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Hydrogen atom

  1. Dec 4, 2007 #1
    [SOLVED] hydrogen atom

    1. The problem statement, all variables and given/known data
    My book says that

    [tex]R_{1,0}(r) = 2(1/a_0)^{3/2} e^{-r/a_0}[/tex]

    is a normalized wavefunction. But if you integrate R_{1,0}(r)^2 over r from 0 to infinity, you do not get 1. What's wrong here?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2007 #2

    Avodyne

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    You need a factor of r^2 that comes from the dx dy dz in spherical coords.
     
  4. Dec 5, 2007 #3

    Gokul43201

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    That's not it - he's got the factor of r^2. LATE EDIT: No, you're right. I misread his (r)^2 as the r^2 from the volume element.

    Ehrenfest: You should not get 1 for that integral. Write down the (general) normalization condition.
     
    Last edited: Dec 5, 2007
  5. Dec 5, 2007 #4
    Now, I am very confused. I did not have the factor of r^2. But I only need that when I am integrating over dtheta dphi.

    Isn't the normalization condition

    [tex]\int_{r=0}^{\infty}|R_{1,0}|^2dr = 1[/tex]

    Obviously that's wrong but why?

    Are you implying that it should really be

    [tex]\int_{all space}|\Psi|^2drd\thetad\phi = 1[/tex]

    But [tex] \Psi = R(r)Y_{l,m}(\theta,\phi)[/tex] and the spherical harmonics are already normalized. So the normalization condition is still [tex]\int_{r=0}^{\infty}|R_{1,0}(r)|^2dr = 1[/tex], right?
     
    Last edited: Dec 5, 2007
  6. Dec 5, 2007 #5
    I edited my last post before I saw yours but here it is again


    Are you implying that it should really be

    [tex]\int_{all space}|\Psi|^2drd\thetad\phi = 1[/tex]

    But [tex] \Psi = R(r)Y_{l,m}(\theta,\phi)[/tex] and the spherical harmonics are already normalized. So the normalization condition is still [tex]\int_{r=0}^{\infty}|R_{1,0}(r)|^2dr = 1[/tex], right?
     
  7. Dec 5, 2007 #6

    Gokul43201

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    Only for a 1D problem.

    That's not it either.

    And that's important. Over what variables are the spherical harmonics normalized? Write down this expression.

    Next:

    [tex]\int d\Omega |\Psi(r,\theta ,\phi )|^2 = 1 [/tex]

    How do you write the volume element [itex]d\Omega[/itex] in spherical co-ordinates?
     
    Last edited: Dec 5, 2007
  8. Dec 5, 2007 #7

    malawi_glenn

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    What do you get? You said in your first post that you did not get 1, what did u get?

    What happens if you insert r^2 and then integrate?

    I have now checked, and I think you are missing the r^2

    [tex] R(r) = \frac{u(r)}{r} [/tex] where u(r) solves the radial shrödringer eq.

    And now the nowmalization becomes:

    [tex]\int_{r=0}^{\infty}|R(r)|^2r^2dr = \int_{r=0}^{\infty}|u(r)|^2dr =1[/tex]

    EDIT: Gokul43201 saw that he made a misstake, now three of us are telling you that you missed the r^2 from the volume element of integration =)
     
  9. Dec 5, 2007 #8
    I think I see the problem.

    We know that

    [tex] \int_{all space}\Psi(r,\phi,\theta)dr r^2 sin(\theta) d\theta d\phi = 1[/tex]

    [tex]\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)[/tex]

    So, [tex]\int_{\Omega}Y(\theta,\phi)d\Omega = \int_{0}^{\pi}\int_{0}^{2\pi} Y(\theta,\phi)d\phi d\theta = 1[/tex].

    Wait, is the first equality in the last equation true, or do I need a factor of sine in there?
     
    Last edited: Dec 5, 2007
  10. Dec 5, 2007 #9

    malawi_glenn

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    need or need.. you are forgetting to square the wavefunction..

    the only problem was that you forgot was the r^2 from the volume element of integration.

    [tex] \int Y^*Y sin(\theta )d\theta d\phi = 1 [/tex] is the normalization condition for the angular part.
     
  11. Dec 5, 2007 #10
    Here is what my last post should have been:

    We know that

    [tex] \int_{all space}\Psi(r,\phi,\theta)*\Psi(r,\phi,\theta)dr r^2 sin(\theta) d\theta d\phi = 1[/tex]

    [tex]\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)[/tex]

    So, [tex]\int_{\Omega}Y(\theta,\phi)*Y(\theta,\phi)d\Omega = \int_{0}^{\pi}\int_{0}^{2\pi} Y(\theta,\phi)*Y(\theta,\phi) sin\theta d\phi d\theta = 1[/tex].

    I have no idea why the first equality in the last equation needs that sin \theta? I have solids angles, I have no idea what they mean any more?
     
  12. Dec 5, 2007 #11
    Never mind. I see. Thanks all. :)
     
  13. Dec 5, 2007 #12
    EDIT: Do not even read this post. It is all wrong. I am not confused any more.

    Now I am really confused. Apparently the normalization condition is really

    [tex]4\pi\int_{r=0}^{\infty}|R(r)|^2r^2dr = 1[/tex]

    If you try this for the wavefunction I have in my first post, you need that factor of 4\pi.

    Which means that [tex] \int \int Y^*Y sin(\theta )d\theta d\phi = 1 [/tex] was not correct!

    But that makes no sense and I am so confused.
     
    Last edited: Dec 5, 2007
  14. Dec 5, 2007 #13

    malawi_glenn

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    [tex] d\Omega = sin\theta d\phi d\theta [/tex]

    =)
     
    Last edited: Dec 5, 2007
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