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S matrix and decaying particles

  1. Nov 21, 2014 #1
    Hi All,

    The S-matrix is defined as the inner product of the in- and out-states, as in Eq. (3.2.1) in Weinberg's QFT vol 1:
    [itex]S_{βα}=(Ψ−β,Ψ+α)[/itex]

    [itex]\Psi_{±α}[/itex] are the eigenstates of the full Hamiltonian with a non-zero interaction term.

    Can [itex]\alpha[/itex] describes a neutron ? Since it is not stable, it is not an eigenstate of the full Hamiltonian, so it should not... But then how can you compute the decay rate of the neutron within the S matrix formalism ?

    I am also wondering how to take neutrinon into account. Supposing [itex]\alpha[/itex] can describe a neutron,
    the neutron may decays as: n → p + [itex]e^−[/itex] + [itex]\overline{\nu}_e[/itex]. But [itex]\overline{\nu}_e[/itex] is not an eigenstate of the full Hamiltonian ; I would rather use the mass eigenstates [itex]\nu_1, \nu_2, \nu_3[/itex]. But then it means that the S matrix formalism is unable to predict that it is really a [itex]\nu_e[/itex] that is created at the time of the interaction, which could lead to false predictions... What do you think ?
     
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  3. Nov 21, 2014 #2

    Orodruin

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    If you have access to Peskin-Schröder, there is a discussion on unstable particles in chapter 7 in relation to the discussion of the optical theorem.

    When it regards neutrinos, you are right. The states should really be the mass eigenstates. However, they remain coherent over pretty large distances due to the very small mass splittings, resulting in neutrino oscillations. If you do not measure the neutrino, you can simply do the total decay rate as a sum of the decay rates into the different mass eigenstates. The result will be essentially ##\Gamma = (|U_{e1}|^2+|U_{e2}|^2+|U_{e3}|^2)\Gamma_0##, where ##\Gamma_0## is the decay rate you would have to one neutrino state if there was no mixing and ##U## is the PMNS matrix. Since the PMNS (as far as we know) is unitary, the sum of the squared PMNS matrix elements equals 1.
     
  4. Nov 21, 2014 #3

    Avodyne

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    Strictly speaking, only stable particles are allowed in the in and out states. The neutron then appears as a resonance in scattering of a neutrino, electron, and proton. The decay rate is identified with the imaginary part of the location of the pole in the scattering amplitude.

    Happily, it turns out that this decay rate is equivalent to what you get by starting with a neutron in the in state, even though you are not really allowed to do this.

    Srednicki's text also has a discussion of this.
     
  5. Nov 23, 2014 #4
    Thanks for your replies. I will have a look to these references
     
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