Bohr–Van Leeuwen theorem

The Bohr–Van Leeuwen theorem applies to an isolated system that cannot rotate. If the isolated system is allowed to rotate in response to an externally applied magnetic field, then this theorem does not apply.[8] If, in addition, there is only one state of thermal equilibrium in a given temperature and field, and the system is allowed time to return to equilibrium after a field is applied, then there will be no magnetization.

The probability that the system will be in a given state of motion is predicted by Maxwell–Boltzmann statistics to be proportional to ${\displaystyle \exp(-U/k_{\text{B}}T)}$, where ${\displaystyle U}$ is the energy of the system, ${\displaystyle k_{\text{B}}}$ is the Boltzmann constant, and ${\displaystyle T}$ is the absolute temperature. This energy is equal to the sum of the kinetic energy (${\displaystyle mv^{2}/2}$ for a particle with mass ${\displaystyle m}$ and speed ${\displaystyle v}$) and the potential energy.[8]

The magnetic field does not contribute to the potential energy. The Lorentz force on a particle with charge ${\displaystyle q}$ and velocity ${\displaystyle \mathbf {v} }$ is

${\displaystyle \mathbf {F} =q\left(\mathbf {E} +\mathbf {v} \times \mathbf {B} \right),}$

where ${\displaystyle \mathbf {E} }$ is the electric field and ${\displaystyle \mathbf {B} }$ is the magnetic flux density. The rate of work done is ${\displaystyle \mathbf {F} \cdot \mathbf {v} =q\mathbf {E} \cdot \mathbf {v} }$ and does not depend on ${\displaystyle \mathbf {B} }$. Therefore, the energy does not depend on the magnetic field, so the distribution of motions does not depend on the magnetic field.[8]

In zero field, there will be no net motion of charged particles because the system is not able to rotate. There will therefore be an average magnetic moment of zero. Since the distribution of motions does not depend on the magnetic field, the moment in thermal equilibrium remains zero in any magnetic field.[8]

So as to lower the complexity of the proof, a system with ${\displaystyle N}$ electrons will be used.

This is appropriate, since most of the magnetism in a solid is carried by electrons, and the proof is easily generalized to more than one type of charged particle.

Each electron has a negative charge ${\displaystyle e}$ and mass ${\displaystyle m_{\text{e}}}$.

If its position is ${\displaystyle \mathbf {r} }$ and velocity is ${\displaystyle \mathbf {v} }$, it produces a current ${\displaystyle \mathbf {j} =e\mathbf {v} }$ and a magnetic moment[6]

${\displaystyle \mathbf {\mu } ={\frac {1}{2c}}\mathbf {r} \times \mathbf {j} ={\frac {e}{2c}}\mathbf {r} \times \mathbf {v} .}$

The above equation shows that the magnetic moment is a linear function of the velocity coordinates, so the total magnetic moment in a given direction must be a linear function of the form

${\displaystyle \mu =\sum _{i=1}^{N}\mathbf {a} _{i}\cdot {\dot {\mathbf {r} }}_{i},}$

where the dot represents a time derivative and ${\displaystyle \mathbf {a} _{i}}$ are vector coefficients depending on the position coordinates ${\displaystyle \{\mathbf {r} _{i},i=1\ldots N\}}$.[6]

Maxwell–Boltzmann statistics gives the probability that the nth particle has momentum ${\displaystyle \mathbf {p} _{n}}$ and coordinate ${\displaystyle \mathbf {r} _{n}}$ as

${\displaystyle dP\propto \exp {\left[-{\frac {{\mathcal {H}}(\mathbf {p} _{1},\ldots ,\mathbf {p} _{N};\mathbf {r} _{1},\ldots ,\mathbf {r} _{N})}{k_{\text{B}}T}}\right]}d\mathbf {p} _{1},\ldots ,d\mathbf {p} _{N}d\mathbf {r} _{1},\ldots ,d\mathbf {r} _{N},}$

where ${\displaystyle {\mathcal {H}}}$ is the Hamiltonian, the total energy of the system.[6]

The thermal average of any function ${\displaystyle f(\mathbf {p} _{1},\ldots ,\mathbf {p} _{N};\mathbf {r} _{1},\ldots ,\mathbf {r} _{N})}$ of these generalized coordinates is then

${\displaystyle \langle f\rangle ={\frac {\int fdP}{\int dP}}.}$

In the presence of a magnetic field,

${\displaystyle {\mathcal {H}}={\frac {1}{2m_{\text{e}}}}\sum _{i=1}^{N}\left(\mathbf {p} _{i}-{\frac {e}{c}}\mathbf {A} _{i}\right)^{2}+e\phi (\mathbf {q} ),}$

where ${\displaystyle \mathbf {A} _{i}}$ is the magnetic vector potential and ${\displaystyle \phi (\mathbf {q} )}$ is the electric scalar potential. For each particle the components of the momentum ${\displaystyle \mathbf {p} _{i}}$ and position ${\displaystyle \mathbf {r} _{i}}$ are related by the equations of Hamiltonian mechanics:

${\displaystyle {\begin{aligned}{\dot {\mathbf {p} }}_{i}&=-\partial {\mathcal {H}}/\partial \mathbf {r} _{i}\\{\dot {\mathbf {r} }}_{i}&=\partial {\mathcal {H}}/\partial \mathbf {p} _{i}.\end{aligned}}}$

Therefore,

${\displaystyle {\dot {\mathbf {r} }}_{i}\propto \mathbf {p} _{i}-{\frac {e}{c}}\mathbf {A} _{i},}$

so the moment ${\displaystyle \mu }$ is a linear function of the momenta ${\displaystyle \mathbf {p} _{i}}$.[6]

The thermally averaged moment,

${\displaystyle \langle \mu \rangle ={\frac {\int \mu dP}{\int dP}},}$

is the sum of terms proportional to integrals of the form

${\displaystyle \int _{-\infty }^{\infty }(\mathbf {p} _{i}-{\frac {e}{c}}\mathbf {A} _{i})dP,}$

where ${\displaystyle p}$ represents one of the momentum coordinates.

The integrand is an odd function of ${\displaystyle p}$, so it vanishes.

Therefore, ${\displaystyle \langle \mu \rangle =0}$.[6]