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.

  • $\begingroup$ There are amorphous ferromagnets (iron-based metglass). $\endgroup$
    – user137289
    Jan 8, 2018 at 16:19

1 Answer 1


The author probably means that, without the anisotropic term with the external magnetic field the average magnetization of a Heisenberg ferromagnetic calculated in the canonical ensemble is zero (the symmetry is not broken). As for "all ferromagnets would be paramagnets, unless some other energy term is not included", you need to use the Heisenberg Hamiltonian to get a ferromagnetic.


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