Phase factors under rotations of strong and weak isospin The strong isospin raising operator changes a $d$ quark into a $u$:
$$
\tau_+ \big|d\big> = \big|u\big>
$$
However, for antiquarks, there is an additional phase factor:
$$
\tau_+ \big|\bar u\big> = - \left|\bar d\right>
$$
This phase factor is the reason the $\pi^0$ wavefunction is proportional to 
$\big|u\bar u\big> - \left|d\bar d\right> $.  But I don't understand why it arises.
The book that I have at hand is Wong's Nuclear Physics.  Given particle creation operators $a^\dagger_{t,t_0}$
and antiparticle creation operators $b^\dagger_{t,t_0}$ for hadrons with strong isospin $t$ and projection $t_0$, Wong states that 
$$
b^\dagger_{t,t_0} = (-1)^{t-t_0}a_{t,-t_0}
$$
because "operators $a^\dagger_{t,t_0}$ and $a_{t,-t_0}$ are not Hermitian conjugate of each other without the factor $(-1)^{t-t_0}$."  There is supposedly  a more detailed argument in Bohr & Mottleson, which I don't have access to at the moment.


*

*Why is this phase factor required?

*Do the same symmetry arguments apply to weak isospin partners?  If so, I need to revise my opinion on this previous question.

 A: An anti-particle transforms in the conjugate transformation $\bar{\mathbf r}$ than the particle ${\mathbf r}$ does under the given symmetry group, $SU(2)$ in your case. No surprise thus in general that the generators $T^i_{\mathbf r}$ and $T^i_\bar{\mathbf r}=-T^{i\,*}_{\mathbf r}$ act differently on ${\mathbf r}$ and $\bar{\mathbf r}$ respectively. 
This is the reason why for e.g. spin-$\frac{1}{2}$ fields, that is linear combinations of particle annihilation operators, $a$, and anti-particles creation operators, $b^\dagger$,  the wave functions $u(p)$ and $v(p)$ in front to $a$ and $b^\dagger$ are not symply related by $p\rightarrow -p$. 
However, for SU(2) doublet transforming with generators $T^i=\sigma^i/2$ (where $\sigma^i$ are the Pauli matrices), it turns out that $\sigma^2 T^i\sigma^2=-T^{i\,*}$ so that $i\sigma^2 \bar{\mathbf{2}}$ transform as a ${\mathbf 2}$, namely 
$$
i\sigma^2\psi^*=\left(\begin{array}{c} \bar{d}\\ -\bar{u}\end{array}\right)\qquad \mbox{and}\qquad \psi=\left(\begin{array}{c} u\\ d\end{array}\right)
$$ 
transform in the same way under $SU(2)$ (and in particular $\tau_+ \bar{u}=-\bar{d}$).
Therefore, if you want to extract  a triplet from $\bar{\mathbf 2}\otimes {\mathbf 2}={\mathbf{1}}+\mathbf{3}$, you just need to take the symmetric combinations of $i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a={\mathbf{3}}$, exactly like if you  wanted to extract the triplet (i.e. the symmetric combination) out of $\mathbf 2\otimes {\mathbf 2}={\mathbf{1}}+\mathbf{3}$. 
Explicitly, ordering the states as $\psi^{a=1}=|up\rangle=u$ and $\psi^{a=2}=|down \rangle=d$ as above (this is the basis where $T^3$ is diagonal with $+1/2$ and $-1/2$ on the upper and lower entry respectively), the symmetric combination with $a=1$ and $c=2$ in $i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a$ is proportional to $\bar{u}u-\bar{d}d$. The relative minus sign that you get comes from the $i\sigma^2$. The other states of the triplet correspond to the choice $a=c=1$ and $a=c=2$, that give $\bar{d}u$ and $\bar{u}d$
$$
\mathbf{3}\propto \frac{1}{2}\left(i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a\right)=\left(\begin{array}{cc}\bar{d}u & \frac{1}{2}(\bar{d}d-\bar{u}u) \\
\frac{1}{2}(\bar{d}d-\bar{u}u) & -\bar{u}d
\end{array}\right)_{ac}
$$
 This are the correct composition indeed, see http://en.wikipedia.org/wiki/Pion to check it out.
If you remove, erroneously,  the $i\sigma$ form the symmetric combination, you get instead $\bar{u}d+\bar{d}u$ (for $a=1$, $c=2$), and $\bar{u}u$ and $\bar{d}d$ for the other choices $a=c$, a result that does not make any sense.
