In Weinberg's book (the quantum theory of fields, volume 1, pag 446, equation (10.4.19) ) is stated that

$\int \ dx \ dy \ dz \ exp(-i p x - ik y + i lz) \langle\Psi_0,T\{J^{\mu}(x)\Psi_n(y)\bar \psi_m(z)\}\psi_0\rangle$

where $J$ is the conserved current in QED, $\Psi_n$ is the spinorial field (electrons and $\Psi_0$ is the exact vacuum of the theory (that is, not the vacuum of the free theory of photons and electrons, but the vacuum of the interacting theory),

may be written as a sum of Feynmann graphs with a outgoing fermion line (with associated the fermion propagator in momentum space), an incoming fermion line (again with associated the propagator) and a photonic line (evidently associated to $J^\mu$) carrying a photon propagator.

Can someone explain me why there is this photonic line? $J^\mu = \bar \Psi \gamma^\mu \Psi$ so I think there should be a (say dashed) line with associated the fourier transform of the "propagator" $[j^\mu(x),\psi_m(y)]$.

(I have used Weinberg notation where upper case field are in Heisenberg picture, lower case in interaction picture)

  • $\begingroup$ it is interacting, right? so this is the interaction term of the photon current with the spinor field, unless the question asks sth different $\endgroup$
    – Nikos M.
    Commented Nov 4, 2014 at 23:58
  • 2
    $\begingroup$ In fact it corresponds to a dressed photon line with a bare photon line stripped away at the external end, see the diagrams I drew in physics.stackexchange.com/q/29890 and in physics.stackexchange.com/q/70882 $\endgroup$
    – Jia Yiyang
    Commented Nov 5, 2014 at 6:53
  • $\begingroup$ @JiaYiyang Thank you very much I was stucked for too long on that! Since you have not post an answer here I will upvote the one in the link. $\endgroup$ Commented Nov 5, 2014 at 21:27


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