# How is ##\Phi \rightarrow K^{+}K^{-}## decay possible?

• I
I know that, for charge conjugation, ##C_{\Phi} = -1##, ##C_{K^{+}} = 0##, ##C_{K^{-}} = 0##, but ##C_{\Phi} \neq C_{K^{+}}C_{K^{-}}##. How ##C_{tot}## is conserved in this interaction?

What do you mean with ##C_{K^{\pm}}=0##?
##K^{\pm}## is not an eigenstate of the charge conjugation operator.

You're right, i've just noticed that. Does it mean i can't check for charge conjugation conservation in this process? If i'm asked to check if this process is forbidden or not, which are the quantum numbers to check? The process is hadronic, and these processes conserve charge conjugation, right?

mfb
Mentor
The hadronic interaction is invariant under charge conjugation, yes. If you replace particles with antiparticles in a possible process the process is still possible. If you replace particles with antiparticles here you get the same process again.
Check the quark flavors and the energy. If they work out the process will be possible in general. It can be very unlikely due to spin/angular momentum, something to check as well.

Does it mean i can't check for charge conjugation conservation in this process?

A single charged kaon is not an eigenstate of the charge conjugation operator. But a ##K^+K^-## system is.

vanhees71