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I have a problem regarding the energy per lattice point in case of the 2-state Potts model. The q=2 state Potts model should correspond to the Ising model. The Hamiltonian for the Ising model is $$H = -J \sum_{\langle s_i s_j\rangle} s_i s_j$$ where an external magnetic field is not present and we set $J =1$.

Let's look at the case of a lattice point $s_i$ with spin $s_i = 1$ and the the four neigbours (up, down, left, right) have spins $(-1,1,-1,-1)$. Our Hamiltonian and therefore the energy at that lattice point is then $$H = -\sum_{\langle s_i s_j\rangle} s_i s_j = -s_i \sum_{\langle s_i s_j\rangle} s_j = -1 \cdot (-1+1-1-1) = -1 \cdot (-2) = 2$$.

Now, for the Potts model we have (according to https://en.wikipedia.org/wiki/Potts_model) the Hamiltonian $$H = -J \sum_{\langle s_i s_j\rangle} \delta_{s_i, s_j}$$ where we again set $J = 1$ and neglect any external magnetic field. Since we have $q=2$ Potts model, we have the possible states $s = 1$ and $s = 2$. Analogous to our Ising model, the lattice point has spin $s_i = 1$ and the neighbours (up, down, left, right) have spins $(2,1,2,2)$. We then have $$H = - \sum_{\langle s_i s_j\rangle} \delta_{s_i, s_j} = - (\delta_{1,2} + \delta_{1,1} + \delta_{1,2} + \delta_{1,2}) = - (0+1+0+0) = -1.$$

This is of course not the same as for the Ising model. But, as far as I understand, it should be equal.

Where am I mistaken? And how does it translate for higher $q$-state Potts models?

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    $\begingroup$ Let us start with the Ising model with Hamiltonian $-J\sum_{i\sim j} \sigma_i\sigma_j$. Note that $\sigma_i\sigma_j = 2\delta_{\sigma_i,\sigma_j} - 1$. We can thus rewrite the Hamiltonian as $-2J\sum_{i\sim j} \delta_{\sigma_i,\sigma_j} + C$, where $C$ is an irrelevant constant (it appears both in the numerator and the denominator of the Gibbs probability measure and thus does not affect the probabilities of configurations). Of course, it can be computed explicitly: $C=J|E|$, where $|E|$ is the number of pairs of nearest-neighbors $\endgroup$
    – Yvan Velenik
    Commented Aug 12 at 7:45
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    $\begingroup$ Observe finally that the new Hamiltonian $-2J\sum_{i\sim j}\delta_{\sigma_i,\sigma_j}$ has Potts form (it does not matter how you label your spins, as long as you use two different labels for the two different spin states). $\endgroup$
    – Yvan Velenik
    Commented Aug 12 at 7:46
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    $\begingroup$ @YvanVelenik thank You very much for the explanation! Is the $-2J \sum_{i \sim j} \delta_{\sigma_i , \sigma_j}$ Hamiltonian also valid for Potts models with $q > 2$ states? $\endgroup$
    – syphracos
    Commented Aug 12 at 7:55
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    $\begingroup$ The Hamiltonian of the Potts model for arbitrary integer $q\geq 2$ is indeed $-K\sum_{i\sim j}\delta_{s_i,s_j}$, where $K$ is the coupling constant (no reason to write it $2J$ as there is no connection to the Ising model anymore). $\endgroup$
    – Yvan Velenik
    Commented Aug 12 at 8:04
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    $\begingroup$ Sure, changing $K$ only changes the temperature scale. $\endgroup$
    – Yvan Velenik
    Commented Aug 12 at 8:09

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