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Proving an eigenfunction (EF) from Lx as combination of EF of Lz

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the eigenfunction of Lx can be written as a combination of eigenfunctions from Lz with the same l but different m. Using the eigenfunction
    [itex]Y_{x} = \frac{1}{\sqrt{2}}(Y^{-1}_{1} - Y^{1}_{1})[/itex] as the eigenfunction of Lx

    2. Relevant equations

    [itex]Y^{m}_{l} = \frac{1}{\sqrt{2 \pi}}e^{im\varphi}[/itex] eigenfunction of Lz
    [itex]L_{z}Y^{m}_{l}= l Y^{m}_{l} [/itex]

    3. The attempt at a solution
    [itex]Y_{x} = \frac{1}{\sqrt{2}}(Y^{-1}_{1} - Y^{1}_{1}) = \frac{1}{\sqrt{4\pi}}(e^{-i\varphi} - e^{i\varphi}) [/itex]

    [itex]\widehat{L}_{x} = i \frac{h}{2\pi}(sin\theta \frac{d}{d\varphi} + cot\theta cos\varphi\frac{d}{d\varphi}) [/itex]

    [itex]\widehat{L}_{x}Y_{x} = i \frac{h}{2\pi}cot\theta cos\varphi\frac{d}{d\varphi}\frac{1}{\sqrt{4\pi}}(e^{-i\varphi} - e^{i\varphi}) [/itex]

    [itex]\widehat{L}_{x}Y_{x} = \frac{h}{2\pi}cot\theta cos\varphi\frac{1}{\sqrt{4\pi}}(e^{-i\varphi} + e^{i\varphi}) ≠ lY_{x} [/itex]

    can't see where I'm going wrong
     
  2. jcsd
  3. Apr 26, 2012 #2

    vela

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    The line above is your problem. ##e^{i m \varphi}## is an eigenfunction of Lz, but the spherical harmonic ##Y^m_l## is an eigenfunction of both Lz and L2.

     
  4. Apr 26, 2012 #3
    Hi, should I be using

    [itex]Y^{m}_{l} \propto P^{m}_{l}(cos\theta)e^{i m \varphi}[/itex] ?
     
  5. Apr 26, 2012 #4

    vela

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    Yes.
     
  6. Apr 26, 2012 #5
    just to confirm is

    [itex]P^{-1}_{1}(cos\theta) = cos\theta[/itex]

    thanks!
     
  7. Apr 26, 2012 #6

    vela

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    No, that's not correct. Don't you have a list of spherical harmonics in your textbook? If not, just google it.
     
  8. Apr 26, 2012 #7
    thank you no I wasn't aware of spherical harmonics, but i am now!
     
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