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
$$|\pi^0\rangle=\frac{1}{\sqrt{2}}\left[ <br /> \frac{1}{\sqrt{2}}\left( |u_\uparrow\bar{u}_\downarrow\rangle - |d_\uparrow\bar{d}_\downarrow\rangle \right)<br /> - \frac{1}{\sqrt{2}}\left( |u_\downarrow\bar{u}_\uparrow\rangle -|d_\downarrow\bar{d}_\uparrow\rangle \right) \right]<br /> =\frac{1}{2}\left[ <br /> |u_\uparrow\bar{u}_\downarrow\rangle - <br /> |u_\downarrow\bar{u}_\uparrow\rangle - <br /> |d_\uparrow\bar{d}_\downarrow\rangle + <br /> |d_\downarrow\bar{d}_\uparrow\rangle \right]$$