# Why do the $u$ and $d$ quark not have an associated quantum number?

All the other quarks ($c$,$s$,$b$ and $t$) have quantum numbers of charmness, strangeness, bottomness and topness that are conserved in strong interactions.

This allows, among other things, flavour changing neutral currents in $K^0$, $B^0$ and $D^0$ mesons as I discussed in this question but prevents them in pions $\pi^0$.

Is there any physical meaning as to why there are no quantum numbers of 'upness' and 'downness'?

• Isotopic spin Apr 12, 2015 at 10:59

They do. It's the third component of isospin, which came about before Murray Gell-Mann's quark model.

\begin{equation} I_3 = \frac{(N_u-N_\bar{u})-(N_d-N_\bar{d})}{2} \end{equation}

All quarks have baryon number 1/3, so that the nucleons can be built up. Baryon number is a conserved quantity in the standard model . In models where the proton can decay, it is not conserved, but no proton decays have been detected up to now.

One has to realize that the quantum numbers S,C,B,T are attributes for all quarks and have a value, even if that value is zero. Zero is a good quantum number.

The heavier quarks , conserving baryon number, can decay through the weak interaction to lower mass quarks. The up and down have no lower to go.

It is just the ordering observed experimentally. It is not simple because of group structures, but the slots in the group representations occupied by the quarks were chosen so as to have correspondence with their quantum numbers.

You certainly can say that up quarks have +1 upness and down quarks have -1 downness. See Griffiths' particle book, 2nd ed., p. 49. It's just not very useful. The only quark with $S = C = B = T = 0$ and electric charge of $+\frac{2}{3}$ is the up quark. The down quark has $-\frac{1}{3}$ charge.

You could also say the up quark has baryon number of $\frac{1}{3}$ and the third component of isospin $+\frac{1}{2}$. You see, there are already plenty of ways to describe the up and down quarks, and these other quantum numbers predated the quark model.