Isospin: how serious must I take it? Superposition of proton and neutron?by nonequilibrium Tags: isospin, neutron, proton, superposition 

#37
Apr312, 07:40 AM

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Hence in the course of beta decay a neutron (isospin down) would continuously transform into a superposition of an isospin up (p+e + antinu) and down (n) state. 



#38
Apr312, 07:57 AM

P: 62

@mr. vodka: sorry once again for my notation; [itex]P^2=P^{\mu}P_{\mu}=h^2\vec{p}^2[/itex] (as operators), as you have written. In my opinion, the reason why we assume only mass eigenstates is that, when the time goes to "infinity", a good approximation of the situation we are considering is that of free particles; now, if I understood correctly, oneparticle irreducible representations of the Poincarč group are "classified" by mass and spin; so this clarifies, in my opinion, why we see, in scattering processes, only mass eigenstates: in few words, because irreducible representations of the Poincarč group are "defined" by mass and spin; you may ask why we build irreducible representations of the Poincarč group; well, this is related to Poicarč invariance; a good reference for this is Weinberg. Moreover, one may ask why one considers also energy eigenstates ("delocalized"); in my opinion, this has to do with formal scattering theory, which allows some kind of manipulations. And, finally, when you ask if you interpreted correctly my point of view, the answer is yes, you interpreted correctly my point of view. @tom.stoer http://www.physicsforums.com/showthread.php?t=591256 Thank you very much! Francesco 



#39
Apr312, 09:55 AM

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#40
Apr312, 10:04 AM

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#41
Apr312, 10:05 AM

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Francesco ps [EDIT]: a question has just occured to my mind: if it is possible to produce in a scattering states which are not mass eigenstates, how do you compute the cross section or the decay rate for such a process? In the phase space factor there i an explicit mass, or am I wrong? 



#42
Apr312, 10:23 AM

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Francesco, in that thread you were citing you wrote:
"The Feynman diagram for every such process is weighted with factors of the PMNS matrix; so why are we speaking about electron neutrinos? The point is that we are not able to measure and observe the mass eigenstates of the neutrinos, our experimental apparatus are not so powerful. What we can say, in my opinion, is that we can treat the system as a statistical mixture of three kind of neutrinos. If we were able to see the mass eigenstates, then we would not observe oscillations." The three mass eigenstates would still form a superposition and not a statistical mixture. The situation is not much different from the emission of radiation by an excited atom. The radiation will not be an energy eigenstate but have a certain linewidth which is due to the exponential decay of the excitation. In the case of photons the measurement devices are so refined that you can detect the phase of the components of the different energy eigenstates. 



#43
Apr312, 10:27 AM

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#44
Apr312, 10:45 AM

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#45
Apr312, 10:52 AM

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Hey guys, we have a thread for oscillations of neutrinos or masseigenstates which are allowed) and we have this thread for superpositions of "charge eigenstates" which are forbidden due to superselection rules. I think here we should focus on the latter one.




#46
Apr312, 11:19 AM

P: 62

Moreover I think I didn't understand your comparison with the emission of radiation: let me be precise and distinguish mass eigenstates from energy eigenstates: if I have understood your comparison, the role of the observable "energy" in your example is played by "mass" in my interpretation; but the two situations are different, from my point of view. In your case you mean that a superposition of energy eigenstates is possible (and I agree with this), but what I stress is that only mass eigenstates can be produced: notice that a linear superposition of energy eigenstates of photons is still a mass eigenstate. Is it wrong or did I misunderstand your comparison? In this case why? One more final question, which might be helpful to clarify my point of view: suppose we have the standard model with a right handed neutrino and we add a Yukawa term analogous to that of the quark. After the electroweak symmetry breaking and diagonalization of the mass matrices, what is the difference between quarks and neutrinos? Nobody have doubts that in calulating effective low energy operators from high energy contribution (a very awful expression to indicate all contribution to hadronic state which are deduced by quark interactions, very very very roughly speaking; e.g. the mixing of the k kbar system already cited) mass eigenstates should be used. What is different in the case of neutrinos? 



#47
Apr312, 01:11 PM

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But I don't think my confusion stems from me misunderstanding (regular) spin (?). 



#48
Apr312, 01:53 PM

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As I said; I guess strangerep wants to discuss the states [itex]l,l_3\rangle[/itex] and (forbidden) superpositions i.e. superselection rules for angular momentum; think about
[tex]1,1\rangle\,+\,1,1\rangle[/tex] [tex]1,1\rangle\,+\,2,2\rangle[/tex] 



#49
Apr312, 02:25 PM

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#50
Apr312, 02:30 PM

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I didn't even know there were any forbidden superpositions when it comes to regular spin...




#51
Apr312, 02:40 PM

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Then how shall
[tex]1,1\rangle\,+\,2,2\rangle[/tex] be realized in nature? A simple definition of a superselection rule are states 1> and 2> for which <1A2>=0 holds for all observables A. I think this is satisfied for the above mention angular momentum eigenstates in QM. 



#52
Apr312, 02:44 PM

P: 1,406

Not to be smart***y, but how would you realize e.g. [itex]1,1 \rangle[/itex] in nature?




#53
Apr312, 02:48 PM

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#54
Apr312, 02:50 PM

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