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Probability Distribution Function, H-atom

  1. Dec 19, 2006 #1
    I'm reading Basdevant/Dalibard on 'Stationary States of the Hydrogen Atom' in preparation for a final this week, and the "Probability distribution function" for finding an electron in a spherical shell of thickness dr in the ground state is given.

    It's not derived, so I was wondering if anyone could explain how to find such a distribution function.

    Momentum, for example. If I wanted to find the probabilty distribution function for momentum, how would I do that?

    I think I've got the wavefunction for the ground state of Hydrogen:
    (using the equation involving spherical harmonics, the radial equation, and n=1, l=0, m=0)
    [tex]|100>=(1/a_{o})^{2/3}e^{-r/a_{o}}\sqrt{1/{4\pi}}[/tex]

    Any insight would be very much appreciated!

    EDIT:
    Oh, to clarify, Basdevant lists this as the answer for the radial probability distribution function:

    [tex]P(r)dr=|\psi_{1,0,0}(r)|^2(4\pi)r^2dr[/tex]

    I just don't know how he got there!
     
    Last edited: Dec 19, 2006
  2. jcsd
  3. Dec 19, 2006 #2

    dextercioby

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    Homework Helper

    Solve the spectral equation for the H atom. Assume discrete spectrum (the continuous one is harder to treat, its treatment is often missing in textbooks) and then write down the eigenvectors of the Hamiltonian. By squaring the modulus of these vectors (position wavefunctions, if you assume that you use the coordinate representation in [itex] L^{2}\left(\mathbb{R}^{3}, d^{3}x\right) [/itex]) you get the probability density function for localizing the electron within an open domain from R^3.

    Fourier transform [itex] \langle r,\theta,\phi|nlm\rangle [/itex] and then consider the square modulus.

     
  4. Dec 19, 2006 #3

    jtbell

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    Staff: Mentor

    You get it from the volume integral of [itex]|\psi_{1,0,0}(r)|^2[/itex] in spherical coordinates, integrating over the angular coordinates but not the radial coordinate:

    [tex]P(r)dr=\int_0^{2 \pi} \int_0^\pi |\psi_{1,0,0}(r)|^2 r^2 \sin \theta dr d\theta d\phi[/tex]

    In case you or anyone else reading this needs reminding, [itex]r^2 \sin \theta dr d\theta d\phi[/itex] is the volume element in spherical coordinates, like [itex]dx dy dz[/itex] is the volume element in rectangular coordinates.
     
    Last edited: Dec 19, 2006
  5. Dec 19, 2006 #4
    Okay, so for momentum, my plan of attack is to FT [tex]\psi_{1,0,0}(r)[/tex] to [tex]\phi(p)[/tex], and then
    [tex]<\phi|p|\phi>=\int(\phi*p\phi)d^3p[/tex].
     
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