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QED photon field

  1. Mar 23, 2009 #1
    This may be a simple question, but I feel as though I don't clearly understand this fully. How come we can take the photon field to be represented by the four potential instead of the E and B fields? Is it equivalent and more convient to do? Also by using the 4 potential do you atomatically have relativity built in instead of using the E and B fields? Thanks to anyone who can clear up this simple issue.
  2. jcsd
  3. Mar 24, 2009 #2


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    E and B are defined in terms of the potentials phi and A, so phi and A are more fundamental I would say. And yes, the four-potential as better covariant properties
  4. Mar 24, 2009 #3
    Using the four-vector potential is equivalent and more convient in many repsects: you atomatically have relativity built in, four-vector is simpler than four-tensor, etc. The Dirac equation that contains this four-vector is similar to the Hamilton-Jacoby classical mechanical equation which is also expressed via potentials rather than field tesnions.

    The Newton and the Hamilton-Jacoby equations give the same classical solutions for particle trajectories despite the "gauge" liberty in choosing the potentials. The same is valid in QED. There are equivalent QED formulations in terms of the field tensions (Hammer C. L., Good R. H. // Ann. of Phys. 1961. V. 12. P. 463., Mandelstam S. // Ann. of Phys. 1962. V. 19. P. 1.)

  5. Apr 26, 2011 #4
    it is just a direct application of Poincaré's Lemma
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