What's is the origin of Orbital Angular Momentum of electrons in atoms?

Consider the Hydrogen 1s electron. We know that, in the quantum picture, the electron isn't orbiting or rotating at all, rather we simply state that the electron is spread over the entire space with the probability of finding it being maximum a radial distance $R$ (=Bohr radius) away from the nucleus. This helps explain why the electron does not radiate EM radiation while in the atom.

But, with this understanding, I do not understand the source of the orbital angular momentum. Is it intrinsic like the spin?

Answer

In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however, it is moving in a very special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense, the orbital is constantly rotating.

To understand precisely what is happening lets calculate some observables. Consider the Hydrogen $1s$ state which is described by

\begin{equation} \psi _{ 1,0,0} = R _1 (r) Y _0 ^0 = R _{1,0} (r) \frac{1}{ \sqrt{ 4\pi } } \end{equation} where $ R _{1,0} \equiv 2 a _0 ^{ - 3/2} e ^{ - r / a _0 } $ is some function of only distance from the origin and is irrelevant for this discussion and the wavefunction is denoted by the quantum numbers, $n$, $ \ell $, and $ m $, $ \psi _{ n , \ell , m } $. The expectation value of momentum in the angular directions are both zero, \begin{equation} \int \,d^3r \psi _{ 1,0,0 } ^\ast p _\phi \psi _{ 1,0,0 } = \int \,d^3r \psi _{ 1,0,0 } ^\ast p _\theta \psi _{ 1,0,0 } = 0 \end{equation} where $ p _\phi \equiv - i \frac{1}{ r } \frac{ \partial }{ \partial \phi } $ and $ p _\theta \equiv \frac{1}{ r \sin \theta } \frac{ \partial }{ \partial \theta } $.

However this is not the case for the $ 2P _z $ state ($ \ell = 1, m = 1 $) for example. Here we have, \begin{align} \left\langle p _\phi \right\rangle & = - i \int \,d^3r \frac{1}{ r}\psi _{ 1,1,1} ^\ast \frac{ \partial }{ \partial \phi }\psi _{ 1,1,1} \\ & = - i \int d r r R _{2,1} (r) ^\ast R _{ 2,1} (r) \int d \phi ( - i ) \sqrt{ \frac{ 3 }{ 8\pi }} \int d \theta \sin ^3 \theta \\ & = - \left( \int d r R _{2,1} (r) ^\ast R _{2,1} (r) \right) \sqrt{ \frac{ 3 }{ 8\pi }} 2\pi \frac{ 4 }{ 3} \\ & \neq 0 \end{align} where $ R _{2 1} (r) \equiv \frac{1}{ \sqrt{3} } ( 2 a _0 ) ^{ - 3/2} \frac{ r }{ a _0 } e ^{ - r / 2 a _0 } $ (again the particular form is irrelevant for our discussion, the important point being that its integral is not zero). Thus there is momentum moving in the $ \hat{\phi} $ direction. The electron is certainly spread out in a "dumbell" shape, but the "dumbell" isn't staying still. Its constantly rotating around in space instead.

Note that this is distinct from the spin of an electron which does not involve any movement in real space, but is instead an intrinsic property of a particle.