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From what I understand, the Ising model is described by the canonical ensemble because it has constant temperature and volume. Why is it the Helmholtz free energy: $$F = \langle E \rangle - TS,$$ that is minimised at equilibrium in canonical ensemble, and not the average energy $\langle E \rangle$.

In more generality, is the equilibrium in the situation defined as the system with the maximum entropy (which leads to minimisation of the free energy) or is it the free energy that is more fundamental?

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You are used to the fact that the internal energy is minimized in equilibrium, but keep in mind that in this case your system is not isolated. Instead, you are dealing with a system in the canonical ensemble, so there is a giant temperature reservoir coupled to your system! The entire system of your spins + the bath is considered to be in the micro canonical ensemble, and their internal energy is minimized in equilibrium. By introducing the Helmholtz free energy

$$F=U-TS$$

you are able to solve the spin system while discarding the bath. Now observe that there is interplay between the internal energy and the entropy. For low temperatures $F\approx U$ so the internal energy is minimized, while for high temperatures $F\approx-TS$ and the systems tends to maximize its entropy. In between the situation is more complex. Thus as you said $F$ is more fundamental.

EDIT 1: Why minimization? Just because entropy is always increasing! If $Q_{\rm bath}$ is the heat flowing into the heat bath then

$Q_{\rm bath}=\Delta U_{\rm bath}=-\Delta U_{}$

since the total energy of the spins + bath is constant, and so the total change in entropy is

$$0\leq\Delta S=\Delta S_{\rm bath}+\Delta S=\frac{Q_{\rm bath}}{T}+\Delta S=$$ $$=\frac{T\Delta S-\Delta U}{T}=-\frac{\Delta\left(U-TS\right)}{T}=-\frac{\Delta F}{T}$$

Thus the change in Helmholtz free energy is

$$\Delta F\leq 0$$

It means that this quantity is decreasing, obtaining a minimum at equilibrium. I should note that I'd assumed that the system does zero work.

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  • $\begingroup$ I understand as much, I am just unsure why the free energy is minimised in equilibrium here. I see that it should be conserved because: $$dU = TdS - PdV = d(TS) - SdT - PdV,$$ therefore: $$d(U-TS) = -SdT - PdV.$$ Which has to be equal to $0$ in the canonical ensemble as $dT = 0$ and $dV = 0$, but why is it minimised? $\endgroup$
    – Akerai
    Commented Nov 17, 2017 at 18:10
  • $\begingroup$ Oh, you're unsure why is this necessarily a minimum? I've edited my answer accordingly. $\endgroup$
    – eranreches
    Commented Nov 17, 2017 at 18:25
  • $\begingroup$ +1 very nice answer but when you write "...internal energy is minimized in equilibrium, but keep in mind that in this case your system is not isolated", it may be misleading. Minimum internal energy is for "closed" systems at fixed entropy: en.wikipedia.org/wiki/Principle_of_minimum_energy and physics.stackexchange.com/q/534173/226902 $\endgroup$
    – Quillo
    Commented Apr 2, 2023 at 17:43

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