# Theoretical prediction for absolute magnetization in 1D Ising model of a ferromagnet

In the 2D Ising model of a ferromagnet, Onsager predicts the absolute magnetization as a function of (unitless) temperature as

$$|M_{2D}(T)|=\left(1-\sinh\left(\frac{\ln\left(1+\sqrt{2}\right)}{T}\right)^{-4}\right)^\frac{1}{8}$$

Does a similar expression for the absolute magnetization exist for the 1D model? I am able to find a theoretical prediction for the non-absolute magnetization in the presence of an external magnetic field $B$ (in unitless variables),

$$M_{1D}(T)=\frac{\exp\left(\frac{1}{T}\right)\sinh(B)}{\sqrt{\left(\exp\left(\frac{2}{T}\right)\sinh^2(B)+\exp{\left(-\frac{2}{T}\right)}\right)}},$$

but this vanishes when $B=0$, due to the symmetry of the system. I have so far been unable to find an analytical expression for $|M_{1D}(T)|$. Does such a thing exist?

• Yes, such an expression exists: it is equal to $0$. (Considering the absolute value does not change anything, once the thermodynamic limit has been taken.) Nov 16, 2017 at 7:06
• I don't understand... at low T, the system is completely magnetized, with equal probability of being up or down. Taking the absolute value (on a per-microstate basis, before averaging) means |M(0)|=1. Perhaps I should specify that I mean to take the absolute value before taking the average, ie, $\langle|M|\rangle$, not $|\langle M\rangle|$ Nov 16, 2017 at 15:41
• No, at low temperature the magnetization density vanishes almost surely (and so does, obviously, its absolute value). It is only at zero temperature that the magnetization is non-zero. This can very easily be proved mathematically. Nov 16, 2017 at 17:52
• I see, so the presence of nonzero magnetization at nonzero temperature in simulation is due to finite size effects? Nov 16, 2017 at 18:05

Let $$m_N = \frac1N \sum_{i=1}^{N} \sigma_i$$ be the (empirical) magnetization density and $\mu_{N;\beta}$ be the Gibbs measure (with periodic boundary condition) at inverse temperature $\beta<\infty$, for a one-dimensional system of $N$ spins $\sigma_1,\ldots,\sigma_N$.
One can then prove the following: For any $\beta<\infty$ and any $\epsilon>0$, $$\lim_{N\to\infty} \mu_{N;\beta}(|m_N| > \epsilon) = 0.$$ (In fact, this probability decays exponentially fast in $N$.) This is standard material (for a proof, see for example Theorem 3.10 in this book). Of course, it follows that the expected value of $|m_N|$ satisfies $$\langle |m_N| \rangle_{N;\beta} \leq \epsilon + \mu_{N;\beta}(|m_N| > \epsilon).$$ In view of the above, this immediately implies that $$\lim_{N\to\infty} \langle |m_N| \rangle_{N;\beta} = 0,$$ for any $\beta<\infty$.
$$M=\tanh\left(\frac{|M|}{T}\right)$$
• Well, again,in dimension $1$, the magnetization density is equal to zero at any positive temperature (in the thermodynamic limit). It is only at $T=0$ that the magnetization is positive (and actually equal to $\pm 1$). (This is an exact mathematical statement, not relying on simulations.) Nov 16, 2017 at 17:54