Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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,


    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]


    [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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook