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In the book of "Introduction To Topological Quantum Computation" by Jiannis K. Pachos, in chapter 5, it tries to explain anyonic exchange.

In the following, the $m$ and $e$ quasi-particles are produced by Pauli matrices $σ_x$ and $σ_z$, and the $ϵ$ is made out of their combination.

In the picture below, which is about exchanging two $ϵ$ quasi-particles, I just cannot figure out why the string (or loop) that resulted in the exchange is constructed like the one which comes in the following explanation of the image. Why is it like that? In particular, why is the first parenthesis different from the second one?

enter image description here

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Hmm it's hard to add drawing, but if you apply those parenthesis from right to left to the given figure, you indeed see two Fermionic excitations are swapped. Note that in that Kitaev model, single application of $\sigma_z$ makes two e anyons(same for $\sigma_x$ and m anyons). Also here $e^2\sim m^2\sim 1$. Therefore, if you apply $\sigma_z$ at a point near an e anyon, it actually moves the anyon by one lattice point.

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  • $\begingroup$ Dear Hwang, my main question is why those parenthesis are not symmetric? What's the reason behind this choice? If we choose the parenthesis to be symmetric, then it will not be Fermionic, am I right? $\endgroup$ – ALirezaSO May 29 '19 at 7:35
  • $\begingroup$ Anyon exchange statistics is defined by swapping the two anyons following through a 'loop'. If the parentheses are symmetric, they do not make a loop, but two anyons penetrate each other, since they follow exactly the same path. This is impossible. Remember it's a 2D physics, lattice-ed version. Asymmetric parenthesis choice given above is a minimal choice for making a loop. If you want to check, you can draw a more clear(larger) loop, so one anyon goes $\rightarrow \downarrow$ and the other one goes $\leftarrow \uparrow$. Although it contains more Pauli matrices, the result will be $-1$. $\endgroup$ – hwang May 29 '19 at 19:56
  • $\begingroup$ Dear hwang, I hope not to waste your time. I just need to understand the orders. If I write the parenthesis (σz1.σx1.σz2.σx2)(σx2.σz2.σx1.σz1) it still returns us the same loop but the value then is 1 instead of -1. Hence, the value is conditional according to the path. If I look at it like a bunch of rotations, I describe the rotation for the first particle as a CCW then a CW rotation and for the second, two CW rotations. If the second's rotations be like the first one, then the value becomes one. Is it right? $\endgroup$ – ALirezaSO Jun 1 '19 at 6:34

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