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Energy levels in Hydrogen - derivation

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    We know that:

    [tex]E=<\psi|H|\psi>[/tex]

    where
    [tex]H=-\frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 r}[/tex]
    and
    [tex]\psi=R(r)Y(\theta, \phi)[/tex]
    with
    [itex]R(r)=\frac{1}{\sqrt{(2n)!}}(\frac{2Z}{na_0})^{3/2}\left(\frac{2Zr}{na_0}\right)^{n-1}e^{-Zr/na_0}[/itex]
    If I want to find the energy, do I just evaluate this integral over all space? I started doing it, but it becomes very untidy at a point - I just wanted to ask if this is the method.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2011 #2

    micromass

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    Mod Note: moved to advanced physics. Please post such a question in that forum.
     
  4. Dec 6, 2011 #3

    dextercioby

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    Essentially yes. However, there are shortcuts you can take to evaluate expectation values of the Hamiltonian. Is psi in an eigenstate (discrete portion of the spectrum), no integral necessary. Likewise for a weighted sum of eigenstates.
     
  5. Dec 6, 2011 #4
    Psi is a wavefunction, and an eigenstate - but I believe it has to be an integral. However, can you tell me what these shortcuts are?
     
  6. Dec 6, 2011 #5

    dextercioby

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    Tabulated (as a result of integrations on which you needn't worry) values for

    [tex]\langle nlm |\frac{1}{r} | nlm \rangle [/tex]


    (nlm are eigenvectors of the Hamiltonian) exist and ease your work.
     
    Last edited: Dec 6, 2011
  7. Dec 7, 2011 #6
    I know these values - for inverse 1st, 2nd and 3rd powers - my concern is the first term - the one involving the differential - it is going to produce 2 integrals on its own.
     
  8. Dec 7, 2011 #7
    NO, this is most definitely not the standard method. Do you know what eigenvalues and eigenfunctions are?
     
  9. Dec 7, 2011 #8
    Yes, I do ofc. I know how to find the energy by requiring the raising operator to terminate the circular orbit wavefunction, I was just wondering is you can do it by directly evaluating this integral.
     
  10. Dec 7, 2011 #9
    This is surely not the standard way. Please do demonstrate.
     
  11. Dec 7, 2011 #10

    vela

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    So what specifically is the problem? These aren't particularly tricky integrals. They're straightforward but tedious. Show us what you've done so far.
     
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