Griffiths in his celebrated book named 'Introduction to Quantum Mechanics' discusses about the total angular momentum of a hydrogen atom on page 187.
He writes:
If a hydrogen atom is in the state $\psi_{nlm}$, the net angular momentum of the electron (spin plus orbital) is $l+1/2$ or $l-1/2$; if you now throw in the spin of the proton, the atom's total angular momentum quantum number is $l + 1$, $l$, or $l-1$.
However, while solving for the hydrogen atom Griffith assumes that the nucleus is stationary.
My question: does the proton have any $l$? If not, then why?
Best Answer
The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation $$ \left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm}, $$ are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical and the quantum versions to factor away the motion of the bigger body (which is close to the centre of mass) and leave an effective one-body problem which is easier to treat.
This means that the orbital angular momentum, with total angular momentum number $l$, is in fact the combined angular momentum of the electron and the proton about their centre of mass. In essence, the proton partakes in part of the orbital motion and takes out some of the angular momentum from the electron. (Note, though, that this is classical language and it explicitly does not hold for the hydrogen atom, where the angular momentum of the motion is essentially indivisible.)
This raises an apparent paradox, which is resolved through the fact that the separation vector obeys dynamics through the reduced mass $\mu=1/\left(\tfrac{1}{m_e}+\tfrac{1}{m_p}\right)\approx\left(1-O\left({m_e\over m_p}\right)\right)m_e\lesssim m_e$ of the system, and this is slightly smaller than the electron mass. This slightly enlarges the orbital radius of the electron (since the Bohr radius is inversely proportional to the mass). The velocity stays constant (at $\alpha c$), which means that the angular momentum $L\sim \mu r v$ stays constant as well.
That said, the proton does have spin angular momentum of its own, but this couples weakly to the electronic motion. This coupling is via the same spin-orbit couplings as the electron, but its much higher moment of inertia means that the relevant energies are much smaller, as are the corresponding hyperfine splittings in the spectrum.