Prove 2 and 3 state clock model is equivalent to Potts model

[Fb.Researcher]

Homework Statement

A p-state clock model is a spin model in which states can be viewed as hands of a clock!
Now knowing the models we should prove these two models are equivalent for $p=q=2$ or $p=q=3$

Homework Equations

Potts model is described by $$ H = -J \sum_{<ij> } \delta_{ \sigma_i \sigma_j} ~ ~ ,~ ~\sigma_i = 0,1 , \dots, q $$
and clock model is described by $$ H = -K \sum_{<ij> } \cos (2 \pi (n_i - n_j ) / p) ~ ~ ~, ~ n_i = 1, 2, ..., p $$ in which $n_i$ is our state. In this models states are vectors in a circle( or hands of a clock ).

The Attempt at a Solution

As far as I know two equivalent model have same Hamiltonian; which means the shape of the sentences in the Hamiltonian should be the same up to constant factor.
for $q = p = 2$ we can say that a two state clock model will have two states with $0 ~ , ~ \pi ~, -\pi$ deference. So $$ \cos (2 \pi (n_i - n_j ) / p) = \begin{cases} 1, & n_i = n_j \\ -1, & |n_i - n_j| = 1\end{cases} $$.
Which is two possible values for the parameter inside the sum. Now I think adding a 1 to the Hamiltonian of clock model with $K = \frac{J}{2}$ plus a constant, which is $\frac{1}{2}$ will produce the same results as Potts model, so these two models are identical.

for $q = p = 2$ with the same procedure I obtained
$$ \cos (2 \pi (n_i - n_j ) / p) = \begin{cases} 0, & n_i = n_j \\ - \frac{1}{2}, & |n_i - n_j| = 2\end{cases} $$.
Which means for $K = -2J$ or $k = 2J$ and a constant factor of $\frac{1}{2}$, these model produce the same results and they are identical.

For higher values of p and q clock model will always produce more than two numbers and we can not say these models are identicalAs you can see I have not find a way to transform delta function to a function similar to the clock model.In fact I do not have a transformation. My solutions does not seem to be professional nor is an advanced solution for a advanced Physics problem.

 

[DeathbyGreen]

I think youre answer is almost there, I worked out a bit of it and got the same thing. I think the next step would be to use an identity of the form

$N\delta_{k,k'} = \sum_1^N e^{i(k-k')x}$

this is an identity commonly used during a Fourier transform. Before using this you need to convert your cosine into exponentials as $cos(\theta) = \frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)$. Obviously you need to manipulate it a bit, but once you get it into a form appropriate to your problem I think it will work out.