I'm studying QFT by David Tong's lecture notes.

When he discusses causility with real scalar fields, he defines the propagator as


then he shows that the commutator of two scalar fields at arbitrary spacetime points $x,y$ is $$[\phi(x),\phi(y)]=D(x-y)-D(y-x),$$

which since the lhs involves operators, I assumed there should also be an implicit identity operator multiplying the rhs.

When he moves on to discuss the propagator of a fermion, however, he defines

$$iS_{\alpha\beta}=\{\psi_{\alpha}(x),\bar{\psi}_{\beta}(y)\},$$ as the propagator, where $\bar{\psi}=\psi^{\dagger}\gamma^{0}$ is the Dirac adjoint.

Could anyone explain to me from where this definition comes from? I find it confusing, since he didn't call the commutator of two scalar fields as the propagator before, he defined it as the vacuum expectation of the field in two different points as $D(x-y)$. Or are the propagators named differently for scalars and spinors? Why does it has to be a matrix instead of a number (meaning, why define it with the adjoint to the right of the anti-commutator instead of the other way around)? Also, why define it with an $i$ factored out?

  • $\begingroup$ Are you asking why it is an anticommutator rather than a commutator? Cf. physics.stackexchange.com/q/17893/2451 $\endgroup$
    – Qmechanic
    Commented Feb 25, 2018 at 15:11
  • 1
    $\begingroup$ Not really. In my understanding it is an anticommutator due to the quantization of the spinor. What troubles me are the questions I wrote in the last paragraph. $\endgroup$ Commented Feb 25, 2018 at 15:17
  • 3
    $\begingroup$ Not related to the question, but I think it's worth mentioning that if you're studying QFT on your own, I wouldn't recommend using lecture notes. If the author is squeezing all that into only 150 pages, the text is bound to leave out important details and explanations. You might end up spending more time than reading a more comprehensive 500+ pages textbook. One suggestion is Srednicki's book. The pre-print is available for free: web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf Also, welcome to Physics Stack Exchange! It's nice seeing a familiar face here! I miss your videos! $\endgroup$
    – Wood
    Commented Feb 25, 2018 at 19:54
  • $\begingroup$ Thank you for your advise. To tell you the truth I took a long time trying to read QFT textbooks, and didn't find any that worked for me. They all seemed formidable and impenetrable. This lecture notes were recommended by my supervisor, and so far I was at least able to make some progress. But I got no illusions about the gaps: I'm not turning into any QFT expert after reading that. Also, thank you for your kind words of welcoming and about my videos :) $\endgroup$ Commented Feb 25, 2018 at 20:48

1 Answer 1


The trouble with saying $A$ is the "definition" of $B$ is that, although we might be able to prove certain definitions for $B$ are equivalent, that doesn't mean they should all be considered "the" definition as some are more useful starting points. I'll try to suggest what the true starting point should be if you want a unified account of propagators, but of course every other useful equation is deducible from that.

Tong introduces each propagator when he needs it for something, and in doing so starts with whichever expression for the propagator suits his present purpose. But I would argue the way one should define a propagator initially is as a Green's function. He does this with $D$, at least if you ignore the matrix element and skip straight to the integral representation, because by inspection $(i\gamma^\mu \partial_\mu-m)D=\delta (x-y)$. (The Feynman propagator of the Klein-Gordon field makes for a conceptually easier starting point, but for some reason Tong gets to that slightly later.)

Why do we care about Green's functions? We can think of the action as a functional-integral generalisation of the matrix identity $\int \exp (-\frac{1}{2}x\cdot Ax) d^n x = (2\pi)^{n/2}\det D,\,D:=A^{-1}$ for an invertible $n\times n$ square matrix $A$. For example, the Klein-Gordon Lagrangian density is (up to a total derivative) $-\frac{1}{2}\phi^\ast A\phi,\,A:=\square + m^2$. Every propagator starts life with this kind of argument; what does a Green's function of $A$ look like? For all its faults, the first few chapter's of Zee's Quantum Field Theory in a Nutshell uses this approach to get the propagators for such a field and also for the more physically relevant problems of electromagnetism and gravity. (I recommend giving it a quick read, if only to see the subtleties involved in dealing with the $A$-needs-to-be-invertible requirement.)


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