〈 Berger | Dillon 〉

The actual wavefunction is a linear combination of the eigenfunctions where the angular dependence goes away. This is intuitive. The atom normally doesn't have the notion of angle.

what do you mean "actual"? Hydrogen can certainly be in an energy eigenstate .. the energy quantum number doesn't depend on azimuthal quantum numbers, but the angular momentum observables do

Take the 2p orbital, for example. The electron is in a superposition of 2px, 2py and 2pz. |ψ2p> = (|2px>+|2py>+|2pz>)/sqrt(3). Now take the pdf <ψ2p|ψ2p> and you'll see the dependence on the angles cancels out. It only depends on the distance to the nucleus.

That's just not true.. a superposition of the 2ps is a superposition of 3 dumbbells each of which has an angular dependence in an independent direction .. it *can* happen that you measure l=0, but it can also happen that you measure l=1

That's what you would measure. But it is hard to imagine a free hydrogen atom being anisotropic...

why is it hard to imagine? .. the point is that the wave function has non-trivial angular dependence and so does the PDF .. i'm sorry if you're upset but it's just the eigensolutions to the schrodinger equation

Not upset at all... But without external field, why would the proton and the electron align in a particular direction?

because energy eigenstates have trivial time evolution .. if you prepare the state in some Ψnlm , then you are simply just more likely to find it in certain places than another, that's like the whole thing .. and you don't even need very exotic states to acheive this at all