baryonsconfinementmesonsquarksstandard-model
I'll be honest, I only have a Highschool education, so there might be something obvious I'm overlooking. However particle physics is of massive Interest to me.
My question is, I know there are unstable quark-antiquark pairs that form Mesons, but are there any Baryons that are not wholly made up of quarks or antiquarks? i.e. "up, up, anti-down"
If the answer is no, why not?
Each quark $q$ and antiquark $\bar{q}$ transform in the fundamental representation ${\bf 3}$ and antifundamental representation $\bar{\bf 3}$ of the color group $SU(3)_C$, respectively.
However, a hadron has to be a color singlet ${\bf 1}$, due to color confinement.
In particular the center $$\mathbb{Z}_3~:=~\mathbb{Z}/3\mathbb{Z}~=~\{1,e^{\pm 2\pi i/3}\} ~\subseteq ~SU(3)_C$$ of the color group $SU(3)_C$ should be in a trivial representation of the cyclic group $\mathbb{Z}_3$. This precisely happens if the quark number is a multiple of 3. This answers OP's title question.
Examples:
A single quark $q$ transforms in the fundamental representation ${\bf 3}$ of $SU(3)_C$, and is hence not allowed. See also related Phys.SE post here.
A diquark $qq$ belongs to the tensor representation ${\bf 3}^{\otimes 2}:={\bf 3}\otimes{\bf 3}\cong\bar{\bf 3}\oplus{\bf 6}_S$, which we have decomposed in irreps of $SU(3)_C$. This contains no singlet ${\bf 1}$, and is hence not allowed. See also related Phys.SE posts here and here about $SU(3)$ tensor representations.
In a meson, the quark-antiquark-pair $q\bar{q}$ belongs to ${\bf 3}\otimes\bar{\bf 3}\cong{\bf 1}\oplus{\bf 8}_M$, which contains a singlet ${\bf 1}$, and is hence allowed.
The third tensor product is ${\bf 3}^{\otimes 3}={\bf 3}\otimes{\bf 3}\otimes{\bf 3}\cong {\bf 1}\oplus 2\cdot{\bf 8}_M\oplus{\bf 10}_S$. In a baryon, the three quarks $qqq$ form a totally antisymmetric representation $\wedge^3 {\bf 3}\cong {\bf 1}$ of $SU(3)_C$, which is isomorphic to a singlet ${\bf 1}$, and is hence allowed. See also related Phys.SE post here. The lightest baryon, the proton, is stable in the standard model due to baryon/quark number conservation. (However, see the hypothetical proton decay.)
The tensor product ${\bf 3}\otimes{\bf 3}\otimes\bar{\bf 3}\cong 2\cdot{\bf 3}\oplus {\bf 6}_M\oplus {\bf 15}_M$ contains no singlet ${\bf 1}$, so the combination $qq\bar{q}$ is not allowed.
The fourth tensor product is ${\bf 3}^{\otimes 4}\cong 3\cdot{\bf 3}\oplus 2\cdot{\bf 6}_M\oplus 3\cdot{\bf 15}_M\oplus{\bf 15}_S$. This contains no singlet ${\bf 1}$, so four quarks $qqqq$ are not allowed.
A "molecule" of mesons and baryons, such e.g. a tetraquark $q\bar{q}q\bar{q}$ or a pentaquark $qqqq\bar{q}$, is also allowed, but is obviously heavier. See also related Phys.SE posts here, here, and here.
TL;DR: The number of quarks minus the number of antiquarks should be divisible by 3.
References:
One is talking quantum mechanics and attributed quantum numbers to elementary particles.
A simple quantum number is charge and it it assigned to quarks ( and antiquarks) as +/-1/3 or +/-2/3 as in the table
Charge is connected with the electromagnetic force.
Flavor is assigned as a quantum number to each quark, and it is connected with the weak interaction.
Each quark at the same time is connected with the strong color force of quantum chromo dynamics. So it can also come in the three color quantum numbers, for identification called red blue and green ( analogous to the weak Strange Charm Bottom Top).
The identifications are not random, they are within the SU(3)xSU(2)xU(1) group representations and algebra of the standard model of particle physics.
The quark forces are attractive only in "colorless" combinations .
Attractive is a wrong attribute. Color is always attractive, but it can be "nullified" in certain color combinations so that stable bound states of quarks appear, as with the rest of the quote.
Flavor characterizes the weak interactions of the quarks. Color the strong ones.
Best Answer
No, a three-quark baryon can not be be made out of two quarks and one anti-quark (and vice versa) as this would necessarily give the particle color.
Each quark carries one of three colors (red, blue, green) and each anti-quark respectively carries anti-color. Color is an additive quantity when constructing particle and the result must be color-neutral, i.e. it either is made out of for example red + anti-red or red + blue + green quarks. Although we can not observe the "color" of the constituents of a particle directly (as all observable particles must be color-neutral), we can measure its effects indirectly via certain cross-sections.
With this being said, we can now clearly see that there is no way in which we can construct particles of two-quarks and one anti-quark as any possible combination would not be color-neutral. Hence, color-neutrality forbids observable three-quark particles composed out of quarks and anti-quarks.
Penta-quarks however can and must contain quarks and anti-quarks as mentioned in a previous answer.