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We calculate the real-space propagator $\Delta(x)$ for a free real scalar field $\varphi(x)$ with mass $m$ by performing the Fourier transform (using sign convention +---)

$$\Delta(x) = \int \frac{d^3k\, d\omega}{(2 \pi)^4} \frac{e^{-i k \cdot x}}{k^2 - m^2}$$

Depending on whether we deform the contour of integration for the $\omega$ integral to the the same or opposite sides of the two poles at $\omega = \pm \sqrt{{\bf k}^2 + m^2}$, we get either the retarded or the Feynman propagator (or the advanced or the anti-time-ordered propagator, but let's ignore those two options).

From the Wikipedia article on propagator, the retarded propagator is

$$G_\text{ret}(x) = i\, \langle [ \varphi(x), \varphi(0) ] \rangle\, \Theta(x_0)= \frac{1}{2 \pi} \delta(\tau^2) - \frac{m\, J_1(m \tau)}{4 \pi \tau}$$

if $x$ is inside the origin's future light cone, and zero otherwise, where $\Theta$ is the step function, $\delta$ the Dirac delta function, $\tau$ the proper time $\sqrt{x \cdot x}$, and $J_1$ the Bessel function. The Feynman propagator is

$$G_F(x) = -i\, \langle \mathcal{T} \varphi(x) \varphi(0) \rangle = \begin{cases} -\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H^{(2)}_1(m \sqrt{s}) \qquad &\text{if } s \geq 0 \\ -\frac{i m }{4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) \qquad &\text{if } s < 0 \end{cases},$$

where $\mathcal{T}$ is the time-ordering symbol, $s$ the spacetime interval $x \cdot x$, $H^{(2)}$ the Hankel function, and $K_1$ the modified Bessel function.

Looking at the operator expectation values, it's clear that the Feynman propagator is the right one to use for calculating probabilities of past-to-future propagation. But looking at the actual functional expressions, $G_\text{ret}$ naively "looks" more causally correct, because it vanishes outside the light cone, as we'd naively expect for a particle propagator. (I know, I know, the spacelike correlations decay exponentially and don't actually violate causality because they can't transmit causal influence, etc. etc.)

Does the retarded propagator $G_\text{ret}$ have any physical significance? I understand why locality requires $G_\text{ret}(x) \equiv 0$ for $x$ outside the lightcone, but it seems a bit strange that once we enforce this, we then completely ignore its value inside the light cone. But I don't know how to intepret the $\langle \varphi(0) \varphi(x) \rangle$ part of $G_\text{ret}$ if $x^0$ is positive.

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The convolution $G_{ret}*f$ of the retarded propagator $G_{ret}$ with a source term $f$ which vanishes sufficiently far in the past is the unique solution of the inhomogeneous Klein-Gordon equation with source term $f$ which vanishes in the far past. It is necessarily a nontrivial superposition of positive and negative energy solutions at all times when it is not identically zero, precisely because it vanishes outside the future light cone of the support of $f$.

The convolution $G_F*f$ of the Feynman propagator $G_F$ with a source term $f$ which vanishes outside a bounded space-time region, on its turn, is a solution of the inhomogeneous Klein-Gordon equation which is of positive energy outside the past light cone of the support of $f$ (i.e. in the far future) and of negative energy outside the future light cone of the support of $f$ (i.e. in the far past). Because of this, it cannot vanish in any nonvoid open region of space-time, unlike $G_{ret}*f$.

The relation between the position-space support and the momentum-space support of $G_{ret}$ and $G_F$ can be seen as a consequence of the uncertainty principle, which restricts the regions where a distribution can vanish in position and momentum space (the latter after a Fourier transform).

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    $\begingroup$ Thank you, this is helpful. But what is the physical (rather than mathematical) interpretation of $G_\text{ret}$ in terms of particle or antiparticle propagation? $\endgroup$
    – tparker
    Commented May 22, 2016 at 23:28
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    $\begingroup$ I find it difficult to offer a physical interpretation of $G_{ret}$ in terms of quantum particles. It is an inherently classical object, in the following sense: if you couple a quantum Klein-Gordon field with a classical external source which vanishes in the far past and uses the vacuum as your "in" state, your state after the source is "turned on" will be a coherent state (hence with an indefnite particle number) such that the one-point function in that state satisfies the inhomogeneous Klein-Gordon equation with the same source. $\endgroup$ Commented May 23, 2016 at 2:17
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    $\begingroup$ @Dr.Yoma you're rather considering the particle interpretation within QFT, which is available in the free case (of course) and, in the interacting case, only asymptotically in the far past or in the far future. That being said, let's address your questions: 1.) you're essentially right regarding the Feynman propagator versus the Wightman propagator $\langle 0|\phi(x)\phi(y)|0\rangle$, in the sense that if you smear the latter with tempered test functions $f,g$ you get the probability transition amplitude for two positive-energy wave packets whose momentum distributions... (to continue) $\endgroup$ Commented Mar 6, 2023 at 19:04
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    $\begingroup$ (continued) ...are given by the respective restrictions of the Fourier transforms of $f,g$ to the positive (squared) mass shell. One must remark that, unlike the Feynman propagator, the Wightman propagator is a bisolution (that is, in $x$ and $y$) of the homogeneous Klein-Gordon equation, that is, without sources. The interpretation of the Feynman propagator is quite different and more closely related to the interacting case (hence the cautionary preamble): if you smear one of its variables with a test function $f$, the result is a solution of the inhomogeneous ...(to continue) $\endgroup$ Commented Mar 6, 2023 at 19:11
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    $\begingroup$ (continued) ...Klein-Gordon equation with source $f$, the result behaves as a negative-energy solution of the homogeneous Klein-Gordon equation in the causal past of the support of $f$ and as a positive-energy solution of the same equation in the causal future of the support of $f$ (in both locations, $f$ should vanish). This essentially means that the Feynman propagator smeared in both variables yields the scattering probability amplitude of a positive-energy "in" particle with a given momentum distribution by a given external source. $\endgroup$ Commented Mar 6, 2023 at 19:19

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