Spin Operator algebra

Question

I am trying to teach myself some QM.

In Christopher J. Cramers textbook Essentials of Computational Chemistry: theory and models, in Appendix C, he goes over Spin algebra.

I am unable to calculate myself that $S^2\alpha = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \alpha$ as well as that $S^2\beta = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \beta$

Where $S^2$ is the spin squared operator and $S = S_x + S_y + S_z$.

This seems like it should be trivial. It certainly is to him, he says that repeated application of the following equations lead to the above

$S_x\alpha = \frac{1}{2}\hbar\beta$

$S_x\beta= \frac{1}{2}\hbar\alpha $

$S_y\alpha = \frac{1}{2}i\hbar\beta$

$S_y\beta = -\frac{1}{2}i\hbar\alpha$

$S_z\alpha = \frac{1}{2}\hbar\alpha$

$S_z\beta = -\frac{1}{2}\hbar\beta$

my understanding is that $S^2=S_x^2 + S_y^2 + S_z^2$. However, when using this operator on $\alpha$ or $\beta$ I am not sure what the eigenvalues would be..., relatedly, I am not sure even how to operate just $S_x^2$ on $\alpha$.

Answer 1

One possible way:

As you say $${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$ Thus \begin{align} {\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2 \left[\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \right] \\ &= \left( \frac{\hbar}{2}\right)^2\left[\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}+\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}+\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \right]\\ &= \left(\frac{3}{4}\right)\hbar^2 \end{align}

Answer 2

Solution

$S^2\alpha = \left(S_x^2 + S_y^2 +S_z^2 \right)\alpha= S_x(S_x(\alpha)) + S_y(S_y(\alpha)) +S_z(S_z(\alpha)) $

from the ''rules", where for instance $S_x$ operating on $\alpha$ gives $\frac{1}{2}\hbar\beta$, we make the three replacements to get

$=S_x(\frac{1}{2}\hbar\beta) +S_y(\frac{1}{2}i\hbar\beta) +S_z(\frac{1}{2}\hbar\alpha) $

Now take the constants outside so that we have

$=\frac{1}{2}\hbar S_x\beta +\frac{1}{2}i\hbar S_y\beta +\frac{1}{2}\hbar S_z\alpha $

We have our familiar Operator_$\phi$ where $\phi = x,y,z$ and we can follow the rules above for how each operator deals with the respective eigenfunction.

This leads to

$=\frac{1}{2}\hbar (\frac{1}{2}\hbar \alpha) +\frac{1}{2}i\hbar (-\frac{1}{2}i\hbar \alpha) +\frac{1}{2}\hbar (\frac{1}{2}\hbar \alpha) $

This can be simplified to (remembering that $i = \sqrt{-1}$ so $ii = -1$,

$=\frac{1}{2}\hbar^2 \alpha +\frac{1}{2}\frac{1}{2}\hbar^2\alpha $

and finally

$S^2\alpha = \frac{1}{2} \left(\frac{1}{2}+1 \right)\hbar^2\alpha$

So, it is indeed trivial once you see it in the correct light.