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Why is the Electromagnetic Field Tensor in the QED Lagrangian?

  1. Jul 29, 2004 #1
    The QED Lagrangian is given by [tex]\mathcal{L}_{\hbox{QED}} = \bar{\psi}(i\partial - m)\psi
    - \frac{1}{4}(F_{\mu\nu})^2 - e\bar{\psi}\gamma^\mu\psi A_\mu[/tex]

    What is the purpose of the middle term. I know that it represents the energy of the E and B fields. However is that due to the external potential A? I am used to thinking about Lagrangians in terms as equal to the kinetic term minus the potential, T-V. This term doesn't seem to be either and nor does the final interaction term. I didn't put the slash in the partial derivative because I didn't know how to. I understand that it is really a sum of three seperate Lagrangians, that is, [tex]\mathcal{L}_{\hbox{QED}} = \mathcal{L}_{\hbox{Dirac}} + \mathcal{L}_{\hbox{Maxwell}} + \mathcal{L}_{\hbox{int}}.[/tex] What does the middle term represent or what does it do for us?

    Thanks a Million.

  2. jcsd
  3. Jul 29, 2004 #2

    Tom Mattson

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    I deleted your other thread, because it is basically the same as this one. If you want to modify your posts, use the Edit feature--no need to start another thread.

    Anyway, here was my response from the other thread.

    Without that term the QED Lagrangian would not be dynamically closed. That is, the photon field would be an external field that interacts with the matter fields. So we include the EM "kinetic energy". The only way to do it that is both Lorentz covariant and gauge invariant is to introduce a massless term that is quadratic in the EM field strength tensor.
  4. Jul 29, 2004 #3
    The middle term is the Lagrangian for the free EM field. By itself, the resulting Euler-Lagrange equations would give the free EM field dynamics. Just like the first term is the free Dirac field Lagrangian. Alone those two terms describe the two systems that are being allowed to interact via the third term. With one or the other of the first terms missing, you know longer have two interacting fields.

    The free field EM term contains the first-derivatives of the A potential. Without it there is no dynamical information about the behavior of the EM field. That is, there is no differential equation to be satisfield by A. I haven't checked, but the A field would probably either be trivial or arbitrary in that case.
  5. Jul 29, 2004 #4
    Ohhh. I think I know what happened. I hit the submit button instead of the preview button. I hit the stop button on my browser, but apparently is must have been too late. Sorry about that.
    Last edited: Jul 29, 2004
  6. Jul 30, 2004 #5


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    It is a point that the free fields in QED ARE the EM and Dirac fields, not some other fields. QED is thus just a minimal extension of Dirac and EM, all it adds is that they interact! And the quantization, of course.
  7. Aug 1, 2004 #6
    Thanks for the wonderful answers, guys. I have a better understanding of this term now.

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