Hamiltonian and eigenstates of rotating cube Suppose we have a rigid cube whose centre is at rest but it can rotate in three dimensions. One might propose for the Hamiltonian and energy eigenstates the following argument:
$$
\hat{H} = 
\frac{\hat{L}_x^2}{2 I_x} +
\frac{\hat{L}_y^2}{2 I_y} +
\frac{\hat{L}_z^2}{2 I_z}
= \frac{\hat{L}^2}{2 I}
$$
using $I \equiv I_x = I_y = I_z$ for coordinate axes aligned with the principle axes of the cube. One then gets energy eigenvalues $L(L+1) \hbar^2/2I$ and each energy eigenstate has degeneracy $2L+1$.
But the above is wrong. It is wrong because it understates the degeneracy of the eigenspace of $\hat{L}^2$ for this case. My question is: how can we see that?
Essentially the same question was already asked in
Quantum rotator and equipartition theorem
and there is an excellent answer there, but it simply states that the eigenspace degeneracy is not $2L+1$ without going into further details on that. I am trying to elucidate how the humble student can see "oh yes, clearly this is wrong because we forgot ..."
 A: The "naive" approach is wrong because it forgets to spell out what the actual space of states is $H$ acts on - sure, you can write down $H = L^2 = L_x^2 + L_y^2 + L_z^2$, but what is the state space of the system we're describing?
That all energy eigenvalues are of the form $\ell(\ell +1)$ is true - but just having the Hamiltonian in hand doesn't tell you what the allowed values of $\ell$ are or how often they are repeated. There is just no basis for the assumption that the space of states should contain the eigenspace of $L^2$ for each $\ell$ only once.
The correct quantization of the rigid rotor can be somewhat subtle because the Euler angles often used as generalized coordinates are $S^1$-valued (or "$2\pi$-periodic", if you prefer), not $\mathbb{R}$-valued, but in any case the $L_i$ are not the canonical momenta (as you can already see from them not commuting with each other!). Regardless of what quantization procedure exactly we implement, our state space however should be wavefunctions of either the coordinates (Euler angles) or the canonical momenta.
Now, if you note that "every integer $\ell$ once" is just the spherical harmonics $Y(\theta,\phi)$ which are functions of two angles, it should seem somewhat intuitive that the idea of each $\ell$-eigenspace occuring only once underestimates the actual state space, which should be functions of three angles.
A: If I remember correctly, the  wavefunctions of the isotropic 3-d quantum rotator are the rotation matrices
$$
D^{l}_{mn}(\theta,\phi,\psi)=\langle l,m|e^{-i\hat L_3\phi}e^{-\hat L_2\theta}e^{-i \hat L_3\psi}|l,n\rangle\\=e^{-im\phi}d^l_{mn}(\theta) e^{-in\psi}, \quad l=1,2,3\ldots
$$
where $\theta,\phi,\psi$ are the Euler angles. The energy is
$E\propto l(l+1)$ and $m,n$ are usual integers $-l\le m\le +l$ so the degeneracy of each level is $(2l+1)^2$.
The basic idea is that the QM of the rotator is the that of free motion on the three-sphere $S^3/Z_2$, which is the group manifold of ${\rm SO}(3)$.
A: Very interesting. I did not know this intriguing issue. Here is my ideas about a suitable setting to accommodate the mathematical description of an inertially isotropic rigid rotor (*), after I have read the other excellent answers ("perched on the shoulders of giants"). It seems to me a nice application of well known results of the theory of compact Lie groups representation.
The configuration space of the isotropic rotor is the group $G=SO(3)$.
That is because the configurations of the system  are one-to-one with the orthogonal triples of axes, with common center, which can be achieved from a given triple with a rigid motion.
It is now natural to define the Hilbert space of the analogous quantised system as ${\cal H}:= L^2(G, \mu)$, where the measure $\mu$ is the unique normalized two-side Haar measure. (It exists because the group is compact and thus unimodular.)
If we choose the Laplace-Beltrami operator (with respect to the unique bi invariant Riemannian metric on the group, the Killing metric), we have that this Hamiltonian also coincide with the familiar Casimir operator $L^2$ of the right-regular action of the group:
$$(U_h\psi)(g):= \psi(gh).$$
There is however still a dimensional  scale factor to be fixed. To interpret the Hamiltonian as the kinetic energy, this factor has to be chosen as $(2I)^{-1}$:
$$H := \frac{L^2}{2I}.$$
At this juncture, we can profitably apply the celebrated Peter-Weyl theorem since the topological (and Lie) group $SO(3)$ is compact and the right representation is strongly continuous.
According to that theorem, the Hilbert space decomposes into an orthogonal Hilbert sum
$${\cal H} = \bigoplus_{j=0}^{+\infty} (2j+1) {\cal H}_{j}\:.$$
The closed subspaces ${\cal H}_{j}$ are unitary irreducible representations of $SO(3)$.  Schur's lemma implies that $L^2$ is constant therein  with (eigen)value $j(j+1)$ (I throughout assume $\hbar=1$). The   dimension, as is well known, is  $2j+1$.
A basis of ${\cal H}_{j}$ is made of the well known $SO(3)$ matrix representations $D^j_{mn}$.
Each such representation takes place once again  exactly $2j+1$ times according to the said theorem. (As a general result of the said theorem, referring to the decomposition if the  right regular representation, the degeneracy of each equivalence class of irreducible reps equals the dimension of the space of the said representation itself.) This is reflected by the factors 2j+1 on the right hand side of the decomposition above.
In summary, the degeneracy of a given $j$ is $(2j+1)^2$.
If the rotor is not the considered one, the configuration space changes. For instance, if there is only one symmetry axis, the configuration space is $SO(3)/SO(2) = S^2$ and  an approach as above should  be still suitable passing to the theory of representation on quotient spaces.

(*) Rotor or Rotator? My Italian mother tongue suggests rotator from Latin and also from the English verb to rotate, but Wikipedia for example uses quantum rotor....
