# Problem in counting bonding pairs (elementary mean-field theory on the Ising Model) [duplicate]

When the Ising model Hamiltonian $$H=-J\sum _{} \sigma _i\sigma _j-H\sum _i \sigma _i$$ is assumed ($$\sum _{}$$ 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, $$H_{MF}= -Jm\sum_{}\sigma _i -H\sum_i \sigma _i.$$ 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 $$\sum _{} =\frac{z}{2}\sum _i$$. 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?

$$\sum\limits_{}\sigma_i = \sum\limits_{j = 1}^{z}\sum\limits_{i}\sigma_i=z\sum\limits_{i}\sigma_i$$, 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 $$\begin{equation*} H = -\frac{J}{2}\sum\limits_{}\sigma_i\sigma_j-H\sum\limits_{i}\sigma_i \end{equation*}$$