I'm new here, loving this website and I'm having some difficulty with the wilson-loop operator in kitaev's honeycomb model.

problem statement The Kitaev model (Kitaev, 2006 is the original paper) consists of spins residing at the lattice sites of a honeycomb lattice with separate nn couplings for the three directions that are identified for the bonds. The wilson-loop operator is $w_p=\sigma^x_1 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6$, where the indices $i \in\{1,...,6\} $ indicate the $6$ lattice sites involved in the hexagonal loop (see picture).

Kitaev honeycomb loop

In Jiannis K. Pachos' book (Introduction to topological quantum computation, 2012) the author states that $(w_p)^2=1$, which I'm trying to find myself. Actually this should be not at all hard, but I'm stuck unfortunately.

attempt at solution I've tried the following $$ (w_p)^2 = \sigma^x_1 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \sigma^x_1 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \\ = -\sigma^x_1 \sigma^x_1 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \\ = -\sigma^y_2 \sigma^y_2 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \\ = + \sigma^z_3 \sigma^z_3 \sigma^x_4 \sigma^y_5 \sigma^z_6 \sigma^x_4 \sigma^y_5 \sigma^z_6 \\ = + \sigma^x_4 \sigma^x_4 \sigma^y_5 \sigma^z_6 \sigma^y_5 \sigma^z_6 \\ = - \sigma^y_5 \sigma^y_5 \sigma^z_6 \sigma^z_6 \\ =-1 $$

Where I've pulled $\sigma_1$ through first, next the $\sigma_2$, etc. And I've used $\{\sigma^\alpha_i , \sigma^\beta_j \}= 2 \delta_{i,j}\delta_{\alpha,\beta} I_2 $ (so that every swapping of unequal $\sigma$'s gives a minus sign and $\sigma^\alpha_i\sigma^\alpha_i =I_2$ ).

So I get $(w_p)^2=-1$, which is not what I wanted to find. All text on the subject state that $w_p$ acting on a lattice configuration yield $w_p=\pm1 $ which can be easily concluded from $(w_p)^2=1$ (the expression I didn't get).

My guess is that my commutation relations are not correct, but I'm unsure. Who can help me out? A big thanks in advance!

Best, L

  • $\begingroup$ $w_p$ is a tensorial product (the site indices are different), not a simple product. $\endgroup$ – Trimok Aug 27 '14 at 15:12
  • $\begingroup$ From your formulas, $\omega_p^2=(-1)^{5+4+3+2+1}=-1$. I wonder whether there is an $i$ missing in your definition of $\omega_p$? $\endgroup$ – innisfree Aug 27 '14 at 16:59

As you have in the commutation relations, $\sigma_i \sigma_j= \sigma_j\sigma_i$ e.g. spin operators on different sites commute, so there is no minus sign to pick up.

  • $\begingroup$ Thanks! Fundamental things like this are what I often forget. $\endgroup$ – L de Pudo Aug 28 '14 at 7:56
  • 1
    $\begingroup$ Is it true? Don't spin-operators on different sites anti-commute? $\{\sigma^\alpha_i,\sigma^\beta_j\}=\delta_{ij}\delta_{\alpha\beta}$. For distinct sites $\alpha\neq\beta$, $\{\sigma^\alpha_i,\sigma^\beta_j\}=0$. The spin operators anti-commute. $\endgroup$ – innisfree Aug 28 '14 at 11:46
  • $\begingroup$ The spin operators anti-commute on the same site, but commute if they're on different sites. If found a quite useful book here that treats the fact. $\endgroup$ – L de Pudo Sep 4 '14 at 13:23

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