# How do I know that gauge fields are bosons?

QED and the Dirac equation have field operators $\psi$ interact with a gauge field $A^{\mu}$.

We identify $\psi$ as a fermionic field and $A^{\mu}$ as a gauge boson - the photon.

Do we or can we know that one is a fermion and the other is a boson?

Or do we get that information from the commutation relations when we quantise the theory?

By the spin-statistics theorem, half integer spin is associated with fermions whereas integer spin is associated with bosons. Gauge fields transform as vectors under Lorentz transformations therefore they have spin one and are bosons.

• This is wrong. The Rarita-Schwinger field is a fermion, and it is a gauge field. More generally, gauge fields do not (necessarily) live in the vector representation of Lorentz. – AccidentalFourierTransform Sep 11 '18 at 19:03
• @AccidentalFourierTransform Alright. Then we should properly define gauge fields. If by gauge field we mean the connection of local symmetries of the Standard Model, then my answer is correct. I know this is a bit restricted definition but it is the one best understood when physicists talk about gauge fields. – Diracology Sep 11 '18 at 20:29
• Gauge field = local section of a fibre bundle (in general, a supervector bundle). The definition is unambiguous, and it does not imply it transforms as a Lorentz vector. You are being too simplistic/reductive: the SM is only a tiny subset of the general setting of gauge theories, and it is most certainly not "the one best understood when physicists talk about gauge fields": tell that to someone working on supergravity/string theory! – AccidentalFourierTransform Sep 11 '18 at 20:42
• @AccidentalFourierTransform I feel "local section of a fibre bundle" is way too permissive. If so, basically ALL fields in physics are gauge fields. – Bence Racskó Sep 11 '18 at 20:44
• @Uldreth Well, in that case, the issue is but a matter of terminology. I would call "gauge field" to any field that is affected by gauge transformations. It seems that what OP and Diracology have in mind is a connection, not (what I call) the gauge field. Still, there is no fundamental reason connections should be Lorentz vectors; it just so happens that in the Standard Model, they are. Oh well, I guess I should have been more clear about what I mean by "gauge field", thank you for pointing it out. – AccidentalFourierTransform Sep 11 '18 at 20:51

The possible Lorentz invariant wave equations is a relatively limited set of equations. Most wave equations one can write will not be Lorentz invariant under any transformation law you can think of.

We know all the possible finite dim Lorentz invariant field equations. One of the properties of these equations is that each of them have a distinct number associated with them that relates to their angular momentum. We need some representation theory to exactly explain what I mean by this, but trust me that there is a number. This number represents how rotations effect the equations, and it is in fact the "Equation's spin". When quantised, this is the particle's spin.

So once you see a particle's wave equation, you can always tell the spin of the particle. From this point you can look at the gauge field's differential equation and immediately tell what is the gauge particle's spin. You can also ask the reverse question - what are the possible wave equations of a spin 7/2 particle?

The main point is, that the possible Lorentz invariant field equations is such a limited set, that we can tell from the wave equations themselves everything there is to know about the particles they represent. If the particle violates these rules, it cannot be Lorentz Invariant.

The kinetic term of the gauge fields, i.e. $$F_{\mu \nu}F^{\mu \nu}$$, is only Lorentz invariant if it has spin 1. You can write different Lagrangians with different kinetic terms, and have gauge fields with different spins.

(The technical name for the "equation's spin" is the total angular momentum of the representation of the Lorentz group of the field)