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?

  • $\begingroup$ $u_{\langle i,j\rangle}$ 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, 2015 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, 2015 at 9:14

1 Answer 1


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.

  • $\begingroup$ Can you explain how to generalize this argument to the honeycomb case? thanks! $\endgroup$
    – Ogawa Chen
    Mar 29, 2021 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.