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Constructing spatial wave function of hydrogen

  1. Mar 25, 2006 #1
    I'm having problem with griffith QM problem 4.43:
    Construct the spatial wave function for hydrogen in the state n = 3, l =2, m = 1. Express your answer as a function of r, [tex]\theta[/tex], [tex]\phi[/tex], and a (the Bohr radius) only.

    My prof. gave hints about radial wave function, but I have no idea how to construct spherical harmonics from scratch. The book only shows spherical harmonic are eigenfunctions of [tex]L^2[/tex] and [tex]L_z[/tex], not how you can generate them with [tex]L_{\pm}[/tex] operators. So is there a way to generate spherical harmonics with operators similar to harmonic oscillator operator method? I guess more specifically, is there a way to get spherical harmonics for [tex]Y_{l=2}^{m=0}[/tex]? I know I can apply [tex]L_{+}[/tex] to get [tex]Y_{l=2}^{m=1}[/tex].

    edit: nevermind, looks like Griffith made generating spherical harmonics into problem 4.22 as oppose to showing how to do it. that was very tricky of my professor to sneak another problem into homework.

    edit2: i'm having problems with 4.22...
    denoting spherical harmonics as [tex]Y_{l}^{m}(\theta,\phi)=f(\theta)g(\phi)[/tex], solve for [tex]g(\phi)[/tex] first.
    [tex]L_{z}Y_{l}^{l}=\hbar l Y_{l}^{l}[/tex]
    [tex]g(\phi)=e^{i l \phi}[/tex] as expected.

    now for [tex]f(\theta)[/tex]
    where [tex]L_{+}=\hbar e^{i \phi}(\partial_{\theta}+i \cot \theta \partial_{\phi})[/tex]
    solving this I get
    [tex]\ln f(\theta) = \frac{l}{\hbar}e^{i \phi} \ln \sin \theta[/tex]
    which doesn't look right... it's an exponential of an exponetial... any clues as to what I did wrong?
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 26, 2006 #2


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    I don't know why you could not simply use Table 4.3 or Equation 4.32 (I am assuming you have the second edition)

    You should not have any exponential of [itex] \phi[/itex] left! You get an equation of the form [itex]L_+ Y_l^l = e^{i(l+1)\phi} (\partial_\theta - l cot \theta) f(\theta) = 0 [/itex]. So you divide by the exponential in phi which goes away and then solve for f(theta).

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