# Nonstandard physics/Quantum Hall effect

It turns out that when a free electron is placed in a magnetic field, its energy levels go from being continuous (by the virtue of being free particles) to being a discrete set. What follows is the calculation of this phenomenom, which is called the Quantum Hall Effect.

## Deduction of the quantization of electrons' energy levels

For a free electrons gas in a magnetic field, the Hamiltonian is:

    $H = \frac{1}{2m} \left|-i \hbar \nabla + \frac{e}{c} \textbf{A}\right|^2$


(Note we will be using Gaussian units.) When the magnetic field is constant, $\textbf{H} = H \textbf{k}$[1], we can choose for our vector potential Landau's gauge:

    $\textbf{A} = - H y \textbf{i}$


This gauge will simplify the resolution of Schrödinger's equation for this Hamiltonian.

Developping the square in the Hamiltonian (do it!) we arrive at:

    $H = \frac{\hbar^2}{2m} \left( -\Delta + K^4 y^2 + 2 K^2 y \cdot i \partial _x \right)$


where we have called $K^2 = \frac{e H}{\hbar c}$. We then, as usual, try to find a base of eigenstates[2] which make Schrödinger's equation separable:

    $\Psi(x,y) = X(x) \cdot Y(y)$

    $H[\Psi(x,y)] = E \cdot \Psi(x,y)$


We have:

    $\partial _x[\Psi] = \frac{X'}{X} \cdot \Psi$

    $\partial _x^2[\Psi] = \frac{X''}{X} \cdot \Psi$


So:

    $H[\Psi] = \frac{\hbar^2}{2m} \cdot \left(- \frac{X''}{X} - \frac{Y''}{Y} + K^4 y^2 + 2 K^2 y \cdot i \frac{X'}{X} \right) \Psi$


The last term inside the parenthesis imposes that the equation will only be separable if X´/X is constant. The only such base that doesn't diverge at infinite is Fourier's base:

    $X(x) = e^{i k \cdot x}$

    $X'(x) = i k \cdot X(x)$

    $X''(x) = -k^2 \cdot X(x)$


We then substitute our solution for X in Schrödinger's equation:

    $\frac{\hbar^2}{2m} \cdot \left(k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y \right) \Psi = E \Psi$


Now the parenthesis is not an operator, but a scalar[3], so we can drop the $\Psi$. Each base function Y must satisfy:

    $\frac{\hbar^2}{2m} \cdot \left(k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y \right) = E$

    $k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y = \frac{2mE}{\hbar^2}$

    $\frac{Y''}{Y} = K^4 y^2 - 2 K^2 k y + \left(k^2 - \frac{2mE}{\hbar^2}\right)$


Which has a smell of Quantum Harmonic Oscillator (id est, Hermite's equation). We can easily transform it algebraically by converting the right hand side to a perfect square (through a change of variable):