Potts model equivalent to Heisenberg model

Question

The mean-field equations for the three-state Potts model $$ \mathcal{H}=-J\sum_{\langle ij \rangle}\delta_{\sigma_i\sigma_j}\:,\qquad\sigma_i=1,\,2,\,3 $$ is equivalent to $$ \mathcal{H}=-J\sum_{\langle ij \rangle}\vec{s}_i\cdot\vec{s}_j $$ with $$ \vec{s}_i= \begin{pmatrix} 1\\0 \end{pmatrix}, \begin{pmatrix} -1/2\\\sqrt{3}/2 \end{pmatrix}, \begin{pmatrix} -1/2\\-\sqrt{3}/2 \end{pmatrix}. $$

This is a problem from J. M. Yeomans book Statistical Mechanics of Phase Transitions and I was wondering how to show this for an arbitrary number of particles.

If we had just two particles it could easily be shown with a small table as follows:

$$ \begin{align} \text{State particle 1}&\quad |\quad\text{State particle 2}&\quad \quad\text{Potts}&\quad |\quad\text{Heisenberg}\\ 1&\quad |\quad 1&\quad 1&\quad |\quad 1\\ 1&\quad |\quad 2&\quad 0&\quad |\quad -1/2\\ 1&\quad |\quad 3&\quad 0&\quad |\quad -1/2\\ 2&\quad |\quad 1&\quad 0&\quad |\quad -1/2\\ 2&\quad |\quad 2&\quad 1&\quad |\quad 1\\ 2&\quad |\quad 3&\quad 0&\quad |\quad -1/2\\ 3&\quad |\quad 1&\quad 0&\quad |\quad -1/2\\ 3&\quad |\quad 2&\quad 0&\quad |\quad -1/2\\ 3&\quad |\quad 3&\quad 1&\quad |\quad 1\\ \end{align} $$

In both models we find the same resulting picture, which becomes more clear or obvious if we would use arrows or something, like when Potts equals to one then $\uparrow$, when Heisenberg is equal to one the $\uparrow$, when Potts equals 0 or Heisenberg $-1/2$ then $\downarrow$.

But I am not even sure if this is what Yeomans had in mind with this problem and how it could be expanded to $N$ particles instead of just two. I think a permutation might do it or that the expectation value of $\mathcal{H}$

$$ \langle\mathcal{H}\rangle=\frac{1}{Z}\sum_{configurations}\!\mathcal{H}\cdot e^{-\beta\mathcal{H}} $$

with $Z$ the partition function and $\beta=\frac{1}{k_BT}$ ($k_B$ the Boltzmann constant) should be the same for both models.

Or is there a more elegant / straightforward solution?

Answer 1

Your first equation is the Potts model hamiltonian, not a mean field equation. Showing that it is equivalent to the second equation is the first part of problem (4.4) in Yeomans. Doing the mean field theory follows afterwards. The motivation for making this transformation is that you can use the same method as the one used in problem (4.3). I'm not going to discuss how to solve either of these problems (that would be off-topic for this site), I just want to make some general points.

Your table of energies is already, almost, a proof of the equivalence. You may need to redefine the energy scale, $J$, since the energy differences between different pair states for the spin version are $3/2$ times those for the original version, as you have written them. An alternative approach is to say that the models are "equivalent" if they are considered at different temperatures (a factor $3/2$ different). The essential point is that the probability of occurrence (in the canonical ensemble) of a given configuration of Potts variables $\{\sigma_i\}$ is the same as that of the corresponding configuration of spins $\{\vec{s}_i\}$.

Note that adding a constant to the total energy makes no difference to the probabilities, so the fact that the lowest pair energies for the spins are $-1/2$ rather than $0$ should not be a problem. Adding the same constant to the energy of every pair is the same as adding a constant to the total energy. You might need to bear it in mind when comparing absolute values of energies and so on, but it is easy to take into account.

The hamiltonian is a sum over all nearest neighbour pairs, so this mapping or equivalence already applies to the full, many-spin system.

There is no need to consider the average energy in order to do this mapping (and incidentally your formula for this is missing a term, the energy of the configuration, inside the sum). You should already be familiar with similar mappings (for example between the Ising model and the lattice-gas model, and the binary alloy lattice model). If not, I advise looking them up, because they give an idea of what is going on here.

I won't add any more, since we do not answer homework-like problems, but the bottom line is that you have done most of the work already.