Conservation of strangeness and eigenstates

[Xico Sim]

Hi, guys.

In Povh's book, page 198, he says: "The strong force conserves the strangeness S and so the neutral kaons are in an eigenstate of the strong interaction."

I do not see why this must be the case. My atempt to understand it:

$$ŜĤ_s |K_0 \rangle = Ĥ_sŜ |K_0 \rangle$$
So
$$Ŝ(Ĥ_s |K_0 \rangle) = -Ĥ_s |K_0 \rangle $$

Since the ket $Ĥ_s |K_0 \rangle$ has strangeness -1, it belongs to the eigensubspace of $Ŝ$ with eigenvalue -1. I don't know how one conclude, from this, that
$$ Ĥ_s |K_0 \rangle \, \alpha \, |K_0\rangle $$
which is what I want to prove.

 

[ChrisVer]

well, abstractly speaking, can't you write $|K_0>$ state in whichever basis you want? and so the strong interactions' basis... the commutation relation you wrote (and I guess you take as given) will hold in any basis of the state $|K_0>$.
So you can define that the $K_0$ represents a strong state.

 

[mfb]

A more experimental approach: neutral kaons cannot decay via the strong interaction (they are the lightest neutral particles with a strange quark and $K^0 \to K^+ \pi^-$ is not possible either), so no matter how you write them as state, they have to be an eigenstate of the strong interaction.

 

[Xico Sim]

So you can define that the $K_0$ represents a strong state.

"So"? i don't see why...

 

[ChrisVer]

whether you define the x-axis showing to your left hand, or showing to your right, if there is a left/right symmetry it doesn't really matter.

If we suppose that K0 in your case is not a strong state, you can "rotate" it into being in the strong state K0'...
THis happens by diagonalizing the Hamiltonian in the strong basis by $H \rightarrow H_{\text{str-diag}}=U H U^\dagger$ and so will the states: $|K_0> \rightarrow U |K_0>$ which will be your states written in the strong interaction basis.
The thing is that U is a unitary matrix (since it only changes the basis) and you of course get the same thing..