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Supersection Rules

  1. Sep 25, 2013 #1
    I was just reading that because of superselection rules we cannot superpose two particles with different electric charges. But when I look in my particles physics books it seems there are decay processes that do this all the time: consider [itex]\pi^+ → μ^+ +\nu_\mu [/itex], for example. Can anyone tell me what's going on here?

    Thanks a lot!
     
  2. jcsd
  3. Sep 25, 2013 #2

    stevendaryl

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    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.
     
  4. Sep 25, 2013 #3
    But the antimuon has positive charge, and the neutrino no charge, no?
     
  5. Sep 25, 2013 #4

    Bill_K

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    You're misunderstanding the meaning of the word "superposition". A state with a muon AND a neutrino is not a superposition. A superposition would be a state that is maybe a muon OR maybe a neutrino.

    As stevendaryl illustrated, the state has probability |α|2 of being |A> and probability |β|2 of being |B>. This can't happen if |A> and |B> have different total charge.
     
  6. Sep 25, 2013 #5
    Thanks. 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?

    And what about when we mix quarks as in the CKM matrix -- is that a mixed state or a superposition also? Thanks a lot!
     
  7. Sep 25, 2013 #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.

    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.
     
  8. Sep 26, 2013 #7
    Thanks Kith -- especially for your explanation of what's going on when we represent the pion decay in that way. Final question: when we represent neutrino mixing as sums of neutrinos of different flavors, are we making a superposition precluded by superselection rules then?

    Thanks again!
     
  9. Sep 26, 2013 #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 specific states of such systems. So in standard QM notation, π+ and ρ+ would be written |π+> and |ρ+> just like we write |1s> and |1p> for hydrogen states.

    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?
     
  10. Sep 27, 2013 #9
    Hmm... I guess I thought that the different neutrinos have different properties, and as such couldn't be superposed any more than two particles with different charges can. But I guess they only differ in their 'generations' (leaving aside any mass differences). So maybe neutrino mass mixing doesn't violate any superselection rules after all. Thanks for you input kith!
     
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