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Orbital angular momentum projection

  1. Nov 18, 2014 #1
    Suppose I have particle in three dimensional space whose position space wavefunction in spherical coordinates is ##\psi(r,\theta,\phi)##. The spherical harmonics ##Y_{\ell,m}## are a complete set of functions on the 2-sphere and so any function ##f(\theta,\phi)## can be expanded as ##f(\theta,\phi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} f_{\ell,m} Y_{\ell,m}(\theta,\phi)##. Since ##\psi## may, in general, have radial dependence, the coefficients ##f_{\ell,m}## will too, i.e., ##f_{\ell,m}(r)##.

    I'm curious, then, what the probability is that the particle will be measured have total orbital angular momentum ##\sqrt{\ell_0(\ell_0+1)}\hbar##. I'm used to seeing expansions of quantum states in which the coefficients are just constants and so I'm a bit thrown by the radial dependence in this case. Intuitively, I'd expect that a total orbital angular momentum measurement would yield ##\sqrt{\ell_0(\ell_0+1)}\hbar## with probability
    ##
    P(\ell_0) = \sum_{m=-\ell_0}^{\ell_0} \int_0^\infty dr |f_{\ell_0,m}(r)|^2
    ##

    Is that right?
     
  2. jcsd
  3. Nov 18, 2014 #2
    Almost.
    Almost. It should be the norm of the part of the wave function with ##\ell = \ell_0##, so

    [tex]P(\ell_0) = \int_0^\infty dr r^2 \int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta \left| \sum_{m = -\ell_0}^{\ell_0} f_{\ell_0, m}(r) Y_{\ell_0, m}(\theta, \phi) \right|^2[/tex]

    You can use the orthogonality of the spherical harmonics to rewrite this as

    [tex]P(\ell_0) = \sum_{m = -\ell_0}^{\ell_0} \int_0^\infty dr r^2 |f_{\ell_0, m}(r) |^2 \int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta \left|Y_{\ell_0, m}(\theta, \phi) \right|^2[/tex]

    If we assume that the spherical harmonics are normalized so that

    [tex]\int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta |Y_{\ell, m}(\theta, \phi)|^2 = 1[/tex]

    Then we get

    [tex]P(\ell_0) = \sum_{m=-\ell_0}^{\ell_0} \int_0^\infty dr r^2 |f_{\ell_0,m}(r)|^2[/tex]
     
  4. Nov 18, 2014 #3
    Interesting, I wouldn't have thought you'd need to do the whole spatial integral, particularly the angular part. I suppose that's just to take the inner product between spherical harmonics?
     
  5. Nov 18, 2014 #4
    Well, you can see that you don't actually have to do any angular integrals in the end. But the norm is defined in terms of the integral over all space, so I started from that.
     
  6. Nov 18, 2014 #5
    Very helpful, thank you.
     
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