Where does phi go in the probability density for hydrogen?

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The analytical solution for the wavefunction of a hydrogenic electron with quantum numbers n, l and m has a spherical harmonic part that involves theta and phi (in spherical coordinates). I was looking in Griffiths, and the spherical harmonics part only has phi as exp(i m phi) where i is the imaginary unit, m is the magnetic quantum number and phi is phi (sorry I didn't use TeX). Phi doesn't show up in any other way, than attached to an i in an exponential...

So the complex conjugate of the wave function multiplied by the wave function itself should kill all the terms with phi. Am I correct?

How does phi factor into the probability density? Isn't it removed in taking psi*psi? I am trying to sketch the probability densities of hydrogen's first few wavefunctions.

Thank you for any help.
 
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Indeed. \varphi only affects phase, and not amplitude, but it is important in other respects. Sometimes it is useful to sketch the phase of the wavefunction using colour.
 
You're right, all eigenstates have the form \psi = A(r,\theta) e^{i m \phi}, so that:

\psi^* \psi = |A(r,\theta)|^2

which doesn't depend on \phi. This is good: just as an energy eigenstate has probabilities that don't depend on time, and a momenutm eigenstate (e^ipx) has probabiliites that don't depend on position, an angular momentum eigenstate should have probabilities that don't depend on \phi (this is related to Noether's theorem, a far reaching result connecting conserved quantities like momentum and energy to symmetries like rotation and time translation).

On the other hand, if we have a state that is not an angular momentum eigenstate, such as:

\psi= A(r,\theta) e^{im\phi} + B(r,\theta) e^{in\phi}

we get:

\psi^* \psi = |A(r,\theta)|^2 + |B(r,\theta)|^2 + 2Re \{ A B^* e^{i(m-n)\phi}\}

or, assuming A and B are real:

\psi^* \psi = A(r,\theta)^2 + B(r,\theta)^2 + 2AB\cos((m-n)\phi)

which should reming you of the oscillation between two energy eigenstates.
 
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