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Where does phi go in the probability density for hydrogen?

  1. May 27, 2008 #1
    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.
     
  2. jcsd
  3. May 27, 2008 #2
    Indeed. [tex]\varphi[/tex] 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.
     
  4. May 27, 2008 #3

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    You're right, all eigenstates have the form [itex]\psi = A(r,\theta) e^{i m \phi}[/itex], so that:

    [tex] \psi^* \psi = |A(r,\theta)|^2[/tex]

    which doesn't depend on [itex]\phi[/itex]. 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 [itex]\phi[/itex] (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:

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

    we get:

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

    or, assuming A and B are real:

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

    which should reming you of the oscillation between two energy eigenstates.
     
    Last edited: May 27, 2008
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