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QM - the Hydrogen atom

  1. Apr 28, 2008 #1
    the Hamiltonian for the hydrogen atom appears as

    H= (Pr)^2/2m+(L^2)/2mr^2-e^2/r

    The Schroedinger equation is


    |E| because we're looking for bound states.

    m - reduced mass.

    the radial equation is


    (h denotes h-bar)

    we change the the dependent variable to u=rR(r)

    and we get


    I don't understand how exactly the last equation is obtained. By the change u=rR, shouldn't we get the factor 1/r inside the brackets?
    Last edited: Apr 28, 2008
  2. jcsd
  3. Apr 28, 2008 #2

    In your first equation (the "Radial Equation"), you have 1/r in the first term - why not multiply both sides (or, just the left side, since the right side is just equal to zero) by r? This way, you get rid of the r in the first term, and when you distribute out the R(r) you can then make the substitution for U(r)=rR(r) (The "Radial Function"). At this point, most authors make the following substitution

    [tex]V_{eff}=\frac{l(l+1)\hbar^2}{2mr^2}-\frac{e^2}{4\pi\epsilon_0 r}[/tex]

    which simplifies things a bit. Of course, you'll have to substitute this back in eventually, however it may make it easier to see things temporarily. Additionally, following this it becomes convenient to make the following substitution


    The math that follows this is really quite messy, however none of it is particularly difficult.
  4. Apr 29, 2008 #3
    thank you(:
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