Quantum energy levels of a point mass rotating about a fixed point The question is: A particle of mass m is attached to a fixed point in space by a massless rigid rod of length a and can freely rotate about this point. Find the quantum energy levels of the system. What is the degeneracy of each energy level?
I used rotational kinetic energy:
$E=\frac{1}{2}I\omega^2=\frac{L^2}{2I}$
and then substituted $I=ma^2$ and $L=\hbar \sqrt{l(l+1)}$ to get:
$E_l=\frac{\hbar^2l(l+1)}{2ma^2}$.
So the energies are quantized as expected. But what is the degeneracy of each level? Plugging in a bunch of values for $l$ doesn't show any $l$s with similar energy so far. Is it correct that the degeneracy of each level is $0$?
 A: For each $l$ there exists $2l+1$ possible values of $m$. Since $m$ must be an integer, and $-l\leq m\leq l$, expanding out the associated Legendre function:
$P_l^m(x)\equiv (-1)^m(1-x^2)^{m/2}(\frac{d}{dx})^mP_l(x)$
where $P_l(x)$ is the $l$th Legendre polynomial in $x$, will show that there is $2l+1$ degeneracies. The solutions to the theta dependence of the angular equation due to separation of variables of the spherical Schrodinger equation are the Legendre polynomials in $\cos(\theta)$.
For example:
$P_0^0=1$,
$P_1^1=-\sin(\theta)$, 
$P_1^0=\cos(\theta)$, etc...
A: As @probably_someone wrote in his comment:

It might be easier to think of the equivalent problem: a particle is constrained to move on a fixed sphere, with no other forces besides the constraint (which is what is meant by "free" in this context). 

Now I guess that the non-degenerate energy levels you wrote down are correct if we confine the rotation to one plane. But there is an infinity of planes in which the rod can rotate, which means there is an infinite degeneracy for all energy levels.
