My QFT knowledge has very much rusted and i got confused by these few lines from Peskin and Schroeder:

p.27: " [..] the amplitude for a particle to propagate from $y$ to $x$ is $\langle 0| \phi(x) \phi(y) |0\rangle $. We will call this quantity $D(x — y)$."

(The relation with the commutator is explicitly calculated at (2.53) p.28, + bottom of p.29: $$ [\phi(x) , \phi(y)] = \cdots = D(x-y)- D(y-x) = \langle 0|[\phi(x) , \phi(y)]|0\rangle$$ the r.h.s. are implicitly understood as being proportional to the $\mathbb{1}$ operator)

Finally the expressions of the retarded and Feynman propagators are given (2.55) p.30

$$D_R := \theta (x^0 -y^0) \langle 0|[\phi(x) , \phi(y)]|0\rangle $$

and (2.60) p.31 (without commutators)

$$D_F := \theta (x^0 -y^0) \langle 0|\phi(x) \cdot \phi(y)|0\rangle + \theta (y^0 -x^0) \langle 0|\phi(y) \cdot \phi(x)|0\rangle $$

which by definition of "propagator" or "Green's function" satisfy $(\square +m^2) G(x,y)= -i\delta^4(x-y)$.

Now my confusion comes from the fact that I remember that propagators had the interpretation of the amplitude of propagation, cf. e.g. wikipedia or Peskin last 2 lines p.82, but this is wrong? (three different functions obviously cannot have the exact same interpretation!)

Remark: I am aware that from the defining relation of a Green's function, one can express the "evolution" of a solution of the (Klein-Gordon) equation, so that in some sense propagators expresses an idea of propagation

The first question is too easy so here is a second: Are propagators always combinations of the $D(x-y)$?

  • for interacting fields
  • for more general PDE?

Remark: I'm not trying to relate interacting theories to the free one, so $D(x-y)$ stands for the amplitude of propagation in each theory not the free one. The underlying idea is that $D(x-y)$ has a clear interpretation while the propagators would not have an easy interpretation on their own unless they are just simple functions of these $D(x-y)$.

  • $\begingroup$ Ok, actually things got clearer as I wrote the question: I'm actually satisfied by the dumbest interpretation possible: the different propagators are explicitly combinations of $\langle 0|\phi(x) \phi(y) |0\rangle$ so the interpretation is that it is the sum of different probability amplitudes whatever that means $\endgroup$ – Noix07 Oct 31 '15 at 16:54
  • $\begingroup$ Thinking of propagators as "propagating amplitude" is sometimes misleading. It provides with some nice intuition, but the actual amplitude is a bit harder to calculate (as you have to take into account eg interactions, through correlators, and LSZ, etc.) $\endgroup$ – AccidentalFourierTransform Oct 31 '15 at 19:47
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    $\begingroup$ In the simple case of one-incoming/one-outgoing particle, then the correlators are simpliy the propagators (plus loops corrections). In general, the correlation functions can always be written (in perturbation theory) as sums over Feynman propagators $D_F$ (together with numeric constants or matrices). $\endgroup$ – AccidentalFourierTransform Oct 31 '15 at 19:52
  • $\begingroup$ Yeah, hmm... I was not trying to relate the interacting theory to the free one but rather the Green's function to the "$D(x-y)$" to get an interpretation. Nevertheless, I indeed now remember that in perturbation theory we could express everthing with $D_F$ whose interpretation is just that of a combination of $D$. (+ vertices or loops) $\endgroup$ – Noix07 Oct 31 '15 at 22:21

Propagators are independent of interactions. They only depend on the free part of the Lagrangian. For example, the KG Lagrangian reads $$ \mathcal L_{KG}=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\mathcal H_I(\phi) $$ where $\mathcal H_I$ could be, eg, $\frac{g}{4!}\phi^4$. There are different definitions of the propagator (all equivalent, of course). For our purposes, the easiest definition is that it is the Green function of the KG PDE. The fastest way to find $D_F$ is to go to momentum space, and "solve" for the propagator: $$ \mathcal L_{KG}(k)=\phi(k)(k^2+m^2)\phi(k)-\mathcal H_I(k)\equiv \phi(k) D_F^{-1}\phi(k)-\mathcal H_I(k) $$ where we find $$ D_F(k)=\frac{1}{k^2+m^2} $$ (there are several conventions on the normalisation of $D_F$, so you can sometimes find some numerical factors in front of the expression above)

On the other hand, its not difficult to prove that $D_F$ agrees with your expression: $$ D_F=\langle 0|\mathcal T\ \phi_0(x)\phi_0(y)|0\rangle $$ where $\phi_0(x)$ is the solution of the free KG equation. If you add interactions, then the amplitude to go from $x$ to $y$ is no longer the propagator. You can call the amplitude $C(x,y)$, and we usually call this new quantity a correlator. In perturbation theory, it is not difficult to prove that $C$ can always be written in terms of the propagator (as an asymptotic expansion over some parameter that weights the interactions). The correlator depends on $\mathcal H_I$, but the propagator does not. You can prove that if $\mathcal H_I=0$, then $C(x,y)=D_F(x,y)$.

With this, we can answer your first two questions: 1) $D_F$ is not the propagation amplitude, but $C(x,y)$ is. They agree when there are no interactions. 2) The propagator is independent of interactions, but the correlator is not. The latter can be written as an asymptotic expansion where the different terms include powers of $D_F$, so that correlators can always we written in terms of propagators.

As for your third question, the propagator does depend on the PDE, as it depends on the free part of the Lagrangian. For example, if you take a different Lagrangian, as Dirac's $$ \mathcal L_D=\bar\psi(i\not\partial-m)\psi-\mathcal H_I(\psi) $$ then using the same line of reasoning as before we find that the propagator is $$ S(k)=\frac{1}{\not k-m} $$

I hope this answers your questions.


Some people make no distinction between the concept of propagator and correlator. They call the former free propagator and the latter exact propagator (or even bare vs dressed propagators). Note that this is not very usual, for most people just call $D_F$ the propagator, and call $C$ correlator (or two-point function). IMO it is very important to understand that these are different objects, so I don't like to use the same word for them.

That being said, it seems that you just call both objects propagator. If this is the case, you should convince yourself that the free propagator is independent of interactions, and the exact one is not. So if you are going to use the same word for both, then at least you should not use the same notation. For example, in Srednicki's book he writes $\Delta$ for the free propagator, and $\boldsymbol \Delta$ for the exact one. By the way, Peskin & Schroeder do call $C$ correlator, and $D_F$ propagator (see page 82).

Anyway, if you see the links I posted, you can see that $D_F(k)=1/k^2+m^2$ (in KG's case), or $D_F(k)=1/\not k-m$ (in Dirac's). Then it is easy to see that there are no coupling constants in these expressions, so they are indeed independent of interactions. By the Dyson series you can calculate the correlators (which do depend on the coupling parameters), which are given, at fist order, by the propagators.

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  • $\begingroup$ Nice try, "$D_F$ is not the propagation amplitude" is what I wanted but I'm afraid I disagree on rest: the propagator is defined as the "inverse" (remains to clarify in which algebra) of the operator so it does depend on interactions $\endgroup$ – Noix07 Oct 31 '15 at 22:40
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    $\begingroup$ @user39158, there is nothing wrong with this answer. Indeed, the propagators are fixed by the kinetic terms (i.e., by the free theory). The dependence on the specific theory shows up only on the vertices of the Feynman diagrams, not in the internal/external lines. $\endgroup$ – Mr. K Nov 1 '15 at 15:09
  • $\begingroup$ Ok maybe, but when I look at Peskin for example (4.10) p.82 it says "two-point Green's function" (by the way = "two-point correlator") for the "$\phi^4$ theory". What perturbation does is to related it to the free field (4.31) p.87 (actually rather the field in "interacting picture", i.e. evolves as the free field) $\endgroup$ – Noix07 Nov 1 '15 at 16:27
  • $\begingroup$ I mean "free propagator" not "field": "relate the interacting propagator to the free propagator" $\endgroup$ – Noix07 Nov 1 '15 at 16:47
  • $\begingroup$ Yes, exactly. It means that the two-point function can be related to the free propagator. The tool for doing so is to expand the $U$ operator through Dyson series. The $U$ operator is given by something like $U\sim \exp(\int \phi^4)$, and when expanded in series we can go back to free propagators by Wick's theorem. You can find the details in the pages 88-90. $\endgroup$ – AccidentalFourierTransform Nov 1 '15 at 19:16

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