# Transverse Ising model with longitudinal field at finite T: magnetisation, susceptibility,

In a Ising model with transversal and longitudinal fields $$H = B_x \sum_i S_x^{(i)} + B_z \sum_i S_z^{(i)} + J \sum_{\langle i,j\rangle} S_z^{(i)}S_z^{(j)}$$ if $B_z=0$, the magnetisation $M$ and susceptibility $\chi$ are defined as $$M = \dfrac{\partial F}{\partial B_x} \qquad \chi = \dfrac{\partial M}{\partial B_x} \;,$$ where $F = -\beta^{-1} \log Z$ is the free energy and $Z$ is the canonical partition function.

But what if $B_z\neq 0$? Should one define a magnetisation vector $\vec{M}$? And for the susceptibility? What is the physical meaning?

• Please pay attention to the interacting term of the Hamiltonian, it seems a bit strange to me since I would expect something like $J\sum_{\left\langle ij \right\rangle} \vec{S}_i \cdot \vec{S}_j$. – Matteo Oct 7 '17 at 9:19
• @Matteo, you can write the interaction you want... what you mentioned is the "Heisenberg type" interaction, which is not what I have. – m137 Oct 8 '17 at 8:05

Of course in this case magnetization $\vec{M}$ is a vector: if you take $\vec{S}_i = S_x \hat{x}+S_z \hat{z}$ it is defined as

$$\vec{M} = \frac{1}{N} \sum_{i=1}^N \vec{S}_i$$

It is easy to prove that:

$$\vec{M} = \frac{\partial F}{\partial \vec{B}} = \frac{\partial F}{\partial B_x}\hat{x} + \frac{\partial F}{\partial B_z}\hat{z}$$

Magnetic susceptibility is now a tensor $\chi_{\alpha \beta}$ (where I used labels $\alpha=x,z$ and $\beta=x,z$) defined as:

$$\chi_{\alpha \beta} = \frac{\partial M_{\alpha}}{\partial B_{\beta}}$$

So you have to deal with four components: $\chi_{xx}$, $\chi_{xz}$, $\chi_{zx}$, $\chi_{zz}$ .