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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?

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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.

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  • $\begingroup$ 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! $\endgroup$ – Kabouter9 Jun 22 at 16:37
  • $\begingroup$ glad my answer helped! $\endgroup$ – Sparsh Mishra Jun 23 at 1:01

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