AHolico
Mar1-04, 12:16 AM
Let \mathbf{S} = \mbox{$\frac {1}{2}$}(\sigma_1 + \sigma_2) be the total spin of a system of two spin-(1/2) particles.
a) Show that \mathbf{P} \equiv (\mathbf{S} \cdot \mathbf{r})^2 / {r^2} is a projection operator
b) Show that tensor operator S_1_2 = 2(3P - \mathbf{S}^2) satisfies S_1_2 ^2 = 4\mathbf{S}^2 - 2S_1_2
c) Show that the eigenvalues of S_1_2 are 0, 2 and -4
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I'm stuck at the first part, and have no idea on how to proceed. I know that if P is a projection operator, then P^2 = P and P^+ = P. So, the first thing I do is expand P to check these properties. First thing I did is getting that r^2 inside of the dot product, leaving me with \mathbf{P} \equiv (\mathbf{S} \cdot \mathbf{n})^2, where n is an unitary vector.
\mathbf{S} = \frac{1}{2} (\sigma_1 + \sigma_2) = \frac{1}{2}\left[ \left(\begin{array}{cc}0 & 1\\1 & 0\end{array}\right) + \left(\begin{array}{cc}0 & -i\\i & 0\end{array}\right)\right] = \frac{1}{2}\left(\begin{array}{cc}0 & 1-i\\1+i & 0\end{array}\right)
\mathbf{n} = \left(\begin{array}{c} cos \theta \\ sin \theta \end{array}\right)
\mathbf{S} \cdot \mathbf{n} = \frac{1}{2}\left(\begin{array}{cc}0 & 1-i\\1+i & 0\end{array}\right)\left(\begin{array}{c} cos \theta \\ sin \theta \end{array}\right) = \frac{1}{2}\left(\begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right)
Now, to square it... should I (a) just multiply that vector by itself, or (b) multiply it complex conjungate and the vector???
a)\frac {1}{4}\left(sin \theta - i sin \theta , cos \theta + i cos \theta \right) \left( \begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right) = \frac{i}{2}\left(\begin{array}{c} -sin \theta \\ cos \theta \end{array}\right)
I thiunk this is not a proyection operator because P^+ = P.
b)\frac {1}{4}\left(sin \theta + i sin \theta , cos \theta - i cos \theta \right) \left( \begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right) = \frac{1}{2}\left(\begin{array}{c} sin^2 \theta \\ cos^2 \theta \end{array}\right)
But clearly this is not a proyection operator because P^2 = P
In some book I found that \vec{\sigma} \cdot \vec{n} = \sigma_1 sin \theta cos \phi + \sigma_2 sin \theta sin \phi + \sigma_3 cos \theta
If I follow this aproach then
\mathbf{S} \cdot \mathbf{n} = \frac {1}{2} \sigma_1 cos \theta + \frac {1}{2} \sigma_2 sin \theta
(\mathbf{S} \cdot \mathbf{n})^2 = \frac {1}{4} \sigma_1^2 cos^2 \theta + \frac {1}{4} \sigma_2^2 sin^2 \theta + \frac {1}{4} \sigma_1 \sigma_2 sin \theta cos \theta + \frac {1}{4} \sigma_2 \sigma_1 sin \theta cos \theta
(\mathbf{S} \cdot \mathbf{n})^2 = \frac {1}{4} \mathbb{I}
The identity matrix is a projection operator, but that constant in front of it is giving me problems. What I'm doing wrong?
a) Show that \mathbf{P} \equiv (\mathbf{S} \cdot \mathbf{r})^2 / {r^2} is a projection operator
b) Show that tensor operator S_1_2 = 2(3P - \mathbf{S}^2) satisfies S_1_2 ^2 = 4\mathbf{S}^2 - 2S_1_2
c) Show that the eigenvalues of S_1_2 are 0, 2 and -4
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I'm stuck at the first part, and have no idea on how to proceed. I know that if P is a projection operator, then P^2 = P and P^+ = P. So, the first thing I do is expand P to check these properties. First thing I did is getting that r^2 inside of the dot product, leaving me with \mathbf{P} \equiv (\mathbf{S} \cdot \mathbf{n})^2, where n is an unitary vector.
\mathbf{S} = \frac{1}{2} (\sigma_1 + \sigma_2) = \frac{1}{2}\left[ \left(\begin{array}{cc}0 & 1\\1 & 0\end{array}\right) + \left(\begin{array}{cc}0 & -i\\i & 0\end{array}\right)\right] = \frac{1}{2}\left(\begin{array}{cc}0 & 1-i\\1+i & 0\end{array}\right)
\mathbf{n} = \left(\begin{array}{c} cos \theta \\ sin \theta \end{array}\right)
\mathbf{S} \cdot \mathbf{n} = \frac{1}{2}\left(\begin{array}{cc}0 & 1-i\\1+i & 0\end{array}\right)\left(\begin{array}{c} cos \theta \\ sin \theta \end{array}\right) = \frac{1}{2}\left(\begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right)
Now, to square it... should I (a) just multiply that vector by itself, or (b) multiply it complex conjungate and the vector???
a)\frac {1}{4}\left(sin \theta - i sin \theta , cos \theta + i cos \theta \right) \left( \begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right) = \frac{i}{2}\left(\begin{array}{c} -sin \theta \\ cos \theta \end{array}\right)
I thiunk this is not a proyection operator because P^+ = P.
b)\frac {1}{4}\left(sin \theta + i sin \theta , cos \theta - i cos \theta \right) \left( \begin{array}{c} sin \theta - i sin \theta\\ cos \theta + i cos \theta \end{array}\right) = \frac{1}{2}\left(\begin{array}{c} sin^2 \theta \\ cos^2 \theta \end{array}\right)
But clearly this is not a proyection operator because P^2 = P
In some book I found that \vec{\sigma} \cdot \vec{n} = \sigma_1 sin \theta cos \phi + \sigma_2 sin \theta sin \phi + \sigma_3 cos \theta
If I follow this aproach then
\mathbf{S} \cdot \mathbf{n} = \frac {1}{2} \sigma_1 cos \theta + \frac {1}{2} \sigma_2 sin \theta
(\mathbf{S} \cdot \mathbf{n})^2 = \frac {1}{4} \sigma_1^2 cos^2 \theta + \frac {1}{4} \sigma_2^2 sin^2 \theta + \frac {1}{4} \sigma_1 \sigma_2 sin \theta cos \theta + \frac {1}{4} \sigma_2 \sigma_1 sin \theta cos \theta
(\mathbf{S} \cdot \mathbf{n})^2 = \frac {1}{4} \mathbb{I}
The identity matrix is a projection operator, but that constant in front of it is giving me problems. What I'm doing wrong?