# QM problem, operators and tensors math

1. Mar 1, 2004

### AHolico

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 $$S_1_2$$$$^2 = 4\mathbf{S}^2 - 2$$$$S_1_2$$

c) Show that the eigenvalues of $$S_1_2$$ are 0, 2 and -4

--------------------------------------------------------------------

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?

Last edited: Mar 1, 2004
2. Mar 1, 2004

### Tom Mattson

Staff Emeritus
Hold on. It appears that you are not writing P as a vector. If you are working in 2D (are you allowed to assume that?), then you should write P as:

P=(1/2)(Pxi+Pyj)
P=(1/2)[(&sigma;1x+&sigma;2x)i+(&sigma;1y+&sigma;2y)j]

But what you have done here is add the &sigma;x and &sigma;y without paying attention to the fact that they are components of a vector.

Edit to add: You also seem to be ignoring the fact that you have a 2-particle system. I think you are getting the labels "1" and "2" mixed up with the components of &sigma;, when in fact they are the labels for the spin operators of particles 1 and 2, respectively.

Last edited: Mar 1, 2004