The partition function for the infinite range Ising model is $$Z(\beta ,h)=\int_{-\infty}^{\infty}dm \frac{1}{\sqrt{2\pi/N\beta J}} e^{-Ng(m)}$$ where $g(m) = \frac{\beta J}{2}m^2 - \ln\left[2\cosh(\beta h + \beta J m)\right]$. Here $J$ is the coupling strength and $h$ is the applied field. To get the analytical partition function one then takes the thermodynamic limit $N\rightarrow \infty$. The magnetization is then defined by $$M = \lim_{h\rightarrow 0} \lim_{N\rightarrow\infty} \frac{1}{\beta N} \frac{\partial \ln Z(\beta,h)}{\partial h}.$$

My question is then following. Why is it important to take the $N\rightarrow\infty$ first and then the $h\rightarrow0$ limit. If we take instead the following quantity $$M_1 = \lim_{N\rightarrow\infty} \lim_{h\rightarrow 0} \frac{1}{\beta N} \frac{\partial \ln Z(\beta,h)}{\partial h},$$ why wouldn't $M_1$ be a valid quantity for magnetization? Since we are taking the $N\rightarrow \infty$ still one should be able to perform the integration by steepest descent method. What stops it to be the magnetization of the system?

  • $\begingroup$ Any physical or mathematical argument is welcome. $\endgroup$ – abhijit975 Sep 22 '19 at 2:19
  • $\begingroup$ It's been awhile since I've looked at this, but is it possible that the derivation that formula for Z is actually only valid in the large N limit? $\endgroup$ – Jahan Claes Sep 22 '19 at 2:30
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    $\begingroup$ The first line (with the integral) is always valid and exact. The steepest descent method is valid only in the $N \rightarrow \infty$ limit. But we are taking that limit anyway in the case of $M_1$ as well but only after taking the limit $h\rightarrow 0$. $\endgroup$ – abhijit975 Sep 22 '19 at 2:45
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    $\begingroup$ The order of the limits is necesary to have a spontaneous magnetization at zero h. If the limit $h \to 0$ is taken first, then by symmetry M must be zero at any finite N. But if the limit $N \to \infty$ is taken first, one obtains a finite value, which goes to a constant as h goes to zero. The limits do not commute. $\endgroup$ – Kuma Sep 22 '19 at 4:19
  • $\begingroup$ @Kuma you said that by symmetry M must be zero at finite N if we take $h\rightarrow 0$ first. How does the symmetry argument change for infinite N? $\endgroup$ – abhijit975 Sep 23 '19 at 15:57

This is an issue of limits not commuting which applies to any version of the Ising model with a phase transition, not just the infinite-range case. I will give a rough version of the explanation found in chapter 2 of Nigel Goldenfeld's book.

For any finite $N$, the partition function is a sum of a finite number of analytic functions (exponentials). Thus, for any finite $N$ the magnetization must be a smooth function, and in particular be continuous. At $h = 0$, symmetry requires that the magnetization $M = 0$. Thus, by this continuity, as $h\to0$ from either side, $M\to0$ smoothly. Therefore, if we take the limit $h \to 0$ at finite $N$ before taking the limit $N \to 0$, we will always find that there is no "spontaneous magnetization" at $h \approx 0$.

On the other hand, the limit of an infinite sum of continous functions need not be continous (consider e.g. the Fourier representation of a square wave). Therefore, taking the limit $N\to \infty$ first allows us to get an $M{\left(h\right)}$ which is not continous at $h = 0$. So although symmetry still requires that $M{\left(0\right)} = 0$, we can have $\lim_{h\to0^+} M{\left(h\right)} > 0$.

This figure from Nigel's book illustrates this discontinuity:

enter image description here


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