If $\phi$ is a quantum field there's a simple interpretation of $$\langle 0 | T \phi(x) \phi(y) | 0 \rangle.$$ This quantity is simply the amplitude to propagate from $y$ to $x$. Diagrammatically, we can compute it by summing all Feynman diagrams with one $\phi$ field at $x$ and one $\phi$ field at $y$.

Now let $j^\mu$ be a conserved current, such as the electromagnetic current in QED. Occasionally, we compute correlators of the form $$\langle 0 | T j^\mu(x) j^\nu(y) |0 \rangle.$$ Is there a simple, possibly diagrammatic interpretation of such a correlation function? How about for a general current in a more complicated theory? How about for $n$ currents?

I know that you can just say what I said in my first paragraph with $\phi$ replaced with $j^\mu$, but I'm looking for something in terms of the fundamental fields.

  • $\begingroup$ The two-point function $\langle 0|T\ \phi(x)\phi(y)|0\rangle$ is not the amplitude to propagate from $y$ to $x$. More generally, correlation functions have no physical interpretation whatsoever. $\endgroup$ – AccidentalFourierTransform Mar 23 '18 at 18:16
  • $\begingroup$ @AccidentalFourierTransform We'll have to agree to disagree again! What I'm saying is approximately true in some limit (tiny coupling, put the system in a box, add a UV cutoff, etc.) so I'm happy to use it if it helps me with intuition. $\endgroup$ – knzhou Mar 23 '18 at 18:20
  • $\begingroup$ What you are saying is not true, not even approximately, not even if you introduce cutoffs and small couplings. There is no meaningful notion of propagation amplitude in a relativistic theory. See e.g. the very pedagogical paper arxiv.org/abs/1712.06605 (in particular, section 2.3). Cheers! $\endgroup$ – AccidentalFourierTransform Mar 23 '18 at 18:30
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    $\begingroup$ @AccidentalFourierTransform I guess I should've added that I'm perfectly willing to do nonrelativistic field theory! Hell, I'll even be happy with an answer in 1+1 dimensions if that helps. $\endgroup$ – knzhou Mar 23 '18 at 18:41
  • $\begingroup$ @knzhou any chance you've come up with an answer to your own question? $\endgroup$ – Dwagg Jan 24 at 20:28

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