I just read a paper by M. Levin and X.-G. Wen (https://arxiv.org/abs/cond-mat/0510613) where they present a general formula to evaluate the topological entanglement entropy $\gamma$. They also prove that $\gamma$ is related to the quantum dimension of the point-like excitations in the theory as follows \begin{equation} \gamma=\log{\sqrt{\sum_a d_a^2}}, \end{equation} where $d_a$ is the quantum dimension of a particle excitation. They derived this formula for a ground state configuration, but i was wondering if it can be applied also to the excited states. Here is my logic.
Consider a gapped Hamiltonian $H_0$ describing a system with a set of point-like excitations $\{ d_a \}$. The ground state of $H_0$ is denoted as $\psi_0$, while $\psi_1$ is the first excited state. $\psi_0$ has topological entanglement entropy $\gamma$.
Now consider a new gapped Hamiltonian $H_1$ describing a system with the same point-like excitations $\{ d_a \}$. Assume that the ground state of $H_1$ is $\psi_1$. Then also $\psi_1$ has topological entanglement entropy $\gamma$, since the excitations $\{ d_a \}$ are the same.
If such $H_1$ exists we conclude that the topological entanglement entropy is the same for $\psi_0$ and $\psi_1$.
Given this logic my question is, is it actually correct to say that $\gamma$ is the same for any state of the system?
