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Though an electrically charged elementary particle is necessarily accompanied by an antiparticle, a particle need not be electrically charged to have an antiparticle. The cleanest example being a neutrino. However, it seems that a particle must be charged under some $U(1)$ group (or its subgroup) to have an antiparticle e.g., lepton number, baryon number, electric charge etc. Let me substantiate my hunch with two examples. Photons and gluons are not charged under any $U(1)$ group. By the above criterion they shouldn't have any antiparticle and they don't. Gluons are charged under color $SU(3)$, a nonabelian group. However, antigluons do not exist. So please comment if it is really necessary for a field to be charged under some $U(1)$ group for antiparticles to exist.

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    $\begingroup$ For me, the simple definition of antiparticle is to see that the identifying quantum numbers of the two candidates add up to zero. see the answer here for gluons quora.com/What-are-the-antiparticles-of-gluons also this talks of anti particles for colored quarks en.wikipedia.org/wiki/Color_charge $\endgroup$ – anna v Oct 30 at 5:43
  • $\begingroup$ @annav I think I am looking for a mathematical reason why gluons don't have antiparticles despite being charged under SU(3) color. $\endgroup$ – mithusengupta123 Oct 30 at 13:26
  • $\begingroup$ The mathematical explanation which I have is: the fields are $\mathbb{Z}_2$-graded algebra with involution valued objects. According to their behavior under involution, they describe particles which are (even) identical to their antiparticles, or not (uneven). For classical fields, the involution is simply complex conjugation. For quantum fields in Fock space, it is Hermitean adjoint. $\endgroup$ – DanielC Nov 5 at 7:41
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The set of all gluons contains the antiparticles of each member of itself. But while there is only one photon which is own antiparticle (it is a singlet), each specific gluon has an antiparticle distinct of itself. But its antiparticle is just another gluon and does not deserve to be called an "antigluon".

It is like saying that the $\pi$ meson is its own antiparticle. In fact this is true of the $\pi_0$ meson only, the antiparticle of the $\pi_+$ meson is the $\pi_-$ meson, but it is not an "anti"-$\pi$ meson, just a different $\pi$ meson ! You have exactly the same situation for the gluons.

To make this more mathematical, a particle which is a singlet for every group, the photon, the graviton, is its own antiparticle.

A non-singlet set invariant under a group (but where each particle is a singlet for all other groups) is such that the antiparticles of each element are within this set and do not deserve the word "anti", but are not identical to their antiparticle.

To make it more explicit the combinations $\bar RB$, $\bar BG$, $\bar GR$, for instance, are color-charged gluons. Their antiparticles $\bar BR$, $\bar GB$, $\bar RG$ are also gluons, not antigluons, and these are in fact all six color-charged gluons.

There are also two (and not three) color-chargeless gluons which are their own antiparticles. They are not exactly the three objects $\bar BB$,$\bar RR$, and $\bar GG$, which in fact do not correctly describe gluons, since there are just two color-chargeless gluons, but this is some complication which is not really useful here.

OK, maybe a few more examples will make this clearer.

The $W_+$ boson has an antiparticle, which is the $W_-$ boson, clearly not identical to it, because of the electrical charge. But the latter is not called anti-$W$ so as a whole the "$W$ boson" as a set of two objects, is its own antiparticle (even though the $SU(2)$ symmetry relating them is far from an exact one, since the $Z_0$ boson, its own antiparticle, has a different name and a very different mass, even though it is somehow the neutral partner of the $W$ in this approximate symmetry). There is however a deep analogy between the set ($W_+, W_-$) and the set ($\pi_+, \pi_-$), but here the neutral particle partner in the (also non-exact) $SU(2)$ isospin symmetry has still a close enough mass to be called $\pi_0$. It is also its own antiparticle, so the full set of the three $\pi$ mesons is, as a set, its own "antiparticle set". But the $\pi_0$ mass is slightly different from the other two $\pi$ mesons and one might have decided at the time of discovery to give it a different name.

In the case of the gluons, because the color $SU(3)$ symmetry is exact, the entire 8-element sets of gluons is its own "antiparticle set", but consists of three pairs of distinct particles which are antiparticle of each other (for instance $\bar RB$ and $\bar BR$), plus two more, each of which is its own antiparticle.

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    $\begingroup$ +1 I like this answer a lot. Let me try to translate this into simple words for those that do not understand much advanced physics: Every particle has an anti-particle (that's pretty much required). This doesn't mean that there must be an even number of every type of particle because some particles are tricky. $\endgroup$ – Jim Nov 4 at 14:12
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All particles have an antiparticle, their transform under C. You may be asking "when is a particle's antiparticle it itself?"

Dirac Fermions cannot be their own antiparticle, as their corresponding fermion number transforms nontrivially under C; this then applies to all types of baryon, lepton, etc numbers. Majorana fermions can. They are their own antiparticles.

Bosons may be their own antiparticle, if they are singlets under C. Work out their charge conjugation. They should be neutral.

So, a photon or a π0, or a $Z^0$ are their own antiparticles. The charged pions, by contrast, are each other's antiparticles. (Since the weak interactions violate C, one normally does not waste breath on $Z^0$ and the Higgs being their own antiparticle...)

But a $\bar R G$ gluon is the antiparticle of a $\bar G R$ gluon, not itself, since SU(3) is a complex group. So, indeed, the antigluons are included in the set of 8 gluons, and, of course, the $\bar R R$ gluon is its own antiparticle, after imposing the tracelessness constraint of color generators.

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  • $\begingroup$ Well, you essentially repeated what I said. Besides there is not thing as a $\bar R R$ gluon ans this is a not orthogonal to the singlet. You need combinations such as $(\bar R R-\bar B B)/\sqrt 2$ or $(2\bar R R-\bar B B-\bar G G)/\sqrt 6$ $\endgroup$ – Alfred Nov 3 at 7:48
  • $\begingroup$ I don’t want to confuse her with tracelessness. But drives home the issue...The fermions matter. $\endgroup$ – Cosmas Zachos Nov 3 at 9:32
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    $\begingroup$ +1 Downvoting correct answers is a persistent problem on Physics SE. In the age of social media of the 21st century, downvoting is a shameful and barbaric practice long abandoned everywhere else. $\endgroup$ – safesphere Nov 4 at 2:07
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Photons are neutral bosons, they are their own antiparticle.

Seen another way, the photon can be considered as its own antiparticle (thus an "antiphoton" is simply a normal photon).

https://en.wikipedia.org/wiki/Photon

Neutrinos are fermions, they do have antiparticles, with opposite lepton number and chirality.

https://en.wikipedia.org/wiki/Neutrino

Gluons do not have EM charge, but they have color charge. It is its own antiparticle. It is because it is adjoint (color charge).

You need charge conjugation to see that the photon and gluon are their own antiparticles.

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