# How the Vortex containing majorana bound state is non-abelian statistics

Recently,I read some papers about non-abelian statistics of majorana fermion, such as: Majorana Returns F. Wilczek http://www.nature.com/nphys/journal/v5/n9/full/nphys1380.html and Non-Abelian Statistics of Half-Quantum Vortices in p-Wave Superconductors D. A. Ivanov http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.86.268.

All of these tell me that the interchanging between two vortexes containing majorana fermion or zero mode will lead to a anyonic matrix phase factor. This conclusion deeply perplex me because the basic fact that the majorana fermion is a fermion which many body non-interacting wavefunction is interchanging anti-symmetry and the resulting phase factor is -1. I don't understand why the vortex containing it is non-Abelian statistics in the way of picture even though I carefully check the mathematics in Ivanov's paper.

I also read the similar question: Majorana particles statistics (Majorana particles statistics) and the answer of prof. Wen.From his answer, I have know that this a confusing. But, I still don't be clear about why the interchanging of vortexes lead to a non-Abelian operation,when these vortexes are represented by majorana field operator $\gamma$ in Ivanov's paper. In my option, at this time, the vortexes are also fermion which only phase factor -1. Which point I was wrong? Please help me, thanks.
This is why people start to abandon calling the zero modes in vortices "Majorana fermions" because they are NOT fermions. $\gamma$ is a Majorana zero mode, which means it always has to pair up with another Majorana zero mode to form a 2-dimensional Hilbert space. Exchanging vortices generates a nontrivial unitary transformation on the degenerate space.
• you mean that the majorana zero mode is not a majorana fermion? It hard to understand that because the field operator $\gamma$ satisfying self-conjugate and the corresponding quasiparticle is fermion(anti-commute). Did you mean there is only the feature of "self-conjugate" (so named "majorana"), but no the feature of fermion? thanks – alxandernashzhang Apr 3 '15 at 1:07
• Self-conjugation is special. Usually fermion operators satisfy $c^2=(c^\dagger)^2=0$. – Meng Cheng Apr 3 '15 at 3:04