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Eigenstates of S^2

  1. Apr 21, 2007 #1
    The problem statement, all variables and given/known data

    I need to show: [tex]S^2 \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)[/tex] has an eigenvalue of zero.

    The attempt at a solution

    [tex]S_1^2 \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)=\hbar^2 \frac{3}{4} \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right) [/tex]

    [tex]S_2^2 \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)=\hbar^2 \frac{3}{4} \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)[/tex]

    [tex]2S_{1z}S_{2z} \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)=-\hbar^2 \frac{1}{2} \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right) [/tex]

    [tex]\left( S_{1+}S_{2+}+S_{1-}S_{2-} \right) \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)=0[/tex]

    As you can see, these values do not add up to zero. [tex]\hbar^2 \frac{3}{4} + \hbar^2 \frac{3}{4} -\hbar^2 \frac{1}{2} =2 \hbar^2 [/tex]

    I am not sure where my calculations went off, if you would like to see more work, please ask.
     
  2. jcsd
  3. Apr 21, 2007 #2

    Meir Achuz

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    S_1x S_2x should equal S_1z S_2z.
    Do it more carefully.
     
  4. Nov 23, 2010 #3
    Your result is wrong because:

    [tex]
    \left( S_{1x}S_{2x}+S_{1y}S_{2y} \right) = \left( S_{1+}S_{2-}+S_{1-}S_{2+} \right)
    [/tex]

    so that
    [tex]
    \left( S_{1+}S_{2-}+S_{1-}S_{2+} \right) \left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)= -\left( \frac{1}{\sqrt{2}} \left[ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right] \right)
    [/tex]

    You also summed in the wrong way all the eigenvalues. You have:
    [tex]
    \hbar^2 \frac{3}{4} + \hbar^2 \frac{3}{4} -\hbar^2 \frac{1}{2} =1 \hbar^2
    [/tex]

    If you sum the new -1 value got from the correct representation of S operators you get 0.


    (Sorry for the top up, just found a wrong answer and felt like to provide the correct solution :))
     
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