Charge conjugation of $|b\bar b\rangle$ states?

I know that for a state of a boson and its anti boson $|b\bar b\rangle$ the charge conjugation is $(-1)^{L+S}$ but I don't understand how this value is arrived at. Wikipedia says that is to do with the fact that the c-parity operating on $|b\bar b\rangle$ is identical to the parity - something that I don't find immediately obvious. Please can someone explain this to me?

(On a side note I think my confusion in part due to the fact I don't know what the charge conjugate does to spins - invert or not?)

• Charge conjugation distinctly commutes with anything spin. It is the same as for positronium. The total C-space-spin wave function for two fermions has to be antisymmetric, so $C (-)^L (-)^{S+1}=-$. – Cosmas Zachos Apr 20 '17 at 19:37

I can only answer partially: $C$-parity on any particle - antiparticle state $| a \bar{a} \rangle$ is identical to parity only if their spin state is the same. Say you have:

• $a$ in spatial state $| 1_{space} \rangle$ and spin state $| 1_{spin} \rangle$ .

• $\bar{a}$ in spatial state $| 2_{space} \rangle$ and spin state $| 2_{spin} \rangle$.

The effect of parity is changing $\textbf{r} \mapsto - \textbf{r}$. At the frame of reference of the center of mass this exchanges the spatial states. But spins remain the same, because angular momentum is a pseudovector. Then you have

• $a$ in spatial state $| 2_{space} \rangle$ and spin state $| 1_{spin} \rangle$ .

• $\bar{a}$ in spatial state $| 1_{space} \rangle$ and spin state $| 2_{spin} \rangle$.

Charge conjugation changes $a \leftrightarrow \bar{a}$, so now you have:

• $\bar{a}$ in spatial state $| 1_{space} \rangle$ and spin state $| 1_{spin} \rangle$ .

• $a$ in spatial state $| 2_{space} \rangle$ and spin state $| 2_{spin} \rangle$.

These are only equivalent if $| 1_{spin} \rangle = | 2_{spin} \rangle$. In the Wikipedia page you linked, the spin state of $\pi^+$ and $\pi^-$ is the same because they have spin $0$.

I also understand that the spatial state exchange contributes a factor of $(-1)^{L_r}$ to both parity and $C$-parity, where $L_r$ is the relative orital angular momentum. But I don't know where the $(-1)^{S}$ factor of the $C$-parity comes from.