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Forming Hydrogen wave functions with real spherical harmonics

  1. May 1, 2013 #1
    Hi, I'm a little confused about how to apply the real spherical harmonics when building a hydrogen wave function.

    I'm doing a computational project, so I want to work with a wave function which is strictly real, and I'm hoping I can do so by building the orbitals using the real spherical harmonics, [itex] p_x, p_y, p_z,... [/itex] etc. Can someone give me a link or reference as to how that works?

    Thanks.
     
  2. jcsd
  3. May 2, 2013 #2

    DrClaude

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    Staff: Mentor

    Not sure exactly what you want, but you construct the angular part of the p orbitals by transforming from the usual basis of spherical harmonics ##Y_{1,1}##, ##Y_{1,0}##, ##Y_{1,-1}## to
    $$
    p_x = \frac{1}{\sqrt{2}} \left( Y_{1,-1} - Y_{1,1} \right) \\
    p_y = \frac{i}{\sqrt{2}} \left( Y_{1,-1} + Y_{1,1} \right) \\
    p_z = Y_{1,0}
    $$
     
  4. May 2, 2013 #3
    Hi, so my question is, does the wave function,

    [tex] \psi_{nlm}(\vec{x}) = R_{n\ell}(r)p_{i}(\theta,\phi) [/tex]

    where [itex] p_{i}(\theta,\phi) [/itex] are the real spherical harmonics, satisfy the same Schrodinger equation as the wave function

    [tex] \psi_{nlm}(\vec{x}) = R_{n\ell}(r)Y_{\ell}^m(\theta,\phi) [/tex] ?

    Thanks.
     
  5. May 3, 2013 #4

    tom.stoer

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    Just act with H, especially with the part containing the phi-derivative, on the new functions and check what happens with m
     
  6. May 3, 2013 #5

    DrDu

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    Yes, the states with different m are degenerate in the hydrogen atom. Hence any linear combination (notably the real ones) of these states is again an eigenstate.
     
  7. May 3, 2013 #6

    DrClaude

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    Staff: Mentor

    To add to what Tom and DrDu wrote, just note that while the real spherical harmonics are eigenfunctions of the Hamiltonian of the hydrogen atom, they are not simultaneously eigenfunctions of ##\hat{l}_z##, as the complex ##Y_{l,m}## are.
     
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