Categories: Physics, Quantum mechanics.

Landau quantization

When a particle with charge $$q$$ is moving in a homogeneous magnetic field, quantum mechanics decrees that its allowed energies split into degenerate discrete Landau levels, a phenomenon known as Landau quantization.

Starting from the Hamiltonian $$\hat{H}$$ for a particle with mass $$m$$ in a vector potential $$\vec{A}(\hat{Q})$$:

\begin{aligned} \hat{H} &= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2 \end{aligned}

We choose $$\vec{A} = (- \hat{y} B, 0, 0)$$, yielding a magnetic field $$\vec{B} = \nabla \times \vec{A}$$ pointing in the $$z$$-direction with strength $$B$$. The Hamiltonian becomes:

\begin{aligned} \hat{H} &= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m} + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m} \end{aligned}

The only position operator occurring in $$\hat{H}$$ is $$\hat{y}$$, so $$[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$$. Because $$\hat{p}_z$$ appears in an unmodified kinetic energy term, and the corresponding $$\hat{z}$$ does not occur at all, the particle has completely free motion in the $$z$$-direction. Likewise, because $$\hat{x}$$ does not occur in $$\hat{H}$$, we can replace $$\hat{p}_x$$ by its eigenvalue $$\hbar k_x$$, although the motion is not free, due to $$q B \hat{y}$$.

Based on the absence of $$\hat{x}$$ and $$\hat{z}$$, we make the following ansatz for the wavefunction $$\Psi$$: a plane wave in the $$x$$ and $$z$$ directions, multiplied by an unknown $$\phi(y)$$:

\begin{aligned} \Psi(x, y, z) = \phi(y) \exp(i k_x x + i k_z z) \end{aligned}

Inserting this into the time-independent Schrödinger equation gives, after dividing out the plane wave exponential $$\exp(i k_x x + i k_z z)$$:

\begin{aligned} E \phi &= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi \end{aligned}

By defining the cyclotron frequency $$\omega_c \equiv q B / m$$ and rearranging, we can turn this into a 1D quantum harmonic oscillator in $$y$$, with a couple of extra terms:

\begin{aligned} \Big( E - \frac{\hbar^2 k_z^2}{2 m} \Big) \phi &= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi \end{aligned}

The potential minimum is shifted by $$y_0 = \hbar k_x / (m \omega_c)$$, and a plane wave in $$z$$ contributes to the energy $$E$$. In any case, the energy levels of this type of system are well-known:

\begin{aligned} \boxed{ E_n = \hbar \omega_c \Big(n + \frac{1}{2}\Big) + \frac{\hbar^2 k_z^2}{2 m} } \end{aligned}

And $$\Psi_n$$ is then as follows, where $$\phi$$ is the known quantum harmonic oscillator solution:

\begin{aligned} \Psi_n(x, y, z) = \phi_n(y - y_0) \exp(i k_x x + i k_z z) \end{aligned}

Note that this wave function contains $$k_x$$ (also inside $$y_0$$), but $$k_x$$ is absent from the energy $$E_n$$. This implies degeneracy: assuming periodic boundary conditions $$\Psi(x\!+\!L_x) = \Psi(x)$$, then $$k_x$$ can take values of the form $$2 \pi n / L_x$$, for $$n \in \mathbb{Z}$$.

However, $$k_x$$ also occurs in the definition of $$y_0$$, so the degeneracy is finite, since $$y_0$$ must still lie inside the system, or, more formally, $$y_0 \in [0, L_y]$$:

\begin{aligned} 0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y \end{aligned}

Isolating this for $$n$$, we find the following upper bound of the degeneracy:

\begin{aligned} \boxed{ n \le \frac{q B L_x L_y}{2 \pi \hbar} = \frac{q B A}{h} } \end{aligned}

Where $$A \equiv L_x L_y$$ is the area of the confinement in the $$(x,y)$$-plane. Evidently, the degeneracy of each level increases with larger $$B$$, but since $$\omega_c = q B / m$$, the energy gap between each level increases too. In other words: the density of states is a constant with respect to the energy, but the states get distributed across the $$E_n$$ differently depending on $$B$$.

1. L.E. Ballentine, Quantum mechanics: a modern development, 2nd edition, World Scientific.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.