Two particle rigid rotor in QM Consider the following example in QM (From Griffiths book 'Introduction to Quantum Mechanics'): If you are given two particles of mass $m$ attached to the ends of a massless rigid rod of length $a$. The system is free to rotate in three dimensions about the center. We want to show that the allowed energies of this rigid rotator are:
$$E_n = \frac{\hbar^2 n(n+1)}{ma^2}~~~\text{for }n=0,1,2,..$$
The following solution is given: First note that classically we have $H = 2(\frac{1}{2}mv^2) = mv^2$ and $|L| = 2(\frac{a}{2}mv) = amv$ then $L^2 = a^2m^2v^2$ and hence $H = \frac{L^2}{ma^2}$. But we know that the eigenvalues of $L^2: \hbar^2l(l+1)$ or labelled with $n:$ $$E_n = \frac{\hbar^2 n(n+1)}{ma^2}~~~\text{for }n=0,1,2,..$$ 
Questions: 
- Why are the eigenvalues given as $\hbar^2l(l+1)$, is this not the eigenvalues for a single particle system? In this case we are considering two particles, should the eigenvalues for the total angular momentum squared not be $2\hbar^2l(l+1)$?
-Lastly, I just want to know if this is the standard approach when dealing with these types of problems. First start with the classic description and then substitute the quantum mechanic analogues of the quantities? 
Thanks for any assistance.
 A: Eigenvalues of angular momentum operators, i.e. operators satisfying commutation relations
$$ [L_i,L_j]= i \epsilon_{ijk} L_k $$
are always $j(j+1)$ with $j$ natural number (or half of natural number if spin is taken into account). Derivation of this fact is universal and doesn't depend on the physical nature of $L$ operator (whether it is angular momentum of one particle or some composite system). You can easily check that if $\vec L_1$ and $\vec L_2$ are two vector operators satisfying angular momentum algebra then $\vec L_1 + \vec L_2$ also is. Therefore if we are not talking about spin here, it has precisely the same set of eigenvalues as $\vec L_1$ and $\vec L_2$. You might be puzzled: why can't we just add eigenvalues? In this case the reason is that angular momenta are vector operators and we need to add them like vectors. Observe that
$$ (\vec L_1 + \vec L_2)^2 = L_1^2 + L_2^2 + 2 \vec L_1 \cdot \vec L_2, $$
where I assumed that $\vec L_1$ and $\vec L_2$ commute. Note that if you consider an eigenstate of both $L_1^2$ and $L_2^2$ it will usually not be eigenstate of $(\vec L_1 + \vec L_2)^2$ at all, because of the dot product term. In general, there are quite some intricacies involved in adding angular momenta. This is due to the fact that not only you need to add them like vectors, but you also need to remember that they are operators and their components don't commute among themselves, which gives rise to some new phenomena not seen in classical mechanics. I'm sure this will be covered in any reasonable QM course you might attend, but if you can't wait I suggest you look into one of the Sakurai books. Now for the second question: Yes, this is the usual approach. This procedure is usually called quantization. Basically we take some classical theory and try to promote classical quantities (momenta of particles, values of electromagnetic field or whatever you like) to quantum operators. Note that even if such procedure is possible, it is to large extent arbitrary and needs to use heuristic arguments. It is quantum theory which is more general and fundamental so it should be in principle possible to derive classical mechanics from quantum, but not the other way around. So quantization of classical theory is an attempt to reproduce reality from partial information.
