6.S: The Hydrogen Atom (Summary)

For the hydrogen atom, we let the nucleus be defined as the center of our coordinate system, and the Hamiltonian becomes:

\[\op{H} = -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\epsilon_0r}, \nonumber \]

where \(r\) is the electron distance, \(\nabla^2\) is in spherical coordinates as before, \(e\) is the charge of the electron, and the Coulombic term is negative to indicate that the proton-electron potential is attractive (i.e., the proton and electron have charges that are opposite in sign). The factor \(4\pi\epsilon_0\) comes from using SI units; the factor disappears if using atomic units. The solution to this equation is tedious, but it turns out that the method of separation of variables can be used to solve this system in terms of radial and angular components:

\[\psi(r,\theta,\phi) = R(r)Y(\theta,\phi). \nonumber \]

The angular solutions are the same as those for the rigid rotor: the spherical harmonics \(Y_l^m(\theta,\phi)\), so the hydrogen atom wavefunctions have quantum numbers \(l\) and \(m_l\). You are already familiar with these from general chemistry, \(l=0\) corresponds to an \(s\) orbital, \(l=1\) is a \(p\) orbital, and so on.

It is found that the energy of the hydrogen atom depends only on a single quantum number \(n\), even though full specification of the wavefunction also requires the quantum numbers \(l\) and \(m_l\):

\[E_n = -\frac{e^2}{8\pi\epsilon_0a_0n^2},\quad n=1,2,\ldots, \nonumber \]

where \(a_0\) is the Bohr radius. The allowed quantum numbers are \(n = 1,2,3,\ldots\), \(0 \leq l \leq n\), and \(0 \leq |m_l| \leq l\).

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