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Box^2 A_μ = J_μ ; [Dirac operator] ψ = 0 or not?

  1. May 1, 2012 #1
    The four vector J_μ is the source of the Electromagnetic potential four vector A_μ. If I wanted a source term for the Dirac equation (not that I think there is such a thing) could it be (must it be) a spinor source term (what ever that means)?

    Related question, consider the two Feynman diagrams below. In the first diagram the current is a source of a virtual photon. In the second a photon is destroyed and a electron positron pair is produced. Can I think of the virtual photon as being a source for the Dirac equation?

    Diagram from,

    http://www.google.com/imgres?hl=en&...49&start=12&ndsp=15&ved=1t:429,r:6,s:12,i:116

    Thanks for any help!
     

    Attached Files:

  2. jcsd
  3. May 2, 2012 #2
    Presumably a source term would replace zero on the right hand side of the Dirac equation:

    [itex](i\gamma^\mu\partial_\mu - m)\psi = 0[/itex]

    Then yes, for the indices to work out, whatever goes on the RHS must be a Dirac spinor.

    The Lagrangian for a Dirac field coupled to the electromagnetic field is

    [itex]\bar{\psi}\{i\gamma^\mu(\partial_\mu - ieA_\mu) - m\}\psi[/itex]

    If you proceed from this Lagrangian to the "classical equation of motion", whatever that means for a spinor field, you get

    [itex]\{i\gamma^\mu(\partial_\mu - ieA_\mu) - m\}\psi = 0[/itex]

    or

    [itex](i\gamma^\mu\partial_\mu - m)\psi = -e\gamma^\mu A_\mu \psi[/itex]

    I guess if you like you can think of the RHS as a source term, but this interpretation is maybe not so clean since the "source" of psi depends on psi.

    In QFT I think we prefer to speak of generalized "interactions" rather than fields being "sources" for one another. As you are seeing, the same interaction term

    [itex]e\bar{\psi}\gamma^\mu A_\mu \psi[/itex]

    can do many things. It can let an electron produce a photon, or let an electron absorb a photon, or destroy an electron-positron pair and create a photon, or destroy a photon and create an electron-positron pair, or destroy a photon, electron, and positron, or create a photon, electron, and positron. Lorentz invariance forces us into a theory where all these processes are essentially equivalent and described by the same term in the Lagrangian.
     
  4. May 3, 2012 #3
    Thanks Duck! So it seems if I take some random Dirac spinor, say (1,0,0,0) and multiply it by δexp(iωt) where δ is some small number and let this source function act on a very small region of space and for a very short time (say one cycle) we can excite the Dirac field? And if ω > m waves will propagate outward? Does the propagator tell us the likely momentum in those outward waves? This is all unphysical but fun to think about?

    Thanks again!
     
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