condensed matter - Ignoring $(sigma_i-M)(sigma_j-M)$ in mean field theory?

A way to do mean field theory for the Ising model is as follows.

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1. First take the Ising Hamiltonian: $$H=-J \sum_{\left} \sigma_i\sigma_j$$


2. 
   
   

   Let $\sigma_i=\sigma_i-M+M$ and likewise for $\sigma_j$ to get: $$H=-J \sum_{\left} (M^2 +(\sigma_i-M)M+(\sigma_j-M)M+\underbrace{(\sigma_i-M)(\sigma_j-M)}_{\bigstar})$$

   
   


3. 
   

   Ignore the stared ($\star$) term.

   
   


4. 
   

   Write down the partition function, apply a self-consistency condition etc.

   
   

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Given that in the Ising model $\sigma_i=\pm1$ thus for any given $i$ and $j$, the ($\star$) term is not going to be small. What is the standard justificiation for then ignoring it?

Answer

Even though $\sigma_i-M$ is not small, expectation value of its square is small as that corresponds to the variance, hence fluctuations, which are assumed to be next order terms in the mean-field-theory. That's why summation $\sum (\sigma_i-M)(\sigma_j-M)$, which is basically autocovariance function along lattice sites, should be small. By the way, here we should have $M=<\sigma_i>$.