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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.

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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.

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  • $\begingroup$ 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)??? $\endgroup$ – annie heart Sep 30 '17 at 23:44
  • $\begingroup$ 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???!!! $\endgroup$ – annie heart Sep 30 '17 at 23:47
  • $\begingroup$ @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. $\endgroup$ – Bruce Lee Oct 1 '17 at 4:27

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