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I try to compute the variational free energy in the Ising Model using the bogoliubov inequality: \begin{equation} \mathcal{F}(\lambda) = F_{0}(\lambda) \ + \ \langle\mathcal{H_{1}(\lambda)} \rangle_{0} \end{equation} Hence, I need the partition function $Z_{0}$ relative to the Hamiltonian $\mathcal{H_{0}}(\lambda) = -\lambda \sum_{i}s_{i}$ , which describes independent spins interacting with an external field. My effort: \begin{equation} \begin{split}Z_{0} & = \sum_{ \{ s_{j} \}} \ e^{-\beta\mathcal{H_{0}(\lambda \ , \ s_{j})}} \\ & = \sum_{s_{1} = \pm1} \dots \sum_{s_{N} = \pm 1} \prod_{i} ^{N}{e^{\beta \ \lambda \ s_{i}}} \end{split} \end{equation} I know that $e^{\beta \ \lambda s_{i}} = e^{\pm\beta \ \lambda}$. Afterwards, I tried using the expression for the number of ways to choose $k$ elements (in this case the number of ways to have a $k$ positive signs in the exponent) out of a set of $N$ elements. Then I summed $k$ from $1$ to $N$ to get the total sum (don't know if this reasoning is correct though): \begin{equation} Z_{0} = \sum^N_{k = 0}\frac{N!}{k!(N-k)!} e ^{k \ \beta \lambda - (N-k) \ \beta \ \lambda} \end{equation} but then I get stuck because it's hard to get the free energy out of it. My calculation does not seem to lead anywhere. Is this the correct way to compute the partition function ?

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    $\begingroup$ Start from the last sum (on $s_N$) and factorize all terms of the product which do not depend on $s_N$. Repeat for the sum on $s_{N-1}$ and so on up to the sum on $s_1$. You'll find that the multiple sum of the product factorizes into a product of sums, each sum being on the two values of $s_i$. And you have done. $\endgroup$ – GiorgioP Dec 15 '19 at 5:48
  • $\begingroup$ Thank you for your answer ! That is indeed a simpler way of getting the answer ($(2\cosh{\beta \lambda})^{N}$). When I put my computed sum into wolfram I also get $(2\cosh{\beta \lambda})^{N}$ , but unfortunately wolfram does not show the steps of the calculation. Just out of curiosity, how can I evaluate my sum to get the simpler answer ? Or is this nearly impossible to get the "nicer" answer ? $\endgroup$ – Einsteinwasmyfather Dec 15 '19 at 16:17
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Okay I found it ! Just use the Binomial theorem: \begin{equation} \begin{split} Z_{0}& = \sum_{k =0}^{N} \frac{N!}{k!(N-k)!}e^{k\beta \lambda - (N-k)\beta \lambda} \\ &= \sum^{N}_{k} {N\choose k} (e^{-\beta\lambda})^{N-k} \ (e^{\beta \lambda})^{k} \\& = (e^{-\beta \lambda} + e^{\beta \lambda})^{N} \\ &= (2\cosh{\beta\lambda})^{N}\end{split}\end{equation} and this is easier to work with :)

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