Eigenstates of S^2: Calculating Eigenvalue for a Quantum System

[PhysicsForums]

Homework Statement

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

The attempt at a solution

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)

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)

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)

\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

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

I am not sure where my calculations went off, if you would like to see more work, please ask.

 

[Meir Achuz]

S_1x S_2x should equal S_1z S_2z.
Do it more carefully.

 

[Dollarius]

Your result is wrong because:

<br /> \left( S_{1x}S_{2x}+S_{1y}S_{2y} \right) = \left( S_{1+}S_{2-}+S_{1-}S_{2+} \right)<br />

so that
<br /> \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)<br />

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

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