# Susceptibility in the paramagnetic Ising model

In the Ising model (2D for simplicity), the magnetic susceptibility casts the form

$$\chi = \beta\left(\langle M^2\rangle - \langle M\rangle^2\right)$$

We know that the susceptibility peaks at the transition temperature between ferromagnetic and paramagnetic phase. However, we can see that in the paramagnetic phase, $$\langle M^2\rangle=\frac{1}{N}\sum_iM_i^2=\frac{1}{N}\sum_i (\pm1)^2=1$$ $$\langle M\rangle^2=\left(\frac{1}{N}\sum_iM_i\right)^2=\left(\frac{1}{N}\sum_i(\pm 1)\right)^2 = 0$$ since $$M_i = \pm 1$$ and there is no net magnetisation in the paramagnetic phase. This means that the susceptibility becomes $$\chi = \beta\sim1/T$$, contradicting to the Curie-Weiss law. Where did I make my mistake?

• Is $\beta=1/(k_BT)$? Jul 28 at 18:26
• Yes, that is correct Jul 28 at 18:29
• For $T>T_c$, we have $\chi \propto (T-T_c)^{-1}$. My focus is only in the paramagnetic phase ($T>T_c$), and gives $\chi=\beta$, clearly contradicts to the Curie-Weiss law. Jul 28 at 21:32
• But the condition 'net magnetisation vanishes' applies for all $T>T_c$. How can you explain this discrepancy between $\chi \propto (T-T_c)^{-1}$ and $\chi \propto T^{-1}$? Jul 29 at 0:59
• Did you check out Wikipedia? This seems very on-point: en.wikipedia.org/wiki/… Jul 29 at 4:11

You are misunderstanding the notation. Let us denote by $$\sigma_i$$, $$i\in V\subset\mathbb{Z}^2$$, the spins in your system.
The magnetization the takes the form $$M_V = \sum_{i\in V} \sigma_i.$$ Therefore (note that the brackets denote expectation with respect to the Gibbs measure, not an average over spins as you seem to be doing), $$\langle M_V^2 \rangle = \Bigl\langle\Bigl( \sum_{i\in V} \sigma_i \Bigr)^2\Bigr\rangle = \Bigl\langle\sum_{i,j\in V} \sigma_i\sigma_j\Bigr\rangle = \sum_{i,j\in V} \langle\sigma_i\sigma_j\rangle.$$ Similarly, $$\langle M_V\rangle^2 = \Bigl( \Bigl\langle\sum_{i\in V} \sigma_i \Bigr\rangle\Bigr)^2 = \Bigl( \sum_{i\in V} \langle\sigma_i\rangle \Bigr)^2 = \sum_{i,j \in V} \langle\sigma_i\rangle\langle\sigma_j\rangle.$$ Its variance is thus given by $$\operatorname{Var}(M_V) = \langle M_V^2 \rangle - \langle M_V \rangle^2 = \sum_{i,j\in V} \bigl( \langle\sigma_i\sigma_j\rangle - \langle\sigma_i\rangle\langle\sigma_j\rangle \bigr).$$ In the thermodynamic limit, using translation invariance, the susceptibility can then be written as $$\chi = \lim_{V\to\mathbb{Z}^2} \frac{\beta}{|V|}\operatorname{Var}(M_V) = \beta \sum_{i\in\mathbb{Z}^2} \bigl( \langle\sigma_0\sigma_i\rangle - \langle\sigma_0\rangle\langle\sigma_i\rangle \bigr).$$ Concerning your computations, note that, when $$i\neq 0$$, $$\langle\sigma_0\sigma_i\rangle\neq 1$$ except at $$T=0$$, while $$\langle\sigma_0\rangle=\langle\sigma_i\rangle = 0$$ if $$T>T_{\rm c}$$.
In fact, the truncated 2-point function $$\langle\sigma_0\sigma_i\rangle - \langle\sigma_0\rangle\langle\sigma_i\rangle$$ decays exponentially fast in $$\|i\|$$ at all $$T\neq T_{\rm c}$$ (in pure phases), but the decay becomes a non-summable power law at $$T_{\rm c}$$, which is the cause of the divergence of the susceptibility $$\chi$$.
• You don't. The Curie-Weiss law only holds in the mean-field approximation, or above 4 dimensions. You can see the list of critical exponents of the Ising model here. The relevant one is $\gamma$. Jul 29 at 8:53
• If your more general question is "how do you obtain the critical behavior of the susceptibility?", then the answer is that this is hard. When $d=2$, you can make explicit computations to obtain the exact critical exponent. When $d=3$, you can obtain a good estimate of the critical exponent using, for instance, conformal bootstrap. The mean-field behavior above $4$ dimension can be rigorously derived, but this is not trivial either. Jul 29 at 9:06