# Spinmodel Statistics

Consider a spin model with the following energy with $$\{\sigma\} =(\sigma_1,\sigma_2,...,\sigma_N)$$ where each $$\sigma$$ can take the values: $$-s,-s+1,...,-1,0,1,...,s-1,s$$ and $$E(\{\sigma \}) = \mu H \sum_{i=1}^{N}\sigma_i$$.

First I had to calculate the magnetization, which is defined as $$\langle M \rangle = \frac{1}{\beta} \frac{\delta Z}{\delta H}$$, So I started by calculating the partition function as always: $$Z =\sum_{\sigma_i} e^{-\beta E(\sigma_i)} = (\sum_{\sigma_i = -s}^{s} e^{-\beta \mu H\sigma_i})^N =(\sum_{k=0}^{2s} e^{-\beta \mu H (k-s)})^N = (e^{\beta \mu H J} \frac{1-e^{-\beta \mu H (2s+1)}}{1-e^{-\beta \mu H}})^N$$. But now I have to differentiate this with respect to $$H$$ which takes quite a lot of time. Also the next question is to calculate the magnetic susceptibility which is defined as $$\chi = \frac{\delta M}{\delta H}\Bigg|_{H=0}$$. So this means that I have to differentiate it again.

I'm sure that I am wrong somewhere, but I don't know where. Can anyone please point out the error?

The partition function that you have calculated is actually correct! but just not convenient. Instead of using the geometric sum, you can use the following identity: $$Exp(a)+ Exp(-a) = 2 Cosh(a)$$ where a is any real number($$a\in \mathbb{R}$$). Then the partition function for just one single spin is: $$z = \sum_{\sigma_i = -s}^{s} Exp(-\beta \mu H \sigma_i) = 1 + \sum_{\sigma_i = 1}^{s} (Exp(\beta \mu H \sigma_i) + Exp(-\beta \mu H \sigma_i)) = 1 + 2\sum_{\sigma_i = 1}^s Cosh(\beta \mu H \sigma_i)$$ Notice the summation indices are different. Now you can take N non-interacting spins which is your problem: $$Z = z^N = \bigl(1 + 2\sum_{\sigma_i = 1}^s Cosh(\beta \mu H \sigma_i)\bigl)^N$$ Now to find the magnetization, it is better perhaps to take the logarithm of this and obtain the Helmholtz free energy. So you obtain: $$F = -(1/\beta)ln(Z) = -(N/\beta)\;Log\bigl(1 + 2\sum_{\sigma_i = 1}^s Cosh(\beta \mu H \sigma_i)\bigl)$$ Now you can take derivatives of this function to find the Magnetisation or the Susceptibility. This type of problem is called noninteracting Ising chain. The link is the reference that I used.

• I was thinkting to do that to, but I'm was not that smart to just to leave that sommation uncalculated. And i didn't know how to calculate a sommation of a cosh so :), and a derivative of a sommation is here the sommation of the derivative. I calculated the magnetization. But if you derivative it once more, to calculate the Susceptibility, a lot is zero because H = 0. :) Thanks! – Kabouter9 Jun 22 at 16:37
• glad my answer helped! – Sparsh Mishra Jun 23 at 1:01