Retarded, Advanced, and Feynman Propagators

Retarded, advanced, and Feynman propagators invert the same free wave operator with different boundary data. The choice follows the physical question: response to a source, a final boundary condition, or a vacuum time-ordered amplitude. This page compares those choices and calculates the homogeneous terms that distinguish them. The scalar contour derivations are given on Klein–Gordon Propagators; the matrix construction is given on Dirac Propagators.

Required background. Klein–Gordon Propagators supplies the prescribed scalar inverses; Dirac Propagators fixes their spinor counterparts. Helpful background. Locality and Causality Warnings explains causal support and spacelike correlations.

Boundary data select an inverse

Use $\hbar=c=1$, $m>0$, $L=\Box+m^2$, and Fourier kernel $e^{-ip\cdot x}$. Throughout the comparison $LG_X=\delta^4$: every scalar object in the table has the same unit source.

Choice	Placement of the positive and negative frequency poles	Boundary question
Retarded $G_R$	Both below the real frequency axis	What field is generated with no response before the source?
Advanced $G_A$	Both above the real frequency axis	What solution obeys the corresponding future boundary condition?
Feynman $G_F$	Positive below, negative above	What inverse has positive frequency forward and negative frequency backward in time?

The retarded solution to a localized source has support only in its causal future. The advanced solution has support only in its causal past. The Feynman solution generally has neither support restriction. These are statements about spacetime support, stronger than merely vanishing at negative or positive time.

The conventional vacuum correlator $D_F=-iG_F$ has the Feynman prescription but satisfies $LD_F=-i\delta^4$. Multiplying it by the appropriate factor corrects its source, not its boundary condition. For Dirac fields the corresponding pair is $S_F=iK_F$ and $DK_F=\delta^4I_4$. Always identify the operator and source before comparing objects called a propagator.

Homogeneous differences live on the mass shell

Set $a=p^2-m^2$. The distributional identity

$$\frac1{a\pm i0} =\operatorname{PV}\frac1a\mp i\pi\delta(a)$$

separates an off-shell principal value from an on-shell term. For the retarded prescription the infinitesimal imaginary part has the sign of $p^0$. Therefore

$$\begin{aligned} \widetilde G_F&=-\operatorname{PV}\frac1a+i\pi\delta(a),\\ \widetilde G_R&=-\operatorname{PV}\frac1a +i\pi\operatorname{sgn}(p^0)\delta(a),\\ \widetilde G_A&=-\operatorname{PV}\frac1a -i\pi\operatorname{sgn}(p^0)\delta(a). \end{aligned}$$

Subtracting gives

$$\begin{aligned} \widetilde G_F-\widetilde G_R &=2\pi i\,\theta(-p^0)\delta(a),\\ \widetilde G_F-\widetilde G_A &=2\pi i\,\theta(p^0)\delta(a),\\ \widetilde G_R-\widetilde G_A &=2\pi i\,\operatorname{sgn}(p^0)\delta(a). \end{aligned}$$

Multiplication by the Fourier symbol $-a$ annihilates every difference. This explicitly verifies that changing the boundary prescription adds a homogeneous solution. It also shows which energy sheet changes. The statement “on shell” refers here to momentum space; it does not mean support only on the spacetime light cone.

The time-symmetric inverse $(G_R+G_A)/2$ has Fourier transform $-\operatorname{PV}(1/a)$. It is not $G_F$: the latter also contains $i\pi\delta(a)$. Averaging causal and anticausal response is therefore not the definition of vacuum time ordering.

What changes for a prescribed source

At fixed spatial momentum let $E=\sqrt{\mathbf p^2+m^2}$. Performing just the on-shell delta integral in the first difference gives

$$g_F(t,\mathbf p)-g_R(t,\mathbf p) =\frac{i}{2E}e^{iEt}.$$

For a smooth time-compact Fourier source $J_{\mathbf p}(t)$, the two source solutions differ by

$$\phi_F(t,\mathbf p)-\phi_R(t,\mathbf p) =\frac{i e^{iEt}}{2E} \int_{-\infty}^{\infty}dt'\,e^{-iEt'}J_{\mathbf p}(t').$$

The integral fixes a negative-frequency homogeneous coefficient. If it is nonzero, the Feynman solution need not vanish before the source begins. No causal-response paradox follows: the two solutions obey different boundary conditions. For an experiment that turns on a source from a specified initial state, the retarded response is the relevant inverse, together with that state’s homogeneous data.

If the source has a smooth Fourier transform that vanishes on both mass-shell sheets, all three particular solutions coincide, subject to the required distributional regularity. Thus the prescription can be irrelevant for a particular source even though the kernels differ. This is a condition on the source, not a general permission to discard $i0$.

Vacuum correlations and measurable response

In a free quantum field vacuum, positive-energy annihilation modes and negative-frequency creation modes combine with time ordering to produce the Feynman inverse. For spinors the exchange of fermionic operators supplies the negative-time minus sign derived on the Dirac-propagator page.

A time-ordered correlation is useful for amplitudes and perturbation theory. A linear response of an observable to a perturbing source is instead built from a retarded commutator, with normalization fixed by the chosen coupling. Local observables commute at spacelike separation. Odd fermion fields satisfy the corresponding anticommutator condition, and the fundamental Dirac retarded kernel uses that graded bracket. An individual time-ordered or positive-frequency correlation need not vanish at spacelike separation. Confusing these objects incorrectly turns a correlation into a signal.

Backgrounds require further care. A stationary problem with a chosen vacuum or frequency splitting may admit a Feynman construction. For a time-dependent background, specifying an in/out state or another state prescription is additional information; a free $i0$ denominator is not a complete definition of every two-point function. The retarded classical inverse is fixed by a well-posed initial-value problem without making that vacuum choice.

Exercises

1. Find $G_F-(G_R+G_A)/2$ in momentum space and verify its homogeneous source equation.

Solution

The difference is $i\pi\delta(p^2-m^2)$. Multiplying by $-(p^2-m^2)$ gives zero. The mean of retarded and advanced inverses lacks this mass-shell term.

2. At fixed spatial momentum, apply a rectangular source pulse: $J_{\mathbf p}(t)=J_0$ for $0<t<T$, and zero otherwise. Find $\phi_F-\phi_R$. For which nonzero pulse durations do the two solutions coincide? Explain whether the common solution is then zero during the pulse.

Solution

The finite time integral is elementary:

$$\phi_F-\phi_R =\frac{J_0}{2E^2}(1-e^{-iET})e^{iEt}.$$

It vanishes when $ET=2\pi n$ for a positive integer $n$: this pulse has zero Fourier component at the mode frequency. For $0<t<T$, the retarded solution is nevertheless $J_0(1-\cos Et)/E^2$. After the pulse it is $J_0[\cos E(t-T)-\cos Et]/E^2$, which vanishes for those durations. Thus equality of the two boundary prescriptions for this source does not mean that the source produced no response. The mode returns to rest after an integer number of oscillation periods.

3. Starting from the scalar differences, how are the unit-source Dirac differences obtained?

Solution

Apply $-(i\gamma^\mu\partial_\mu+m)$ to each scalar difference. For example, $K_F-K_R=-(i\gamma^\mu\partial_\mu+m)(G_F-G_R)$. Its Dirac source vanishes because the scalar difference is homogeneous. Multiplying by $i$ would instead compare the corresponding correlator-normalized kernels.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill, 1964. Green-function boundary conditions.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940. Propagators, distributions, and time ordering.
• Tong, David. Lectures on Quantum Field Theory. University of Cambridge, 2006, sections 2.6–2.7 and 5.4–5.5. Scalar causality and propagators; spinor propagators.