Phase diagram of the Ashkin-Teller model for unequal intraplane couplings

The 2d Ashkin-Teller model is a $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ symmetric model consisting of two planes of Ising models on the square lattice with an interplane four-spin coupling:

$$H = -\sum_{} J_s \sigma_i \sigma_j + J_t \tau_i \tau_j + J_i \sigma_i \sigma_j \tau_i \tau_j$$

I am trying to understand the phase diagram of this model in the ferromagnetic case $$J_s,J_t,J_i>0$$ for the regime $$J_s>J_t$$ and $$J_i$$ is small relative to both $$J_s$$ and $$J_t$$. What does the phase diagram look like? In particular, what are the universality classes of the transitions?

Here is what I suspect will happen. When $$J_s - J_t >> J_i$$, on tuning temperature, I think there will be two distinct phase transitions. The higher temperature phase transition will be when $$T^s_c = O(J_s)$$, below which the more strongly coupled $$\sigma$$ spins will be ferromagnetically ordered. The lower temperature phase transition will occur around $$T^t_c= O(J_t)$$ where the $$\tau$$ spins finally order below that temp.

I expect both of these phase transitions to be in the Ising universality class described by $$c=1/2$$ conformal field theory. I would appreciate confirmation for this fact.

However, I'm less confident about the case where $$J_i$$ is on the order of $$J_s - J_t$$, or when $$J_i$$ is much larger than the difference.

The (ferromagnetic, planar) Ashkin-Teller model undergoes two phase transitions, except when $$J_s=J_t\geq J_i$$, in which case there is a unique phase transition.

The existence of multiple phase transitions in this model was first proved (in parts of the phase diagram) rigorously in Journal of Statistical Physics volume 29, pages 113–116 (1982).

In the special case $$J_s=J_t$$ (the so-called isotropic case), the rigorous proof that there are two phase transitions when $$J_i>J_s=J_t$$ and a single phase transition (occurring on the self-dual curve) when $$J_i\leq J_s=J_t$$ was given recently in this paper.

The phase transitions are expected to be in the Ising universality class as long as $$J_s\neq J_t$$; see Communications in Mathematical Physics volume 256, pages 681–735 (2005) for rigorous results. Concerning the isotropic case $$J_s=J_t$$, let me cite from the latter paper:

The isotropic case was studied by Kadanoff who, by scaling theory, conjectured a relation between the critical exponents of isotropic AT and those of the Eight vertex model, which had been solved by Baxter and has nonuniversal indexes. Further evidence for the validity of Kadanoff’s prediction was given by Pruisken and Brown (using second order renormalization group arguments) and by [Luther and Peschel, den Nijs] (by a heuristic mapping of both models into the massive Luttinger model describing one dimensional interacting fermions in the continuum). Indeed nonuniversal critical behaviour in the specific heat in the isotropic AT model, for small λ, has been rigorously established in [Mastropietro].

• +1 Thank you for the thorough answer. I think the only part of the phase diagram that is still a little less clear to me is $J_i >> J_s > J_t$. I'll try to summarize my haphazard thoughts below. Commented Oct 15, 2023 at 20:48
• This is the rough picture I'm thinking in my head: let $J_s > J_t$ and $J_i$ much smaller than either and $J_i$ much smaller than $J_s - J_t$. Then the phase transitions correspond to ordering $J_s$ first and then ordering $J_t$ on lowering the temperature. Now consider an opposite limite where $J_i >> J_s = J_t$. Then there are again two transitions, but on lowering the temperature the composite spin $\sigma \tau$ orders first before the individual planes order simultaneously. Are all four of these transitions Ising-type? Commented Oct 15, 2023 at 20:59
• Then the case that I think I'm a little turned around about is $J_i >> J_s > J_t$. I'm roughly expecting that this case will have similarities to $J_i >> J_s = J_t$. That is, the composite spin $\sigma \tau$ orders first on lowering the temperature. I suppose then that on further lowering the temperature, the two planes will eventually start to order simultaneously despite $J_s > J_t$. Commented Oct 15, 2023 at 21:04
• Note that by setting $\theta_i=\sigma_i\tau_i$ and expressing the Hamiltonian in terms of $\sigma$ and $\theta$ variables, you recover an Ashkin-Teller model in which you have swapped the roles of $J_t$ and $J_i$ (of course, you can do the same for $J_s$ and $J_i$). Commented Oct 16, 2023 at 7:47
• Thanks, that duality answers all my questions in the comments above! Commented Oct 16, 2023 at 21:38