagnetic field is somehow treated as part of the system (so we really have a gauge theory, $p+ip$ coupled to a $Z_2$ gauge field), then the vortices become true anyonic excitations of the system. This is what happens in Kitaev's honeycomb lattice model (the gapped B phase), and is then similar to the $\nu=5/2$ Moore-Read state, although not quite the same topological order.

Regarding the ground state on torus, as you observed, naively one can have four ground states for chiral p-wave superconductor/superfluid, corresponding to periodic/anti-periodic boundary conditions for fermions. Physically, however the twisted boundary condition induces a finite gradient energy from the stiffness of the superconductor (as you already observed, boundary conditions can be modified by dragging a vortex around a non-contractible loop), and therefore the four states do not have the same energies. Again, if the gauge field is dynamical they will become nearly degenerate , and one of the sector has an odd number of fermions, which must be thrown away from the gauge theory (since in gauge theory the physical state must be gauge-invariant). So one obtains three degenerate ground states, agreeing with Ising anyon model.

The Moore-Read state has a different topological order than the Ising anyons. One can think of the MR state as the $p+ip$ superconductor of composite fermions, but there is also a charged sector.

A very insightful discussion of these issues can be found in http://arxiv.org/abs/1212.6395.

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  • $\begingroup$ Thank for the nice answer and the reference (it looks indeed very useful). There is one thing which I have trouble to understand: how you get to the number 3 (ground state degeneracy) -> how you can say which states has an odd number of fermions? $\endgroup$ – nesseril Jan 28 '16 at 18:35
  • $\begingroup$ This is a simple exercise: take your favorite lattice model of $p+ip$ superconductor, determine the fermion parity of the ground state on a torus with periodic/anti-periodic boundary conditions. You go to momentum space, and observe that except four high symmetry points ( $(0,0), (0, \pi), (\pi,0), (\pi,\pi)$, all other momentums are paired up, so we only need to look at these four points. The pairing vanishes at these four points, so we just need to find whether these four points are occupied or not. $\endgroup$ – Meng Cheng Jan 28 '16 at 19:14

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