Quantum Mechanics/The Hydrogen Atom

The Hydrogen Atom

Introduction

By now you're probably familiar with the Bohr model of the atom, which was a great help in classifying the position of fundamental atomic specta lines. However, Bohr lucked out in more ways than one. The hydrogen atom turns out to be one of the few systems in Quantum Mechanics that we are able to solve almost precisely. This has made it tremendously useful as a model for other Quantum Mechanical systems, and as a model for the behavior of atoms themselves.

Fundamentals

We can assume that the hydrogen atom is governed by the Coulomb potential, namely:

$V(r)=-{\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{r}}$

such that, $H\Psi =-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\Psi }{dx^{2}}}-{\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{r}}$

Obviously, simply by inspection, we can see that the Hydrogen Atom is a spherical system. Hence it makes more sense to deal with the Hydrogen atom in spherical coordinates. One should remember at this point that, via Separation of variables, you can obtain the solution to the spherical Laplacian in three-dimensional space:

$\nabla ^{2}f={1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial f \over \partial r}\right)+{1 \over r^{2}\sin \theta }{\partial \over \partial \theta }\left(\sin \theta {\partial f \over \partial \theta }\right)+{1 \over r^{2}\sin ^{2}\theta }{\partial ^{2}f \over \partial \phi ^{2}}$

the solutions to this function when we use separation of variables inside a Hamiltonian gives us two different functions, the Radial Wave functions (not useful now, but good to know):

$A\cdot J_{l}\left(Z_{nl}{\frac {r}{a}}\right)$

where $J_{l}$ are the spherical Bessel functions of type l, and $Z_{nl}$ are the zeroes of said Bessel functions.

The other component, the angular component are the Spherical harmonics which are explored in detail on Wikipedia.

Calculations

Essentially, we're ahead of the game at this point. We already know the answers. The hydrogen wave function will have to involve the spherical solutions to the Laplacian and will be related to both the angular and radial components. This is most fortunate; for us to attempt to solve the Laplacian while doing the hydrogen atom would be difficult. However, we have some tasks left. The situation must be normalized, and we must deal with the fact that our potential has a pesky r dependence.

Results

We end up with the results that you knew we were going to. We can write down the hydrogen wave equation as:

$\Psi _{nlm}(r,\phi ,\theta )=R_{nl}(r)Y_{m}^{l}(\theta ,\phi )$

where $Y_{m}^{l}(\theta ,\phi )$ are the spherical harmonics.