It took me some time after my Group Theory lecture to really start to understand how to apply all the mathematics in physics, in particular with respect to the quark model. I am currently reading the book "An Introductory Course of Particle Physics" by Palash B. Pal (pp. 201-227 for isospin symmetry and pp. 254-297 for the quark model) and it cleared up a lot of questions I had for quite some time. However, there are still certain points I am terribly confused about and I can't seem to find a satisfying explanation anywhere.
For later reference, I already here quote the free, non-interacting Lagrange density of QCD with six flavors, $u$, $d$, $s$, $c$, $b$, and $t$: $$ \mathcal{L}^\text{free}_\text{QCD} = \sum_{\substack{\mathrm{f} = u,d,s,\\ \quad c,b,t}}{\overline{q}_\mathrm{f}(x)\left(\mathrm{i} \gamma^\mu \partial_\mu - m_\mathrm{f}\right)q_\mathrm{f}(x)}.$$ At this, $q_\mathrm{f}(x)$ are the "usual" quark spinor fields and $\overline{q}_\mathrm{f}(x)$ are their Dirac adjoints, $\overline{q}_\mathrm{f}(x) \equiv q_\mathrm{f}(x)^\dagger \gamma^0$.
There are mainly two things I am struggling with:
As far as I understand, all representations of SU(2) are self-conjugate. Concerning the fundamental representation, the $\mathfrak{su}(2)$ generators are conventionally given by the Pauli matrices $\tau_{i=1,2,3}$ and implicate a conjugate representation according to $\tilde{\tau}_i \equiv - \tau_i^*$. Since we can find a similarity transformation $S=\mathrm{i} \tau_2$ such that $\tilde{\tau}_i = S \tau_i S^{-1}$, the two representations account for the same content in different bases. So far so good regarding the generators. The space of states in the $\mathbb{2}_\mathrm{f}$ representation is given by two-dimensional column vectors which the generators act on. We say the up- and down-quark form a doublet in the $\mathbb{2}_\mathrm{f}$ according to $$q = \begin{pmatrix} u \\ d \end{pmatrix}.$$ In the aforementioned book, it is shown on pp. 220-221 that the antiquarks then form a doublet according to $$\overline{q} = \begin{pmatrix} \overline{d} \\ -\overline{u} \end{pmatrix},$$ see also Quark composition of the neutral pion. Now, if the representation is equivalent to the conjugate representation, why are the state vectors different then? I do understand the proof given in the book, but somehow I can not figure out the connection to the representation being self-conjugate. Naively, I would have expected the states to simply be given by $$\overline{q} = \begin{pmatrix} \overline{u} \\ \overline{d} \end{pmatrix},$$ since they reside in the conjugate representation, which, however, is equivalent to the representation itself. Adding to this confusion is the fact that for some reason, this is exactly what the author does for the $\mathbb{3}_\mathrm{f}$ of $\mathrm{SU}(3)$ on p.271: $$q = \begin{pmatrix} u \\ d \\ s \end{pmatrix}, \qquad \overline{q} = \begin{pmatrix} \overline{u} \\ \overline{d} \\ \overline{s} \end{pmatrix}.$$ Should the antiquark triplet not take a different form then as well -- if so, what form? Furthermore, it appears to me that I am not even able to recover the (two flavor) QCD Lagrange density with former two doublets: $$ \mathcal{L}^\text{free}_\text{QCD, u, d} \neq \overline{q}\left(\mathrm{i} \gamma^\mu \partial_\mu - M\right)q, \qquad q = \begin{pmatrix} u \\ d \end{pmatrix}, \qquad \overline{q} = \begin{pmatrix} \overline{d} \\ -\overline{u} \end{pmatrix}, \qquad M = \mathrm{diag}(m_u,m_d),$$ right?
The second thing I am confused about is, to some degree, actually connected to the first question. I always believed that the Dirac adjoint spinor fields $\overline{q}_\mathrm{f}(x)$ are (interpreted as) the spinor antifields, i.e. antiquarks in the QCD Lagrange density above. I feel reconfirmed in this belief by the answer from @ACuriousMind in Antiparticles vs. conjugate particles. Am I assuming right that the quark spinor fields $u(x)$, $d(x)$, and $s(x)$ are exactly the $u$, $d$, and $s$ appearing in the quark doublets and triplets above? Because if so: let's say my Lagrange density obeys a $\mathrm{SU}(2)$ isospin symmetry among the $u$- and $d$-quark fields, thus in particular $m_u = m_d$. In order to check that the Lagrange density indeed respects that symmetry, I would merely do the following: $$q = \begin{pmatrix} u \\ d \end{pmatrix} \rightarrow U \begin{pmatrix} u \\ d \end{pmatrix},$$ where $U \in \mathrm{SU}(2)$ and which implies $$\overline{q} \rightarrow \overline{q} U^\dagger.$$ The unitary matrices cancel out in the QCD Lagrange density and it immediately follows what I just claimed. How is latter transformation justified, given that the antiquarks should actually be given by what I stated in my first question?
