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Consider the Kitaev honeycomb model: $\quad -J_x\sum_{x\; links} S_i S_{i+x}- J_y\sum_{y\; links} S_i S_{i+y}- J_z\sum_{z\; links} S_i S_{i+z}$.

From Lieb's theorem, the ground state is given by, $w_p=1$ ($w_p$ being the Wilson-loop operator). Now, $w_p=\prod_{\langle j,k \rangle \in \partial p} \hat{u}_{\langle j,k \rangle}$. Here $\hat{u}_{\langle j,k \rangle}$ is the link operator and $\partial p$, path around a plaquette.

My question is the following. Since $\hat{u}_{\langle j,k \rangle}=\pm 1$ there are eight ways to get $w_p=1$. Therefore does the ground state of Hamiltonian has a degeneracy of eight?

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  • $\begingroup$ $u_{\langle i,j\rangle}$ are not physical observables (i.e. not gauge invariant). In fact, all configurations of $u$ with the same values of $w_p$ are equivalent to each other through gauge transformations. So no, there are no additional degeneracies associated with different $u$ corresponding to the same $w$. $\endgroup$ – Meng Cheng Dec 14 '15 at 22:40
  • $\begingroup$ Thanks. Could you explain (or give a reference on) how one can prove that all configurations of $u$ with the same values of $w_p$ are equivalent to each other through gauge transformations? $\endgroup$ – user101331 Dec 29 '15 at 9:14
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The ground state degeneracy depends on the manifold where you put the model on.

If it's on ${\mathbb R}^2$ (open boundary condition), the ground state is unique. If it's on torus $T^2$ (periodic boundary condition), there is 4-fold groundstate degeneracy.

To illustrate this point, we use square lattice for simplicity. Similar argument works for honeycomb lattice. For open boundary condition, consider a configuration of spins on each edge with $w_p = \prod_{e \in \partial p} s_e = +1$ for each plaquette $p$. You can apply gauge transformation (flipping 4 spins on adjacent edges of a vertex $v$) properly and make $s_e=+1$ for $e$ being the bottom horizontal edges of the lattice. Then, we apply gauge transformation to make $s_e=+1$ for $e$ being the vertical edges of the bottom row. Since gauge transformation doesn't affect $w_p$, we still have $w_p=+1$ for all $p$. Since bottom horizontal edges and vertical edges are all $+1$, the second horizontal edges from bottom are $+1$ automatically. We can keep doing this to fix all vertical edges to be $+1$ and horizontal edges will become $+1$ automatically. We can connect to the configuration with all $s_e=1$. Therefore, there is only one state corresponding to $w_p=1$.

The story is different for torus $T^2$ since there is no "bottom row" for torus. There are 4 different states which can't be connected to each other by gauge transformation. For the details, please see this section 3.1 of Kitaev's paper (https://arxiv.org/pdf/0904.2771.pdf). The basic idea is that the product of spins along two nontrivial cycles on torus $T^2$ can be $\pm 1$. Therefore, there are $2 \times 2=4$ ground states.

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  • $\begingroup$ Can you explain how to generalize this argument to the honeycomb case? thanks! $\endgroup$ – Chuan Chen Mar 29 at 13:22

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