Editing Quantum Hall effect (section)

=== Landau levels ===
{{Main|Landau levels}}
In two dimensions, when classical electrons are subjected to a magnetic field they follow circular cyclotron orbits. When the system is treated quantum mechanically, these orbits are quantized. To determine the values of the energy levels the Schrödinger equation must be solved.

Since the system is subjected to a magnetic field, it has to be introduced as an electromagnetic vector potential in the [[Schrödinger equation]]. The system considered is an electron gas that is free to move in the x and y directions, but is tightly confined in the z direction. Then, a magnetic field is applied in the z direction and according to the [[Landau gauge]] the electromagnetic vector potential is <math>\mathbf{A} = (0, Bx, 0)</math> and the [[scalar potential]] is <math>\phi=0</math>. Thus the Schrödinger equation for a particle of charge  <math>q</math> and effective mass <math>m^*</math> in this system is:
: <math>\left \{ \frac{1}{2m^*} \left [ \mathbf{p}-q\mathbf{A}\right ]^2 + V(z) \right \}\psi(x,y,z)=\varepsilon\psi(x,y,z)</math>
where <math>\mathbf{p}</math>  is the canonical momentum, which is replaced by the operator <math>-i\hbar\nabla</math> and <math>\varepsilon</math> is the total energy.

To solve this equation it is possible to separate it into two equations since the magnetic field just affects the movement along x and y axes. The total energy becomes then, the sum of two contributions <math>\varepsilon = \varepsilon_z + \varepsilon_{xy}</math>. The corresponding equations in z axis is:
: <math>\left [- \frac{\hbar^2}{2m^*} {\partial^2\over\partial z^2} + V(z) \right ]u(z)=\varepsilon_zu(z)</math>
To simplify things, the solution <math>V(z)</math> is considered as an infinite well. Thus the solutions for the z direction are the energies <math display="inline">\varepsilon_z=\frac{n_z^2\pi^2\hbar^2}{2m^*L^2}</math>, <math>n_z=1,2,3...</math>  and the wavefunctions are sinusoidal. For the <math>x</math> and <math>y</math> directions, the solution of the Schrödinger equation can be chosen to be the product of a plane wave in <math>y</math>-direction with some unknown function of <math>x</math>, i.e., <math>\psi_{xy}=u(x)e^{ik_yy}</math>. This is because the vector potential does not depend on <math>y</math> and the momentum operator <math>\hat p_y</math> therefore commutes with the Hamiltonian. By substituting this Ansatz into the Schrödinger equation one gets the one-dimensional [[harmonic oscillator]] equation centered at <math display="inline">x_{k_y}=\frac{\hbar k_y}{eB}</math>.

: <math>\left [- \frac{\hbar^2}{2m^*} {\partial^2\over\partial x^2}+\frac{1}{2}m^*\omega_{\rm c}^2(x-l_B^2k_y)^2\right ]u(x)=\varepsilon_{xy}u(x)</math>
where <math display="inline">\omega_{\rm c}=\frac{eB}{m^*}</math> is defined as the cyclotron frequency and <math display="inline">l_B^2=\frac{\hbar}{eB}</math> the magnetic length. The energies are:
: <math>\varepsilon_{xy}\equiv\varepsilon_{n_x}=\hbar \omega_{\rm c} \left ( n_x+\frac{1}{2} \right )</math>,             <math>n_x=1,2,3...</math>

And the wavefunctions for the motion in the <math>xy</math> plane are given by the product of a plane wave in <math>y</math> and [[Hermite polynomials]] attenuated by the gaussian function in <math>x</math>, which are the wavefunctions of a harmonic oscillator.

From the expression for the Landau levels one notices that the energy depends only on <math>n_x</math>, not on <math>k_y</math>. States with the same <math>n_x</math> but different <math>k_y</math> are degenerate.