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A Definition of a decay process

  1. Dec 5, 2016 #1
    Consider the following momentum-space Feynman diagram

    https://upload.wikimedia.org/wikipedia/commons/b/b5/Lepton-interaction-vertex-eeg.svg

    This Feynman diagram is the leading-order contribution to the any of the following processes:

    1. ##e^{-} \rightarrow e^{+} + \gamma##
    2. ##e^{+} \rightarrow e^{-} + \gamma##
    3. ##\gamma \rightarrow e^{+} + e^{-}##

    The process in 3 is clearly the decay of a photon to an electron-positron pair.

    Why can't the processes in 1 and 2 be considered to be decay processes of the electron and positron respectively?
     
  2. jcsd
  3. Dec 5, 2016 #2
  4. Dec 5, 2016 #3
    What if I replace the photon by a massive scalar boson?
     
  5. Dec 5, 2016 #4
  6. Dec 5, 2016 #5
    Ok, so why is the process ##h \rightarrow h + h## allowed?

    In the rest frame, the total energy is the rest energy of ##h##, but the outgoing particles have twice the rest energy, so energy is not conserved.
     
  7. Dec 5, 2016 #6
    Sorry, this doesn't make any sense to me, I better let the pros take this one.
     
  8. Dec 5, 2016 #7

    Orodruin

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    It is not allowed for free Higgses.

    Neither is any of your reactions listed in the OP.
     
  9. Dec 5, 2016 #8
    Ah! I see!

    But consider the following Lagrangian

    $$\mathcal{L} = \bar{\psi}_{e}(i\gamma^{\mu}{\partial_{\mu}}-y_{e}\nu)\psi_{e}-y_{e}\bar{\psi}_{e}h\psi_{e}+\frac{1}{2}(\partial_{\mu}h)(\partial^{\mu}h)-\frac{1}{2}\left(2|\kappa^{2}|\right)h^{2}-\frac{\lambda}{6}\nu h^{3}-\frac{\lambda}{24}h^{4}$$

    which has the intreaction term ##-y_{e}\bar{\psi}_{e}h\psi_{e}##. This term leads to an interaction vertex of the form given in the diagram:

    https://i.imgsafe.org/64394a297d.jpg

    This diagram describes the leading-order Feynman diagram of the process ##h \rightarrow e^{-} + e^{+}##.

    Can we have a process like ##h \rightarrow h##, where the electron and anti-electron are in a loop? Can we call this process the decay of ##h##?
     
  10. Dec 5, 2016 #9

    Orodruin

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    No, the ##h## can decay to the electron positron pair if its mass is large enough. The ##h \to h## diagram is not a decay, it is a contribution to the self energy of the ##h##.
     
  11. Dec 5, 2016 #10
    Ok, so (assuming that the mass of ##h## is greater than the mass of ##e^{-}##) apart from ##h \rightarrow e^{-} + e^{+}##, what other decays of ##h## are possible?
     
  12. Dec 5, 2016 #11

    Orodruin

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    The mass needs to be greater than two electron masses for the decay to occur.

    What other decays that will occur depends on what else you stuff into your theory.

    As given, none (apart from possible higher order decays to several electron-positron pairs, but those are suppressed).
     
  13. Dec 5, 2016 #12
    Why are higher-order decays suppressed?
     
  14. Dec 5, 2016 #13

    Orodruin

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    Because they contain more vertices and are therefore only available at higher order in perturbation theory. If the total mass of the decay products is close to that of the decaying particle, you will also get additional phase space suppression.
     
  15. Dec 5, 2016 #14
    But you can't deny that these higher-order decay processes still exist!

    It's just that they are suppressed since they contain more vertices, right?
     
  16. Dec 5, 2016 #15

    Orodruin

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    Where did I do that?

    Also, they will not exist unless the sum of the masses of the outgoing particles is smaller than the ##h## mass.

    This is what I said.
     
  17. Dec 5, 2016 #16
    Is the image in the following link the leading order Feynman diagram for the (higher-order) decay process ##h \rightarrow e^{-} + e^{+} + e^{-} + e^{+}##

    https://i.imgsafe.org/650622580e.jpg?
     
  18. Dec 5, 2016 #17

    Orodruin

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    No. It has three electrons and three positrons in the out state. Also, there are several different diagrams contributing.
     
  19. Dec 5, 2016 #18
    How do I understand that there are several different diagrams contributing?

    By several different diagrams, do you mean several interaction vertices?
     
  20. Dec 6, 2016 #19

    Orodruin

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    Because you can draw them while respecting the Feynman rules.

    No.
     
  21. Dec 6, 2016 #20
    Okay, so if instead, we had the Lagrangian

    $$\mathcal{L} = \bar{\psi}_{e}(i\gamma^{\mu}{\partial_{\mu}}-y_{e}\nu)\psi_{e}-y_{e}\bar{\psi}_{e}h\psi_{e}+\bar{\psi}_{\mu}(i\gamma^{\nu}{\partial_{\nu}}-y_{\mu}\nu)\psi_{\mu}-y_{\mu}\bar{\psi}_{\mu}h\psi_{\mu}+\frac{1}{2}(\partial_{\mu}h)(\partial^{\mu}h)-\frac{1}{2}\left(2|\kappa^{2}|\right)h^{2}-\frac{\lambda}{6}\nu h^{3}-\frac{\lambda}{24}h^{4}$$

    we can write down processes like

    $$h \rightarrow e^{-} + e^{+}, \qquad\qquad h \rightarrow e^{-} + e^{+} + e^{-} + e^{+}, \dots$$

    $$h \rightarrow \mu^{-} + \mu^{+}, \qquad\qquad h \rightarrow \mu^{-} + \mu^{+} + \mu^{-} + \mu^{+}, \dots$$

    $$h \rightarrow e^{-} + e^{+} + \mu^{-} + \mu^{+}, \dots $$

    I was wondering if a particle (e.g. muon) must always be produced with its corresponding antiparticle (e.g. antimuon).

    Also, I was wondering if there are only a finite number of processes for this reaction.
     
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