# Does the Bohr van Leeuwen Theorem also apply to ferromagnetism?

I know that the Bohr-van Leeuwen theorem shows that there could be not consistent pure classical explanation of dia- and paramagnetism.

Does the same theorem also rule out a consistent classical theory of ferromagnetism?

Do you have any realiable references for this?

• Comment to the question (v2): Wikipedia in the first paragraph claims it is the case and refers to Amikam Aharoni, Introduction to the Theory of Ferromagnetism, 1996. – Qmechanic Jun 30 '14 at 19:29
• "the Bohr-van Leeuwen theorem shows that there could be not consistent pure classical explanation of dia- and paramagnetism" this is often said but it is misleading. There is diamagnetism in classical physics - free charged particle in external magnetic field moves in such a way as to produce magnetic moment opposing the magnetic field: en.wikipedia.org/wiki/Diamagnetism#Langevin_diamagnetism – Ján Lalinský Jul 1 '14 at 10:35
• @JánLalinský the Bohr-van Leeuwen theorem applies to systems in thermal equilibrium (for instance, plasmas are apparently able to classically produce magnetic effects because they are not in equilibrium). – Danu Jul 1 '14 at 15:41
• @Danu, I know that the theorem is derived for equilibrium. Still the common statement I quoted is misleading. Partly because there is classical diamagnetism in non-equilibrium, partly because the derivation assumes only one very special situation from classical physics, not all of them. – Ján Lalinský Jul 1 '14 at 16:32

The Bohr–van Leeuwen theorem is a theorem in the field of statistical mechanics. The theorem states that when statistical mechanics and classical mechanics are applied consistently, the thermal average of the magnetization is always zero. This makes magnetism in solids solely a quantum mechanical effect and means that classical physics cannot account for diamagnetism, paramagnetism or ferromagnetism.

The theorem applies to any form of magnetism. Continuing to read the same wikipedia page, which supplies an intuitive as well as a more formal proof, one sees that the argument formally boils down to showing that the thermal average of the magnetic moment $\mu$ is zero:

$$\langle \mu \rangle=0$$

This is done without any assumptions on the origin of the magnetic moment $\mu$. I will now reproduce another proof, which is found in several textbooks.

Consider an $N$-particle system with only particles with charge $e$ and mass $m$ (the proof is easily generalized). We define the (thermal average of the) magnetization as

$$\langle \mu \rangle=\left\langle -\frac{\partial F}{\partial B}\right\rangle$$ Where $F=-T\ln Z$ is the free energy, and $Z$ is the partition function. The classical partition function is $$Z=\int d\vec{p}_1\dots d\vec{p}_N\int d\vec{r}_1 \dots d\vec{r}_N\ e^{-\beta H}$$ Where $H$ is the classical Hamiltionian. In the presence of a magnetic field, we have $$H=\frac{1}{2m}\sum_{i=1}^{N}\biggl(\vec{p}_i-\frac{e}{c}\vec{A}_i\biggr)^2+eV(\vec{r}_1,\dots\vec{r}_N)$$ By making the substitution $\vec{p}_i\to \vec{p}_i-\frac{e}{c}\vec{A}_i$ in each integral over the momenta we can completely eliminate the dependence of $Z$ on $\vec{A}_i$ and therefore on the magnetic field $\vec{B}=\vec{\nabla}\times\vec{A}$. Therefore, $F$ doesn't depend on $B$ either, and $$\langle \mu \rangle= \left\langle -\frac{\partial F}{\partial B}\right\rangle=0$$

In conclusion, the Bohr-van Leeuwen theorem shows that magnetism cannot be accounted for classically, independent of the origin of the magnetization. When applying a magnetic field and allowing a solid to reach thermal equilibrium, there can be no net magnetization (classically). In particular, it also rules out classical ferromagnetism.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Mar 22 at 18:41

I hope that doubts about the change of integration variables which eliminates vector potential have been cleared up. The key point is that, for each coordinate $${\bf x}_i$$, a change of momentum variable $${\bf p^{\prime}}_i= {\bf p}_i -{\bf A({\bf x}_i)}$$ makes the vector potential term disappearing, due to the unbounded integration limits on momenta.

However, the recent bounty on this question draws the attention on some incompleteness of the existing answer with respect to the original question.

As far as the formal proof of the consequences of BvL theorem, everything is fine. Where the answer is not satisfactory is when the the consequences of the theorem are extended to every form of magnetism. In particular to ferromagnetism (which was the subject of the original question).

It is true, and it remains true that real ferromagnetism cannot be explained in a self-consistent way using only classical mechanics, electromagnetism and classical statistical mechanics. But the reason has nothing to do with BvL theorem.

In order to understand this point is necessary to recall the modern (quantum mechanical) explanation of ferromagnetism. It is based on two key ingredients:

1. electrostatic interaction;
2. antisymmetry of the electronic wavefunction (i.e. Pauli's principle).

Therefore, it is clear that ingredient n.$$2$$ is missing in any purely classic explanation, accordingly with the conclusion that in a purely classical world it would be impossible to provide a consistent explanation of ferromagnetism.

It is also clear that, since the origin of the electronic spin magnetic momentum is not the orbital motion, and then its origin has nothing to do with a Hamiltonian of a spinless point particle like that present in Danu's answer, the BvL theorem does not say anything about ferromagnetism.

A final word of caution is in order. The result of a careful analysis of the origin of magnetic effects in condensed matter calls for quantum mechanics, as key ingredient to provide a consistent explanation of the origin and behavior of magnetic effects. However, as often in physics, one could be interested in modeling the behavior of magnetic systems, without any request of describing at the same time the origin of magnetism. And indeed, classical statistical mechanics can routinely applied to describe para/ferromagnetic transitions, on the base of purely classical statistical mechanics (think for example to the prototype of all the magnetic models, the Ising model).

## protected by ACuriousMind♦Jun 16 '17 at 15:57

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