# Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?

I simply have no idea how this result is found $$I=\hbar \sqrt{\ell(\ell+1)}.$$ The result seems to just be dumped in textbooks rather than explained. I can get the result that $I_z=\hbar m_j$. Please explain as clearly as possible how this is derived.

Also, any pointers towards any books/videos that may help me learn quantum mechanics would be appreciated.

• – Qmechanic Apr 3 '15 at 20:16
• That's actually a bit of a thorny problem, but the gist has to be something like, "we get this equation for $\Phi(\phi)$, we can assume $\Phi(\phi) = Z(\cos(\phi))$ because on this range $\cos$ is invertible; that gives the associated Legendre equation in $z = \cos\phi$, then to get a proper solution we need (at least) $\Phi'(0) = \Phi'(\pi) = 0$, and possibly for other derivatives to also vanish." Those boundary conditions are pretty much the only things that can quantize the associated number to $\ell (\ell + 1).$ – CR Drost Jul 18 '15 at 22:14
• ACuriousMind's answer is correct, but is not going to demystify this issue for people who are not fluent in group theory. For a very rough, intuitive, visual argument, see section 14.2.4 of my book Simple Nature: lightandmatter.com/area1sn.html . – user4552 Apr 21 '18 at 19:15

This comes from the representation theory of the rotation group $\mathrm{SO}(3)$.

Quantum mechanics takes place in a vector space, and observables are operators on this space. The total amount of angular momentum is obtained from the angular momenta $L_x,L_y,L_z$ about the three axes of space as $$L = \sqrt{L_x^2+L_y^2+L_z^2}$$ since that is the length of the vector $\vec L = (L_x,L_y,L_z)^T$.

The representation theory of the rotation group now tells you that the only possible values for $L$ on so-called irreducible representations, to which the states with $I = \sqrt{l(l+1)}$ belong, are restricted to $L = \sqrt{l(l+1)}$ with $l$ an integer. You cannot make $L_x,L_y,L_z$ behave as angular momentum operators (i.e. as operators which generate the rotations in the sense that they form the Lie algebra belonging to the rotation group) without $L$ taking these integer values. The proof of this is technical and found e.g. on the Wikipedia page I linked in the beginning, or, in another approach, found in my answer here.

The prefactor $\hbar = \frac{h}{2\pi}$ for $\sqrt{l(l+1)}$ is found by dimensional analysis and comparing to experimental results.

• It's probably not as general, but there's also the fact that if you're solving Schrodinger's equation for a rotational system you'll end up with Legendre polynomials (or, in general, spherical harmonics), and if you don't quantize $\ell$ you won't get well-behaved solutions. – zeldredge Apr 3 '15 at 19:03
• Is it not possible to get this result from the Schrodinger equation? – RobChem Apr 3 '15 at 19:06
• @zeldredge: Actually, the spherical harmonics(link to relevant part of the Wiki article) with fixed $l$ are all the irreducible representations of the rotation group, so I believe that is a fully equivalent way if you have a rotationally invariant Hamiltonian. – ACuriousMind Apr 3 '15 at 19:07
• Oh, I think you're right. Sorry, my representation theory isn't as good as it could be and I don't have good intuition for how it enters into things. – zeldredge Apr 3 '15 at 19:09
• @RobChem: It is possible e.g. if you look at the hydrogen atom. You find that the solutions to the angular dependent part, the spherical harmonics, exhibit precisely the $l(l+1)$, but there it just kinda falls out of the structure of the solutions with little insight to it. The representation theory is the underlying reason. – ACuriousMind Apr 3 '15 at 19:11