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

by nonequilibrium
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nonequilibrium
#1
Apr2-12, 05:17 PM
P: 1,412
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 [itex]\hat A[/itex]) 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 [itex]\hat A[/itex]? 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 [itex]| K_1 \rangle := \frac{1}{\sqrt{2}} \left( |K^0 \rangle - | \overline K^0 \rangle \right)[/itex] and analogously [itex]| K_2 \rangle[/itex] since if the weak force conserves CP, then kaons can only interact with the weak force in the forms of the (only) CP eigenstates [itex]|K_1 \rangle[/itex] and [itex]|K_2\rangle[/itex] (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 [itex]K_1[/itex]:
[itex]|K_L \rangle = \frac{1}{\sqrt{1+\epsilon^2}} \left( |K_2 \rangle + \epsilon |K_1 \rangle \right)[/itex]
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?
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strangerep
#2
Apr3-12, 04:38 AM
Sci Advisor
P: 1,898
Quote Quote by mr. vodka View Post
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
#3
Apr3-12, 12:33 PM
P: 1,412
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 [itex]\hat A[/itex] can experience that interaction force? (and apparently the other way around is also true, and that would answer my revised question (B))

samalkhaiat
#4
Apr3-12, 05:14 PM
Sci Advisor
P: 892
2 questions on symmetries: conserved in interaction => eigenstate in interaction ?

Quote Quote by mr. vodka View Post
A) Say observable A (with operator [itex]\hat A[/itex]) 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 [itex]\hat A[/itex]? 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 [itex]\hat{A}[/itex] as conserved “charge”, then only charged particles, i.e., the eigenstates of the conserved charge, will experience the interaction described by [itex]\hat{A}[/itex].

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 [itex]| K_1 \rangle := \frac{1}{\sqrt{2}} \left( |K^0 \rangle - | \overline K^0 \rangle \right)[/itex] and analogously [itex]| K_2 \rangle[/itex] since if the weak force conserves CP, then kaons can only interact with the weak force in the forms of the (only) CP eigenstates [itex]|K_1 \rangle[/itex] and [itex]|K_2\rangle[/itex] (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 [itex]K^{0}-\bar{K}^{0}[/itex] mixing, CP violation and strangeness oscillations can be found in chapter 9 of the following text book;

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

Sam


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