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I have that the energy of my model is such that every spin interact with each other:

\begin{equation} E = -\frac{J}{2N}\left(\sum_{i}\sigma_i \right)^2 - h\sum_i\sigma_i \end{equation}

And I have to prove that the partition function $Z$, that is defined as,

\begin{equation} Z = \sum_{\sigma_1,\dots,\sigma_N}e^{-\beta E} \end{equation}

Can be put in the form

\begin{equation} Z = \frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^2}\left[2\cosh \left(\beta h + 2x\sqrt{\frac{\beta J}{2N}}\right)\right]^N\mathrm {dx} \end{equation}

Where we use the Gaussian relation

\begin{equation} e^{y^2} = \frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^2 + 2yx}\mathrm {dx} \end{equation}

I tried to put $y = \sum_i \sigma_i$ working the partition function. If you put that energie in the partition function then $Z$ become

\begin{equation} Z = \sum_{\sigma_1,\dots,\sigma_N}e^{-\beta E} = \sum_{\sigma_1,\dots,\sigma_N}\exp \left\{\frac{\beta J}{2N}\left (\sum_i \sigma_i\right)^2 + \beta h \sum_i \sigma_i \right\} \end{equation}

\begin{equation} = \sum_{\sigma_1,\dots,\sigma_N}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^2}\exp\left\{2x\sqrt{\frac{\beta J}{2N}}\sum_i \sigma_i\right\}\mathrm{d}x \exp\left\{\beta h \sum_i \sigma_i\right\} \end{equation}

Where I just used the relation above. What I cannot understand is from where the $\cosh$ came from: The sigmas sum and the other some can be put together in the form \begin{equation}\label{equação seis} Z = \frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^2}\left\{\sum_{\sigma_1,\dots,\sigma_N}\exp\left(\left(2\sqrt{\frac{\beta J }{2N}}+\beta h\right)\sum_i \sigma_i\right)\right\}\mathrm{d}x \end{equation}

trivially. Then what I want to know is that how can I prove now that that part will became equal to the $\cosh$ part.

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closed as off-topic by Norbert Schuch, AccidentalFourierTransform, Jon Custer, user36790, Gert Dec 6 '16 at 4:04

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  • $\begingroup$ Just apply the Gaussian relation to the first part of the Boltzmann weight (corresponding to the quadratic part of the Hamiltonian). If you can't do it, this is explained in many textbooks (just google Hubbard-Stratonovich). $\endgroup$ – Yvan Velenik Dec 5 '16 at 7:36
  • $\begingroup$ I will try do this. $\endgroup$ – user78217 Dec 5 '16 at 14:39
  • $\begingroup$ Worked but now I ended up having to show that equality $\endgroup$ – user78217 Dec 7 '16 at 16:00
  • $\begingroup$ Do you mean the Gaussian relation? Just bring the $e^{y^2}$ to the other side and observe that the exponent becomes $-(x+y)^2$. Change variables to $z=x+y$ and the result follows immediately. $\endgroup$ – Yvan Velenik Dec 7 '16 at 16:34
  • $\begingroup$ Sorry, I did not realize you had changed your question. Well, now you just have to sum over the spin values. Note that the function you are summing can be factorized: $e^{A\sum_i \sigma_i} = \prod_i e^{A\sigma_i}$.This means that you can sum over each spin separately. $\endgroup$ – Yvan Velenik Dec 8 '16 at 12:15