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

+ 2 like - 0 dislike
650 views

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

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

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

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
Possibly related.

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

+ 5 like - 0 dislike

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
answered Sep 30, 2017 by (50 points)
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
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
@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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification