I am familiar with the expression for spin 1/2 but haven't seen one for spin 1.
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1$\begingroup$ Could you explain what you mean exactly? What is the expression you're familiar with for spin 1/2? $\endgroup$– fqqJan 17, 2014 at 16:13
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$\begingroup$ Using the same method as this uvm.edu/~kspartal/EPR_Simulation/AboutSpins.pdf which is for the spin 1/2 particle and apply it to the spin 1 particle $\endgroup$– user32462Jan 19, 2014 at 17:21
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$\begingroup$ If you plug the spin-1 representation for $S_x$, $S_y$ and $S_z$ (which I posted below) into the very first equation of this pdf (i.e. into the definition of $S_r$) you can go through the steps of this document and obtain the respective expressions for spin-1. The difference being, that you are now dealing with $3\times 3$-matrices instead. $\endgroup$– AndréJan 26, 2014 at 11:03
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$\begingroup$ Already done it. The problem just came when normalising my eigenvectors using the mathematica function I had some degeneracies but calculating it by hand sorted it out. $\endgroup$– user32462Jan 26, 2014 at 13:31
1 Answer
I'm not entirely sure if I'm getting your question correctly, but the general spin operator is given by \begin{equation} \widehat{S} = a\cdot \widehat{S}_x + b\cdot \widehat{S}_y + c\cdot \widehat{S}_z \end{equation} where $a,b,c \in \mathbb{R}$ such that $\vec{n} \equiv (a\; b\; c)$ is the direction in which you measure the spin. Note that so far, no particular spin (1/2, 1, 3/2 etc.) has been specified. As you will probably know, the operators $\widehat{S}_x, \widehat{S}_y$ and $\widehat{S}_z$ are required to satisfy a very specific algebraic relation, namely \begin{equation} [\widehat{S}_\alpha, \widehat{S}_\beta] = i\hbar\ \varepsilon_{\alpha\beta}^{~~~~~\gamma}\ \widehat{S}_\gamma \qquad \forall\, \alpha,\beta,\gamma \in \{x,y,z\}\ . \end{equation} Now, to put it in "simple words", you go looking for matrices that satisfy this relation (i.e. representations of the algebra). If you look for $2\times 2$-matrices you may find that \begin{equation} \widehat{S}_x \stackrel{\cdot}{=} \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix},\ \widehat{S}_y \stackrel{\cdot}{=} \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix},\ \widehat{S}_z \stackrel{\cdot}{=} \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}\ \end{equation} will do the trick (recognize the Pauli matrices). This representation by $2\times 2$ matrices is called a spin-1/2 representation. Now, you may go on and look for a representation by $3\times 3$ matrices and find \begin{equation} \widehat{S}_x \stackrel{\cdot}{=} \frac{\hbar}{\sqrt{2}}\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\, ,\ \widehat{S}_y \stackrel{\cdot}{=} \frac{\hbar}{\sqrt{2}}\begin{pmatrix}0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix}\, ,\ \widehat{S}_z \stackrel{\cdot}{=} \hbar\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}\ . \end{equation} This is a spin-1 representation. Note that conventionally the basis is chosen in such a way that $\widehat{S}_z$ is a diagonal matrix and the entries on its main diagonal correspond to the possible outcomes of your measurements, i.e. ($+\frac{\hbar}{2}$, $-\frac{\hbar}{2}$) for the spin-1/2 case and ($+\hbar$, $0$, $-\hbar$) for the spin-1 case.
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3$\begingroup$ This is fine for non-elementary particles, as well as for atoms with angular momentum $L$, but beware of photons, which don't have a rest frame, and therefore only helicity. $\endgroup$ Jan 17, 2014 at 17:03
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$\begingroup$ That is absolutely true. As I said, I'm not sure if I'm getting the question correctly. $\endgroup$– AndréJan 17, 2014 at 17:09
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$\begingroup$ My best interpretation of the question is the literal one, so I would say a Stern-Gerlach apparatus, heh $\endgroup$ Jan 18, 2014 at 11:29