User:Jorge~enwikiversity/Quantum Hall effect

This page is about the quantum Hall effect. It builds upon the knowledge of:
 * what the Hamiltonian of a classical charged particle subject to a magnetic field is,
 * what Schroedinger's Equation is,
 * the technique of Separation of Variables to solve Differential Equations,
 * the solution of the Quantum Harmonic Oscillator
 * and what the Fourier Transform is.

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, $$\textbf{H} = H \textbf{k}$$, 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 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, 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):