Correlation Functions and Propagators

Two-point distributions can encode state fluctuations, field commutators, causal response, or in–out propagation. They share the Klein–Gordon operator but are not all inverses of it: Wightman functions and the commutator are homogeneous bisolutions, whereas retarded, advanced, and Feynman functions acquire contact sources. Their different state, ordering, support, and pole prescriptions are the mathematical content of the different physical questions.

Required background. Quantum Mechanics supplies expectation values and operator ordering; Classical Field Theory supplies the Klein–Gordon operator and source equation; Special Relativity supplies causal separation and Lorentz covariance. Helpful background. Calculus and Linear Algebra supplies the calculus and linear algebra used in Fourier representations; Waves, Green Functions, and Geometric Optics supplies the Green-function viewpoint. The retarded and advanced Green-function construction is the closest classical analogue.

Quantum correlations and Green functions

Choose the object from the question before doing algebra. Use $G^+$ for state fluctuations and positive-frequency content, $C$ for causal commutators, $G_{\mathrm R}$ for the response to a source with no effect before the source acts, and $G_{\mathrm F}$ for vacuum time ordering in an in–out perturbative amplitude. $G_{\mathrm A}$ is the future-boundary counterpart of $G_{\mathrm R}$.

For the free scalar field of Free Fields, Creation Operators, and Fock Space, use

$$K\equiv\Box-m^2, \qquad \omega_{\mathbf k}=\sqrt{\mathbf k^2+m^2}.$$

The Minkowski-vacuum Wightman function is

$$G^+(x,y) =\langle0|\hat\phi(x)\hat\phi(y)|0\rangle =\int\frac{d^3k}{(2\pi)^3\,2\omega_{\mathbf k}}\, e^{-i\omega_{\mathbf k}(t_x-t_y) +i\mathbf k\cdot(\mathbf x-\mathbf y)}.$$

It is a state-dependent distribution and a homogeneous bisolution,

$$K_xG^+(x,y)=K_yG^+(x,y)=0,$$

including distributionally at coincidence. It is singular there, but that singularity is not a Green-function delta source. Reversing the operator order gives $G^-(x,y)=G^+(y,x)$.

Let

$$C(x,y)=\langle0|[\hat\phi(x),\hat\phi(y)]|0\rangle =G^+(x,y)-G^+(y,x).$$

For a free field this commutator is a state-independent c-number, obeys $K_xC=K_yC=0$, and vanishes at spacelike separation. Its equal-time data, derived from the ladder algebra, are

$$C(0,\mathbf r)=0, \qquad \partial_\tau C(0,\mathbf r) =-i\delta^{(3)}(\mathbf r),$$

where $\tau=t_x-t_y$ and $\mathbf r=\mathbf x-\mathbf y$. Our mostly-plus conventions define

$$G_{\mathrm R}(x,y) =-i\,\theta(t_x-t_y)C(x,y), \qquad G_{\mathrm A}(x,y) =+i\,\theta(t_y-t_x)C(x,y),$$

so that

$$K_xG_{\mathrm R}(x,y) =K_xG_{\mathrm A}(x,y) =\delta^{(4)}(x-y).$$

$G_{\mathrm R}$ has support only when $x$ lies in or on the causal future of $y$; $G_{\mathrm A}$ has the time-reversed support. The retarded function is therefore the inverse used for a causal initial-value response. Its sign changes if one defines the source term or Klein–Gordon operator differently.

For clarity, the equation $K\phi=J$ follows from adding $-\int d^4x\,J\phi$ to the scalar action. The corresponding Hamiltonian perturbation is $+\int d^3x\,J\phi$. The linear-response page instead writes $H'=-fX$ and obtains the kernel $+i\theta\langle[X(t),X(0)]\rangle$. For $X=\phi$, identifying $f=-J$ makes that kernel $-G_{\mathrm R}$ and gives the same response $\delta\langle\phi\rangle=G_{\mathrm R}*J$. Here the star denotes integration over the source point. The apparent sign difference comes from the named source, not from a different causal law.

The contact normalization follows rather than being assumed. Because $K=-\partial_\tau^2+\boldsymbol\nabla_{\mathbf r}^2-m^2$ and $C(0,\mathbf r)=0$, no $\delta'(\tau)$ term survives. Distributional differentiation gives

$$\begin{aligned} K\!\left[\theta(\tau)C(\tau,\mathbf r)\right] &=-\delta(\tau)\partial_\tau C(0,\mathbf r) =+i\delta^{(4)}(x-y),\\ K\!\left[\theta(-\tau)C(\tau,\mathbf r)\right] &=+\delta(\tau)\partial_\tau C(0,\mathbf r) =-i\delta^{(4)}(x-y). \end{aligned}$$

Multiplying the first line by $-i$ and the second by $+i$ proves $KG_{\mathrm R}=KG_{\mathrm A}=\delta^{(4)}$. This jump calculation is the independent sign and normalization check.

Pole prescriptions and time support

The same support information appears as a boundary value in the complex $p^0$ plane. With $e^{ip\cdot(x-y)}=e^{-ip^0\tau+i\mathbf p\cdot\mathbf r}$,

$$\begin{aligned} G_{\mathrm R}(x-y) &=\lim_{\epsilon\to0^+} \int\frac{d^4p}{(2\pi)^4}\, \frac{e^{ip\cdot(x-y)}} {(p^0+i\epsilon)^2-\omega_{\mathbf p}^2},\\ G_{\mathrm A}(x-y) &=\lim_{\epsilon\to0^+} \int\frac{d^4p}{(2\pi)^4}\, \frac{e^{ip\cdot(x-y)}} {(p^0-i\epsilon)^2-\omega_{\mathbf p}^2}. \end{aligned}$$

Both retarded poles, $p^0=\pm\omega_{\mathbf p}-i0$, lie below the real axis. For $\tau\lt 0$ the contour closes above, encloses no pole, and gives zero; for $\tau\gt 0$ it closes below and gives $-iC$. The advanced prescription puts both poles above, so the support is reversed.

The Feynman function is instead time ordered:

$$G_{\mathrm F}(x,y) =\langle0|T\hat\phi(x)\hat\phi(y)|0\rangle =\theta(t_x-t_y)G^+(x,y) +\theta(t_y-t_x)G^+(y,x).$$

In the Fourier convention $e^{ip\cdot(x-y)}$ with $p\cdot x=-p^0t+\mathbf p\cdot\mathbf x$,

$$G_{\mathrm F}(x-y) =\lim_{\epsilon\to0^+} \int\frac{d^4p}{(2\pi)^4}\, \frac{-i\,e^{ip\cdot(x-y)}} {p^2+m^2-i\epsilon}, \qquad K_xG_{\mathrm F}=i\delta^{(4)}.$$

The $i\epsilon$ displaces the positive- and negative-energy poles on opposite sides of the contour. It implements vacuum time ordering, not retarded support.

More explicitly,

$$p^2+m^2-i\epsilon \simeq- \left(p^0-\omega_{\mathbf p}+i0\right) \left(p^0+\omega_{\mathbf p}-i0\right).$$

The positive-energy pole is below the real axis and the negative-energy pole is above it. For $\tau\gt 0$, closing below selects the positive-energy Wightman term; for $\tau\lt 0$, closing above selects the reversed-order term. This contour argument derives

$$G_{\mathrm F} =\theta(\tau)G^+ +\theta(-\tau)G^-.$$

Finally, the derivative jump is $\partial_\tau G^+(0,\mathbf r)-\partial_\tau G^-(0,\mathbf r)=\partial_\tau C(0,\mathbf r)=-i\delta^{(3)}(\mathbf r)$. Therefore time ordering gives $KG_{\mathrm F}=i\delta^{(4)}$, explaining why its source normalization differs by $i$ from the retarded and advanced inverses. Tong 2006–2007, §2.6.1 and §§2.7–2.7.2, printed pp. 36–41, PDF derives the commutator support and these Green-function boundary values.

A compact comparison

Object	Extra input	Primary use
$G^+$	state and operator order	vacuum or state fluctuations; detector spectra
$C$	canonical algebra and dynamics	microcausality; spectral information
$G_{\mathrm R/A}$	future/past support	causal response and initial-value evolution
$G_{\mathrm F}$	state, time ordering, $i\epsilon$	perturbative amplitudes and generating functionals

The table is not a list of interchangeable names. For example, a source switched on near $y$ produces a classical response governed by $G_{\mathrm R}$, whereas an internal line in an in–out scattering amplitude uses $G_{\mathrm F}$. Neither object by itself is an observable.

Spectral and curved-spacetime lessons

In an interacting vacuum, the two-point function admits a spectral representation: an isolated pole can mark a stable one-particle state, while positive continuous spectral weight describes multiparticle channels. The exact propagator is not generally the reciprocal of the classical quadratic polynomial; self-energy corrections alter its analytic structure. Srednicki, Chapter 13, Eqs. (13.1)–(13.17), printed pp. 106–108, PDF derives the pole-plus-continuum form; his propagator includes one extra overall factor of $i$ relative to $G_{\mathrm F}$ on this page.

On a globally hyperbolic curved spacetime, normally hyperbolic free-field equations retain unique retarded and advanced inverses tied to causal support. A preferred vacuum Wightman or Feynman function generally does not follow from the geometry alone. This is why causal Green functions can be geometrically robust even when particle language is observer- or state-dependent. Wald 1995, printed pp. 6–10, especially pp. 8–9, PDF explains the missing preferred Hilbert-space construction. The geometric continuation appears on Advanced, Retarded, and Causal Propagators, while Quasifree States and n-Point Functions develops the state-dependent side.

Checks and failure modes

• $G^\pm$ and $C$ are homogeneous distributions even at coincidence. $G_{\mathrm R}$, $G_{\mathrm A}$, and $G_{\mathrm F}$ are homogeneous only away from their contact source.
• A denominator without its contour or boundary value does not specify a Lorentzian propagator. The Fourier phase convention is part of the prescription.
• Spacelike commutator vanishing does not mean the Wightman function vanishes. Vacuum correlations are not controllable signals.
• A Euclidean Green function becomes a Lorentzian correlator only after a specified analytic continuation and boundary prescription.
• Retarded propagation does not mean support only on the light cone. Massive fields, and massless fields in many curved spacetimes, also develop interior “tails.”

Exercises

Build a causal source response

Let $J$ be a smooth source of compact spacetime support and define

$$\phi_J(x)=\int d^4y\,G_{\mathrm R}(x,y)J(y).$$

Show that $K\phi_J=J$ and that $\phi_J(x)$ receives no contribution from source points outside the causal past of $x$.

Solution

Acting with the differential operator under the integral gives

$$K_x\phi_J(x) =\int d^4y\,K_xG_{\mathrm R}(x,y)J(y) =\int d^4y\,\delta^{(4)}(x-y)J(y) =J(x).$$

By definition, $G_{\mathrm R}(x,y)$ vanishes unless $x$ lies in or on the causal future of $y$. Thus only $y\in J^-(x)$ contributes. This support statement is stronger than merely requiring $t_y\lt t_x$ because it also excludes spacelike-separated source points.

Infer support from pole placement

For fixed spatial momentum, evaluate the support of

$$I(\tau)=\lim_{\epsilon\to0^+} \int_{-\infty}^{\infty}\frac{dp^0}{2\pi}\, \frac{e^{-ip^0\tau}} {(p^0+i\epsilon)^2-\omega^2}.$$

Identify the prescription and give $I(\tau)$ explicitly for $\omega\gt 0$.

Solution

Both poles, $p^0=\pm\omega-i0$, lie below the real axis. For $\tau\lt 0$, the exponential decays on the upper semicircle; closing there encloses no pole, so $I(\tau)=0$. For $\tau\gt 0$, closing below is clockwise and gives $-i$ times the sum of the two residues:

$$I(\tau) =-i\,\theta(\tau) \frac{e^{-i\omega\tau}-e^{i\omega\tau}}{2\omega} =-\theta(\tau)\frac{\sin(\omega\tau)}{\omega}.$$

This is the retarded boundary value for the fixed-momentum operator $-\partial_\tau^2-\omega^2$. Its derivative jumps by $-1$ at the origin, so applying that operator gives $\delta(\tau)$.

Continue to Interactions, Time Ordering, and Perturbation Theory to see why $G_{\mathrm F}$ organizes the expansion around a free vacuum.

References

• Mark Srednicki, Quantum Field Theory, official author draft of the Cambridge University Press text, Chapter 13, printed pp. 106–108. Author-draft PDF.
• David Tong, Quantum Field Theory, University of Cambridge Part III lecture notes, Michaelmas Terms 2006 and 2007, §2.6.1 and §§2.7–2.7.2, printed pp. 36–41. Official author PDF.
• Robert M. Wald, “Quantum Field Theory in Curved Spacetime,” arXiv:gr-qc/9509057 (1995), printed pp. 6–10. Author-manuscript PDF.