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Eigenfunction of Angular Momentum Squared Operator

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    The square of the angular momentum operator is (in SPC);

    http://img6.imageshack.us/img6/67/54712598.png [Broken]

    Show that Y([itex]\theta[/itex],[itex]\phi[/itex]) = C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]

    Is an eigenfunction. C is a constant.

    2. Relevant equations



    3. The attempt at a solution

    I am not able to get it in the form of (A number) times Y([itex]\theta[/itex],[itex]\phi[/itex]) = C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]. No matter what, I always end up with;

    -2C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex] - 4C[itex]e^{2i\phi}[/itex] - and I don't know what to do with that.

    This is clearly not what I want, there is an extra term that I just can't seem to eradicate. What am I doing wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 29, 2011 #2

    fzero

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    You might want to show some of the earlier steps. You've made a mistake combining the first 2 terms.
     
  4. Nov 29, 2011 #3
    Hmm, well with

    [itex]\frac{\delta^{2}}{\delta\theta^{2}}[/itex] of Y([itex]\theta[/itex],[itex]\phi[/itex]) I get

    (having taken the constant C out)

    [itex]-2e^{2i\phi}[/itex]([itex]sin^{2} - cos^{2}[/itex])


    With [itex]cot\theta \frac{\delta}{\delta\theta}[/itex]

    I get

    [itex]-2e^{2i\phi}[/itex][itex]\frac{cos\theta}{sin\theta} sin\theta cos\theta[/itex] = [itex]-2e^{2i\phi}[/itex][itex]cos^{2}\theta[/itex]

    and lastly with

    [itex]\frac{1}{sin^{2}\theta}\frac{\delta^{2}}{\delta {\phi}^{2}}[/itex]

    I get [itex]-4e^{2i\phi}[/itex]

    Adding all the terms gives me;

    [itex]-2e^{2i\phi}[/itex][itex]sin^{2} \theta[/itex] [itex]-4e^{2i\phi}[/itex]

    Have I done something wrong?
     
  5. Nov 29, 2011 #4

    fzero

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    You've factored out the [itex]-\hbar^2[/itex] so there's no minus sign in front of this term.
     
  6. Nov 29, 2011 #5
    [itex]\frac{\delta}{\delta\theta}[/itex] of

    [itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]

    is

    [itex]2sin\theta(-cos\theta)e^{2i\phi}[/itex] = [itex]-2sin\theta cos\theta e^{2i\phi}[/itex]

    surely?

    So times by [itex] cot\theta [/itex]

    is

    [itex]-2 cos^{2} \theta e^{2i\phi}[/itex]

    ?
     
  7. Nov 29, 2011 #6

    fzero

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    No,

    [itex] \frac{d}{d\theta}\sin\theta = \cos\theta,~~~\frac{d}{d\theta}\cos\theta = -\sin\theta.[/itex]

    You'll want to doublecheck the first term as well then, but you had the overall sign correct there.
     
  8. Nov 29, 2011 #7
    Oh my good lord.

    This is just embarrassing. It's not like I'm even tired or anything, I just continuously made an elementary mistake.

    Thanks. I'll be handing in my University resignation form tomorrow ._.
     
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