What is the CP operator on pion combinations?

[rubenvb]

Hi,

I have a question regarding the CP operator on pion systems.
1) CP \mid \pi^0 \rangle
2) CP \mid \pi^+ \pi^- \rangle
3) CP \mid \pi^0 \pi^0 \rangle

I'd like to solve this in the above ket notation and apply the operators as is on the different parts of the represented wave function. My solution for 2) is:
CP \mid \pi^+ \pi^- \rangle
= C \mid \pi^- \pi^+ \rangle (switch pions physically in e.g. x-coordinate)
= \mid \pi^+ \pi^- \rangle (invert charges)
Thus CP is +1 for \mid \pi^+ \pi^- \rangle. This does not seem to work for 1). Note I have somehow lost the notion of (-)^l that should be present somewhere :S

Any help is appreciated.

 

[ansgar]

write the state for the pi0 in terms quarks

 

[rubenvb]

write the state for the pi0 in terms quarks

OK, this is where I get:
CP \mid \pi^0 \rangle = CP \frac{\mid u \bar{u} \rangle - \mid d \bar{d} \rangle}{\sqrt{2}}
= \frac{ CP \mid u \bar{u} \rangle - CP \mid d \bar{d} \rangle}{\sqrt{2}}
= \frac{C \mid \bar{u} u \rangle - C \mid \bar{d} d \rangle}{\sqrt{2}}
= \frac{ \mid u \bar{u} \rangle - \mid d \bar{d} \rangle}{\sqrt{2}} = \mid \pi^0 \rangle
This would mean CP for a \pi^0 is +1, while I kind of remember it being -1... What do I do wrong?

 

[humanino]

sorry for my first message, I did not pay attention.[/size]

Together with the flavor wavefunction \frac{1}{\sqrt{2}}\left( |u\bar{u}\rangle - |d\bar{d}\rangle \right)
one has to take into account spin \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle \right) as well
|\pi^0\rangle=\frac{1}{\sqrt{2}}\left[ <br /> \frac{1}{\sqrt{2}}\left( |u_\uparrow\bar{u}_\downarrow\rangle - |d_\uparrow\bar{d}_\downarrow\rangle \right)<br /> - \frac{1}{\sqrt{2}}\left( |u_\downarrow\bar{u}_\uparrow\rangle -|d_\downarrow\bar{d}_\uparrow\rangle \right) \right]<br /> =\frac{1}{2}\left[ <br /> |u_\uparrow\bar{u}_\downarrow\rangle - <br /> |u_\downarrow\bar{u}_\uparrow\rangle - <br /> |d_\uparrow\bar{d}_\downarrow\rangle + <br /> |d_\downarrow\bar{d}_\uparrow\rangle \right]<br />