I have a one-dimensional system of two Spins ( $\sigma_1$, $\sigma_2$) in a magnetic field B with energy: \begin{equation} E = -J \sigma_1 \sigma_2 -B (\sigma_1 + \sigma_2) \end{equation} each $\sigma$ can be $\pm 1$. I want to compute the partition function and afterwards from that, the free energy $F$. My approach by going through all the possible configurations of $\sigma_1$ and $\sigma_2$ is: \begin{align} Z &= \sum_n \exp{(-\beta E_n)} \\ &= \exp{(- \beta[-J-2B])} + \exp{(- \beta [-J+2B])} + 2 \exp{(-\beta J)} \\ &= 2 \exp{(\beta J)}\cosh{(2 \beta B )} + 2\exp{(-\beta J)} \end{align}

I found online that there's a general form for Z for N Spins (https://www.uni-muenster.de/Physik.TP/archive/typo3/fileadmin/lehre/teilchen/ws0910/IsingI.pdf), which would give me the following expression for 2 Spins: \begin{equation} Z_N = E_+^N + E_-^N \\ = 2\exp{(2 \beta J)} \cosh{(2 \beta B)} - 2 \exp{(-2 \beta J)} \end{equation} Which is similar but not equal to my solution, but I can't figure out where my calculation mistake is; and furthermore what is the right solution, because in each case the expression for the free energy would be pretty ugly.

  • $\begingroup$ I think you should explain better how you obtained the last expression. $\endgroup$ – valerio Mar 20 '18 at 15:12
  • $\begingroup$ The general formula you refer to corresponds to periodic boundary condition, while you consider free boundary condition. This explains the fact that it has $2\beta J$ rather than your $\beta J$. The minus sign in the last expression is probably a mistake on your part (putting $\beta=0$ in the formula yields $0$, which is wrong). $\endgroup$ – Yvan Velenik Mar 20 '18 at 15:51

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