# Interpretation of the electromagnetic field strength tensor as a spin-1 field

If I understand correctly the electromagnetic field strength tensor $F_{\mu\nu}$ could be considered as a spin-1 field. In that case, what can one say about the total spin and the $z$-component of the spin for this field? Also, how is $F_{\mu\nu}$'s spin related to that of the photon field ($A_{\mu}$)?

The spin of the electromagnetic field tensor $F_{\mu\nu}$ is best understood by writing it as a spinor. A spin 1 field is a represented by a symmetric spinor $\xi^{AB}$ or by a dotted symmetric spinor $\eta_{\dot{A}\dot{B}}$. In order to get the field transforming correctly under parity, the electromagnetic field has to be a direct sum using the symmetric spinor and it's complex conjugate dotted spinor. \begin{equation} F_{\mu\nu} \sim \xi^{AB}\oplus [\xi^{*}]_{\dot{A}\dot{B}} \end{equation} The symmetric spinor $\xi^{AB}$ has three independent complex components $\xi^{11},\xi^{12}=\xi^{21},\xi^{22}$. Linear combinations of these components correspond to the three complex components of the electromagnetic field $B^{r}+iE^{r}$. The source free Maxwell equations are obtained by acting on the symmetric spinor with the Hermitian momentum operator $\hat{p}^{\dot{A}}_{\ B}$ \begin{equation} \hat{p}^{\dot{A}}_{\ B}\xi^{BC}=0 \end{equation} This equation is similar to the Dirac equation for a massive spin 1/2 field. The photon is massless, so it has helicity = $\pm 1$ instead of spin (essentially spin 1 along or against the direction of flight). The photon has two helicity degrees of freedom, but the spin 1 field $F_{\mu\nu} \sim \xi^{AB}\oplus [\xi^{*}]_{\dot{A}\dot{B}}$ has six real components. The Maxwell equations $\hat{p}^{\dot{A}}_{\ B}\xi^{BC}=0$ project the spinor onto a two-dimensional subspace.
This is as far as I know how to answer the question at present. The gauge field $A^{\mu}$ is a four vector so it ought to be a Hermitian spinor field of type $X^{\dot{A}}_{\ B}$ which is the tensor product of two spin 1/2 fields. It has four components, so the gauge fixing must come in to reduce four to the two helicity degrees of freedom.