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In the Ising model the mean value of any particular spin is:

$$ m = \left<s_i\right> = \frac{ \sum_{s_i}e^{-\frac{H}{T} }s_i} { \sum_{s_i} e^{-\frac{H}{T} } } .$$

I read in several sources that due to "translational invariance" $m$ is "independent of the site label $i$". That means that all spins have the same average value. My question is whether this is an approximation or this is an actual fact? Why does translational invariance imply that all spins have the same average value? If we assume that there are no fluctuations then $m$ would be a constant throughout the lattice by assumption, not by "translational invariance" right?

Again, I'm not sure what is meant by translational invariance here.

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    $\begingroup$ Translational invariance, here, means that the law of physics do not change by the transformation $i \rightarrow i+n$, with $n$ positive or negative integer. So, all sites are equivalent. $\endgroup$ – Trimok Jul 29 '13 at 16:29
  • $\begingroup$ right i think i'm just overthinking this right $\endgroup$ – Timtam Jul 29 '13 at 16:35
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This is an assumption, but in the case of the of the Ising model it is an obviously correct one. The statement that $m$ is the same at all sites is the assumption that the state of the system respects translational invariance. In general a system may spontaneously break translational symmetry, just as it breaks the $s\rightarrow-s$ symmetry. So when you do the usual mean field theory you are assuming that the system maintains the symmetry.

Take for example the Ising model on a square lattice but with an antiferromagnetic interaction, so that neighboring spins want to anti-align. In this case, at zero temperature, the ground state has $s=+1$ on half the sites and $s=-1$ on the other half. This spontaneously breaks the translational symmetry.

Let's say you didn't notice this and proceeded with the derivation as usual. So you assume the magnetization is equal on every site and went through the mean field theory. Nothing would break, mathematically speaking. However, you would find that there is no spontaneous symmetry breaking and that $m=0$. This is wrong. You get the wrong answer because you tried to approximate the system with a ferromagnetic state, when you should have used an anti-ferromagnetic.

Moral: All other problems aside, mean field theory is only as good as the state you are using. Even worse, if you use the wrong state, the mean field theory will give you no idea that you made the wrong choice.

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