I am following a course on QFT that is based upon canonical quantization and not path integrals.

When calculating scattering amplitudes we compute, at the relevant order, the corresponding matrix element, which might look like

$$\int \textrm{d}^4x\textrm{d}^4y\,\langle 0|\hat{a}_p\hat{a}_q T\{\bar{\psi}_x\phi_x\psi_x\bar{\psi}_y\phi_y\psi_y\}\hat{a}^\dagger_r\hat{a}^\dagger_s|0\rangle .$$

In order to compute this matrix element, we simply replace the time ordered product by the normal order of contractions, and then we see we need to fully contract the fields with the external states, as is explained in Peskin and Schroeder and in Computing S-Matrix Elements from Feynman Diagrams.

We then come up with Feynman Rules, define propagators as vacuum expectation values and retrieve the expression for the amplitude.

However, I have just read that both propagators and Feynman Rules can be recovered from the Lagrangian of the theory. How is this related to the usual definition of propagators $$\langle 0|T\{\phi_x\phi_y\}|0\rangle=D(x-y)$$ and the Feynman rules derived from the Wick's theorem?

Also, please, could you provide me with a good reference to derive Feynman Rules from Wick's Theorem? What I have found so far is just a list of rules to apply, but I would like to understand where they come from.

Edit: Just to clarify the question. I am familiar with the definition of propagator $\langle 0|T\{\phi_x\phi_y\}|0\rangle=D(x-y)$. How is this related to the fact that propagator can also be retrieved from second order terms in the lagrangian? View for instance https://cds.cern.ch/record/319569/files/AT00000309.pdf.

Second part of the question is: since Feynman Rules are just a way to calculate Wick contrations in the reasoning I exposed in the first part of my question, I do not see why they can also be derived from the Lagrangian, as explained in the same link. In P&S, Feynman Rules are obtained as a way to compute the matrix element of the fully contracted fields and initial/final states.


1 Answer 1

  1. The propagator $\langle 0\vert T\phi(x)\phi(y)\vert 0\rangle$ is a Green's function for the free equation of motion of the field $\phi$. If you write the quadratic term in the Lagrangian as $\phi D \phi$ for some differential operator $\phi$, then this means the propagator is the inverse of $D$ (where "inverse" means it is the Green's function for this operator), i.e. you can tell the propagator for a field by looking at its quadratic term in the Lagrangian.

  2. "The Feynman rules" is a bit of a vague term that often encompasses not only the graphical rules for how to associate Feynman diagrams to contractions of operators and how to evaluate those diagrams, but also how to know which diagrams to draw to compute the S-matrix elements/scattering amplitudes for a theory with a given Lagrangian. E.g. "the only internal vertices allowed are those with four lines attached" for $\phi^4$ theory is "a Feynman rule" in this latter, broader sense.

    The LSZ formula says that S-matrix elements are of the form $\langle \Omega \vert T\prod_i \phi(x_i)\vert \Omega \rangle$ where $\lvert \Omega\rangle$ is the interacting vacuum and then a computation in the interaction picture shows that those can be computed by looking at $$ \langle 0\vert T\prod_i \phi_I(x_i) \mathrm{e}^{-\mathrm{i}\int H_I(t)\mathrm{d}t}\vert 0\rangle,$$ where $H_I$ is the interacting part of the Hamiltonian. It is here that the Lagrangian implicitly enters - for the standard cases, this $H_I$ is just the potential in the Lagrangian. A Taylor expansion of the exponential in terms of some parameter in the potential then gives us expressions of the form $$\langle 0\vert T \prod_i \phi_I(x_i) \left(\int V(\phi_I(y))\mathrm{d}y\right)^n\vert 0\rangle$$ and it is to these expressions that we now apply Wick's theorem/the "generic" Feynman rules. So, knowing this, you can "read off" the diagrams that will occur in this expansion just by looking at the potential term in the Lagrangian, and that's what people mean when they say you can read off "the Feynman rules" from the Lagrangian.

  • $\begingroup$ Thank you very much for your answer, I believe it is complete and fully addresses my question. Regarding the propagator, is it a coincidence that it is a Green's Function for the differential operator? I mean, why is this fact related to a time-ordered vacuum expectation value? $\endgroup$ Feb 3, 2022 at 9:41
  • $\begingroup$ @Elementarium If you're asking about any other way to see this except just explicitly computing that it indeed is the Green's function, I don't know of one off the top of my head. $\endgroup$
    – ACuriousMind
    Feb 3, 2022 at 10:07

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