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

  1. First take the Ising Hamiltonian: $$H=-J \sum_{\left<i,j\right>} \sigma_i\sigma_j$$

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

  3. Ignore the stared ($\star$) term.

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

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?

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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 $M=<\sigma_i>$.

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  • $\begingroup$ Good answer! I just want to add something: ignoring fluctuations in mean field approximation is not that good because it makes you predict a phase transition at finite temperature even in 1D Ising model $\endgroup$
    – Matteo
    Commented Apr 14, 2018 at 7:13
  • $\begingroup$ Thanks. MFT is supposed to be used in higher dimensions though, it is especially, and ironically, efficient for $d>4$ as far as I remember. $\endgroup$
    – SonerAlbayrak
    Commented Apr 14, 2018 at 8:02
  • $\begingroup$ You say that the expected value of the square should be small - I agree with this. But this does not mean that the sum of the squares should be small. I cannot see how you go from this observation to concluding that $\sum (\sigma_i-M)(\sigma_j-M)$ should be small? $\endgroup$
    – Quantum spaghettification
    Commented Apr 14, 2018 at 8:54
  • $\begingroup$ I am assuming that we are working in a homogeneous lattice (as is the case for the usual Ising model, which is also compatible with your equation) for which the expectation value of the operator is its mean value in the lattice. Hence that summation is exactly the autocovariance function, which measures the variance. The thing is not only that the variance should be small, but also that it should be small with respect to the mean. $\endgroup$
    – SonerAlbayrak
    Commented Apr 14, 2018 at 9:04
  • $\begingroup$ Another way of seeing this is as follows: In your equation, take everything in $M^2$ paranthesis. First three terms are of order $\left(\delta\sigma/M\right)^0$ or $\left(\delta\sigma/M\right)^{-1}$ whereas the last term is of the order $\left(\delta\sigma/M\right)^{-2}$. And we are assuming that variance is suppressed with respect to the mean in MFT, so higher orders should be relatively suppressed by assumption. This assumption is justified to a degree for large systems, as mean grows linearly with the system size whereas standard deviation grows as square root of the system size. $\endgroup$
    – SonerAlbayrak
    Commented Apr 14, 2018 at 9:07
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I could not find any references that explains this in detail so if you think it is wrong please let me know.

This issue is a lot more subtle then it first appears - especially in the case where the external magnetic field, $h=0$, like I have put in the question.

Why is it a subtle issue? Well for $h=0$ our Hamiltonian has symmetry under $\sigma_i \leftrightarrow -\sigma_i$, meaning for any configuration $\{\sigma_i\}$ the configuration $\{-\sigma_i\}$ has exactly the same energy and therefore contributes the same waiting to the partition function. If the $\star$ term is small for $\{\sigma_i\}$ then it clearly won't be for $\{-\sigma_i\}$ (in general it will be large and positive).

To see this think about the case where $M$ is close to $+1$, for the configuration $\{\sigma_i=1\;\forall i\}$ this last term will indeed be small but for $\{\sigma_i=-1\;\forall i\}$ it won't be.

Thus for any case except $M=0$ ignoring the $\star$ term dramatically decreases the probability of the flipped spins for the case of $h=0$.

Therefore to make the mean field theory valid in this case and for the argument as presented in Soner's answer to work we must a spontaneous symmetry breaking field present (or at least have it's present in the back of our minds). This means that the inverted configurations already have much smaller probability then those with mean value near $M$ and as such ignoring $\star$ does not have a big effect on our partition function.

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    $\begingroup$ Yes, my answer was assuming a lot in the background. For example, as far as I know, MFT is a saddle-point approximation in the sense that we only take the contribution of "classical path" in the path integral and ignore other contributions. To me, that means we already choose a vacuum and ignore the effects of other vacuums (say soliton/instanton) and we also ignore the fluctuations around chosen vacuum. $\endgroup$
    – SonerAlbayrak
    Commented Apr 15, 2018 at 17:13
  • $\begingroup$ For $h\ne 0$, it is clear which minimum is true vacuum (hence even though the fluctuations in the other minima is not "small" as you point out, this is not relevant). I honestly do not know the case $h=0$ in an obvious manner, but if I remember correctly we can obtain it as $h\to 0$ in the end (after taking thermodynamic limit) so it should not be too incorrect to see it as a special limit of the general case. Well, this is the situation as far as I know. $\endgroup$
    – SonerAlbayrak
    Commented Apr 15, 2018 at 17:19

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