# Quantum numbers

1. Aug 29, 2010

### Gavroy

hey,

i asked myself, how one could derive the relations for the quantum numbers...
so why is:
l always: 0<l<n-1
from what follows this relation for l
does anyone know this?

2. Aug 29, 2010

### nnnm4

I believe 0<l<n-1 for convergence of the Frobenius solution to the angular momentum eigenvalue equation.

3. Aug 29, 2010

### peteratcam

Yes, the "reason" is that n is an accidental quantum number in a sense, because the hydrogen atom has an extra symmetry that makes lots of states degenerate which you wouldn't expect them to be.

For any spherically symmetric problem, the obvious quantum numbers to use are the number of radial nodes, and then the angular momentum quantum numbers. Imagine a piece of graph paper with number of radial nodes along the x-axis, and angular momentum quantum number along the y-axis. Then at every position with non-negative integer coordinates there are (2l+1) degenerate states. Then you have a nice classification of all states of hydrogen, labial by the number of radial nodes, and the angular momentum.

The extra symmetry of hydrogen means that states on diagonals are degenerate, and so the energy quantum number is an equivalent way to label the states, but the n quantum number is constrained since (#radialnodes+ l +1 = n).