# Is the fully connected Potts model exactly solvable?

Suppose that we have "spins" $$\sigma_1,\dots,\sigma_N$$, with $$\sigma_i\in\{1,\dots,q\}$$, for $$i=1,\dots,N$$, and that our Hamiltonian is $$H = -\frac{J}{N} \sum_\stackrel{i,j=1}{i\ne j}^N \delta(\sigma_i,\sigma_j) - h \sum_{i=1}^N \delta(\sigma_i,1),$$ in which $$\delta$$ is a Kronecker delta: $$\delta(\sigma_i,\sigma_j)=1$$ if $$\sigma_i=\sigma_j$$, and $$\delta(\sigma_i,\sigma_j)=0$$ if $$\sigma_i\ne \sigma_j$$.

This is a fully connected Potts model, defined on the complete graph, where each spin interacts with every other spin in the system.

Define the model partition function as $$Z = \sum_{\sigma_1=1}^q \dots \sum_{\sigma_N=1}^q \exp(-\beta H),$$ in which $$\beta=1/(k_B T)$$.

Direct computation of $$Z$$ is not feasible for any reasonable $$N$$, since it involves the sum of $$q^N$$ terms.

Is there in the literature a known way to perform the above summations and find a "manageable" expression $$f(J,h,\beta,N$$) for the partition function $$Z$$?

I'm looking for an exact result holding for every $$N\geq 2$$, and finite $$J$$, $$h$$ and $$\beta$$, which can be computed in polynomial time.

• @Mathphysmeister : The OP is interested in the model on the complete graph, not the one-dimensional model. I don't think that one can get a general closed form expression for the partition function in that case. One can, however, get some expression for the free energy density in the thermodynamic limit. See, for instance, this paper. Commented Jul 14, 2020 at 9:46
• The grand partition function looks like it would be easier.
– Buzz
Commented Jul 14, 2020 at 16:34
• Typically models of this kind can be solved exactly by "mean-field" in the limit $N\to\infty$.
– lcv
Commented Jul 14, 2020 at 20:38
• @Zen it's very unlikely that the exact solution at finite $N$ exist. Sometimes similar (i.e. mean field) models can be solved by Bethe Ansatz at finite size, but even then you have to solve the equations for the rapidities and these can never be solved explicitly. I think the thermodynamic limit is the best you can get. I would look into the reference of Yvan or similar ones.
– lcv
Commented Jul 15, 2020 at 16:35
• @Zen : Is the Hamiltonian in your latest version really what you wanted to write? I see 2 issues with it. First, $J$ should really be $J/N$ if you want a meaningful behavior in the thermodynamic limit (of course, this does not matter too much as long as you work with finite $N$, but still...). Second, your "magnetic field" term is not correct for Potts spins. Don't you want rather a term of the form $-h \sum_i \delta(\sigma_i,1)$ for instance (that would be a field favoring spins of type $1$, other "magnetic field"-like terms are possible, of course). Commented Jul 16, 2020 at 10:26

You should not expect close form expressions for the finite-$$N$$ partition functions.

In fact, this is already the case when $$q=2$$. The latter is equivalent to the Curie-Weiss model, in which the spins $$\sigma_1,\dots,\sigma_N$$ take values in $$\{-1,1\}$$ and the Hamiltonian takes the form $$H = -\frac{J}{N}\sum_{i,j=1}^N \sigma_i\sigma_j - h \sum_{i=1}^N \sigma_i.$$ Note that I don't impose that $$i\neq j$$ in the first sum. This only shifts the energy by $$J$$ and thus plays no role, while slightly simplifying the exposition.

Let me describe the best you can hope for in this model. Introducing the magnetization $$M=\sum_{i=1}^N \sigma_i$$, the Hamiltonian can be reexpressed as $$H = -\frac{J}{N} \Bigl( \sum_{i=1}^N \sigma_i \Bigr)^2 - h M = -\frac{J}{N} M^2 - h M.$$ From this observation, one can proceed in two different ways, both providing an expression for the partition function.

The first way is combinatorial. Just observe that $$Z_N = \sum_{k=0}^N \binom{N}{k} \exp\Bigl( \frac{\beta J}{N} (2k-N)^2 + \beta h (2k-N) \Bigr), \tag{1}$$ where the sum is over the number $$k$$ of spins $$\sigma_i$$ such that $$\sigma_i=1$$ (in particular, $$M=k-(N-k)=2k-N$$). This is the first "explicit" expression for the partition function. It reduces the partition function from a sum over $$2^N$$ configurations to a sum over the $$N+1$$ possible values of the magnetization.

The second approach is via the Hubbard–Stratonovich transformation, which implies that $$\exp\bigl( \frac{\beta J}{N} M^2 \bigr) = \sqrt{\frac{N}{\pi\beta J}} \int_{-\infty}^{+\infty} \exp \bigl( - \frac{N}{\beta J} x^2 + 2 M x \bigr) \, \mathrm{d}x.$$ From this, we can write \begin{align} Z_N &= \sum_{\sigma_1,\dots,\sigma_N} \sqrt{\frac{N}{\pi\beta J}} \int_{-\infty}^{+\infty} \exp \bigl( - \frac{N}{\beta J} x^2 + (2 x + \beta h) M \bigr) \, \mathrm{d}x \\ &= \sqrt{\frac{N}{\pi\beta J}} \int_{-\infty}^{+\infty} \exp \bigl( - \frac{N}{\beta J} x^2 \bigr) \prod_{i=1}^N \underbrace{\sum_{\sigma_i=\pm 1} \exp \bigl( (2 x + \beta h) \sigma_i \bigr)}_{=2\cosh(2x + \beta h)} \, \mathrm{d}x \\ &= \sqrt{\frac{N}{\pi\beta J}} \int_{-\infty}^{+\infty} \exp \bigl( - N \varphi(x) \bigr) \, \mathrm{d}x , \tag{2} \end{align} with $$\varphi(x) = \frac{1}{\beta J} x^2 - \log\cosh(2x + \beta h) - \log 2$$. This is the second "explicit" expression for the partition function. It reduces the partition function from a sum over $$2^N$$ configurations to an integral.

It does not look likely that one can explicitly evaluate the sum in (1) or the integral in (2). So I believe that this is the best you can hope for. Of course, both (1) and (2) can be used to extract a lot of information on the model. For instance, a saddle-point analysis of (2) would yield sharp approximations of the partition function for finite (but large) values of $$N$$.

For larger values of $$q$$, one can obtain expressions similar to those in (1) and (2) (instead of considering the magnetization $$M$$, one should consider the vector $$(N_1,\dots,N_q)$$ where $$N_k$$ is the number of spins taking value $$k$$).

What was proposed in (1) of a previous answer can be generalized to compute $$Z$$ exactly in polynomial time for $$q\ge2$$. The key is that, as in the $$q=2$$ case, the Hamiltonian only depends on the number of spins taking each of the values.

Define $$n_\sigma$$ to be the number of spins taking value $$\sigma\in\{1,2,\ldots,q\}$$. Formally, $$n_\sigma=\sum_{i=1}^N\delta(\sigma_i, \sigma)$$. Then observe that \begin{align} \sum_{i,j}\delta(\sigma_i,\sigma_j) &= \sum_i \sum_{j}\delta(\sigma_i,\sigma_j)\\ &= \sum_{\sigma=1}^q \sum_{i}\delta(\sigma_i,\sigma)\sum_{j}\delta(\sigma,\sigma_j)\\ &= \sum_{\sigma=1}^q n_\sigma^2 \end{align} where to get the second equality one groups together spins with the same value $$\sigma$$.

Therefore the Hamiltonian reads $$H = -\frac JN\sum_{\sigma=1}^q n_\sigma^2-hn_1$$

Now the partition function be computed by grouping together all configurations sharing the same $$\{n_1,n_2,\ldots,n_q\}$$ and counting how many there are \begin{align} Z_N &= \sum_{\sigma_1,\sigma_2,\ldots,\sigma_N} \exp\left(-\beta H\right)\\ &= \sum_{n_1,n_2\ldots,n_q} \exp\left(-\beta H\right)\sum_{\sigma_1,\sigma_2,\ldots,\sigma_N} \mathbf{1}\left[n_\sigma=\sum_{i=1}^N\delta(\sigma_i, \sigma)\;\forall\sigma\right]\\ &= \sum_{n_1,n_2\ldots,n_q}\binom{N}{n_1 n_2 \cdots n_q} \exp\left(\frac{\beta J}{N}\sum_{\sigma=1}^q n_\sigma^2+\beta hn_1\right) \end{align}

which takes $$\mathcal O (N^q)$$: you need $$q$$ nested loops, one for each of the $$n_\sigma$$'s.

It is implicit that the sum only runs over allowed sets $$\{n_1,n_2,\ldots,n_q \:|\: n_\sigma \in \{0,1,\ldots,N\} \forall \sigma, \sum_{\sigma=1}^q n_\sigma=N\}$$.