# Double semion model on a square lattice

We consider the double semion model proposed in Levin and Wen's paper

http://arxiv.org/abs/cond-mat/0404617

http://journals.aps.org/prb/abstract/10.1103/PhysRevB.71.045110

In their paper, the double semion model is defined on a honeycomb lattice.

Now I am trying to study the same model on a square lattice.

Question 1: Is the following Hamiltonian correct?

$$H=-\sum_{\textrm{vertex}} \prod_{k \in \textrm{vertex}}\sigma_{k}^{z} + \sum_{\textrm{plaquette}} \left[ \prod_{j \in \textrm{legs}} i^{(1-\sigma_{j}^{z})/2} \right] \prod_{k \in \textrm{plaquette}} \sigma_{k}^{x}.$$ On the figure there are totally 8 green legs around each plaquette. As shown in Levin and Wen's paper, the ground state of the double semion model is the equal-weight superposition of all close loops, and each loop contributes a minus sign. Given a loop configuration, the wave function component is given by $(-1)^{\textrm{number of loops}}$. If we have even (odd) number of loops, the wave function component of this configuration is $+1$ ($-1$). On the honeycomb lattice everything looks fine. But I am confusing about the state on the square lattice when the strings are crossing.

Question 2: For the following two configurations, should we regard them as one loop or two loops? Do they have the same amplitude in the ground state wave function?

Here we consider a $3 \times 3$ torus, i.e., we have periodic boundary conditions on both directions. The red line denotes the string, i.e., the spin is $\left| \downarrow \right\rangle$ on each red link. This is configuration I. This is configuration II.

• First thing to check is to make sure that all terms in your Hamiltonian commute with each other. Then to check whether the ground state wavefunction is a double semion, let us consider one of the plaquette term acting on a state with no strings(i.e. $\sigma_z=-1$ everywhere). The term creates a closed string along the plaquette, but the phase factor $\prod_{j\in\text{legs}}i^{(1-\sigma^z_j)/2}=i^8=1$ (while on a honeycomb lattice it is $i^6=-1$ ). So it does not seem to work out. – Meng Cheng Jan 28 '15 at 4:38
• @MengCheng I think I am using a different notation. string means $\sigma_{z}=-1$, and no string means $\sigma_{z}=+1$. If we create a one-plaque string from the no-string state, the phase factor is $i^0=1$ for both square and honeycomb lattice. In this case, is the Hamiltonian correct? Thanks! – No. 9999 Jan 28 '15 at 4:51
• @No.9999 But if you create a single loop, the sign should change! --- This also tells you that you cannot just change your convention for what you call "string" in the double semion model. (This is indeed different for the Toric Code.) – Norbert Schuch Jan 28 '15 at 11:27
• @NorbertSchuch I think we can still use this convention. Please note that I put a plus sign before the plaquette terms, while for the toric code model we usually put a minus sign at the same place. Thanks to your answer below, I have found the correct Hamiltonian by counting the phase factor in a completely different way (but still use my old convention). – No. 9999 Jan 28 '15 at 20:32
• Hi @No.9999, I'm interested in the correct form of the double semion Hamiltonian on the square lattice. May I know what it is? Thanks. – nervxxx Mar 24 '15 at 18:19