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The expectation value for the radial part of the wavefunction of Hydrogen.

  1. Sep 29, 2011 #1
    The wavefunction of hydrogen is given by


    \psi_{nlm}(r, \theta, \phi) = R_{nl}(r)Y_{lm}(\theta, \phi)

    If I am only given the radial part, and asked to find the expectation value of the radial part I integrate the square of the wavefunction multiplied by r cubed allowing r to range from 0 to infinity. I don't understand where the extra factor of r squared comes from? I suspect it has something to do with multiplying by a volume element, but it is unclear to me why the factor of 4 pi that would normally come with spherical integration that depends on r alone disappears. I missed this on a test, recently, and was hoping someone could explain.

  2. jcsd
  3. Sep 29, 2011 #2


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    You're integrating in spherical coordinates so the volume element is [itex]dV=r^2sin\theta dr d\theta d\phi[/itex]
    Last edited: Sep 29, 2011
  4. Sep 29, 2011 #3


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    The 4pi is normalized away by the spherical harmonics. You essentially already integrated over a spherical shell, and over that entire spherical shell, you should get the integral to be 1.
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