# 3-dimensional 3-state Potts model critical temperature

I was given that the free energy per lattice site of the 3-dimensional 3-state Potts model in the mean field approximation is $$f\equiv \frac{F}{\text{# sites}} = \sum_{k=0}^{2}(-3Kx_k^2 + \frac{1}{\beta}x_k \ln x_k\Big)$$ where here $$K$$ is from $$H = -K \sum_{\langle ij\rangle} \delta_{\sigma_i\sigma_j}$$ and $$x_k$$ are the order parameters, the proportion of lattice sites in the state k. To find the temperature where a phase transition occurs, I figured that we take derivatives: $$\frac{\partial }{\partial x_{k_0}}f = -4Kx_{k_0} + \frac{1}{\beta}\ln x_{k_0} + \frac{1}{\beta} = 0\Rightarrow 4K\beta x_{k_0}-\ln x_{k_0}=1$$ but this seems difficult so solve. It's also not clear to me what to do since we have three order parameters.

Context: Consider a system on lattice sites $$i \in \mathbb{Z}^d$$, where the state $$\sigma_i$$ of a site can take values in $$\{1,\dots, q\}$$. We have a Hamiltonian on this system given by $$H = -K \sum_{\langle ij\rangle} \delta_{\sigma_i\sigma_j}$$ where the sum is over pairs $$i,j$$ of neighboring lattice sites and $$K>0$$.

Small caveat: if you’re doing mean field, you lose any notion of dimension, you effectively replaced your lattice by a complete graph. It is therefore contradictory to talk about mean field and dimensionality (unless you are over the upper critical dimension, which is not your case).

You need to be careful when calculating the partial derivatives. The three order parameters are not independent, but satisfy: $$\sum x_i=1$$ To find the stationary points, you therefore need to introduce a Lagrange multiplier, so the criterion is not that the partial derivatives are zero, but rather that they have a common value determined by the normalization constraint.

It turns out that that you don’t need 2 independent order parameters, you can assume that two are equal and that will be enough (this is general for the $$q$$ state Potts model). I don’t think that you’ll be able to calculate the critical temperature analytically. You can visualize the phase transition though by graphing $$f$$ as a function of $$\beta$$ using the single order parameter. Note that the transition is not continuous, unlike Ising ($$q=2$$ Potts model), the order parameter jumps at the transition. In particular, it is not detectable by solely investigating the local stability of the ordered state like you can do for Ising.

Hope this helps.

• Thanks for the response! Maybe I'm missing something, but how does the Lagrange multiplier method not give a trivial answer here? Commented Dec 7, 2023 at 0:07
• You can localize the disordered state pretty easily. However locating the spontaneously broken state isn’t computable analytically
– LPZ
Commented Dec 7, 2023 at 8:45
• Okay, I think I’m seeing that. I see where our third order parameter $x_3$ is redundant because of the constraint. However, why can we take $x_1=x_2$? Commented Dec 7, 2023 at 15:05