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2 questions on symmetries: conserved in interaction => eigenstate in interaction ?by nonequilibrium
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#1
Apr212, 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 Knaught particles are not CPeigenstates, 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) 


#2
Apr312, 04:38 AM

Sci Advisor
P: 1,898

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...) 


#3
Apr312, 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)) 


#4
Apr312, 05:14 PM

Sci Advisor
P: 892

2 questions on symmetries: conserved in interaction => eigenstate in interaction ?
I tell people to think of particle as a set of charges (a collection of real numbers); Spacetime charges (mass and angular momentum) related to the Poincare symmetry group. These put restriction on possible motion in spacetime. Internal charges (electric, isospin, strangeness, ...) related to certain groups of “internal” symmetry. These put restrictions on possible “motions in the internal space” i.e., interactions. Particle Physics B. R. Martin, G. Shaw John Wiley & Sons 1992. Sam 


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