# What is the only quantum number that does not derive from the Schrodinger equation?

Jul 31, 2017

That would be the spin quantum number, ${m}_{s}$. It is purely a property of a fermion, such as an electron, that its spin is $\pm \frac{1}{2}$.

Here's where the rest of them came from. The wave function $\psi$ is justifiably assumed that it can be separated into three functions of one variable (i.e. we tried a trick and it worked, and it is no longer a trick):

${\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right) = {R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$

$= {R}_{n l} \left(r\right) \Theta \left(\theta\right) \Phi \left(\phi\right)$

• The magnetic quantum number ${m}_{l}$ came from solving the $\phi$ part of the angular portion, i.e. for $\Phi \left(\phi\right)$, and using a cyclic boundary condition.
• The azimuthal quantum number $l$ came from solving the $\theta$ part of the angular portion, i.e. for $\Theta \left(\theta\right)$, using the associated Legendre polynomials, which are functions of $\cos \theta$.
• The principal quantum number $n$ came from solving the radial portion, i.e. for ${R}_{n l} \left(r\right)$, but it is one of the more difficult and esoteric parts of solving the Schrodinger equation for the hydrogen atom, and few textbooks show it in full.

It basically comes from a convergence condition when solving for a coefficient in a power series, though I've never had to know that myself.

I did show a good portion of solving the Schrodinger equation for the hydrogen atom here, if you want more context:
https://socratic.org/questions/solve-schr-dinger-equation-for-hydrogen-atom

Or, a more condensed version is found here:
http://users.aber.ac.uk/ruw/teach/237/hatom.php