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$\newcommand{\ket}[1]{| #1 \rangle}$

The general spin formulation is the same as the cinetic momentum :

\begin{equation} [S_{x},S_{y}]=i\hbar S_{z},\quad [S_{y},S_{z}]=i\hbar S_{x},\quad [S_{z}, S_{x}]=i\hbar S_{y} \end{equation} with other spin operators : \begin{equation} S^{2} = S_{x}^{2}+S_{y}^{2}+ S_{z}^{2}\quad,\quad S_{+} = S_{x} + i S_{y}\quad,\quad S_{-} = S_{x} - i S_{y} \end{equation} They do not commute except : \begin{equation} [S^{2}, S_{z}] = 0 \end{equation} which means one can find an orthonormal basis of the Hilbert space out of common eigenvectors : \begin{equation} S^{2}\ket{sm} = \hbar^{2}s(s+1)\ket{sm},\quad S_{z}\ket{sm} = \hbar m\ket{sm} \end{equation} and the action of $S_{+}$ and $S_{-}$ operators on $\ket{sm}$ eigenvalues are : \begin{equation} S_{+}\ket{sm} = \hbar\sqrt{(s-m)(s+m+1)}\ket{s(m+1)}\\ S_{-}\ket{sm} = \hbar\sqrt{(s+m)(s-m+1)}\ket{s(m-1)} \end{equation} Vector normalization imply $-s\le m\le s$.

So for spin 1 particle like photons, there are three values possible for $m\in \{-1, 0, 1\}$, meaning that the dimension of the Hilbert space is $3$.

Then, if I order eigenvectors as follows (for $s=1$) :

m $-1$ $0$ $+1$
$\ket{sm}$ $\ket{-1}$ $\ket{0}$ $\ket{1} $

I can write the matrix representation of previous spin operators in the basis $\{\ket{-1},\ket{0},\ket{1}\}$ :

\begin{equation} S^{2} = 2\hbar^{2} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix},\quad S_{z} = \hbar \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \end{equation}

then \begin{equation} S_{ + } = \hbar \sqrt{2} \begin{pmatrix} 0 &0 & 0 \\ 1 & 0 &0 \\ 0 & 1 & 0 \\ \end{pmatrix},\quad S_{-} = \hbar \sqrt{2} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \end{equation} and finally \begin{equation} S_{x} = \frac{S_{ + } + S_{ - }}{2} = \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 &1 & 0 \\ 1 & 0 &1 \\ 0 & 1 & 0 \\ \end{pmatrix},\quad S_{y} = \frac{S_{ + } - S_{ - }}{2i} = \frac{i\hbar}{\sqrt{2}} \begin{pmatrix} 0 &1 & 0 \\ -1 & 0 &1 \\ 0 & -1 & 0 \\ \end{pmatrix} \end{equation}

The point is that photon spin cannot have the state $s=0$. Then, the matrix operation of the photon spin operator should be of dimension $2\times 2$. Is it possible to reflect in previous $3\times 3$ operators the fact that one state is forbidden and come to a $2\times 2$ representation ?

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    $\begingroup$ The physical problem is that the analysis you're showing here implicitly assumes that the particle is at rest, but a photon has no rest frame. More technically, to describe the spin states of a particle, one needs to look at representations the "little group" of the Poincaire group, or the subgroup that leaves a particle's momentum invariant -- massive and massless particles have different little groups, and you're studying the little group for a massive particle, while the photon is massless. A proper description of this topic can be found, eg, in Vol 1, Chapter 2 of Weinberg's QFT textbook. $\endgroup$
    – Andrew
    Mar 20 at 12:11
  • $\begingroup$ Thanks for the reference, it seems it is needed to tackle quantum field theory to fully understand the photon spin and the spin in general. But it is complex notations and I quickly lose myself, I would interested if someone knows a friendly introduction to this $\endgroup$
    – deb2014
    Mar 20 at 13:33
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    $\begingroup$ Yes, unfortunately to rigorously understand photon spin, you do need a pretty solid grasp of group theory, special relativity, and non-relativistic quantum mechanics. $\endgroup$
    – Andrew
    Mar 20 at 13:55
  • $\begingroup$ @deb2014 the only friendly introduction to QFT is no introduction to QFT, for most students of physics. $\endgroup$
    – JEB
    Mar 21 at 1:09

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