Comprehensive Study Notes

The wave function specifying the state of an electron depends not only on the coordinates $x, y$, and $z$ but also on the spin state of the electron. What effect does this have on the wave functions and energy levels of the hydrogen atom?

To a very good approximation, the Hamiltonian operator for a system of electrons does not involve the spin variables but is a function only of spatial coordinates and derivatives with respect to spatial coordinates. As a result, we can separate the stationary-state wave function of a single electron into a product of space and spin parts:

\(
\psi(x, y, z) g\left(m_{s}\right)
\)

where $g\left(m{s}\right)$ is either one of the functions $\alpha$ or $\beta$, depending on whether $m{s}=\frac{1}{2}$ or $-\frac{1}{2}$. [More generally, $g\left(m{s}\right)$ might be a linear combination of $\alpha$ and $\beta ; g\left(m{s}\right)=c{1} \alpha+c{2} \beta$.] Since the Hamiltonian operator has no effect on the spin function, we have

\(
\hat{H}\left[\psi(x, y, z) g\left(m{s}\right)\right]=g\left(m{s}\right) \hat{H} \psi(x, y, z)=E\left[\psi(x, y, z) g\left(m_{s}\right)\right]
\)

and we get the same energies as previously found without taking spin into account. The only difference spin makes is to double the possible number of states. Instead of the state $\psi(x, y, z)$, we have the two possible states $\psi(x, y, z) \alpha$ and $\psi(x, y, z) \beta$. When we take spin into account, the degeneracy of the hydrogen-atom energy levels is $2 n^{2}$ rather than $n^{2}$.