I'm going to give a slightly different example of why this is useful in Quantum Mechanics and how from it we can arrive at Quantum Numbers. A quick background:
- In Q.M a complete description of the state of a physical system is given by a
normalized vector $|\phi>$ in the Hilbert space appropriate to the system
- Observable quantities (energy, position,momentum etc.) are represented by hermitian operators. Our goal is to have a hermitian operator which we can diagonalise to form an eigenbasis of the Hilbert space.
Now given an observable $Q$ represented as an operator we have that our eigenvalues $q_j$ will label our eigenvectors as $|q_j>$. Now suppose that the eigenvalues are not all distinct: then the eigenvectors will not give us a complete orthogonal basis of the Hilbert space. In this case, we use additional labels or 'quantum numbers' corresponding to a different observable $R$.
E.g:
$$
|q_j,1>,|q_j,2>\\
Q|q_j,i>=q_j|q_j,i>\\
R|q_j,i>=r_i|q_j,i>
$$
And then the e-vectors can be labelled
$$
|q_j,r_i>
$$
Thus we can see that having non-distinct eigenvalues of an operator leads to the idea of a quantum number.
As an aside, when does this happen?
Given two operators $A,B$,that give a complete basis, we require that:
$$
A|a_i,b_j>=a_i|a_i,b_j>\\
B|a_i,b_j>=b_j|a_i,b_j>
$$
Now let $[A,B]$ be the commutator of two matrices then:
$$
[A,B]|a_i,b_j>=(a_ib_j-b_ja_i)|a_i,b_j>=0
$$
Now as the vectors form a complete basis they are nonzero thus $[A,B]=0$.
Similarly the converse can be shown: $[A,B]=0 \implies $complete set of common eigenvectors.