Physics of tagging at B factories At some B factories, mesons carrying $b/\bar{b}$ quarks are created by $e^-e^+$ collisions at $\gamma(4S)$ resonance. $\gamma(4S)$ decays into antisymmetric wavefunction given by 
$$ \frac{1}{\sqrt{2}}(\left|B_0(\vec{k})\right>\left|\bar{B}_0(-\vec{k})\right>-\left|\bar{B}_0(\vec{k})\right>\left|B_0(-\vec{k})\right>) $$
Following this, I have two questions: 


*

*This wavefunction clearly violates Bose statistics. So the full wavefunction has to be symmetric. What is the other part of this wavefunction? (spin is zero)

*The fact that it is an antisymmetric wavefunction helps in tagging flavours of particles. I didn't understand this statement. Please explain how. Also, what if its a symmetric wavefunction(like in case of $\gamma(5S)$). Can't we tag flavours there?
 A: This is only an answer to the first question. When you invoke the spin statistic theorem (i.e. the fact that identical particles, here bosons must have a symmetric wave function) for the system made of $B^0$ and $\bar{B}^0$ mesons, you have to consider:


*

*spin exchange 

*position exchange

*Charge conjugation 


in order to build the whole wave function. 
1) has no incidence here since $B^0$ has a spin 0. 
2) Doing $\vec{k} \leftrightarrow -\vec{k}$ introduces a minus sign. In other words your system has an odd parity. 
3) has also to be considered since $B^0$ and $\bar{B}^0$ are not strictly speaking the same particles, one being the antiparticle of the other. Changing in your wave function the $B^0$ by $\bar{B}^0$ but keeping the original $\vec{k}$ (or $-\vec{k}$) introduces another minus sign. 
So, globally the changes $\vec{k} \leftrightarrow -\vec{k}$ followed by $B^0 \leftrightarrow \bar{B}^0$ don't change the sign of the wave function as it should for this system.
Remark: point 2 and 3 can be checked by looking at the quantum numbers of the $\Upsilon(4s)$ which has $J^{PC}=1^{--}$ meaning an odd parity and an odd C-parity. 
