This is a problem that's been bothering me for a while. For electrons "in" hydrogenic atoms, do the electrons in states for which l=0 actually orbit the nucleus, or would they only ever be observed to be moving radially (assuming they could be directly observed in that way).(adsbygoogle = window.adsbygoogle || []).push({});

Here's my logic. I understand that the solutions of the Schrodinger equation for the hdyrogenic atom are also eigenfunctions of the total angular momentum squared operator, [itex]L^2 \psi = \hbar^2 l(l+1) \psi[/itex]. The eigenvalue is [itex]\hbar^2 l(l+1)[/itex]. As a result the expectation value for the magnitude of the angular momentum of the electron about the nucleus is [itex]\hbar \sqrt{l(l+1)}[/itex].

So, if I understand eigenvalues and eigenfunctions, which I probably don't, that means the angular momentum for states where [itex]l=0[/itex] the angular momentum will be observed to be zero, the electron will never be "orbiting". Even though there is probability density all around the nucleus. So it could be observed to change angular position, but it would never observed to be moving angularly. This seems peculiar to me. Please help.

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# Angular momentum and the hydrogenic atom

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