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---
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.
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