How can a torus admit half a flux quantum, and why does a vortex induce an AB phase?

There is an issue that I have with the argument given in “Topological Degeneracy of non-Abelian States for Dummies” http://arxiv.org/abs/cond-mat/0607743 , regarding the ground state degeneracy of the Pfaffian state on a torus: This is They argue that adiabatic pairwise annihilation of quasiparticles around the $x$-direction (implemented by $T_x$) should be equivalent to inserting half a flux quantum into the y-direction “hole” of the torus, i.e.

$$T_x=F_y^2.$$

Explicitly, this is described in the following paragraph, on page 9, Section 4: Consistency between non-Abelian statistics and charge fractionalization:

The issue I have with this is that the magnetic field in a vortex is localized at a point, and is not like the uniform field described by "insertion of half a flux quantum into a hole", and so the authors still have to prove that a quasiparticle encircling the appropriate fundamental cycle of the torus still picks up the same Aharonov Bohm phase, despite the fact that the magnetic fields are different in each case.

Moreover, I do not believe that it is possible for for half a flux quantum to exist: then the Chern number of the electromagnetic $U(1)$-bundle would then be 1/2, not an integer, which violates mathematics!!!

What's going on here?

• Your last objection has a well-known resolution: all that means is that fractional value of Hall conductance (the "Chern number") requires ground state degeneracy. So when you adiabatically insert $2\pi$ $U(1)$ flux, you actually move to a different ground state. No mathematics is violated. – Meng Cheng Dec 9 '15 at 1:27
• "actually move to a different ground state". - Could you elaborate on that point? Also, we're talking about half a flux-quantum, so that would be $\pi$ flux. – David Roberts Dec 9 '15 at 1:58
• Sorry, I was thinking about a unit flux quanta. Please discard my previous comment since it was probably irrelevant in this context. – Meng Cheng Dec 9 '15 at 2:18

The existence of half flux quanta is special to Moore-Read state (the paper cited in the OP called it the Pfaffian state). Let us take the simplest example, the Moore-Read state at $\nu=1/2$, corresponding to $M=1$ in the paper. The quasiparticles are the following: the trivial electrons, a charge-$e/2$ quasiparticle, a neutral fermion, and charge-$\pm e/4$ non-Abelian quasiparticle (often called the quasiholes, and the paper called them the vortex/anti-vortex). It is useful to use the composite fermion picture, where the composite fermion is one electron plus two units of flux quanta, and the Moore-Read state can be thought of as the $p+ip$ superconductor of the composite fermions. Of course there is also the charge sector, which is very important in this discussion. More formally, we can construct the Moore-Read state by a parton approach, write the electron operator $c$ as $c=\psi b$ where $\psi$ is the neutral composite fermion and $b$ is a charge-$e$ boson. We then put $\psi$ fermions into a $p+ip$ superconductor, and $b$ into a $\nu=1/2$ bosonic Laughlin state. It is not hard to see that the resulting wavefunction is exactly the same as the Moore-Read state. The charge-$e/4$ quasiparticle is then the vortex for the composite fermion.
Now we can understand the effect of external U(1) flux insertion. Inserting a $2\pi$ U(1) flux nucleates the Abelian charge-$e/2$ quasiparticle, as one would expect for a $\nu=1/2$ bosonic Laughlin state. Now the question is whether a $\pi$ U(1) flux is allowed. The reason that usually we do not allow such non-unit fluxes is because they introduce a "branch cut" into the wavefunction of electrons, so can not represent a quasi-particle type excitation of the system, which is the objection raised by the OP near the end (although phrased differently). However, what happens in the Moore-Read state is that when a $\pi$ U(1) flux is inserted, the $b$ boson sees it as a $\pi$ flux, but at the same time a vortex in the $p+ip$ supercondutor is also created, so the $\psi$ fermion also sees a $\pi$ flux (from the vortex). The electron thus sees no effectively no flux, so there is no "branch cut". Therefore, half flux quanta insertion creates a charge-$e/4$ "vortex" quasiparticle.