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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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