Landau Diamagnetism

In this appendix we derive the expression for Landau diamagnetism of free electrons. Our approach will be to derive the result first for 2-$d$ from which we will easily be able to obtain the 3-$d$ result by integrating over $k_z$.

Firstly, we remind ourselves of some statistical mechanics. It is convenient to calculate the grand partition function, ${\cal Z}$ where

$${\cal Z} = {\rm Tr} \exp[\beta(\mu N - {\cal H})] ,$$ (49)

where $N$ is the number of particles and ${\cal H}$ is the Hamiltonian. The trace is over all states for a given $N$ and over all $N$. This ensemble, where both $N$ and $E$ are allowed to vary, is called the ``grand canonical ensemble''. ${\cal Z}$ is related to the ``grand potential'', $\Omega$, by

$$\Omega = -k_B T \log {\cal Z} .$$ (50)

The mean number of particles is given by

$$N = -{\partial \Omega \over \partial \mu} ,$$ (51)

which implicitly determines $\mu$. $\Omega$ is related to the free energy $F \equiv U - T S$ by

$$F = \Omega + \mu N .$$ (52)

Note that $\Omega$ is considered as a function of $\mu$ and so differentiating the right hand side of Eq. (A4) with respect to $\mu$ (keeping $N$ constant) gives

$${\partial F \over \partial \mu} = 0.$$ (53)

This will be useful later. Note that a lot of confusion in statistical mechanics comes from a lack of understanding of what is being kept constant in partial derivatives. Once the correct value of $\mu$ has been determined for a given $N$, we determine the magnetization from

$$M = - {\partial F \over \partial H} ,$$ (54)

where $H$ is the magnetic field and the derivative is at fixed $N$. Note that in general we need to determine $M$ from $- (\partial F / \partial H)_N$ rather than $-(\partial \Omega / \partial H)_\mu$, because the number of particles is kept constant when the field is applied, not the chemical potential. However, as we shall see, they both give the same result to lowest order in $H$, the case of interest here, because of Eq. (A5). The susceptibility is then found from

$$\chi = {\partial M \over \partial H} = -{\partial^2 F \over \partial H^2} ,$$ (55)

so we have to compute the $H^2$ term in the free energy.

A useful feature of the grand canonical ensemble is that, for non-interacting particles, the grand potential factorizes into a product of grand potentials for each single particle state, and so the grand potential is a sum of grand potentials for single particle states. Hence, for free electrons,

$$\Omega = -k_B T \sum_k 2 \ln[ 1 + e^{\beta(\mu - \epsilon_k)} ] ,$$ (56)

which can be conveniently be expressed in terms of the density of states, $g(\epsilon)$ (for both spin species) by

$$\Omega = -k_B T \int_0^\infty g(\epsilon) \ln[ 1 + e^{\beta(\mu - \epsilon_k)} ] \, d\epsilon .$$ (57)

Consider first $H=0$. We are interested in the low-$T$ limit where, as discussed in class, the difference between $\mu$ and its $T=0$ limit, $\epsilon_F$, is negligible. As discussed above $g(\epsilon)$ is a constant $A \, m/ (\pi \hbar^2)$. and so, for small $T$ and $H=0$

$$\Omega(H=0) = -k_B T\, A{ m \over \pi \hbar^2} \int_0^\infty \ln[ 1 + e^{\beta(\mu - \epsilon)} ] \, d\epsilon$$ (58)

$$\simeq -k_B T\, A{ m \over \pi \hbar^2} \int_0^{\epsilon_F} \beta (\epsilon_f - \epsilon) \, d\epsilon$$ (59)

$$= -A{ m \over \pi \hbar^2} {\epsilon_F^2 \over 2} = -{N \epsilon_F \over 2},$$ (60)

where the last equality is from Eq. (14). For $T\to 0$, where $T S$ is negligible, $\Omega$ is just $U + \mu N$ with $\mu = \epsilon_F$, and so

$$U = {N \epsilon_F \over 2}$$ (61)

in agreement with Eq. (15), which gives the energy per electron rather than the total energy as here.

Now we add a field, which changes $\Omega$ in two ways. Firstly it changes the density of states, as we have discussed in detail in the main part of the text. Secondly it changes the chemical potential, so $\mu(H) = \mu(0) + {\rm const.} H^2 +O(H^4)$. However, we shall see that the change in $\Omega$ due to the change in $\mu$ with $H$ does not affect the free energy $F$, so we just focus here on the change in $\Omega$ due to the modification of the density of states.

In the presence of a field $H$, the energy levels take the discrete values $\epsilon_n \equiv (n + \mbox{\small$1 \over 2$}) \hbar \omega_c$ and so

$$\Omega(H) = -k_B T {A m \over \pi \hbar^2} \hbar \omega_c \sum_n \ln[ 1 + e^{\beta(\mu - \epsilon_n)} ] ,$$ (62)

which is to be contrasted with Eq. (A10) for $H=0$. The difference is that Eq. (A14) can be though of as a discretized approximation to Eq. (A10) of the sort that is often used in numerical analysis. The integral over a range of width $\hbar \omega_c$ is replaced by $\hbar \omega_c$ times the value of the the integrand at the midpoint of the interval. The difference between the integral and the approximation to it using this ``midpoint rule'' is well known (see e.g. Numerical Recipes in C (or in Fortran) by Press et al, Eq. (4.4.1)), and can be easily derived by replacing the function in each interval by a polynomial, doing the integral with the first few terms of the polynomial, and summing over intervals. The result can be expressed as

$$\int_{x_0}^{x_N} f(x) \, dx = h [ f_{1/2} + f_{3/2} + \cdots... ...N-1/2} ] + h^2 {(f^\prime_N - f^\prime_0) \over 24} + O(h^4) ,$$ (63)

where $x_n = x_0 + n h$ and $f_n = f(x_n)$. This approximation is good provided the function $f(x)$ varies smoothly over a single interval of width $h$. For the present problem, $h$ is replaced by $\hbar \omega_c$, and the integrand varies rapidly on a scale of $k_B T$. Hence use of Eq. (A15) will be valid for $k_B T \gg \hbar \omega_c$. In this situation, many Landau levels are partially occupied and the oscillations found in the main part of the text are washed out. In our case the integral goes to infinity but both the function and its derivative vanish in this limit, and so, evaluating the derivative of the integrand at $\epsilon = 0$, we get, to order $H^2$,

$$\Omega(H) = \Omega(0) + {e^2 \over 24 \pi m c^2} H^2 + {\partial \Omega \over \partial \mu} \delta \mu(H),$$ (64)

where $\Omega(0)$ corresponds to the integral in Eq. (A15) and $\Omega(H)$ to the sum, and, from now on, we work per unit area. The last term in Eq. (A16) is the change in $\Omega$ from $\delta \mu(H)$, the change in the chemical potential due to the field, which is also of order $H^2$.

Substituting Eq. (A16) into in Eq. (A4) gives

$$F(H) = F(0) + {e^2 \over 24 \pi m c^2} H^2 ,$$ (65)

where, from Eq. (A3), we have noted that the contribution from $\delta \mu(H)$ cancels. This cancellation occurs because $\partial F / \partial \mu = 0$, as noted in Eq. (A5).

The magnetization is given by $M = - \partial F / \partial H$ which yields $M = \chi_{dia}^{(2d)} H$ with $\chi_{dia}^{(2d)}$, the diamagnetic susceptibility of free electrons per unit area in two dimensions, given by

$$\chi_{dia}^{(2d)} = -{e^2 \over 12 \pi m c^2} .$$ (66)

To get the corresponding result in 3-$d$ we need to add the motion in the $z$-direction, specified by $k_z$, and integrate $k_z$ from $-k_F$ to $k_F$. This is easy because Eq. (A18) is a constant independent of the (2-$d$) density of electrons (which would vary with $k_z$) and hence the diamagnetic susceptibility of free electrons in 3-$d$ per unit volume is given by

$$\chi_{dia}^{(3d)} = -{e^2 \over 12 \pi m c^2} \int_{-k_F}^{k_F} {dk_z \over 2\pi} = -{e^2 k_F \over 12 \pi^2 m c^2} .$$ (67)

This is the expression first found by Landau. It is equivalent to Eqs. (31.72) and (31.69) of Ashcroft and Mermin.