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Calculating the Invariant Matrix

  1. May 29, 2014 #1

    ChrisVer

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    I am having one question... If we know the form of the effective Lagrangian, let's say the form:

    [itex]L= g (\bar{\psi}_{e} \gamma^{\mu} P_{L} \psi_{\nu})(\bar{\psi}_{p} \gamma_{\mu} P_{L} \psi_{n}) [/itex]

    How can someone calculate the spin averaged invariant matrix [itex]\large M[/itex]?
    I mean I can do the whole calculation if it's to have the [itex]u,v[/itex] in place of [itex]\psi[/itex]. I am having also a problem of seeing when this is done in QED scatterings too, since we know that [itex]M= j^{\mu}j_{\mu}[/itex] with [itex]j^{\mu}= \bar{\psi} \gamma^{\mu} \psi [/itex]...
    but I don't know if I have the whole Dirac spinor [itex]\psi[/itex] what someone is supposed to do?

    In most cases for the weak interaction, I'm seeing [itex]M[/itex] given by [itex]u,v[/itex] (like in Halzen & Martin)...

    Thanks
     
  2. jcsd
  3. May 29, 2014 #2
    It's exactly the calculation you are used to in terms of the u and v spinors. What appears in the Lagrangian (for example the Fermi Lagrangian that you wrote) is the Dirac field which can be expressed as a sum of Dirac spinors, u and v.

    So suppose you want to describe the effective muon decay, [itex]\mu^-\to e^-+ \nu_\mu +\bar\nu_e[/itex]. Then your effective Lagrangian is going to be:
    $$
    L=\frac{G_F}{\sqrt{2}}\left(\bar\psi_\mu\gamma^\alpha P_L\psi_{\nu_\mu}\right)\left(\bar \psi_{\nu_e}\gamma_\alpha P_L\psi_e\right).
    $$
    However, when computing the matrix element using the Feynman diagram this Lagrangian leads to:
    $$
    \mathcal{M}=\frac{G_F}{\sqrt{2}}\left(\bar \nu_\mu\gamma_\alpha P_L\mu\right)\left(\bar e\gamma^\alpha P_L\nu_e\right).
    $$
    Now in the matrix element you have the spinors appropriate for each particle and thus you can use all the well-known trace technology to compute the square amplitude averaged over the spins.
     
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