May 26, 2016

Yes.

#### Explanation:

Hadrons can divided into two sub-groups, Baryons and Mesons. Examples of Baryons are Protons and Neutrons, and for Mesons, Kaon and Pion particles.

In most simple cases, the Baryons have a three-quark structure and the Mesons and a quark-antiquark structure. For 'antibaryons', this can be found by taking a three-antiquark structure.

For example, the proton (denoted $p$) has the quark structure $u u d$ (where $u$ denotes the up quark with positive $\frac{2}{3}$ charge and $d$ denotes the down quark with negative $\frac{1}{3}$ charge).

An important characteristic to remember is that Baryon number is conserved within an interaction. An ordinary Baryon has a Baryon number of $1$ and ordinary quarks have a baryon number of $\frac{1}{3}$, such that the $u u d$ combination gives $1$ as the baryon number of a proton. Similarly, all antibaryons and antiquarks have the opposite baryon number value from their counterparts (using negatives where positives where) This is why an antiproton (denoted $\overline{p}$) has the quark structure $\overline{u} \overline{u} \overline{d}$, which if you notice is the complete opposite from the proton structure. The antiproton will therefore have the opposite baryon number ($- 1$) and charge ($- 1$).

Things get a bit strange with Mesons, such that some particles like the ${K}^{0}$ is its own antiparticle. I'll explain how this can happen next:
Firstly, because Mesons have the quark-antiquark structure, it's baryon number is $0$ (since recall that a quark has $\frac{1}{3}$ and an antiquark has $- \frac{1}{3}$). We also need to introduce the strange quark, which has a charge of $- \frac{1}{3}$ and a baryon number of $\frac{1}{3}$. (Look up strange number conservation, namely not occurring in weak interactions, for further extension here). Kaons have the superscript value of its charge such that ${K}^{0}$ has a charge of $0$. So we can 'make' it from the quark structure $s \overline{u}$. Consider that the so-called antiparticle of this is going to be $\overline{s} u$, but if we look at the charges this will also produce the ${K}^{0}$, therefore the ${K}^{0}$ is its own antiparticle. The antiparticle of ${K}^{+}$ is ${K}^{-}$ too.

This answer is very long but hopefully it gives some insight into antiparticles within the hadron group.