After about 20 years I'm going back and brushing up on quantum. I am confused about an electron in an s-orbital having zero angular momentum as well as how angular momentum in general relates to Heisenberg uncertainty. I probably have a few misconceptions as well as missing some key pieces. Any help you can provide to clear up my confusion is greatly appreciated.(adsbygoogle = window.adsbygoogle || []).push({});

If an electron occupying an s-orbital has zero angular momentum wouldn't that infer that it has zero angular velocity? If so wouldn't this mean that the electron is at rest and now localized in space? This qualitatively seems troubling to me. I also seem to remember someone mentioning that electron spin is a relativistic effect and due to the electron moving at a velocity near the speed of light. If so, then how can an electron in an s-orbital have a +1/2 or -1/2 spin if it is stationary?

A related but more general question about angular momentum, I have heard arguments using Heisenberg uncertainty to explain 1) why an electron doesn't crash into and stay localized at the nucleus and 2) why the ground energy state can't be equivalent to the zero point energy. It is argued that if the electron is stationary at the nucleus or had zero point energy one would know both the exact velocity (ie zero) and the exact position, and in turn this violates Heisenberg uncertainty. But for angular momentum if one localizes the electron to a specific orbital (doesn't have to be an s-orbital),[tex]\Delta[/tex]L = 0. The angular position [tex]\phi[/tex] of the electron can be anywhere between 0 and 2[tex]\Pi[/tex] but I thought I remember that in this case [tex]\Delta\phi[/tex] = [tex]\pi[/tex]/[tex]\sqrt{}3[/tex]. In this case the products of the uncertainties equals zero. Doesn't this violate Heisenberg uncertainty?

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# S-orbitals, angular momentum and Heisenberg uncertainty

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