The Bohr–Van Leeuwen theorem states that when statistical mechanics and classical mechanics are applied consistently, the thermal average of the magnetization is always zero.^[1] This makes magnetism in solids solely a quantum mechanical effect and means that classical physics cannot account for paramagnetism, diamagnetism and ferromagnetism. Inability of classical physics to explain triboelectricity also stems from the Bohr–Van Leeuwen theorem.^[2]

History

What is today known as the Bohr–Van Leeuwen theorem was discovered by Niels Bohr in 1911 in his doctoral dissertation^[3] and was later rediscovered by Hendrika Johanna van Leeuwen in her doctoral thesis in 1919.^[4] In 1932, J. H. Van Vleck formalized and expanded upon Bohr's initial theorem in a book he wrote on electric and magnetic susceptibilities.^[5]

The significance of this discovery is that classical physics does not allow for such things as paramagnetism, diamagnetism and ferromagnetism and thus quantum physics is needed to explain the magnetic events.^[6] This result, "perhaps the most deflationary publication of all time,"^[7] may have contributed to Bohr's development of a quasi-classical theory of the hydrogen atom in 1913.

Proof

Statistical mechanics
Thermodynamics Kinetic theory
Particle statistics Spin–statistics theorem Indistinguishable particles Maxwell–Boltzmann Bose–Einstein Fermi–Dirac Parastatistics Anyonic statistics Braid statistics
Thermodynamic ensembles NVE Microcanonical NVT Canonical µVT Grand canonical NPH Isoenthalpic–isobaric NPT Isothermal–isobaric
Models Debye Einstein Ising Potts
Potentials Internal energy Enthalpy Helmholtz free energy Gibbs free energy Grand potential / Landau free energy
Scientists Maxwell Boltzmann Bose Gibbs Einstein Ehrenfest von Neumann Tolman Debye Fermi
v t e

An intuitive proof

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]

A more formal proof

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

${\displaystyle dP\propto \exp {\left}d\mathbf {p} _{1},\ldots ,d\mathbf {p} _{N}d\mathbf {r} _{1},\ldots ,d\mathbf {r} _{N},}$

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,

${\displaystyle \mathcal{H}=\frac{1}{2m_{\text{e}}}\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{({\mathbf{p}}_i-<!--\; -\; -->\frac{}{}}^{}}$Zdroj:https://en.wikipedia.org?pojem=Bohr–Van_Leeuwen_theorem
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