6
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jan 28, 2015 at 4:38
  • $\begingroup$ @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! $\endgroup$ Commented Jan 28, 2015 at 4:51
  • $\begingroup$ @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.) $\endgroup$ Commented Jan 28, 2015 at 11:27
  • $\begingroup$ @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). $\endgroup$ Commented Jan 28, 2015 at 20:32
  • $\begingroup$ 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. $\endgroup$ Commented Mar 24, 2015 at 18:19

1 Answer 1

Reset to default
4
$\begingroup$

The defining property of the double semion model is the nature of the ground state as a superposition of loop pattern with alternating signs, and not the form of its Hamiltonian. As you noticed, it is not clear how to count loops on a square lattice. As far as I see, this is one reason why the string-net models are defined on honeycomb lattices, since it allows to count loops unambiguously. (In fact, any trivalent graph would do.)

If you want to define a way to count loops on a square lattice, one way is to "decorate" it such that it becomes a trivalent lattice, this is, you replace every four-valent vertex by two three-valent vertices with an edge inbetween. The state of the extra edge is uniquely determined by the state on the surrounding edges, and thus, this gives you a way to count loops on the square lattice. In the same way, you can map the honeycomb Hamiltonian to a new Hamiltonian on the square lattice. Note, however, that this mapping will necessarily break some lattice symmetry.

Your Hamiltonian is rotationally invariant, so I suspect it is not the correct Hamiltonian. I have not analyzed it carefully, but you could try to exactly diagonalize it on a 4x4 lattice and check the ground subspace. Alternatively, you can study different moves to go from one configuration to another and check if they all give the same phase (I suspect that no, and there will be cancellations). For this, of course, you first have to pick a convention of how to count loops.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks a lot! Indeed, we can get the correct state and the correct Hamiltonian by mapping the honeycomb lattice to the square lattice. The correct Hamiltonian is not rotationally invariant any more. @Norbert Schuch $\endgroup$ Commented Jan 28, 2015 at 20:16

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.