# Gauge fermions versus gauge bosons

Why are all the interactions particle of a gauge theory bosons. Are fermionic gauge particle fields somehow forbidden by the theory ?

• Comment on question (v1): isn't the phrase "fermionic gauge boson" a bit oxymoronic? – joshphysics Feb 24 '14 at 5:51
• Well that is the point of my question: why is it a oxymoron? – Anne O'Nyme Feb 24 '14 at 5:55
• Josh is saying that "fermionic gauge particle" would be a more applicable name. To answer your question: this is disallowed by local quantum field theory. Look up the spin-statistics theorem in Weinberg or Srednicki. – Matthew Feb 24 '14 at 5:59
• fermionic boson is like married bachelor. – Alfred Centauri Feb 24 '14 at 14:09

The reason that the gauge particle must be a spin 1 gauge boson is because there aren't any renormalizable alternatives. To see this consider the Dirac Lagrangian:

$$\bar{\psi} i \gamma ^\mu \partial _\mu \psi$$ This term is not gauge invariant under the transformation, $\psi \rightarrow e ^{ i T ^a \theta ^a (x) } \psi$, because of the derivative spoils the desired transformation of $\partial _\mu \psi$. To fix this we must add a contribution that transforms in the same way as the derivative, i.e., transforms as a vector. In other words we modify the derivative such that, $$D _\mu \psi \rightarrow e ^{ i T _a \theta _a (x) } D _\mu \psi$$

The question is what to add to $D _\mu$. We can potentially add spin $0, \frac{1}{2} , 1 , \frac{3}{2} ,$ and $2$ particles to fix this. We go case by case.

There is no combination of spin zero fields that transform as a vector without adding derivatives (adding derivative to fix the derivative covariance would take you in circles), thus we can't have a spin zero gauge boson.

Next consider adding a spin $1/2$ gauge boson we could write ($\psi _a$ is a gauge particle, not $\psi$), $$D _\mu = \partial _\mu + \sum _a T ^a \left( g\bar{\psi} ^a \gamma _\mu \psi ^a + g ' \bar{\psi} ^a \gamma _\mu \gamma ^5 \psi ^a \right)$$ However, this would give an interaction $$\sum _a i\left[ \bar{\psi} \gamma ^\mu\psi \right] \left[ \bar{\psi} _a \gamma _\mu \psi _a \right]$$ and similarly for the $\gamma ^5$ term. These interactions are non-renormalizable as they involve four fermions. Non-renormalizable interactions arise from effective field theories and are suppressed by the scale at which they arise. This would make the gauge interactions non-fundamental but instead involve a massive vector particle integrated out. For the integrated out interaction to be renormalizable it must be between two fermions and a spin $1$ field. This brings us back to the usual case.

The spin $1$ field works well and exists in the SM. I'm not sure about the spin $3/2$ field as I have no experience with working with such fields however, I presume it won't work for similar reasons. I also know that spin $2$ fields must mediate gravitational fields and thus would give a nonsensible result.

• Thanks for your answer. Is this answer somehow related to "local quantum field theory" (as mentioned in the comment by Matthew)? – Hunter Feb 24 '14 at 14:12
• @Hunter: I think what Matthew was referring to (which I completely omitted since I don't know too much about it), is that any particle with a Lorentz index $W_\mu$, must be a spin 1 particle. So if I understand correctly, he took my answer for granted - That any new gauge field will be of the form $W_\mu$. – JeffDror Feb 24 '14 at 14:21
• Ahh ok, well, I wasn't aware of your answer, so thanks! – Hunter Feb 24 '14 at 14:31
• Non-renormalizability is not a very strong argument (think Effective Field Theory). Though I agree that the argument "that's what works in Nature" is a good answer to "why not gauge fermion ?", it does not tell if it's in principle not possible for some good reason (unitarity ?). I'm no expert in SUSY, but isn't this an example where the local transformation is mediated by fermionic operator, implying fermionic gauge particle ? – Adam Feb 24 '14 at 14:39
• Excuse me, how to see the spin 1/2 gauge boson covariant derivative works? I mean from your first equation, we expect something like $$-\bar{\psi} \gamma^{\mu} T^a \partial_{\mu} \theta^a \psi$$ generated from a local phase transformation, how does it cancelled by the spin-$1/2$ covariant derivative? – user26143 Feb 24 '14 at 14:50

A more intuitive answer that doesn't involve (non-)renormalizability:

Any gauge theory couples particles to each other. Some particle with some spin that couples to the gauge theory (take, for instance, a spin-$1/2$ particle) must couple to itself by an identity operator, which is the singlet generator included by definition in every group theory. The identity operator doesn't change any property of the particle, as it is basically a 'do nothing'-operator, without any physical meaning. This operator doesn't change spin, meaning $\Delta m=0=m_{s}$ at all times, and it therefore must have integer spin. We don't exclude $m_s= \pm 1,\pm 2,...,$ we only state that the identity operator with $m_s=0$ must exist. Since all elements of some group theory must have the same spin, a gauge theory can not contain half-integer spin gauge particles (generators).

In this explanation I only spoke about whether the gauge particle can or cannot have half-integer spin, not about whether they can be bosons or fermions. I did this on purpose, since there is no mathematical reason as to why the generators should have bosonic statistics. In some BSM (beyond the Standard Model) theories, in fact, there exist fermionic vector (spin-$1$) particles. These vector-like fermions could be the gauge particles of some new gauge theory we don't know of yet.

Some nuances have to be made to the reasoning given here:

-If conservation of spin is not exact, the reasoning is invalid. There might be some physics beyond the standard model that doesn't conserve angular momentum, resulting in exotic phenomena.

-SUSY (super symmetric) gauge theory might provide a way out by considering some SUSY transformation which gauges away the spin problem (Wess-Zumino gauge).

I hope this helps.