Need to write up the lectures I gave on kaons. May not get to it this semester but it can be found in any particle physics book.
Consider a states _{} and some symmetry operator _{} such that:
So both states are eigenvectors of the operator but with different eigenvalues. Clearly one can define a state _{}.
The state _{} is not an eigenstate of the operator _{}.
If Atransformations form a group and the Lagrangian is invariant under this group, the decay of the state _{} must preserve the value _{}, the observable associated with _{}. If the state _{} decays then the decay is really the independent decay of the two states _{}. The decay channels split into two independent paths. The result is simply the weighted combination of the separate decay processes.
When studying parity the initial state is examined to see if it is a parity eigenstate. Kaons are all psuedoscalar mesons with definite parity of 1. Also the original parity violating system was a nucleus with spin and also a state of definite parity. Under these conditions the final state should also be of definite parity. The decay of the kaon into three pions violates the conservation of parity. If one chooses to examine or construct states of mixed parity and parity is conserved the final states will also be of mixed parity. The initial and final states would be better treated separately because they cannot mix or interfere as a result of the observable parity which therefore distinguishes the final states.
When symmetries are examined the states considered may or may not be states with definite values for the symmetry operation. The K^{+}, for example, is a state of definite parity but not a state of definite charge conjugation because
_{}
_{}^{} 
_{} 
_{} 
_{} 
psuedoscalar 
_{} 
_{} 
_{} 
1 
psuedoscalar 
_{}^{} 
_{} 
_{} 
1 
psuedoscalar 
_{} 
_{} 
_{} 
1 
psuedoscalar 
_{} 
_{} 
_{} 
1 
psuedoscalar 
_{} 
_{} 
_{} 
1 
psuedoscalar 
1 _{} _{} 
_{} 
C=+1 for _{} 
P=1 
psuedoscalar 
2 _{} _{} sstate 

C=+1 
P=+1 
scalar 
3 _{} _{} sstate 

C=+1 for _{} 
P=1 

The kaon belongs to meson octet with the pion and the eta. It is a J^{P}=0^{1} meson.
wikipedia 
CPT requires particle and antiparticle to have identical mass, and lifetime. Thus the strong kaon states can be separated into particle antiparticle but the _{},_{} cannot be particle antiparticle pairs (each is its own anti particle as shown above). The is also a mass difference and lifetime difference between these two particles.
The results at this point leave CP as a good symmetry. When the _{} was observed to
decay into 2 pions, evidence for CP violation was
obtained. To add this to the theory weak
interaction must mix the _{},_{}. This mixture is represented as another set of kaon states _{},_{}. The important fact
is that CP is indeed violated but CPT is not.
More specific details are for the interested student but not required.