Retarded vs Feynman Klein-Gordon Propagators

Question

Although I follow all the manipulations -- Green's functions, choice of contour/i$\epsilon$ prescription, etc -- I seem to be struggling with too many trees. The forest remains blurry. In particular: the retarded Klein-Gordon propagator $$D_R(x-y)=\Theta(x_0-y_0)\langle 0|[\phi(x),\phi(y)]|0\rangle$$ (Peskin eqn 2.55) is both Lorentz invariant and causal. But so too is the Feynman propagator $$D_F(x-y) =D_R(x-y) + D_R(y-x).$$ What I don't understand must be so basic as to be embarrassing: $D_R$ seems to clearly forbid spacelike transition probabilities -- so then why is $D_F$ the "go to" propagator?

Answer 1

All different Green's functions satisfy the same equation, but it is the boundary conditions that fix the "liberty". In QM we have the initial and "future" data, not just the initial ones.

Answer 2

In typically scattering amplitudes, we compute the $S$ matrix, which encodes probabilities for in states in the distant past to evolve into out states in the distant future. This requires the use of the time ordered product \begin{equation} S = \mathcal{T} e^{i \int_{-\infty}^\infty {\rm d} t H(t)} \end{equation} where $\mathcal{T}$ is the time ordering operator. The fact that we need time ordering, means that we are usually interested (in scattering theory) in time ordered correlation functions, like \begin{equation} G_F(x,y) = \langle 0 | \mathcal{T} \phi(x) \phi(y) | 0 \rangle \end{equation} This is the Feynman Green's function, $G_F(x,y)$. Obviously I have left out a lot of steps, but this is a high-level overview of where $G_F$ comes from.

You are right to be worried about causality; it is not obvious in this formalism. One way to express it is to note that the retarded propagator is related to the Feynman propagator like so: \begin{equation} G_R(x,y) = G_F(x,y) - G_F^\star(y,x) \end{equation} To quote from Section David Tong's QFT notes:

There are words you can drape around this calculation. When $(x- y)^2 < 0$, there is no Lorentz invariant way to order events. If a particle can travel in a spacelike direction from $x \rightarrow y$, it can just as easily travel from $y \leftarrow x$. In any measurement, the amplitudes for these two events cancel.

With a complex scalar field, it is more interesting. We can look at the equation $[ \phi(x)^\dagger, \phi(y)] = 0$ outside the lightcone. The interpretation now is that the amplitude for the particle to propagate from $x \rightarrow y$ cancels the amplitude for the antiparticle to travel from $y \rightarrow x$. In fact, this interpretation is also there for a real scalar field because the particle is its own antiparticle.

In general, I highly recommend Tong's notes for learning QFT.

Answer 3

It just hit me: The causality of $G_R$ is consistent with the sharp boundary of the light cone -- as one demands of a classical relativistic particle. But this is not appropriate for a quantum system, since a sharp boundary at the light cone is not consistent with Heisenberg uncertainty. So another contour is necessary to define a Green's function which can respect both relativistic causality and quantum uncertainty. And that leads to $G_F$.