# Magnetic susceptibility in zero external field?

I want to know why the magnetic susceptibility in zero external field is :

(the fluctuations in magnetization)

$$\chi = \frac{1}{k_bT} \left( \langle M^2 \rangle - \langle M \rangle^2 \right)$$

can any one give a derivation?

Second question(if any one is familiar): I want to calculate magnetic susceptibility in The Ising Model using Wolff Cluster algorithm, this is the formula I should use?!!

Define partition function in the presence of external magnetic field (source field-$$h$$) as, $$Z[h]=\sum_{\{\sigma_{i}^{}\}\in\text{Phase space}}^{} e^{-\beta\left[H[\{\sigma_{i}\};\{J_{ij}\}]-h \hat{M}\right]},$$ where $$H[\{\sigma_{i}\};\{J_{ij}\}]$$ is a generic (classical) Hamiltonian model of a magnet with magnetic moments, $$\{\sigma_{i}\}$$ on the lattice interacting with each other by generic exchange couplings $$\{J_{ij}\}$$, and the total magnetic moment (defined as $$\hat{M}=\sum_{i\in \text{Lattice site}}^{}\sigma_{i}^{}$$) is coupled to an external magnetic field, $$h$$.
$$M[h]=\langle\hat{M}\rangle=\frac{1}{\beta}\frac{\partial}{\partial h}\ln[Z[h]],$$ where $$\langle \hat{O} \rangle=\frac{}{Z[h]}\sum_{\{\sigma_{i}\}\in \text{Phase space}}^{} \hat{O} e^{-\beta\left[H[\{\sigma_{i}\};\{J_{ij}\}]-h \hat{M}\right]}.$$
As the magnetic susceptibility (at the zero external magnetic field) is defined as $$\chi=\frac{\partial M[h]}{\partial h}\Big|_{h \to 0}^{}.$$
Use the above, $$\begin{eqnarray} \chi&=&\frac{\partial M[h]}{\partial h}\Big|_{h \to 0}^{}\\ &=&\beta[\langle\hat{M}^2_{}\rangle-\langle\hat{M}\rangle_{}^2].\end{eqnarray}$$
• The sign discrepancy is due to the extra minus sign in the partition function. Usually the Ising hamiltonian is written as $-J\sum_{ij} S_iS_j - h\sum_i$, so there is no need for the extra $-\beta$. Or equivalently you have to add two minus signs in the hamiltonian. For this derivation it's actually sufficient to only change the sign of $h$. This convention changes the meaning of magnetic field or the type of coupling with a spin. Commented Oct 19, 2021 at 17:43
• @AnotherUser Thanks. I will update the answer by changing the sign of $H$. Commented Oct 20, 2021 at 18:44