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Thanks a lot!

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- Thread starter metroplex021
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- #1

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Thanks a lot!

- #2

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Thanks a lot!

I thought that the rule was that there cannot be a superposition [itex]|\Psi\rangle = \alpha |A\rangle + \beta |B\rangle[/itex] with the total charge in [itex]|A\rangle[/itex] unequal to the total charge in [itex] |B\rangle[/itex]. A decay such as the one in your example doesn't violate that.

- #3

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But the antimuon has positive charge, and the neutrino no charge, no?

- #4

Bill_K

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As stevendaryl illustrated, the state has probability |α|

- #5

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And what about when we mix quarks as in the CKM matrix -- is that a mixed state or a superposition also? Thanks a lot!

- #6

kith

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Schematically, if you have a pion state [itex]|\pi^+\rangle[/itex] and let it evolve in time, you get something like [itex]α(t)|\pi^+\rangle + β(t)|μ^+\rangle⊗|\nu_\mu\rangle + γ(t)|e^+\rangle⊗|\nu_e\rangle + ...[/itex] If you perform a measurement on this superposition state, your final state will be one of the terms. Each of them is a definite state, not a mixture. For example, the term you were discussing is a state where you have exactly one anti-muon and one myon neutrino.So in the decay above -- which I've written down just as you see it in the textbooks -- represents a mixed state, not a coherent superposition?

If you apply a matrix to a state vector you get a new state vector. You can't get statistical mixtures this way. In order to get them you need a so-called super operator: an operator which takes a state operator (density matrix) to another state operator. Such super operators become necessary only in open quantum systems.And what about when we mix quarks as in the CKM matrix -- is that a mixed state or a superposition also?

- #7

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Thanks again!

- #8

kith

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I've often found it useful to relate things from particle physics and QFT as much as possible to what I already knew from ordinary QM. A very simple but nevertheless important realization was that things like the pion are not physical systems like a hydrogen atom, but specificThanks Kith -- especially for your explanation of what's going on when we represent the pion decay in that way.

No. Ordinary selection rules are often only approximately true because we use "wrong" idealized observables and neglect more subtle effects. Superselection rules on the other hand have to be strictly obeyed. Which rule do you think is violated by neutrino mixing?Final question: when we represent neutrino mixing as sums of neutrinos of different flavors, are we making a superposition precluded by superselection rules then?

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