# 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?

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

$$\psi _{ 1,0,0} = R _1 (r) Y _0 ^0 = R _{1,0} (r) \frac{1}{ \sqrt{ 4\pi } }$$ 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, $$\int \,d^3r \psi _{ 1,0,0 } ^\ast p _\phi \psi _{ 1,0,0 } = \int \,d^3r \psi _{ 1,0,0 } ^\ast p _\theta \psi _{ 1,0,0 } = 0$$ 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.

• (no need for all those integrals because I do know QM very well, but thank you) "such that the wave that it forms around the nucleus keeps the shape of the orbital" Agreed. "In some sense the orbital is constantly rotating". How can you prove this? – student1 Mar 30 '14 at 22:14
• @student1: It is proven above. The wavefunction has a nonzero expectation value for the momentum in the $\hat{\phi}$ direction at all times. – JeffDror Mar 30 '14 at 22:25
• You are correct, I will mark this as the correct answer. However, I think that the interpretation is not necessarily "the orbital is constantly rotating". – student1 Mar 31 '14 at 5:46
• @student1: I am not sure how else it can be interpreted-if we found there is momentum in the $\hat{\phi}$ direction then something needs to be moving in that direction at all times and hence you must have a rotating orbital. – JeffDror Apr 2 '14 at 12:36
• Rotating orbital could only be correct if you consider the states with $m\ne0$. Moreover, for the states with $|m|\ne l$ the rotation rate doesn't really reflect the magnitude of angular momentum. You should be looking at the $\exp(i(m\phi-\omega t))$ factor in the wavefunction to see how the orbital "rotates". What is lost in such a description is the standing rotational wave in the direction of change of $\theta$ angle, which also has its share of angular momentum, although doesn't make the orbital appear rotating. – Ruslan Mar 9 '17 at 14:35

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.

• But still you did not tell me where does this "something" came from, if it is not intrinsic. – student1 Mar 30 '14 at 5:19
• @student1 Why the quotes around "something?"; do you somehow find the (standard) expression I gave for the orbital angular momentum operator to be ambiguous? The point is that it's precisely the same physical quantity as in classical mechanics, it's simply promoted (as all observables are) to be an operator in quantum mechanics, and the possible values of its measurement are quantized. If this is not what you mean by "where it comes from" then please clarify. You may also find en.wikipedia.org/wiki/… useful. – joshphysics Mar 30 '14 at 5:22
• Sorry if there was a misunderstanding. If we just promote X,P to be operators, then the idea of an orbiting particle should still remain the same, but this is precisely what QM says is wrong. – student1 Mar 30 '14 at 5:48
• @student1 "If we just promote X,P to be operators, then the idea of an orbiting particle should still remain the same" how so? The whole machinery of quantum mechanics assumes that states of the system are elements of a Hilbert space and that observables are operators on this space. When applied to orbital angular momentum, this is no way implies that the "idea of the orbiting particle remains the same." That idea depends on the evolution of the particle state being a path through space which is absent from the quantum model. – joshphysics Mar 30 '14 at 6:30
• @student1 By the way, what I've said above is just a mathematically explicit formulation of PhotonicBoom's answer. Every book on quantum mechanics covers this, and so does wikipedia, in case you are looking for references. – joshphysics Mar 30 '14 at 6:32

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.)

• Thanks. Well, we can always suffice by saying "This property has no exact counterpart in CM so you should not think about...", and this is indeed the case for say the spin, but I thought in the case of Orbital Angular Momentum we could still find an explanation. – student1 Mar 30 '14 at 22:10
• Do not try to restrain people from applying familiar concepts. The same “average momentum is 0, but the angular momentum can be non-zero” reasoning Robin Ekman applied to an unspecified orbital (presumedly a p orbital) is applicable to a rotating disc. This ${\hat x}×{\hat p}$ depends on ℓ like have we $\int x×p\,d^3x ≠0$ for a rotating disc and 0 for a solid ball at rest. The main difference: for a ball at rest $p=0$ everywhere, whereas for an s orbital (ℓ = 0) we have ${\hat p}≠0$ but $-i\frac\partial{\partial x}$ is collinear with $x$ on an eigenfunction, so ${\hat x}×{\hat p}=0$. – Incnis Mrsi Aug 20 '14 at 14:59

I think the answers did not cover some important things:

1. 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)

2. 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

3. 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

• ... and yet all of those features are consistent with the vanishing $L=0$ of $s$ states. – Emilio Pisanty Apr 13 '18 at 10:06