Why are all force particles bosons?

Question

All of the force-particles in the standard model are bosons, now my question is pretty short, namely:

Why are all force particles bosons?

This can't be a coincidence.

Answer 1

That's an interesting question, even though it might be biased by the definition of forces, and on what particles they apply. For instance, if you want to describe the force that exists between photon (even though direct photon-photon scattering has not been observed yet), it is mainly due to electron loops, so in that case the `force' is fermionic.

On a more fundamental level, all the forces are related to gauge invariances of different symmetries. These invariances are implemented via usual matrices $\psi\to U^{-1} \psi U$ where $U$ depend on the gauge field. Because these matrices have bosonic properties (they are not spinors), they are thus described by bosonic fields (and not fermions).

The question then boils down to "why are all symmetries `bosonic' ?". One answer could be : because Nature says it is. Supersymmetry tells you that you can have fermionic symmetries, that will then be described by fermions gauge field (even though I'm not expert in supersymmetry, so don't trust me too much on that point).

Answer 2

The simplest Feynman diagram for an interaction between two particles looks like a letter "H". The cross-bar is a force-carrier being exchanged. At each vertex, you have a particle either emitting or absorbing a force-carrier. If the force-carrier has a half-integer spin, then you can't emit or absorb it without violating conservation of angular momentum. For example, an electron can't emit a spin-1/2 particle, because you can't couple spin 1/2 and spin 1/2 to make spin 1/2.

Answer 3

There is a simple way to see that, without no much mathematics.

Fundamental matter particles are spin one-half fermions (neutrinos, electrons, quarks). Each particle corresponds to several degrees of freedom, say $2$. Now, let us see these 2 degrees of freedom as a complex $2$-"vector" (in fact, it is not a Lorentz vector, it is a Weyl spinor, but this is not important here).

Imagine now a $3$-vertex interaction, with $2$ matter particles, and an other particle, which we expect to be called force-particle.

We want to write this interaction $\mathcal{L}_I$ like a Lorentz-invariant quantity from these $2$ "vectors" and an other mathematical object representing the "force particle", with only one power of these quantities.

Let $ \psi_1$ and $ \psi_2$ represent matter particles, and a mathematical object $A$ representing the force particle.

Suppose $A$ is a vector, we may try $\mathcal{L}_I = \psi_1.A$, but there would be no $\psi_2$, and this would be a $2$-vertex interaction, and we are looking for a $3$-vertex interaction.

We may try $\mathcal{L}_I = (\psi_1.A) (\psi_2.A)$, but we would have $2$ powers of $A$, meaning a $4$-vertex interaction with 2 matter particles and 2 force-particles.

We see that $A$ cannot be a "vector", and that it should be a "matrix", with an interaction like $\mathcal{L}_I = \psi_1.A\psi_2$

The lesson of this is that mathematical quantities representing force particles are different of the mathematical quantities representing matter particles.

While Weyl spinors (our "vectors") like $\psi_\alpha$ or $\psi^ \dot\alpha$ represent matter particles, force particles are represented by quantities $A^\alpha _\dot\alpha$, which act as "matrices" on the $\psi_\alpha, \psi^ \dot\alpha$

One may recover the usual $A_\mu$ quantities identifying spin-one particles, by : $A_\mu = (\sigma_\mu)^\dot\alpha _\alpha A^\alpha _\dot\alpha$, where the $\sigma_\mu$ are the Pauli matrices.