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 is partly due to the fact I don't know what the charge conjugation does to spins:   does it invert them, or not?)
 A: 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.
A: You are misreading WP. C=C-parity  is not linked to P. It also leaves spins alone.
It is stated clearly there that, for fermion constituents,
$$C= (-1)^{L+S}.$$
The reason is that C flips a fermion into an antifermion and multiplies it by a minus sign, but, doing that, also introduces a minus sign if the two fermions are in an antisymmetric state. This happens only if they are in an even L, $(-)^L=+$, and an  even spin combination (singlet), $(-)^{S+1}$;  or  an odd L state and  odd spin combination (triplet). In these cases, C=+.
Otherwise C=-.
So, the spin singlet parapositronium $^1S_0$ has C=+  and decays to 2γ, while the spin triplet orthopositronium, $^3S_1$ has C=- and decays to 3γ.
Now contrast the two lowest pseudoscalar Υs, so with the same parity:
$$
^1S_0 ~~~9.39GeV, ~~~~C=+ \\
^3S_1 ~~~9.46 GeV, ~~~~ C=- ~~.
$$
