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.
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