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This question already has an answer here:

When the Ising model Hamiltonian $$H=-J\sum _{<ij>} \sigma _i\sigma _j-H\sum _i \sigma _i$$ is assumed ($\sum _{<ij>}$ is the summation over all the bonds or adjacent pairs of sites, $\sum _i$ is the summation over all the sites), I think it is known that the self-consistent equation in mean-field theory is given as $$m=\tanh (\beta Jzm)$$ where $\beta$, $z$ are the inverse temperature and the number of adjacent bonds, respectively, and $m$ is the mean magnetization $\frac{1}{N}\sum _i\sigma _i$.

But I'm confused about its derivation.

  1. When I replace $\sigma _j$ by $m$ for the mean-field approximation, $$H_{MF}= -Jm\sum_{<ij>}\sigma _i -H\sum_i \sigma _i.$$ Since the number of bonds is $\frac{zN}{2}$ (ignoring boundary conditions) and $\sigma _i$ in the first term doesn't depend on $j$, we can use $\sum _{<ij>} =\frac{z}{2}\sum _i$. Thus we obtain, $$H_{MF}= -(\frac{1}{2}Jzm+H)\sum_i\sigma _i =-A\sum _i \sigma _i.$$ Using this $H_{MF}$, $\langle\sigma _i\rangle$ is calculated as below. $$\langle\sigma _i\rangle= \frac{\mathrm{Tr}[\sigma _i \exp (\beta A\sigma _i)]}{\mathrm{Tr}[\exp (\beta A\sigma _i)]}=\tanh (\beta A).$$ Imposing $\langle\sigma _i\rangle=m$, the self-consistent equation is found to be $$m=\tanh \left(\beta (\frac{1}{2}Jzm+H)\right).$$ I suppose it contradicts the known result.
  2. When I replace $\sigma _i$ by $m+(\sigma _i-m)$ and the ignore square of the fluctuation term, the same calculation yields the self-consistent equation $m=\tanh (\beta Jzm)$.

Are there any mistakes in the above discussion?

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marked as duplicate by user197851, Community Jan 21 at 12:02

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$\sum\limits_{<ij>}\sigma_i = \sum\limits_{j = 1}^{z}\sum\limits_{i}\sigma_i=z\sum\limits_{i}\sigma_i$, no need for that $\frac{1}{2}$. If you want to say that there is an energy $J$ per bounding then you need to change that directly in the Hamiltonian \begin{equation*} H = -\frac{J}{2}\sum\limits_{<ij>}\sigma_i\sigma_j-H\sum\limits_{i}\sigma_i \end{equation*}

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