If one considers an ideal gas of non interacting charged particles of charge $q$ in a uniform magnetic field $\mathbf{B} = \mathbf{\nabla} \wedge \mathbf{A}$, then the classical partition function in the canonical ensemble reads (in SI units):

$Q(\beta,V,N,\mathbf{B}) = \frac{1}{N!}q(\beta,V,\mathbf{B})^N$

where $q(\beta,V,\mathbf{B}) = \int \frac{d\mathbf{p} d \mathbf{r}}{h^3}\:e^{-\frac{\beta}{2m}(\mathbf{p}-q\mathbf{A}(\mathbf{r}))^2}$

If we integrate first with respect to momenta over all possible values from $-\infty$ to $+\infty$ for each component, a simple change of variable leads to

$q(\beta,V,\mathbf{B})=\frac{V}{\Lambda^3}$ which is the ideal gas result and where $\Lambda$ is the thermal de Brooglie wavelength.

If one then wants to get the magnetization per particle $\mathbf{\mu}$ induced by the field $\mathbf{B}$ it is simply:

$\mathbf{\mu} = -\frac{\partial \langle \epsilon \rangle}{\partial \mathbf{B}} = \frac{\partial }{\partial \mathbf{B}}\left( \frac{\partial \ln(q(\beta,V,\mathbf{B}))}{\partial \beta} \right) = \frac{\partial }{\partial \beta}\left( \frac{\partial \ln(q(\beta,V,\mathbf{B}))}{\partial \mathbf{B}} \right) = \mathbf{0}$

This is one way to state the Bohr-van Leeuwen theorem.

Now, I physically understand this result as coming from some symmetry associated with the momenta (it is as likely to go to the right as it is to go to the left) and the fact that the boundaries of the integral over the momenta are infinite.

If the problem is treated quantum mechanically, the eigenstates of one charge particle are discretized Landau levels with a typical spacing between two neighbouring levels that is $\hbar \omega_c$ where $\omega_c = qB/m$ is the cyclotron frequency and one finds that the sum over these states depends on the magnetic field $\mathbf{B}$.


I am lost in my interpretation of the quantum to classical limit for this system...so far I thought that the quantum -> classical limit for the statistical properties of an individual particle was related to the way of counting the number of states for this particle i.e. whether we consider the set of states as a continuum or as a discrete set. This analogy seems to work in this case as well since the classical limit arises if $k_B T \gg \hbar \omega_c$. However two major points differ from what I am used to:

  • The quantum treatment of this system yields a non zero magnetic moment (although it vanishes at infinite temperatures) in the limit where $k_B T \gg \hbar \omega_c$ while the classical treatment gives strictly zero.
  • I do not understand how does the left-right symmetry argument used in the classical partition function disappear in the quantum treatment to yield a partition function that depends on $\mathbf{B}$.
  • Is there any classical way to assess that quantum corrections will be of order $\mathcal{O}(\Lambda/R_c)$ where $R_c \sim \sqrt{m k_B T}/(qB)$ is the typical size of the radius of the helical paths taken by a charged particle?

Sorry if my questions seem confused, I will try to improve them if they are not clear enough.

EDIT: I realize that one of my points is not very clear and shall explain it with the example of a true harmonic oscillator. If I consider classical statistical mechanics, I know that $\langle \frac{1}{2}m\omega^2 x^2 \rangle = \frac{1}{2}k_B T$. This tells me that the typical uncertainty on the position of my particle is $\sigma_x = \sqrt{k_B T/(m\omega^2)}$. Incindently this length is also the typical confinement length scale owing to the harmonic potential. One way to semi-classicaly probe the validity of the classical limit is to imagine the particle as a non dispersive wave packet of width $\Lambda = h/\sqrt{2\pi m k_B T}$ and to realize that interferences (ultimately leading to quantization) are not important if $\Lambda \ll \sigma_x$. This is very appealing because one can then probe the validity of a classical approximation using a $\sigma_x$ that comes from a classical treatment.

My biggest problem with a charged particle in a magnetic field is that the Bohr-van Leewen theorem apparently prevents this typical length scale (that I know for sure is $R_c$) to be found with a classical statistical treatment.

  • $\begingroup$ I think the reason of non zero magnetic moment is non commutativity of momentum and vector potential. $\endgroup$
    – WInterfell
    Apr 8 '13 at 19:50

Landau diamagnetism takes place because of the noncommutativity of $\mathbf{p}$ and $\mathbf{A}(\mathbf{r})$. In the classical treatment no such noncommutativity exists, thus the magnetic susceptibility is identically zero.

One way to appreciate the dependence of the result on $\hbar$ and at the same time perform the full quantum treatment is to perform the computation in the coherent state basis. In this basis, the Landau Hamiltonian has the form of an isotropic two dimensional harmonic oscillator with the Larmor frequency as a natural frequency, (for some detail, please see the following question .

$H = \hbar \omega_c (a^{\dagger}a + \frac{1}{2})$

Due to the nonvanishing commutation relation between the creation and anihilation operators, the exponential of the Hamiltonian operator is given by:

$\exp(-\beta H) = \exp(-\frac{1}{2}\beta \hbar \omega_c) \exp(-\beta \hbar \omega_ca^{\dagger}a(1-e^{-\beta \hbar \omega_c}))$

In the coherent space basis, the partition function is given by:

$Z = \omega_c \int d^2\alpha \exp(-\frac{1}{2}\beta \hbar \omega_c) \exp(-\beta \hbar \omega_c\bar{\alpha}\alpha(1-e^{-\beta \hbar \omega_c})) = \omega_c (1-e^{-\beta \hbar \omega_c})^{-1}$

(The multiplicative factor proportional to $\omega_c$ is the Jacobian of the transformation of the phase space volume). (I am being careless in the immaterial constant terms).

This partition function gives the correct magnetic susceptibility. In addition, In the small \hbar limit the partition function becomes a constant, thus giving the Bohr-van Leewen theorem.

  • $\begingroup$ Thanks for the reply. I didn't know this calculation with coherent states. I am afraid, however, that this doesn't fully reply to my concerns. One problem I have is that, as it is explained in the wiki, the Bohr-van Leewen theorem arises from the fact that the mean angular momentum is the integral of an odd function and is therefore zero. Technically it seems very similar to computing the mean momentum for a particle in a box of an oscillator, we find zero by symmetry [to be continued] $\endgroup$
    – gatsu
    Apr 3 '13 at 20:35
  • $\begingroup$ This zero value is true whether we deal with the classical or the quantum version of the problem for a free particle in a box of a "true" harmonic oscillator. That is why I don't understand what is the physical difference in this case of a magnetic field. $\endgroup$
    – gatsu
    Apr 3 '13 at 20:40
  • $\begingroup$ It is also strange that a charged particle in a magnetic field is the only case I know where one cannot "rigorously" point out a classical characteristic length to compare with the thermal wavelength. It seems obvious that this characteristic length is the typical radius $R_c$ I gave above and if $\Lambda > R_c$ then the wavefunction will be as large as the "closed" trajectories in the xy plane and this effective confinement will lead to quantization of the energy levels that are the Landau levels $\endgroup$
    – gatsu
    Apr 3 '13 at 20:45
  • $\begingroup$ Sorry, but what is the proportionality dimensional factor in the front of Your expression for partition function (it has to be dimensionless)? $\endgroup$
    – Name YYY
    Jun 2 '16 at 15:59
  • $\begingroup$ @ YYY Thank you for your remark, I usually like to keep all units and constants and believe that they enhance the physics insight. This is why I'll update this answer and write the full prefactors. The explanation why a factor proportional to $\omega_c$ is present is because the transformation Jacobian between the coherent coordinates and the momenta includes a factor $2 e B \hbar$ which is inversely proportional to the cyclotron frequency. (The integration in the partition function is over all phase space). $\endgroup$ Jun 3 '16 at 6:13
  • The quantum treatment of this system yields a non zero magnetic moment (although it vanishes at infinite temperatures) in the limit where $k_B T \gg \hbar \omega_c$ while the classical treatment gives strictly zero.
  • I do not understand how does the left-right symmetry argument used in the classical partition function disappear in the quantum treatment to yield a partition function that depends on $\mathbf{B}$.

The derivation of the zero magnetic moment (the so called Bohr-van Leeuwen theorem) from the canonical probability distribution is mathematically correct. The reason the magnetic moment obtained is zero is the use of canonical probability distribution, which says the particle has constant position probability distribution inside the box and all directions of velocity are equally probable everywhere.

The equality of all directions of motion is reasonable for a system in a box, because the particle cannot go past the walls and as a result, the particle cannot exercise its natural circular motion if close enough to the wall. If the particle hits the wall, it gets reflected and this makes it reasonable to assume all directions are equally probable, even if particle is close to the wall.

If the walls are removed, charged particle exercises circular motion without encountering obstacles and this results in magnetic moment pointing in the direction determined by the magnetic field (magnetic moment will oppose magnetic field - this is called diamagnetism). It is quite easy to calculate this magnetic moment as a function of particle's energy. When many such particles are allowed to move without this restriction, large net magnetic moment may be obtained. This is possible because the velocity probability distribution at the edges of the system is not isotropic anymore.

Canonical distribution is thus inappropriate for calculating magnetic effects - being a function of energy only, it cannot capture the fact the velocity distribution at the edges of the system is not isotropic, but prefers the direction of circulation of the current determined by the external magnetic field. It assumes magnetized state of matter is what should happen when single non-interacting charged particle is put in a box, but that is physically wrong.

Frequent use of this calculation as an example of inadequacy of non-quantum physics for magnetism is thus misconceived right from the start. In contrast to other uses of an imaginary box in calculations of statistical physics, for magnetic effects of external field the box's effect on the system cannot be ignored. Magnetic moment of uniformly magnetized body can be entirely cancelled out with the right surface current. In this case, reflections from the walls of the box provide such cancelling current.

The quantum calculation is very different. First, eigenvalues of the Hamiltonian are found and then box is introduced as well. This time, however, it is used only to limit the position of the center of the wave function mentally, not all of its support throughout the space. There is no physical interaction of the walls with the system involved that would make it possible for the velocity probability distribution to be isotropic everywhere. Also the partition function is calculated very differently - as a sum over quantum numbers, instead of an integral over phase space. It is thus not that much surprising that the calculation leads to non-trivial magnetic properties - in the quantum calculation, the box is "not really there".

  • $\begingroup$ Thanks for the reply. The point about the walls is interesting and I have encounter it sometimes (the first time in Peierls' book on "Surprises in Theoretical Physics"). The thing is that this heuristic argument is sufficient but not necessary to get the Bohr-von Leeuwen result. In fact, as I have explained in my post even before integrating over the positions one can integrate over the momenta by making a change of variable $p \rightarrow v = p-qA$ without affecting the bounds of integration. This is enough to yield the theorem. $\endgroup$
    – gatsu
    Jun 14 '15 at 9:40
  • $\begingroup$ Moreover, as I also pointed out the corrections to the zero classical result are of the order $\Lambda/R_c \sim \hbar \omega_c/k_BT$ and do not involve at all the volume of the box. I would therefore suspect that the boundaries play in fact a second order role in the emergence of diamagnetism in quantum (statistical) mechanics (and its disappearance in classical mechanics). $\endgroup$
    – gatsu
    Jun 14 '15 at 9:57
  • $\begingroup$ @gatsu, I've modified my answer. Indeed the canonical distribution already leads to zero magnetism. The canonical distribution is a reasonable tool when the system is in a box. Here, however, the canonical distribution fails and it is natural to see one reason for this in the fact that diamagnetism does not happen to particles in a box, but to particles that move freely in magnetic field. The box and the canonical distribution should thus be avoided. The quantum calculation does this - the eigenvalues are usually found for particle in infinite space, not for particle in a box. $\endgroup$ Jun 14 '15 at 15:09
  • $\begingroup$ @gatsu, it does not care for calculation of eigenvalues, but this calculation is correct only if there is no box (the system is infinite). To calculate degeneracy, the box is essential, but it is used only to count, the physical interaction of the system with the box is not taken into account. $\endgroup$ Jun 14 '15 at 21:47
  • $\begingroup$ This thread of comments is quite long already. I think you do not understand me, but that's OK, if you wish to pose some other questions, please post them in new thread and I can try to answer. $\endgroup$ Jun 15 '15 at 18:42

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