quantum field theory - Propagator and probability amplitude that a particle propagates

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 $(\square +m^2) G(x,y)= -i\delta^4(x-y)$.

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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

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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 $D(x-y)$.