---
title: "Landau quantization"
sort_title: "Landau quantization"
date: 2021-07-01
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

When a particle with charge $$q$$ is moving in a homogeneous
[magnetic field](/know/concept/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](/know/concept/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.