# Why do ferromagnetic crystals need to be anisotropic?

In the book Introduction to the theory of Ferromagnetism by Amikam Aharoni, it is said (pages 83-84) that in order to have a real ferromagnet, it is necessary to have another energy term (the anisotropic ones) other than the Heisenberg hamiltonian.

Why is this so?

Following his demonstration, we start with a ferromagnetic particle of magnetic moment $\vec\mu$ at an angle $\theta$ w.r.t. a fixed magnetic field $\vec H$. This gives an energy interaction $E=-\mu H \cos(\theta)$. At thermal equilibrium the probability of having a particular angle at temperature T is proportional to $\exp[\frac{\mu H \cos(\theta)}{k_B T}]= \exp[x \cos(\theta)]$. So the average for an ensemble of particles is $$\langle \cos(\theta)\rangle= \frac{\int_0^{2\pi}\int_0^\pi \cos(\theta)e^{x\cos(\theta)}\sin(\theta)d\theta d\phi}{\int_0^{2\pi}\int_0^\pi e^{xcos(\theta)}\sin(\theta)d\theta d\phi}=\coth(x)-\frac{1}{x}=L(x)$$ where $L(x)$ is the Langevin function.

But $\langle \cos(\theta)\rangle$ is just the component parallel to $\vec H$ of the normalized magnetization vector: $$\langle \cos(\theta)\rangle = \frac{M_H}{M}=L\Big(\frac{\mu H}{k_B T}\Big)$$ this function is the same obeyed, classically, by paramagnets, so this proves that all ferromagnets would be paramagnets, unless some other energy term is not included.

Now, I don't get the meaning of consider an ensemble of particles instead of just one single particle.

EDIT I'm not interested in amorphous ferromagnetic materials.

• There are amorphous ferromagnets (iron-based metglass). – Pieter Jan 8 '18 at 16:19