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I have a question on the terminology being used to compute results on the Ising Model. Every explanation I've seen defines the magnetisation of the system as some variation of:

$$ M = \frac{1}{N}\sum_i \langle\sigma_i\rangle = \frac{\langle \sum_i \sigma_i \rangle}{N} $$

Where $\langle . \rangle$ is the ensemble average. This formulation is then used to derive important results, such as the mean field approximations that derives a self-consistency:

$$ M = \tanh(\beta(h+qJM)) $$

Where $\beta$ is the inverse temperature, $h$ the external field, $q$ is the number of neighbours per site, $J$ the coupling constant. (Taken from here).

However this is leading to two questions for me which I haven't seen an obvious answer for:

  1. How is $M$ ever non-zero for $h=0$? By the symmetry of the possible Hamiltonians, any positive value has an equal weight that is negative. The terms will always cancel, so surely the ensemble average must be zero.

  2. Relatedly, in the ferromagnetic phase ($T<T_c$), how does one interpret the non-zero steady state solutions? When such results are discussed as observed magnetisations I don't quite follow as $M$ is defined as an ensemble average so it is never "observed" in that sense. Or are they referring to the a specific configuration quantity $M = \frac{\sum_i \sigma_i}{N}$ and the terminology is just a bit inconsistent?

I should note I'm not looking for a strict physical interpretation (it makes perfect sense, as the answers for this question lay out), my confusion is with the precise mathematics of the situation. I'm finding it uncomfortable to manipulate $M$ as the central quantity in such derivations when I don't have a clear idea of what it is consistently referring to. For the purposes of calculation, what is the appropriate definition of $M$?

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    $\begingroup$ I already answered similar questions several times. Some of the relevant answers: here, here, here and here. $\endgroup$
    – Yvan Velenik
    Nov 17, 2019 at 9:30

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You can calculate the magnetization in case of spontaneous symmetry breakdown using the method of quasi-averages : calculating magnetization using the formula from your question for non-zero magnetic field $h$ and taking the limit as $h$ approaches zero from above.

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  • $\begingroup$ This approach makes sense physically but it's not quite what I'm asking when it comes to computation (I linked in the answer to another question with similar responses). I suppose I'm asking something simpler: how do I interpret M? To put it differently you've outlined a physical reason one might observe non-zero "magnetisation" (non-ensemble) by rendering a specific configuration through a process applied to the system. Makes perfect sense - but the derivations keep referring to M as an ensemble average? How can it take on a single value, let alone a non-zero one? $\endgroup$
    – Sue Doh Nimh
    Nov 17, 2019 at 3:58
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    $\begingroup$ @SueDohNimh : So you asked: "For the purposes of calculation, what is the appropriate definition of $M$?" I suggest that you define $M$ as a quasi-average. $\endgroup$
    – akhmeteli
    Nov 17, 2019 at 4:15
  • $\begingroup$ Thank you - so to make sure I understand: what you've outlined gets me to the self-consistent expression. Then we take the h limit which allows for a continuum of non-zero magnetisations which all makes sense. And then would it be possible to clarify at the limit h=0 - the mathematical expression for M as defined still results in perfect symmetry. Any explicit computation of this will still always return 0. Is this just a convention to in some sense "ignore" the original definition once we've computed a solution (context: I am not a physicist so maybe a bit unfamiliar with conventions) $\endgroup$
    – Sue Doh Nimh
    Nov 17, 2019 at 4:34
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    $\begingroup$ @SueDohNimh : "And then would it be possible to clarify at the limit h=0 - the mathematical expression for M as defined still results in perfect symmetry" According to the quasi-average recipe, one takes the limit of $M$ for $h$ approaching zero from above, so you don't get zero magnetization in case of spontaneous symmetry breakdown (the ordinary limit does not even exist in this case).. $\endgroup$
    – akhmeteli
    Nov 17, 2019 at 4:43

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