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 \(B\).

References

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

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