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2 questions on symmetries: "conserved in interaction => eigenstate in interaction"? |
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| Apr2-12, 05:17 PM | #1 |
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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 [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) |
| Apr3-12, 04:38 AM | #2 |
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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...) |
| Apr3-12, 12:33 PM | #3 |
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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)) |
| Apr3-12, 05:14 PM | #4 |
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Recognitions:
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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); 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. Particle Physics B. R. Martin, G. Shaw John Wiley & Sons 1992. Sam |
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