Phase transition in Ising Model with local $\mathbb{Z}_2$ symmetry

Question

I am studying the Ising model with a local $\mathbb{Z}_2$ gauge symmetry

\begin{equation} \mathcal{H} = -\sum_{\text{plaquettes}} \sigma^z(\vec{x}, \vec{\mu})\sigma^z(\vec{x}+\vec{\mu}, \vec{\nu})\sigma^z(\vec{x}+\vec{\mu}+\vec{\nu}, -\mu)\sigma^z(\vec{x}+\vec{\nu},-\vec{\nu}) \end{equation} where $\mu,\nu$ are the unit vectors along the $x$ and $y$ directions respectively. Note that the spins are place on the bonds between two sites instead of the site itself. For the unfamiliar reader, this paper might be of use. Another resource is Field Theories of Condensed Matter Physics by E. Fradkin. This model has a local gauge symmetry which corresponds to flipping the spins emanating from a site.

This model has a phase transition but does not have a local order parameter. I ran a Monte Carlo simulation on the above Hamiltonian but the result was inconclusive as I did not see any formation of domains as in the case of the on-site Ising Model. Did I make a mistake or this is the expected behavior?

This is the final picture after 1000 Monte Carlo sweeps. The initial picture looks similarly grainy. Should it not be the case that the picture of spins gives hint towards a phase transition unlike here? I am confused.

Answer 1

The model you wrote is called the $\mathbb{Z}_2$ lattice gauge theory (LGT). Two important notes:

1. In two spatial dimensions (which your simulation appears to be in), the $\mathbb{Z}_2$ LGT does not have a phase transition at finite temperatures. It is always disordered. There are a number of ways to see this, but the simplest is to just calculate the partition function -- it is exactly solvable.

2. In a $d \geq 3$-dimensional system, when the model does have a phase transition, it will not be characterized by the formation of domain walls. This is what occurs in an Ising model with spontaneous symmetry breaking, but not in a LGT. Instead, the two phases are characterized by the behavior of Wilson loops. For a set of links $\ell$ forming a closed loop $\gamma$ in the lattice, you should find: $$ \left\langle \prod_{\ell \in \gamma} \sigma_{\ell} \right\rangle \sim \begin{cases} e^{-f(T) \text{Area}(\gamma)}, & T > T_c \\ e^{-g(T) \text{Perim}(\gamma)}, & T < T_c \end{cases} $$ where $f(T)$ and $g(T)$ are smooth functions of $T$, $\text{Area}(\gamma)$ gives the two-dimensional area enclosed by $\gamma$, and $\text{Perim}(\gamma)$ gives the perimeter of $\gamma$. These are called the area law and perimeter law, respectively.

Answer 2

I'm actually not sure if this is the $\mathbb{Z}_2$ lattice gauge theory (LGT). This is because $\mathbb{Z}_2$ LGT needs to have spins on top of the edges of the square lattice and not the vertices. These give you different models. The way it's explained in the original question \begin{equation} \sigma^z(\vec{x},\vec{\mu})\sigma^z(\vec{x}+\vec{\mu},\vec{\nu}) \sigma^z(\vec{x}+\vec{\mu}+\vec{\nu},-\mu) \sigma^z(\vec{x}+\vec{\nu},-\vec{\nu}) \end{equation} contains at least one mistake ($\mu$ without the $\vec{}$ sign), and seems a bit off. If we are using the usual square lattice and the lattice points can be written as $(x,y)$ with $x,y\in\mathbb{Z}$, then something like \begin{equation} \sigma^z(x,y)\sigma^z(x+1,y) \sigma^z(x,y+1) \sigma^z(x+1,y+1) \end{equation} would totally make sense. This corresponds to having spins on top of the vertices of the square lattice and for each plaquette you have a four-body term. The $\mathbb{Z}_2$ LGT on the other hand, will have terms that look like \begin{equation} \sigma^z(x+1/2,y)\sigma^z(x+1/2,y+1) \sigma^z(x,y+1/2) \sigma^z(x+1,y+1/2) \end{equation} where the $+1/2$ corresponds to having spins in the middle of the edges instead of vertices.

If you were actually interested in the former, that is called the plaquette Ising model (in $2d$), and it has subsystem symmetry, which is different from the gauge symmetry $\mathbb{Z}_2$ LGT has. A quick demonstration of this fact is that if you flip 4 spins on the four edges coming out of a single vertex in LGT, that will give you exactly the same energy. This is exactly what local/gauge symmetry is in this model, and if you have $N$ vertices, you have $O(2^N)$ different states that give you exactly the same energy. This is like the $\mathbb{Z}_2$ symmetry the original Ising model has, but in steroids (there you only had 2 in general). On the other hand, the plaquette Ising model doesn't have that, but you can still flip $L=\sqrt{N}$ spins that are on a single $x$ axis. This gives you $O(2^{2L})$ degeneracy, which is intuitively something between global symmetry and local symmetry, named subsystem symmetry. You can search for papers that talk about plaquette Ising model and they show first-order transitions, but you need to define different order parameters.

Judging from your picture, I'm actually thinking that maybe you were considering the LGT, and not PIM. In that case, I guess I just got confused from the way you wrote the Hamiltonian.