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 SchrÃ¶dinger 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 $B$.

## References

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