# 1D Ising model thermodynamic limit

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?

• Any physical or mathematical argument is welcome. – abhijit975 Sep 22 '19 at 2:19
• 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? – Jahan Claes Sep 22 '19 at 2:30
• 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 $h\rightarrow 0$. – abhijit975 Sep 22 '19 at 2:45
• 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. – Kuma Sep 22 '19 at 4:19
• @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? – abhijit975 Sep 23 '19 at 15:57

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 $$\lim_{h\to0^+} M{\left(h\right)} > 0$$. 