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The isospin doublet consisting of $u$ and $d$-quark is defined as $$ \begin{pmatrix} u\\ d \end{pmatrix}. \tag{1} $$ But the isospin doublet consisting of the antiquarks, $\bar{u}$ and $\bar{d}$, is given by $$ \begin{pmatrix} -\bar{d}\\ \bar{u} \end{pmatrix} \quad \text{but not}\quad \begin{pmatrix} \bar{u}\\ \bar{d} \end{pmatrix}. \tag{2} $$

The particle-to-antiparticle transformation involves a complex conjugation! Therefore, if $(u,d)^T$ transforms like the fundamental ($2$), the entity: $$ \begin{pmatrix} \bar{u}\\ \bar{d} \end{pmatrix}\equiv C\gamma^0 \begin{pmatrix} {u}^*\\ {d}^* \end{pmatrix} = \begin{pmatrix} C\gamma^0u^*\\ C\gamma^0d^* \end{pmatrix} $$ transform as $2^*$ (the antifundamental i.e. the conjugate of $2$). Let me explain. To elaborate, if $D(g)$ is a representation, $$D(g_1)D(g_2)=D(g_1\circ g_2)$$, then complex conjugation gives, $$D^*(g_1)D^*(g_2)=D^*(g_1\circ g_2)$$ i.e. $D^*$ is also a representation, called the conjugate representation. So if $$ \begin{pmatrix} u\\ d \end{pmatrix}\to \underbrace{e^{i(\sigma_a/2)\theta_a}}_{D-\text{representation}}\begin{pmatrix} u\\ d \end{pmatrix}, $$ then $$ \begin{pmatrix} u^*\\ d^* \end{pmatrix}\to \underbrace{e^{i(-\sigma^*_a/2)\theta_a}}_{D^*-\text{representation}}\begin{pmatrix} u^*\\ d^* \end{pmatrix}, $$ which proves that if $(u,d)^T$ transforms in the fundamental $(2)$, then $(u^*,d^*)^T$ transforms in the antifundamental $(2^*)$.

The matter should have ended here. Why do we need further construction? I have shown that if $(u,d)$ transform like the fundamental ({2}) and $(u^*,d^*)$ transforming as the antifundamental ${ 2}^*$. Is there anything wrong till this point? If not, sadly I do not see why we need to define the anti-quark doublet as $(-\bar{d},\bar{u})^T$ instead of working with $(u^*,d^*)^T$ or, $(\bar{u},\bar{d})^T$.

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1 Answers1

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I would like to post an answer based on my current understanding. It has been shown (see the question) that the $(u^*,d^*)^T$ transforms in the antifundamental representation of SU(2) i.e. $$\begin{pmatrix} u^*\\ d^* \end{pmatrix}\to \underbrace{e^{i(-\sigma^*_a/2)\theta_a}}_{D^*-\text{representation}}\begin{pmatrix} u^*\\ d^* \end{pmatrix} $$ By performing a similarity transformation with $S=-i\sigma_2$, we can make a change of basis, to write, $$-i\sigma_2\begin{pmatrix} u^*\\ d^* \end{pmatrix}\to \left\{(-i\sigma_2)e^{i(-\sigma^*_a/2)\theta_a}(-i\sigma_2)^{-1}\right\}(-i\sigma_2)\begin{pmatrix} u^*\\ d^* \end{pmatrix} $$ which simplifies to $$ \begin{pmatrix} -d^*\\ u^* \end{pmatrix}\to e^{i(\sigma_a/2)\theta_a}\begin{pmatrix} -d^*\\ u^* \end{pmatrix} $$ where we made use of the fact that $\sigma_2\sigma_a^*\sigma^{-1}_2=-\sigma_a$. Thus both $(u^*,d^*)^T$ and $(-d^*,u^*)^T$ belong to the same representation (the antifundamental), except written in a different basis. The reason we prefer the construction $(-d^*,u^*)^T$ over $(u^*,d^*)^T$ is that we can combine quark and anti-quark states using the same Clebsch-Gordon coefficients (Halzen and Martin, Section 2.7).

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