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Next: Two dimensions, Bloch electrons Up: magnetic_field Previous: Introduction

Two dimensions, free electrons

It is easier to consider first the case of two dimensions, where the electrons are confined to the $x$-$y$ plane in a perpendicular field. Hence there is no motion in the $z$ direction so the term in Eq. (2) involving $k_z^2$ does not occur. You showed in Qu. 3 of Homework 3 that the density of states in two dimensions is a constant,
\begin{displaymath}
g(\epsilon) = A {m \over \pi \hbar^2} ,
\end{displaymath} (6)

in the absence of a field. Here we give the total density of states, rather than the density of states per unit area, so a factor of $A$ appears. In a field, the allowed energy values are discrete with a constant spacing between levels of $\hbar \omega_c$. Now the number of zero field states in an interval $\hbar \omega_c$ is given by
$\displaystyle g(\epsilon) \hbar \omega_c$ $\textstyle =$ $\displaystyle A {m \over \pi \hbar^2} \hbar {e H \over mc}$ (7)
  $\textstyle =$ $\displaystyle {2 e \over h c } A H$ (8)
  $\textstyle =$ $\displaystyle {\cal N} ,$ (9)

the degeneracy of a Landau level. In other words, the field bunches the states into discrete levels, but the total number of states in a region much larger than the Landau level spacing is unchanged by the field. The density of states, both with and without a magnetic field is shown in Fig. 1.

Figure: The dashed line shows the density of states of the two dimensional free electron gas in the absence of a magnetic field. It has the constant value $g(\epsilon) = A m / \pi\hbar^2$. In the presence of a magnetic field the energy levels are bunched into discrete values $(n_L + 1/2)\Delta \epsilon $ where $n_L = 0, 1, 2, \cdots$, and $\Delta \epsilon = \hbar \omega_c$, where $\omega _c = eH/mc$ is the cyclotron frequency. Hence the density of states is a set of delta functions, shown by the vertical lines. The weight of the delta function is equal to the zero field density of states times $\Delta \epsilon $, so the energy levels are just shifted locally, the total number of states in a region comprising a multiple of $\Delta \epsilon $ being unchanged.
\begin{figure}\begin{center}
\leavevmode
\epsfysize =6cm
\epsfbox{fig1.eps}\end{center}\end{figure}

At zero temperature, as we increase the magnetic field the number of occupied Landau levels will change, since the number of electrons is fixed but the degeneracy of each Landau level changes with $H$. This will lead to an oscillatory behavior of the energy as a function of the magnetic field, that we discuss below. At finite temperature these oscillations will be washed out if $k_B T \gg
\hbar \omega_c$, because then many Landau levels will be partially filled. In order to see the oscillations, we therefore need

\begin{displaymath}
\hbar \omega_c = {e \hbar \over m c } H \gg k_B T ,
\end{displaymath} (10)

which requires large fields and very low temperatures, of order a few K. At higher temperatures, there is a smooth change in the energy with $H$ which leads to a small diamagnetic response, see Appendix A, Ashcroft and Mermin (AM) p. 664, and Peierls, Quantum Theory of Solids, pp. 144-149.

It is also important to realize that if there are impurities, then the electron states will have a finite lifetime, $\tau$, which will broaden the levels by an amount $\hbar / \tau$. This will also wash out the oscillations if the level broadening is greater than the Landau level splitting. Hence, to observe the oscillations we also need a second condition,

\begin{displaymath}
\omega_c \tau =
{e H \over m c } \tau \gg 1 ,
\end{displaymath} (11)

which requires very clean samples.

Let us now determine the change in energy of our two-dimensional model at $T=0$ as a function of field. Let $n = N/A$ be the number of electrons per unit area. Hence, for $H=0$ the Fermi energy is determined from

\begin{displaymath}
N = \int_0^{\epsilon_F} g(\epsilon) \, d\epsilon,
\end{displaymath} (12)

or
\begin{displaymath}
N = g(\epsilon) \epsilon_F = A {m \over \pi \hbar^2} \epsilon_F ,
\end{displaymath} (13)

which gives
\begin{displaymath}
\epsilon_F = n {\pi \hbar^2 \over m} .
\end{displaymath} (14)

Hence the ground state energy per electron in zero field is
\begin{displaymath}
\epsilon(H=0) \equiv {E(0)\over N} = \int_0^{\epsilon_F} \ep...
...\epsilon, = {\epsilon_F \over 2} = n {\pi \hbar^2 \over 2 m} ,
\end{displaymath} (15)

using Eq. (14).

In the presence of the field the levels, $n_L = 0, 1, 2, \cdots, p -1$ will be fully occupied with ${\cal N}$ electrons and level $ p$ will be partially occupied with $\lambda {\cal N}$ electrons, where $0 \le \lambda <
1$.

Counting up electrons one has

\begin{displaymath}
N = {\cal N} \nu
\end{displaymath} (16)

where
\begin{displaymath}
\nu = p + \lambda
\end{displaymath} (17)

is called the filling factor of the Landau levels. Note that $\lambda$ and $\nu $ take a continuous range of values, whereas $ p$ is an integer. From Eq. (4) we have
\begin{displaymath}
\nu = n { \phi_0 \over H} = {H_1 \over H} \left( = {\epsilon...
...c} = n {\pi \hbar^2 \over m} {1\over \hbar \omega_c} \right) ,
\end{displaymath} (18)

where the last equality is from Eq. (14), in which
\begin{displaymath}
H_1 = n \phi_0
\end{displaymath} (19)

is the field required to put all the electrons in the lowest Landau level. Consequently
\begin{displaymath}
p = \left[ { H_1 \over H} \right] , \qquad \lambda = { H_1 \over H} -
\left[ {H_1 \over H} \right]
\end{displaymath} (20)

where $[x]$ means the largest integer less than or equal to $x$. Note that $\lambda$ is the fractional filling of the last Landau level.

The energy per electron is then just obtained by summing the energies of each Landau level, i.e.

$\displaystyle \epsilon(H)$ $\textstyle =$ $\displaystyle \left\{ (0+\mbox{\small$1 \over 2$}) + (1+\mbox{\small$1 \over 2$...
...+ \mbox{\small$1 \over 2$}) \lambda \right\}
{{\cal N} \over N } \hbar \omega_c$ (21)
  $\textstyle =$ $\displaystyle \left\{ { p^2 \over 2} + (p + \mbox{\small$1 \over 2$})\lambda \right\}
{\hbar \omega_c \over \nu }$ (22)
  $\textstyle =$ $\displaystyle \left\{ { (p + \lambda)^2 \over 2} +
{\lambda - \lambda^2 \over 2} \right\}
{\hbar \omega_c \over \nu } .$ (23)

Eliminating $p + \lambda \equiv \nu$ in favor of $n$ using Eq. (18) one finds
\begin{displaymath}
\epsilon(H)
= {\pi \hbar^2 \over 2 m} n
+ {\hbar \omega_c \over 2\nu}
\lambda (1 - \lambda)
\end{displaymath} (24)

which, from Eqs. (3), (15) and (18)-(20), can be conveniently written as
\begin{displaymath}
{\epsilon(H) \over \epsilon(0)} =
1 + \left( {H \over H_1} \right)^2 \lambda (1 - \lambda) .
\end{displaymath} (25)

Eq. (25) is our main result. It gives the field dependence of the ground state energy (per electron) of free electrons in two-dimensions.

Figure: The energy of the two dimensional electron gas at $T=0$ according to Eq. (25), as a function of $H/H_1$ where $H_1 = n\phi _0$ is the field at which all the electrons are in a completely filled lowest Landau level. Note that the overall size of the energy change varies as $H^2$ in addition to the oscillations. The oscillations get closer together for small $H$. In fact, apart from the smooth $H^2$ variation, the data is a periodic function of $1/H$, see Figs. 3 and 4.
\begin{figure}\begin{center}
\leavevmode
\epsfysize =6cm
\epsfbox{fig2.eps}\end{center}\end{figure}

A sketch of the of energy as a function of $H$, according to Eq. (25), is shown in Fig. 2. One clearly sees oscillations whose amplitude increases smoothly which $H$, actually as $H^2$. The period of the oscillations gets smaller with decreasing $H$, since the energy is actually a periodic function of $1/H$ not $H$ (apart from the $H^2$ variation in amplitude). This is seen in Fig. 3 below. Although the energy is a continuous function of $H$ the derivative is discontinuous when the field is such that the Landau levels are completely filled (remember that $\lambda$ is the fractional filling of the last Landau level). Physically, this is because if one decreases $H$ below $H_{p} = H_1 / p$, say, then it is Landau level $ p$ which starts to be occupied, whereas if one increases $H$ above $H_{p}$ then it is Landau level $p-1$ which starts to be depleted.

Fig. 2 is to be contrasted with the situation at a temperature where $k_B T \gg
\hbar \omega_c$, where thermal fluctuations average over the oscillations. Then one has only the smooth increase in the energy proportional to $H^2$. Since the magnetization is given by $M = -
\partial E /\partial H$, where $E$ is the energy per unit volume, one then finds a small diamagnetic (i.e. negative) susceptibility, $\chi_{\rm dia}$, where $M = \chi_{\rm dia} H$, see Appendix A, AM p. 664, and Peierls, Quantum Theory of Solids, pp. 144-149. This was first calculated by Landau in 1930.

Figure 3: The energy of the two dimensional electron gas at $T=0$ as a function of $ \nu = H_1/H$ where $\nu $ is the filling factor (the number of filled Landau levels) and $H_1 = n\phi _0$ is the field at which all the electrons are in a completely filled lowest Landau level. For metals, $H_1$ is much greater than a typical laboratory field so in experimental situations the filling factor is large. Note that the behavior is periodic apart from a gentle overall decrease of the amplitude which is negligible for $\nu \gg 1$.
\begin{figure}\begin{center}
\leavevmode
\epsfysize =6cm
\epsfbox{fig3.eps}\end{center}\end{figure}

Note from Eq. (20) that $\lambda$ is a periodic function of $1/H$ and hence, apart from the slowly varying factor of $H^2$, the energy, given by Eq. (25), is an oscillatory function of $1/H$. This is clearly seen in Fig. 3.

The oscillation in the ground state energy give rise to oscillations in the magnetization per electron $M$, given by

\begin{displaymath}
M = - {\partial \epsilon \over \partial H} ,
\end{displaymath} (26)

which, from Eqs. (5), (15), (17), (18), (20) and (25) is given by
\begin{displaymath}
{M \over \mu_B} =
1- 2\lambda - {2 \lambda (1 - \lambda) \over p + \lambda},
\end{displaymath} (27)

where
\begin{displaymath}
\mu_B = {e \hbar \over 2 m c}
\end{displaymath} (28)

is the Bohr magneton. For $p \gg 1$, which is generally the case of interest, this is given, to a good approximation, by
\begin{displaymath}
{M \over \mu_B} = 1- 2\lambda .
\end{displaymath} (29)

Since $\lambda$ is periodic in $1/H$, see Eq. (20), it follows that $M$ is also periodic in $1/H$, see Fig. 4.

Figure 4: The magnetization of the two dimensional electron gas at $T=0$ (in units of $\mu _B$) as a function of $ \nu = H_1/H$ where $\nu $ is the filling factor and $H_1 = n\phi _0$, obtained from Eq. (27). Note that the magnetization is a periodic function of $1/H$.
\begin{figure}\begin{center}
\leavevmode
\epsfysize =8cm
\epsfbox{fig4.eps}\end{center}\end{figure}

The main conclusion so far is that $M$ is a periodic function of $1/H$ with period given by

\begin{displaymath}
\Delta \left({1\over H}\right) = {1\over H_1} = {1 \over n \phi_0} .
\end{displaymath} (30)

From Eq. (14) it follows that the density is related to the Fermi wave vector $k_F$ by

\begin{displaymath}
n = {k_F^2 \over 2 \pi} ,
\end{displaymath} (31)

and so, using Eq. (5),
\begin{displaymath}
\Delta \left({1\over H}\right) = {2 \pi e \over \hbar c} {1\over A_F} ,
\end{displaymath} (32)

where
\begin{displaymath}
A_f \equiv \pi k_F^2
\end{displaymath} (33)

is the area of the Fermi surface. In Sec. III we shall show that Eq. (32), which relates the periodicity to just fundamental constants and the area of the Fermi surface, is true quite generally, not just for free electrons. In Sec. IV we will also discuss the implications of this result for de Haas-van Alphen experiments which use oscillations in the magnetization to map out Fermi surfaces of metals, see e.g. Ashcroft and Mermin, Ch. 12.

Figure 5: The solid circles are the semi classical orbits of free electrons in two dimensions in a magnetic field. Each circle corresponds to a particular Landau level. The electrons follow orbits around these circles with a period $T = 2\pi / \omega _c$, where $\omega _c$ is the cyclotron frequency. As the strength of the field is increased the radii increase and the orbits ``pop out'' of the Fermi circle, denoted by a dashed circle. This leads to oscillatory behavior in the energy and magnetization provided $k_B T$ is much less than the spacing of the Landau levels, $\hbar \omega_c$.
\begin{figure}\begin{center}
\leavevmode
\epsfysize =6cm
\epsfbox{fig5.eps}\end{center}\end{figure}

It is interesting to see how Eq. (32) emerges from the semi-classical picture. The semi classical orbits for free electrons are circles, on which the energy stays constant. Denoting by $k_p$ the magnitude of the wave vector corresponding to Landau level $ p$, then equating $\hbar^2 k_p^2 / 2 m$ to $(p + \mbox{\small$1 \over 2$}) \hbar \omega_c$ one gets

\begin{displaymath}
k_p^2 = (p + \mbox{\small$1 \over 2$}) {2 e H \over \hbar c}
\end{displaymath} (34)

and so the area in $k$-space swept out by an electron in Landau level $ p$ is
\begin{displaymath}
A_p = \pi k_p^2 = (p + \mbox{\small$1 \over 2$}) {2 \pi e H \over \hbar c} ,
\end{displaymath} (35)

which can also be written as
\begin{displaymath}
{1 \over H} = \left(p + \mbox{\small$1 \over 2$}\right) {2 \pi e \over \hbar c} {1 \over
A_p}.
\end{displaymath} (36)

Hence, in the semi-classical picture of free electrons is a magnetic field, the allowed states in $k$-space form circles whose radius increases with $H$, see Fig. 5. As the states go through the Fermi wave vector and become de-populated this gives rise to oscillations in the energy. From Eq. (36) the change in $1/H$ for one Landau level to go through the Fermi surface (i.e. let $p \to p - 1$ with $A_L = A_F$, the area of the Fermi surface) is given by

\begin{displaymath}
\Delta \left({1\over H}\right) = {2 \pi e \over \hbar c} {1\over A_F} ,
\end{displaymath} (37)

in agreement with Eq. (32).


next up previous
Next: Two dimensions, Bloch electrons Up: magnetic_field Previous: Introduction
Peter Young 2002-10-31