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I am revisiting Statistical Mechanics to better understand models of spin glass and was wondering to what extent axiomatics of Stat.Mech. applies to an ensamble of spin configurations. In particular, how big should the system be to even be statistical, hundreds, thousands of spins? If one goes beyond near-neighbor interactions, how far can one go with the interaction strength to still preserve the statistical independence axiom? In other words, all those papers, where one talks about fully-connectted spin glass with 50 spins and then applies Statistical Mechanics to analyze them, make me feel awkward.

Any logical reasoning is appreciated!

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  • $\begingroup$ You can apply statistical physics to any number of degrees of freedom, but success in reproducing thermodynamic facts is not guaranteed if the number is low. To get results consistent with macroscopic thermodynamics, one usually analyzes the limit $N\to\infty$. For example, faithful description of sharp phase boundaries and related behaviour during a phase transition can be recovered from statistical physics only in this limit. For finite $N$, the statistical physics description may have unphysical features, such as no sharp distinction between gas and liquid, or no magnetic domain walls. $\endgroup$
    – Ján Lalinský
    Commented Nov 18, 2023 at 2:56
  • $\begingroup$ @JánLalinský "You can apply..., but success is not guaranteed." - this kind of statement can be made about anything. Stat. Mech. has been developed for large systems near their equilibrium. Those system, by def., have lots of particles and those particles do not exhibit long-range interactions, otherwise it wouldn't have been an equilibtium. All-to-all Ising graph of ~50 spins violates both. You rightfully brought up the limit of $N \rightarrow \infty$. But nowhere in the whole Stat.Mech will you see the limit of $N=1$ or even a countable number of particles. $\endgroup$
    – MsTais
    Commented Nov 18, 2023 at 3:13
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    $\begingroup$ I agree that for statistical physics to work, interaction has to fade away fast enough. Some analysis is in Landau&Lifshitz, if I remember correctly, central static field of force must fade away faster than $1/r^3$, otherwise homogeneous system energy is not proportional to volume. $\endgroup$
    – Ján Lalinský
    Commented Nov 19, 2023 at 17:39

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Statistical physics tells you something about the quantity you are expected to measure in situation where you can apply the Boltzmann distribution. It doesn't matter how many states the system may have. It can have as few or many as you like. It works for macroscopic and microscopic states. When a system is coupled so strongly that it effectively has only one state, then this framework doesn't do any averaging and simply gives you the measurement that belongs to this state -- and that's the correct result. The only question is: Is it ok to use this probability distribution? You need:

  • Always: thermal equilibrium.
  • Sometimes: $k_B T \ll \min (\Delta E)$
  • Nice to have: Applicability of the approximations used in the model (such as Stirling's formula, where you need $(12r)^{-1}$ elements to get a relative error < $r$).
  • etc.
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  • $\begingroup$ Ok, I will rephrase my question: can I apply Boltzmann distribution to a system with 10 particles, 50 particles, 1000 particles? Also, it seems to me that you use terms "particles" and "states" interchangeably. One state can be represented by N particles, and one particle can be in N states (even at once, but in quantum). $\endgroup$
    – MsTais
    Commented Nov 18, 2023 at 1:57
  • $\begingroup$ @MsTais I didn't use the word "particle". A system has N elements (such as spins or people) and you can map them to $\Omega (N)$ states. For example the memory in your computer has $2^N$ states, where $N$ is your memory capacity in bits. As I said, you can apply the Boltzmann distribution to only one bit or one particle or one person. It's totally ok. Its only task is to associate states with probabilities. It doesn't matter how many $N$s or $\Omega(N)$s there are. $\endgroup$
    – secr
    Commented Nov 18, 2023 at 2:17
  • $\begingroup$ How can you even infer the distribution from only one data point (one particle)? Also, do you work with quantum systems more? Cuz there it is a bit different cuz of entanglement and superposition, and you can actually play this game sometimes. In classical - no. $\endgroup$
    – MsTais
    Commented Nov 18, 2023 at 3:09
  • $\begingroup$ @MsTais You have no data points at all. You have a model that promises probabilities. At some point in your life you do an experiment where you get a measurement. This will give you a data point. Then you do some other stuff. Then you repeat the experiment. Or maybe someone else is repeating the experiment somewhere else. This will give you a second data point etc. $\endgroup$
    – secr
    Commented Nov 18, 2023 at 4:20
  • $\begingroup$ Ok, I think we are stuck here. I would advise you to go and consult the first half a chapter of Statistical Mechanics in any textbook. My choice this time was Landau Lifshitz. If you prefer Western school, then Feynman is good. What you described in your latter response is not a axiomatic framework of statistical mechanics. $\endgroup$
    – MsTais
    Commented Nov 20, 2023 at 1:42

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