Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
High Energy, Nuclear, Particle Physics
Open Questions about Neutrinos Today
Reply to thread
Message
[QUOTE="Vanadium 50, post: 6435610, member: 110252"] Much of the confusion about oscillation vs. interference comes about from thinking of the flavor states as the "real particles" that somehow don't have a well-defined mass. If I say "I produce an electron neutrino", I really mean "I produce a particular admixture of mass eigenstates". If the three mass states were far apart - far enough apart that we could tell them apart - in mass, one would see three entries in the PDG: [LIST] [*][itex]\pi \rightarrow \mu \nu_1[/itex] [*][itex]\pi \rightarrow \mu \nu_2[/itex] [*][itex]\pi \rightarrow \mu \nu_3[/itex] [/LIST] with appropriate branching fractions. There would be no oscillations: you'd produce a ν[sub]2[/sub] or whatever and that would be the end of it. Of course you could detect an electron in a beam of pion-produced neutrinos: ν[sub]2[/sub] has a (fixed) probability of producing an electron. However, for neutrinos in our universe, we cannot distinguish which mass eigenstate was in flight, so QM says we have to add amplitudes. And thus the neutrinos[I] interfere[/I]. This is exactly the same as in the double-slit experiment. Slits far apart and you have no interference. Slits close together so you can't tell which slit the light wave went through, and you do. Now, you might say "This is all well and good with muon decay, where the Michel electron has a range of energies. But if I make my neutrinos via pion decay, it's a two-body decay. I make my muon beam monochromatic, and I measure the energy of the decay muon amd now I know the 4-momentum of the neutrino: contract it and I know the mass. It doesn't matter that this is impractical. It's possible, and that's all I need to add intensities and not amplitudes." There are several reasons this line of reasoning doesn't work, all of them subtle. I think the easiest to understand is the relation E[sup]2[/sup]-p[sup]2[/sup]=m[sup]2[/sup] is a statement about plane waves. But the pion and muon are not exactly plane waves. The neutrinos are produced in a decay pipe, so the muon and pion don't have plane-wave wavefunctions. They have particle-in-a-box wavefunctions. (Because they are particles in a box.) And while the deviation from E[sup]2[/sup]-p[sup]2[/sup]=m[sup]2[/sup] is small because they are in a n equals a zillion state, it's still large enough to make the mass of the neutrino produced uncertain. And so you need to add amplitudes, not intensities. So [itex]\pi \rightarrow \mu \nu \implies \nu + N \rightarrow N + e [/itex] is actually the interference of three processes: [itex]\pi \rightarrow \mu + \nu_1 \implies \nu_1 + N \rightarrow N + e [/itex], [itex]\pi \rightarrow \mu + \nu_2 \implies \nu_2 + N \rightarrow N + e [/itex] and [itex]\pi \rightarrow \mu + \nu_3 \implies \nu_3 + N \rightarrow N + e [/itex]. The process appears to evolve with time just as a two-slit interference pattern appears to evolve in space. PS In a decay pipe, the particle isn't even in an energy eigenstate. It's in a mixture. But all have quantum numbers in the zillions. [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
High Energy, Nuclear, Particle Physics
Open Questions about Neutrinos Today
Back
Top