The $SU(2)$ isospin for quark v.s. anti-quark: Fix $\bar{d}=-|1/2,1/2>$?

Question

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

Comments

Possibly related.

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

Answer 1

If you start from the quarks transforming under fundamental representation of $SU(2)$ $$\psi'_i = U_{ij}\psi_j$$

and complex conjugate both sides, you get

$${\psi'}_i^* = U_{ij}^* \psi_j^*$$

which transforms under the anti-fundamental rep of $SU(2)$.

For $U \in SU(2)$, there exists an $S \in SU(2)$ such that $S^{-1}US = U^*$. Note that this is a special property restricted to $SU(2)$ matrices only and doesn't generalize to $SU(n)$. The previous equation in matrix form therefore becomes: \begin{align} \psi'^* = (S^{-1}US)\psi^* \implies S\psi'^* = U(S\psi^*) \end{align} So $S\psi^*$ transforms as $\psi$. It turns out that in the Pauli representation that $S = i\sigma^2$, and that is the reason for the minus sign in your equations as $\sigma^2$ has a minus sign in one of the components and a plus in the other. In other words

$$i \sigma^2 = \begin{bmatrix} 0 &1 \\ -1 & 0 \end{bmatrix}$$

and therefore the minus sign.

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

Comments

Sorry I am confused now, so cannot accept you as the answer. Did you say $U$ is traceless Hermitian, or do you really mean the generator $T^a$ in Lie algebra is traceless (Pauli matrices)???

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

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In that sense, In p.169 of Griffiths's book, he said it is not essential issue, is actually very important minus sign, in my opinion???!!!

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

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@annieheart yup, that was a typo, thanks for pointing it out. regarding griffiths' statement, maybe he doesn't need it in any further calculation that's why he is saying it not to be an essential issue.

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