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The expectation of 'z' and 'x+iy'

  1. May 6, 2008 #1
    can anyone give me any ideas on how to evaluate this:

    <z>=<[tex]\Phi[/tex]1|z|[tex]\Phi[/tex]2>

    (for say hydrogen wavefunctions). Similarly

    <x+iy>=<[tex]\Phi[/tex]1|x+iy|[tex]\Phi[/tex]2>

    FYI, I'm trying to understand how radiation is polarised (an external B field polarises radiation, so we must consider the dipole transition matrix thus:

    <r>=<[tex]\Phi[/tex]1|r|[tex]\Phi[/tex]2>

    so I am simply resolving 'r' into two components (in the xy plane and z axis).
     
  2. jcsd
  3. May 6, 2008 #2
    The only hard part in the calculation is knowing [tex]\left\langle r \right\rangle[/tex],
    from which
    [tex]x = r \sin\theta \cos\varphi[/tex]
    [tex]y = r \sin\theta \sin\varphi[/tex].
    Can you think of a spherical harmonic equal to [tex]\sin\theta e^{i\varphi}[/tex]? Then you need to know how to compute [tex]\left\langle Y^\ell_m\right\rangle[/tex], which is easy enough for simple cases, but if you want you can consult a table of 3-j symbols.

    That should be all that you need.

    Good luck.
     
  4. May 7, 2008 #3
    aah, that's clever. thanks Ibrits.

    for hydrogen like wavefunctions, x+iy=[tex]\sin\theta e^{i\varphi}[/tex]=Y(l=1,m=1) so yeah its just the expectation of that.

    what about the expectation of z?
     
  5. May 7, 2008 #4
    I assume you know the representation of [tex]z[/tex] in spherical coordinates. I also assume you have a table of spherical harmonics handy. It shouldn't be hard to figure the rest out =)
     
  6. May 7, 2008 #5
    oh i see, i was being really stupid (as per usual)... z=rcos(theta), so its just <r><Y1,0>. thanks:)
     
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