QM problem, operators and tensors math

[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 - 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 approach 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?

 

[quantumdude]

Originally posted by AHolico
\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?

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)(P_xi+P_yj)
P=(1/2)[(σ_1x+σ_2x)i+(σ_1y+σ_2y)j]

But what you have done here is add the σ_x and σ_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 σ, when in fact they are the labels for the spin operators of particles 1 and 2, respectively.

 

[R0man]

Any help will be appreciated.

There are a few mistakes in your calculations. First, the definition of \mathbf{P} should be (\mathbf{S} \cdot \mathbf{r})^2 / r^2, not (\mathbf{S} \cdot \mathbf{n})^2 as you have written. This is because we want \mathbf{P} to be a function of the position vector \mathbf{r}, not just a unit vector \mathbf{n}.

Next, your calculation for \mathbf{S} \cdot \mathbf{n} is incorrect. Remember that \mathbf{S} is a matrix, so we can't just multiply it by a vector like \mathbf{n}. Instead, we need to use the dot product formula \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z. In this case, we have \mathbf{S} = \frac{1}{2}(\sigma_1 + \sigma_2) and \mathbf{n} = (sin \theta cos \phi, sin \theta sin \phi, cos \theta). Plugging these into the dot product formula gives

\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} sin \theta cos \phi\\ sin \theta sin \phi \end{array}\right) = \frac{1}{2}\left(\begin{array}{c} sin \theta sin \phi - i sin \theta cos \phi\\ sin \theta cos \phi + i sin \theta sin \phi \end{array}\right)

Squaring this vector gives

(\mathbf{S} \cdot \mathbf{n})^2 = \frac{1}{4}\left(sin^2 \theta sin^2 \phi + sin^2 \theta cos^2 \phi - 2sin \theta cos \theta sin \phi cos \phi \right) = \frac{1}{4}sin^2 \theta

This is not a projection operator because it is not equal to its own square.

To find the correct projection operator, we need