The isospin for two quarks $u,d$, it is chosen that

$$
u=|1/2,1/2>, d=|1/2,-1/2>,
$$
in fundamental Rep of $SU(2)$.

the anti-quarks have anti-fundamental Rep of $SU(2)$ [thus same as the fundamental Rep of $SU(2)$],
$$
\bar{d}=-|1/2,1/2>, \bar{u}=|1/2,-1/2>,
$$

be aware that there is a minus sign in front of the $\bar{d}=-|1/2,1/2>$.

Somehow the minus sign is crucial, to get the triplet and singlet state wavefunction correct. namely, we have

$$
2 \otimes 2 = 3 \oplus 1
$$
where the 3 is the triplet (3 pions) and 1 is the singlet (another meson), for the pseudo-scalar mesons.

Why do we know that we should choose $\bar{d}=-|1/2,1/2>$? In p.169 of Griffiths's book, he said the minus sign "is a technical detail, but it does not affect the result essentially."

I find that the minus sign is very crucial to get the correct wavefunction of pion $\pi^0$ to be one of the triplet

$$
u\bar{u}-d\bar{d}
$$
$$
=|1,0>=|1/2,1/2>|1/2,-1/2>+|1/2,-1/2>|1/2,1/2>
$$

instead of being a singlet

$$
u\bar{u}+d\bar{d}
$$
$$
=|0,0>
$$

so why do we know the minus sign is there and is it key to the physics or not? I am asking a deeper reason behind it, because I know the minus sign there makes 100% sense for the pion wavefunctions.

This post imported from StackExchange Physics at 2020-10-29 20:02 (UTC), posted by SE-user annie marie heart