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.


3 Answers 3


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).

  • $\begingroup$ About the supersymmetry point I might be able to second that in about 3 months (at the end of my SUSY-course). But indeed there are fermionic force lines, but they always come in pairs, doesn't that also give a kind of bosonic character ? $\endgroup$
    – Nick
    Commented Oct 20, 2013 at 11:26
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    $\begingroup$ I think it depends how you define a force. If it is only processes that contributes to the scattering of the same kind of particles, then it will always be bosonic (c.f. Ben Crowell argument). But if you consider electron-photon scattering, then the force carrier is an electron at tree level. Maybe you can have the same kind of processes in SUSY, but I'm no expert. $\endgroup$
    – Adam
    Commented Oct 20, 2013 at 16:25

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.

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    $\begingroup$ I think this is misleading. There do exist internal lines in feynman diagrams that are fermionic. you mean conservation of angular momentum? $\endgroup$
    – anna v
    Commented Oct 20, 2013 at 4:26
  • $\begingroup$ @annv, but don't fermionic lines also always come in pairs, wich adds the total momentum to an integer ? $\endgroup$
    – Nick
    Commented Oct 20, 2013 at 11:17
  • $\begingroup$ @annav: Thanks for the correction re momentum->angular momentum. Re internal fermionic lines, I didn't make any claims in general about internal fermionic lines. I made a much more specific claim. $\endgroup$
    – user4552
    Commented Oct 20, 2013 at 14:30
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    $\begingroup$ What about particle-antiparticle annihilation into two photons? That's an interaction between two particles in which the exchanged virtual particle is fermionic. Are you only counting interactions that preserve particle species? $\endgroup$
    – tparker
    Commented Aug 29, 2018 at 2:30
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    $\begingroup$ I'm sure this answer is in good faith but for the readers please notice that Feynman diagrams do not describe interactions: they are just pieces of an infinite Taylor expansion. $\endgroup$
    – gented
    Commented Apr 30, 2019 at 14:44

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.

  • $\begingroup$ nice comment (and don't worry, I'm quite aware of the QFT scheme so you don't need to dumb down comments to much ;) ). But how can you then identify A with spin-1 particles ? I can't see that immediately since for me the Pauli-matrices are used in electrodynamics to discribe 1/2-spin systems. I do know that they are also used in Clifford-Algebra's, but that's a chunck of theory that I dind't get to go trough yet :( $\endgroup$
    – Nick
    Commented Oct 20, 2013 at 20:25
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    $\begingroup$ In terms of representations of the Lorentz group, a spinor $\psi_\alpha$ is a representation $(\frac{1}{2},0)$ , a spinor $\psi^ \dot\alpha$ is a representation $(0,\frac{1}{2})$ , and $A^\alpha _\dot\alpha$ is a representation $(\frac{1}{2},\frac{1}{2})$. This last representation is also called vector representation. The Pauli matrices $\sigma_\mu$ just make the link between the $A^\alpha _\dot\alpha$, and the more used notation $A_\mu$. $\endgroup$
    – Trimok
    Commented Oct 21, 2013 at 16:36
  • $\begingroup$ .....In fact, the Lorentz group is $SL(2, \mathbb C) \sim SO(3,1)$. $A^\alpha _\dot\alpha$ could be seen as a representation $(\frac{1}{2},\frac{1}{2})$ of $SL(2, \mathbb C)$, while $A_\mu$ may be seen as a fundamental representation of $SO(3,1)$ $\endgroup$
    – Trimok
    Commented Oct 21, 2013 at 16:36

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