# User:Jorge~enwikiversity/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, ${\textbf {H}}=H{\textbf {k}}$ , we can choose for our vector potential Landau's gauge:

    ${\textbf {A}}=-Hy{\textbf {i}}$ This gauge will simplify the resolution of Schrödinger's equation for this Hamiltonian.

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

    $H={\frac {\hbar ^{2}}{2m}}\left(-\Delta +K^{4}y^{2}+2K^{2}y\cdot i\partial _{x}\right)$ where we have called $K^{2}={\frac {eH}{\hbar c}}$ . 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:

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

    $X(x)=e^{ik\cdot x}$ $X'(x)=ik\cdot X(x)$ $X''(x)=-k^{2}\cdot X(x)$ We then substitute our solution for X in Schrödinger's equation:

    ${\frac {\hbar ^{2}}{2m}}\cdot \left(k^{2}-{\frac {Y''}{Y}}+K^{4}y^{2}-2K^{2}ky\right)\Psi =E\Psi$ Now the parenthesis is not an operator, but a scalar, so we can drop the $\Psi$ . Each base function Y must satisfy:

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