# 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 ($A_{\mu}$)?

• It's the gauge field $A_{\mu}$ that has spin 1. (The gauge bosons have spin one.) The field strength tensor itself is not a gauge field! – psm Sep 24 '15 at 8:07
• Don't forget pair production or atomic orbitals where electrons "exist as standing waves". See this gif and try to envisage a wave in a tight closed path. IMHO you should think of the electron's Fµv field as a spin-½ standing-wave "spinor" configuration of the spin-1 Aµ photon wave. – John Duffield Sep 24 '15 at 13:35

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. $$F_{\mu\nu} \sim \xi^{AB}\oplus [\xi^{*}]_{\dot{A}\dot{B}}$$ 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 $\hat{p}^{\dot{A}}_{\ B}$ $$\hat{p}^{\dot{A}}_{\ B}\xi^{BC}=0$$ 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.
• Hi Stephen. Thanks for your kind response. So, it seems one can not easily visualize how the spin of $F_{\mu\nu}$ is related to that of $A_{\mu}$. However, I'm interested to know how does the spin of $F_{\mu\nu}$ couples to that of charged fermions in a magnetic interaction: $\bar\psi\sigma^{\mu\nu}F_{\mu\nu}\psi$ (where $\sigma^{\mu\nu}$ is the Lorentz generator). In the case of charged fermion and photon interaction: $\bar\psi\sigma^{\mu}A_{\mu}\psi$ the spins of the fermions add up to give the spin of the photon. Is this the case for the magnetic interaction involving $F_{\mu\nu}$? – physics_2015 Sep 26 '15 at 4:23
• @physics_2015 : $F_{\mu\nu}$ is a pure spin 1 field. $A_{\mu}$ is a tensor product of two spin 1/2 fields so it is a direct sum of a spin 0 and a spin 1. The spin 0 field must (I guess) be removed by the gauge fixing. I'm interested in these questions as well. I don't understand the way you couple the fermion field to the gauge field, I would write it $e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$. Would you explain this to me? – Stephen Blake Sep 26 '15 at 5:59
• Hi Stephen, I was using the Weyl representation notation. So I was referring to the following coupling $\chi^{\dagger}\sigma^{\mu}A_{\mu}\chi$ where $\chi$ is a Weyl spinor. – physics_2015 Sep 27 '15 at 23:26