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?
 A: 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 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.
A: 
I do not understand the source of the orbital angular momentum. Is it intrinsic like the spin?

No.
Despite the fact that the picture of the orbiting electron is not quantum-mechanically correct, orbital angular momentum is still written in terms of position and momentum of the electron.  More precisely, orbital angular momentum is represented as a triple of operators $\mathbf L = (L_1, L_2, L_3)$ which are defined in terms of the electron position and momentum operator components $X^i$ and $P^i$ in the following way:
\begin{align}
  L_i = \epsilon_{ijk}X_jP_k
\end{align}
So when one makes a measurement of the orbital angular momentum of the electron, one is still measuring something relating to the position and momentum of the electron, not an intrinsic property of the electron.
A: You should not try to rephrase things in terms of some fuzzy smeared out averaged classical physics, or that the electron "really" is moving in an orbit or something like that.
What is angular momentum? Angular momentum is the quantity that is conserved in a rotationally symmetric system. You can work out that in quantum mechanics the (orbital) angular momentum operator takes the form $\hat L = \hat x \times \hat p$ where $\hat p, \hat x$ are the momentum and position operators
$L$ is an observable so you can talk about the angular momentum of any state, but $L$ commutes with neither $x$ not $p$, so it's harder and not very useful to try to think of it in terms in some definite motion. Indeed the average momentum in a hydrogen orbital is 0, but the angular momentum can be non-zero. (And you should not think of this as "when the electron is at $x$ it's moving in the opposite direction as when it is as $-x$". Nothing about this sentence makes sense in QM.)
A: I think the answers did not cover some important things:


*

*the Heisenberg uncertainty principle states that the more we know the location of the electron, the less we know it's momentum (the higher it will be)

*since the electron is in a closed space according to QM in a stable energy level around the nucleus, it's momentum must be higher

*the closer we know it's position, so the more it is in a smaller closed space around the nucleus, the higher it's momentum will be, so on a lower level orbit, it will have a higher momentum
