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 $\alpha$.

  • 2
    $\begingroup$ The operators work one after the other just as if you are differentiating and the order of doing this matters, so if $S_x^2$, say, operates on $\alpha$ then $S^2_x\alpha\equiv S_x(S_x\alpha) = S_x(\hbar/2) \beta=(\hbar/2)^2\alpha$. The $S^2$ operator has 9 terms as $S_xS_y$ is not necessarily the same as $S_yS_x$. $\endgroup$
    – porphyrin
    Commented Mar 13, 2020 at 8:23
  • $\begingroup$ I have to go back a step. Particle in a box has mathematical eigenfunctions that can clearly be differentiated. I am not seeing what the mathematical form for $\alpha$ is. Hence, I can't differentiate anything. I am missing a pre-requisite piece of info $\endgroup$
    – B. Kelly
    Commented Mar 13, 2020 at 14:09
  • $\begingroup$ $S$ is not equal to $S_x+S_y+S_z$, but rather $(S_x,S_y,S_z)$ or a vector of the spin in each direction. $S^2$ is actually the dot product of this vector with itself, which is why you get $S^2=S_x^2+S_y^2+S_z^2$. Generally, we don't worry about the mathematical form of $|\alpha\rangle$ or $|\beta\rangle$ and just treat the equations you listed as rules for how they behave. So porphyrin is right that you should just apply your operators one after another and follow the rules you have listed. $\endgroup$
    – Tyberius
    Commented Mar 13, 2020 at 14:33
  • $\begingroup$ I answered it myself. It is trivial, I was starting on the wrong foot when I looked at it. $\endgroup$
    – B. Kelly
    Commented Mar 13, 2020 at 16:19
  • $\begingroup$ I would put your solution as a self answer rather than editing it into the question. $\endgroup$
    – Tyberius
    Commented Mar 13, 2020 at 16:50

2 Answers 2


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}

  • 1
    $\begingroup$ This is correct, but doesn't answer the question. I appreciate the effort though. $\endgroup$
    – B. Kelly
    Commented Mar 13, 2020 at 16:22


$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.


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