Must Particles Be Eigenstates of Conserved Quantities in Interactions?

[nonequilibrium]

2 questions on symmetries: "conserved in interaction => eigenstate in interaction"?

Hello, I'm currently taking an introductory course in elementary particles (level: Griffiths) and I have 2 questions that are severely bothering me; all help is appreciated! They are related to Griffiths' "Introduction to Elementary Particles".

A) Say observable A (with operator \hat A) is conserved in, say, the strong interaction, then why must any particle interacting with the strong force (incoming or outgoing) be in an eigenstate of \hat A? For example in the strong interaction (which conserves S) particles must be in an S eigenstate, or in the presumption that the weak force conserves CP, the particles would have to be in a CP eigenstate to partake in weak decay. Why?

B) After establishing that the K-naught particles are not CP-eigenstates, Griffiths construes eigenstates | K_1 \rangle := \frac{1}{\sqrt{2}} \left( |K^0 \rangle - | \overline K^0 \rangle \right) and analogously | K_2 \rangle since if the weak force conserves CP, then kaons can only interact with the weak force in the forms of the (only) CP eigenstates |K_1 \rangle and |K_2\rangle (cf A). He then mentions that CP is not conserved, and finally claims (p147)

Evidently, the long-lived neutral kaon is not a perfect eigenstate of CP after all, but contains a small admixture of K_1:
|K_L \rangle = \frac{1}{\sqrt{1+\epsilon^2}} \left( |K_2 \rangle + \epsilon |K_1 \rangle \right)

But how does this follow? I have no clue! In Griffiths it was proven that K_1 and K_2 are CP eigenstates, so what is he saying?

 

[strangerep]

In Griffiths it was proven that K_1 and K_2 are CP eigenstates, so what is he saying?

Let's review some stuff about ordinary spin again. A spin-1/2 particle can have either spin-up or spin-down along an arbitrary direction. Let's say we pick the z direction. The operator for the z-component of the angular momentum vector is denoted $J_z$. Now suppose we send a beam of such particles through a Stern-Gerlach setup which outputs two beams corresponding to the 2 possible orientations of $J_z$. Now we pick one of those output beams and try to measure the x component of spin $J_x$. What happens? We find that half have $J_x$ along the +x direction and the rest in the -x direction. This is because the $J_z$ and $J_x$ operators don't commute. Hence they do not have simultaneous eigenvectors. (That can only happen if the operators commute). A specific eigenvector of $J_z$ is in general a superposition of eigenvectors of $J_x$.

The same essential principle underlies what Griffiths is saying about the K states.
If K_1 and K_2 are CP eigenstates then an arbitrary superposition of them is not (in general).

(Not sure how much of this will make sense, but I guess you'll tell me...)

 

[nonequilibrium]

Oh my, I'm sorry, first of all it seems I had misread Griffiths' quote! I think I read that K_L (defind as such) was an eigenstate of CP, which is of course ridiculous, hence my confusion. I'm sorry for wasting your time. So to be clear: the things you said are familiar to me.

However, even reading the above Griffiths quote correctly, I still don't understand it (as in: I don't see how he draws that conclusion), but I think I would understand it if I understood my question (A), so everything comes back to my first question: how does it follow that if an interaction conserves observable A, only eigenstates of \hat A can experience that interaction force? (and apparently the other way around is also true, and that would answer my revised question (B))

 

[samalkhaiat]

A) Say observable A (with operator \hat A) is conserved in, say, the strong interaction, then why must any particle interacting with the strong force (incoming or outgoing) be in an eigenstate of \hat A? For example in the strong interaction (which conserves S) particles must be in an S eigenstate, or in the presumption that the weak force conserves CP, the particles would have to be in a CP eigenstate to partake in weak decay. Why?

You will understand this better when you study Noether theorem; if you think of \hat{A} as conserved “charge”, then only charged particles, i.e., the eigenstates of the conserved charge, will experience the interaction described by \hat{A}.

I tell people to think of particle as a set of charges (a collection of real numbers);
Space-time charges (mass and angular momentum) related to the Poincare symmetry group. These put restriction on possible motion in space-time.
Internal charges (electric, iso-spin, strangeness, ...) related to certain groups of “internal” symmetry. These put restrictions on possible “motions in the internal space” i.e., interactions.

B) After establishing that the K-naught particles are not CP-eigenstates, Griffiths construes eigenstates | K_1 \rangle := \frac{1}{\sqrt{2}} \left( |K^0 \rangle - | \overline K^0 \rangle \right) and analogously | K_2 \rangle since if the weak force conserves CP, then kaons can only interact with the weak force in the forms of the (only) CP eigenstates |K_1 \rangle and |K_2\rangle (cf A). He then mentions that CP is not conserved, and finally claims (p147)
But how does this follow? I have no clue! In Griffiths it was proven that K_1 and K_2 are CP eigenstates, so what is he saying?

An elementary, yet extremely accurate, treatment of K^{0}-\bar{K}^{0} mixing, CP violation and strangeness oscillations can be found in chapter 9 of the following textbook;

Particle Physics
B. R. Martin, G. Shaw
John Wiley & Sons 1992.

Sam