Spectral Representation and Dispersion Integrals

The previous lesson on causality, commutators, and support explained why the commutator of two local scalar fields vanishes outside the light cone, even though the Feynman propagator itself does not. This page turns that causal statement into an analytic one. The same two-point function that describes propagation in spacetime becomes, after Fourier transform, a function of complex energy or complex invariant momentum. Poles, branch cuts, and $i0$ prescriptions are not optional decorations; they encode the spectrum of the theory.

The main lesson is that an exact two-point function is not just a corrected free propagator. It is a spectral average of free propagators with all possible invariant masses. A stable one-particle state gives an isolated pole. Multiparticle states give a continuum. The discontinuity across the continuum cut is a positive spectral density, and analytic functions with such cuts obey dispersion relations: their real parts are reconstructible from their imaginary parts up to subtraction constants.

This is the bridge from the causal spacetime picture to the momentum-space picture used in scattering theory and renormalization. It also prepares Thresholds, Cuts, and Imaginary Parts, where thresholds and imaginary parts are read directly from the phase space of intermediate states.

Here is the reader’s dictionary for this page:

Spectral feature	Momentum-space singularity	Physical meaning
$Z\delta(s-M^2)$	simple pole at $p^2=M^2$	stable one-particle state created by the field
$\rho_{\rm cont}(s)\theta(s-s_{\rm th})$	branch cut beginning at $s_{\rm th}$	continuum of multiparticle states
$\operatorname{Disc}D(s)$	jump across the cut	on-shell intermediate states
subtraction polynomial	analytic ambiguity	local counterterms or renormalization conditions

The table is deliberately phrased in both analytic and physical language. A spectral representation is useful precisely because it lets one translate between them.

Wightman functions as boundary values

This lesson preserves the spectral-to-dispersion sequence of the course notes. For a detailed construction of exact two-point functions, see Spectral Decomposition of Two-Point Functions. Lorentzian Boundary Conditions and iε fixes the prescription conventions, while Subtracted Dispersion Relations develops the additional hypotheses required for scattering-amplitude applications.

Upper-rim boundary value and discontinuity sign

For a scalar two-point function we write the physical boundary value as

$$\widetilde G_F(p)=iD(p^2+i0),$$

so a free particle has

$$D(z)=\frac{1}{z-m^2},\qquad z\notin[0,\infty),$$

with $D(p^2+i0)=1/(p^2-m^2+i0)$ on the upper rim of the real axis. With this convention, the analytic function and its physical boundary value are

$$D(z)=\int_0^\infty ds\,{\rho(s)\over z-s}, \qquad D(p^2+i0)=\int_0^\infty ds\,{\rho(s)\over p^2-s+i0}.$$

and therefore, on the cut,

$$\operatorname{Im}D(s+i0)=-\pi\rho(s), \qquad \operatorname{Disc}D(s)=-2\pi i\rho(s).$$

Many books absorb different factors of $i$ into the definition of the propagator. The positivity of $\rho$ and the location of the cuts are invariant statements; the sign of $\operatorname{Im}D$ depends on this convention.

Start with the free real scalar field

$$\phi(t,\mathbf x) = \int{d^3p\over(2\pi)^3}{1\over\sqrt{2\omega_{\mathbf p}}} \left( a_{\mathbf p}e^{-i\omega_{\mathbf p}t+i\mathbf p\cdot\mathbf x} +a^\dagger_{\mathbf p}e^{i\omega_{\mathbf p}t-i\mathbf p\cdot\mathbf x} \right), \qquad \omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

The positive-frequency Wightman function is

$$D^+(t,\mathbf x) = \langle0|\phi(t,\mathbf x)\phi(0,\mathbf0)|0\rangle = \int{d^3p\over(2\pi)^3}{1\over2\omega_{\mathbf p}} e^{-i\omega_{\mathbf p}t+i\mathbf p\cdot\mathbf x}.$$

This integral is not an ordinary function everywhere; it is a distribution. The correct definition is obtained by giving time a small negative imaginary part,

$$t\mapsto t-i0,$$

so that positive-energy modes are slightly damped:

$$D^+(t,\mathbf x) = \lim_{\epsilon\downarrow0} \int{d^3p\over(2\pi)^3}{1\over2\omega_{\mathbf p}} e^{-i\omega_{\mathbf p}(t-i\epsilon)+i\mathbf p\cdot\mathbf x}.$$

The reversed ordering is the opposite boundary value:

$$\langle0|\phi(0,\mathbf0)\phi(t,\mathbf x)|0\rangle =D^+(-t,-\mathbf x),$$

which is obtained with $t+i0$. Thus the two Wightman functions are boundary values of the same analytic object from opposite sides.

Positive-energy propagation selects a boundary value in complex time. The ordering $\phi(t)\phi(0)$ is obtained by $t\to t-i0$, while the reversed ordering is obtained by $t\to t+i0$. The time-ordered propagator glues these boundary values together.

The time-ordered two-point function is therefore

$$G_F(t,\mathbf x) = \theta(t)D^+(t,\mathbf x)+\theta(-t)D^+(-t,-\mathbf x).$$

Equivalently,

$$G_F(x) = \int {d^4p\over(2\pi)^4}\,{i\,e^{-ip\cdot x}\over p^2-m^2+i0}.$$

The $+i0$ in momentum space is the same boundary-value information as the $t\mp i0$ prescription in coordinate space. Positive-energy modes propagate forward in the time-ordered product, while negative-energy poles are placed on the opposite side of the contour.

The massless coordinate-space singularity

For a massless scalar in four spacetime dimensions, the positive-frequency integral can be evaluated explicitly:

$$D^+(t,\mathbf x) = \int{d^3p\over(2\pi)^3}{1\over2|\mathbf p|} e^{-i|\mathbf p|(t-i0)+i\mathbf p\cdot\mathbf x} = {1\over4\pi^2}\,{1\over \mathbf x^2-(t-i0)^2}.$$

The reversed ordering is

$$D^+(-t,-\mathbf x) = {1\over4\pi^2}\,{1\over \mathbf x^2-(t+i0)^2}.$$

The light cone is the singular surface. For $t\neq0$,

$$\mathbf x^2-(t\mp i0)^2 = \mathbf x^2-t^2\pm i0\,\operatorname{sgn}(t).$$

The elementary distribution identity

$${1\over z+i0}-{1\over z-i0}=-2\pi i\delta(z)$$

then explains why the massless commutator is supported on the light cone. The singularity is not a bug in the propagator. It is the analytic shadow of sharp relativistic propagation.

In $d>2$ spacetime dimensions, dimensional analysis gives the leading massless short-distance behavior

$$D^+(x) \sim {1\over\left(\mathbf x^2-(t-i0)^2\right)^{(d-2)/2}}.$$

Equivalently, a free scalar field has engineering dimension

$$[\phi]={d-2\over2}.$$

In $d=2$ the exponent vanishes and this formula must be replaced by a logarithm; the massless scalar also has an infrared zero-mode ambiguity. The engineering dimension is still zero, but dimensional analysis alone does not determine the logarithm.

This short-distance power will later reappear in renormalization: loop divergences are the momentum-space version of coincident-point singularities.

Inserting a complete set of states

The free result is only the beginning. Let $\Phi(x)$ be a Hermitian scalar Heisenberg operator in an interacting theory, and remove any vacuum expectation value by defining

$$\widehat\Phi(x)=\Phi(x)-\langle0|\Phi(0)|0\rangle.$$

Translation invariance gives

$$\widehat\Phi(x)=e^{iP\cdot x}\widehat\Phi(0)e^{-iP\cdot x}.$$

Insert a complete set of exact energy-momentum eigenstates between the two fields:

$$\begin{aligned} \langle0|\widehat\Phi(x)\widehat\Phi(0)|0\rangle &=\sum_{n\ne0} \langle0|\widehat\Phi(x)|n\rangle\langle n|\widehat\Phi(0)|0\rangle \\ &=\sum_{n\ne0} |\langle0|\widehat\Phi(0)|n\rangle|^2 e^{-ip_n\cdot x}. \end{aligned}$$

Every term has positive energy, $p_n^0\ge0$, and invariant mass squared $p_n^2\ge0$. Lorentz invariance allows us to group the states by their invariant mass squared $s=p_n^2$. This defines a nonnegative spectral density $\rho(s)$ such that

$$\boxed{ \langle0|\widehat\Phi(x)\widehat\Phi(0)|0\rangle = \int_0^\infty ds\,\rho(s)\,D^+_s(x), }$$

where $D^+_s(x)$ is the free Wightman function for a scalar particle of mass $\sqrt{s}$:

$$D^+_s(t,\mathbf x) = \int{d^3p\over(2\pi)^3}{1\over2\sqrt{\mathbf p^2+s}} e^{-i\sqrt{\mathbf p^2+s}\,t+i\mathbf p\cdot\mathbf x}.$$

The positivity is crucial:

$$\rho(s)\ge0.$$

It follows directly from the squared matrix elements $|\langle0|\widehat\Phi(0)|n\rangle|^2$. This is why the spectral representation is more than a formal integral transform. It expresses unitarity. Without the vacuum subtraction, a nonzero one-point function would add a disconnected constant in position space, or a term proportional to $\delta^{(d)}(p)$ in momentum space; it is not part of the propagating spectral measure discussed here.

One should read $\rho(s)$ as a positive measure, not necessarily as a smooth function. Isolated particles appear as delta functions in this measure. Continua appear as ordinary functions beginning at thresholds. Writing both with the same symbol is what makes the pole-plus-cut structure look so simple.

The time-ordered version is the Källén–Lehmann representation:

$$\boxed{ G_F(x) = \langle0|T\widehat\Phi(x)\widehat\Phi(0)|0\rangle = \int_0^\infty ds\,\rho(s)\,\Delta_F(x;s), }$$

where

$$\Delta_F(x;s) = \int{d^4p\over(2\pi)^4} {i\,e^{-ip\cdot x}\over p^2-s+i0}.$$

In momentum space,

$$\boxed{ \widetilde G_F(p) =i\int_0^\infty ds\,{\rho(s)\over p^2-s+i0}. }$$

This formula says that the exact propagator is a superposition of free propagators. The interactions have not disappeared; they are encoded in $\rho(s)$.

Poles, continua, and the meaning of the spectral density

Suppose $\widehat\Phi$ has nonzero overlap with a stable one-particle state of physical mass $m_{\rm phys}$. Then the spectral density contains a delta function:

$$\rho(s)=Z\delta(s-m_{\rm phys}^2)+\rho_{\rm cont}(s).$$

Use the standard relativistic one-particle normalization

$$\langle\mathbf p|\mathbf q\rangle =2E_{\mathbf p}(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q).$$

Then the matrix element

$$\langle0|\widehat\Phi(0)|\mathbf p\rangle=\sqrt Z$$

defines the pole residue; Lorentz invariance makes this scalar matrix element momentum independent. The number $Z$ measures how strongly the field creates the physical one-particle state from the vacuum. The propagator contains

$$\widetilde G_F(p) \supset {iZ\over p^2-m_{\rm phys}^2+i0}.$$

The continuum part begins at the lightest multiparticle threshold accessible to $\Phi$. For example, in a theory with a $\phi\mapsto-\phi$ symmetry, the field $\phi$ has odd parity under this symmetry and can couple to one-particle states and to odd-particle continua. If the lightest particle has mass $m$, the first odd multiparticle continuum begins at

$$s_{\rm th}=(3m)^2,$$

provided the one-particle state is stable and the symmetry forbids two-particle states in this channel. Without such a selection rule, a two-particle threshold at $s_{\rm th}=(2m)^2$ is typical.

For a canonically normalized elementary field, the equal-time commutator fixes the sum rule

$$\int_0^\infty ds\,\rho(s)=1, \qquad 0\le Z\le1.$$

The inequality follows because the remaining continuum weight is nonnegative. A rescaled field or a composite operator has no unit normalization sum rule. Moreover, time-ordered correlators of composite operators can require local contact terms, which appear as subtraction polynomials in momentum space in addition to their spectral part.

A stable particle gives an isolated delta function in $\rho(s)$ and an isolated pole of the propagator. Multiparticle states give a continuous spectral density and a branch cut beginning at the threshold $s_{\rm th}$.

The pole is a particle. The cut is a continuum of possible intermediate states. A spectral density is therefore a bookkeeping device for what the field can create.

Discontinuities and imaginary parts

Let the analytic function off its cut be

$$D(z)=\int_0^\infty ds\,{\rho(s)\over z-s}.$$

Away from the real positive axis, $D(z)$ is analytic. On the cut,

$$D(s+i0)-D(s-i0)=-2\pi i\rho(s).$$

Thus

$$\rho(s)=-{1\over2\pi i}\operatorname{Disc}D(s) =-{1\over\pi}\operatorname{Im}D(s+i0).$$

This is the simplest form of the rule that an imaginary part measures the density of physical intermediate states. The minus sign comes directly from approaching the kernel $1/(z-s)$ from the upper half-plane:

$${1\over x+i0}=\operatorname{PV}{1\over x}-i\pi\delta(x).$$

For the full Feynman propagator itself,

$$\widetilde G_F(p)=iD(p^2+i0),$$

the entire spectral measure—isolated delta functions and continuum density—appears distributionally in its real part:

$${i\over x+i0}=i\,\operatorname{PV}{1\over x}+\pi\delta(x).$$

Multiplying by the extracted factor $i$ rotates the imaginary part of $D$ into the real part of $\widetilde G_F$. This is one reason signs in spectral formulas can look different across textbooks. The invariant content is the same: the discontinuity across the physical cut is fixed by the positive spectral density.

Dispersion integrals

The spectral representation is a special case of a more general complex-analysis statement. Let $F(z)$ be analytic in the complex $z$-plane except for a cut beginning at $s_{\rm th}$, and suppose $F(z)$ is real analytic:

$$F(z^*)=F(z)^*.$$

If $F(z)$ falls sufficiently fast at infinity, Cauchy’s theorem gives

$$\boxed{ F(z)={1\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s-z}. }$$

This is an unsubtracted dispersion relation. The discontinuity on the cut determines the whole function.

Cauchy’s theorem expresses an analytic function in terms of its discontinuity along the physical cut. If the large circle does not vanish, one subtracts enough polynomial terms to make the contour integral convergent.

Many QFT functions do not fall fast enough at infinity. Then one subtracts at a point $z_0$ away from the cut:

$$F(z)=F(z_0)+{z-z_0\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over(s-z_0)(s-z)}.$$

If $s_{\rm th}>0$, the origin is an allowed subtraction point and this becomes

$$F(z)=F(0)+{z\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s(s-z)}.$$

With $N$ subtractions at the origin (again assuming it lies off the cut),

$$\boxed{ F(z)=P_{N-1}(z)+{z^N\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s^N(s-z)}. }$$

Here $P_{N-1}(z)$ is a polynomial of degree at most $N-1$. In a renormalizable QFT, this polynomial ambiguity is exactly what one expects from local counterterms. For a two-point function, constant and linear subtraction terms correspond to local terms such as

$$\phi^2, \qquad \partial_\mu\phi\partial^\mu\phi.$$

So the dispersion relation packages two ideas together:

1. the nonlocal part of the amplitude is controlled by physical intermediate states;
2. the remaining ambiguity is local and must be fixed by renormalization conditions.

This is often the cleanest way to separate trustworthy information from convention-dependent information. The discontinuity is fixed once the spectrum and matrix elements are fixed. The subtraction polynomial depends on how the local parameters of the theory are defined.

Self-energy and spectral cuts

For a scalar field, the exact propagator can also be written in Dyson form,

$$\widetilde G_F(p) ={i\over p^2-m_0^2-\Sigma(p^2)+i0},$$

where $\Sigma$ is the self-energy. The spectral representation applies to the full propagator, while dispersion relations apply to $\Sigma$ or to any other 1PI function with the appropriate analytic properties.

At low order, a self-energy graph becomes nonanalytic when its internal lines can simultaneously go on shell. This is the origin of branch cuts. In a $\phi^4$ theory, the tadpole shifts the mass but has no external-momentum cut. The first momentum-dependent scalar self-energy appears in the two-loop sunset diagram, schematically

$$\Sigma(p) \sim \int {d^4k_1\over(2\pi)^4}{d^4k_2\over(2\pi)^4} {i\over k_1^2-m^2+i0} {i\over k_2^2-m^2+i0} {i\over (p-k_1-k_2)^2-m^2+i0}.$$

The denominator can become singular when the three internal particles are simultaneously on shell and carry total momentum $p$. This is the three-particle threshold.

A self-energy diagram develops a discontinuity when the dashed cut can pass through internal lines that are all on shell. The cut is the diagrammatic image of the multiparticle continuum in the spectral representation.

The time-domain picture makes the same point. If a field can create intermediate states with energies $E_n(\mathbf p)$, then its correlator contains terms of the form

$$e^{-iE_n(\mathbf p)t}.$$

A single isolated energy gives a pure exponential and hence a pole after Fourier transform. A continuum of energies gives an integral over exponentials, and that integral produces a branch cut. Thresholds, Cuts, and Imaginary Parts uses this idea to extract threshold behavior from nonrelativistic phase space near the onset of a multiparticle continuum.

Euclidean viewpoint

For spacelike momentum, write

$$p^2=-Q^2, \qquad Q^2>0.$$

The Euclidean two-point function is conventionally written with a positive denominator,

$$D_E(Q^2) =\int_0^\infty ds\,{\rho(s)\over Q^2+s}.$$

With the earlier Lorentzian definition $\widetilde G_F=iD(p^2+i0)$, this Euclidean object corresponds to the positive continuation $D_E(Q^2)=-D(-Q^2)$ after stripping the overall Lorentzian factor. This minus sign is pure bookkeeping; the useful fact is that $D_E(Q^2)$ is real and positive if $\rho(s)\ge0$.

The Euclidean form shows why Euclidean correlation functions are often cleaner: the poles and cuts are not crossed, and the spectral density appears through a smooth kernel.

At large Euclidean momentum,

$${1\over Q^2+s} ={1\over Q^2}\left(1-{s\over Q^2}+{s^2\over Q^4}-\cdots\right),$$

provided the moments are sufficiently well behaved. This expansion is a first glimpse of the short-distance logic behind the operator product expansion: high Euclidean momentum probes moments of spectral data and local short-distance structure.

Summary

The two-point function of an interacting scalar field has three equivalent descriptions. In spacetime, it is a boundary value of an analytic function, with $i0$ prescriptions selecting how singularities are approached. In the Hilbert space, it is a sum over exact intermediate states. In momentum space, it is an analytic function with poles and cuts.

The Källén–Lehmann representation is the cleanest synthesis:

$$\widetilde G_F(p) =i\int_0^\infty ds\,{\rho(s)\over p^2-s+i0}, \qquad \rho(s)\ge0.$$

A stable particle is an isolated pole. A multiparticle continuum is a branch cut. The discontinuity across the cut is the spectral density. Dispersion relations then reconstruct analytic functions from their discontinuities, up to subtraction polynomials. Those subtraction polynomials are not mysterious; in QFT they are the analytic counterpart of local counterterms.

Common pitfalls

Confusing the Feynman propagator with a positive spectral function. The positivity belongs to $\rho(s)$, which comes from squared matrix elements. The time-ordered propagator contains extra factors of $i$ and principal values.

Forgetting the boundary value. Expressions such as $1/(x^2)$ or $1/(p^2-m^2)$ are incomplete as distributions. The $i0$ prescription says which side of the singular surface is meant.

Thinking every imaginary part means instability. A cut in a two-point function may simply mean that the field can create multiparticle states. An unstable particle is more subtle: its pole moves away from the real axis on a second sheet.

Ignoring subtraction constants. Dispersion integrals often require subtractions. The subtraction constants are physical renormalization data, not optional nuisances.

Treating $\rho(s)$ as smooth by default. A stable particle contributes a delta function, not a narrow smooth bump. A resonance is different: it is associated with a pole on another sheet and appears as structure in the continuum.

Exercises

Exercise 1

Evaluate the massless positive-frequency Wightman function in four spacetime dimensions:

$$D^+(t,\mathbf x) =\int{d^3p\over(2\pi)^3}{1\over2|\mathbf p|} e^{-i|\mathbf p|(t-i\epsilon)+i\mathbf p\cdot\mathbf x}, \qquad \epsilon>0.$$

Show that

$$D^+(t,\mathbf x) ={1\over4\pi^2}\,{1\over\mathbf x^2-(t-i0)^2}.$$

Solution

Let $r=|\mathbf x|$. The angular integral is

$$\int d\Omega\,e^{i\mathbf p\cdot\mathbf x} =4\pi{\sin(pr)\over pr}.$$

Thus

$$D^+(t,r) ={1\over(2\pi)^3}\int_0^\infty dp\,{p^2\over2p} 4\pi{\sin(pr)\over pr}e^{-ip(t-i\epsilon)}.$$

So

$$D^+(t,r) ={1\over4\pi^2r}\int_0^\infty dp\,\sin(pr)e^{-ip(t-i\epsilon)}.$$

Using

$$\sin(pr)={e^{ipr}-e^{-ipr}\over2i},$$

we get

$$\int_0^\infty dp\,\sin(pr)e^{-ip(t-i\epsilon)} ={1\over2i}\left[ {1\over \epsilon+i(t-r)}-{1\over \epsilon+i(t+r)} \right].$$

This simplifies to

$${r\over r^2-(t-i\epsilon)^2}.$$

Therefore

$$D^+(t,r) ={1\over4\pi^2}\,{1\over r^2-(t-i\epsilon)^2}.$$

Taking $\epsilon\downarrow0$ gives the stated boundary value.

Exercise 2

Let $\Phi(x)$ be a Hermitian scalar operator and $\widehat\Phi=\Phi-\langle0|\Phi|0\rangle$. Starting from the connected Wightman function

$$W_c(x)=\langle0|\widehat\Phi(x)\widehat\Phi(0)|0\rangle,$$

insert a complete set of energy-momentum eigenstates and explain why the spectral density in the Källén–Lehmann representation is nonnegative.

Solution

Translation invariance gives

$$\widehat\Phi(x)=e^{iP\cdot x}\widehat\Phi(0)e^{-iP\cdot x}.$$

Insert the identity $1=\sum_n |n\rangle\langle n|$, with $P^\mu|n\rangle=p_n^\mu|n\rangle$:

$$\begin{aligned} W_c(x) &=\sum_{n\ne0} \langle0|\widehat\Phi(x)|n\rangle\langle n|\widehat\Phi(0)|0\rangle\\ &=\sum_{n\ne0} |\langle0|\widehat\Phi(0)|n\rangle|^2e^{-ip_n\cdot x}. \end{aligned}$$

The coefficient of each term is an absolute square. Grouping states by invariant mass squared $s=p_n^2$ defines $\rho(s)$. Since $\rho(s)$ is built from sums and integrals of nonnegative quantities,

$$\rho(s)\ge0.$$

This positivity is a direct consequence of Hilbert-space unitarity.

Exercise 3

Let $F(z)$ be analytic except for a cut $[s_{\rm th},\infty)$ with $s_{\rm th}>0$, and real analytic, $F(z^*)=F(z)^*$. Suppose $F(z)/z\to0$ as $|z|\to\infty$. Derive the one-subtracted dispersion relation

$$F(z)=F(0)+{z\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s(s-z)}.$$

Solution

Apply the unsubtracted dispersion relation to

$$H(z)={F(z)-F(0)\over z}.$$

By assumption, $H(z)$ vanishes fast enough at infinity. Its discontinuity is

$$\operatorname{Im}H(s+i0) ={\operatorname{Im}F(s+i0)\over s},$$

because $F(0)$ is real. Therefore

$$H(z)={1\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s(s-z)}.$$

Multiplying by $z$ gives

$$F(z)-F(0) ={z\over\pi}\int_{s_{\rm th}}^\infty ds\, {\operatorname{Im}F(s+i0)\over s(s-z)}.$$

This is the desired one-subtracted dispersion relation.

Exercise 4

Assume an exact scalar propagator has spectral density

$$\rho(s)=Z\delta(s-m^2)+\rho_c(s)\theta(s-s_{\rm th}).$$

Write the Euclidean propagator $D_E(Q^2)$ and identify the large-distance behavior of the Euclidean coordinate-space correlator.

Solution

The Euclidean momentum-space propagator is

$$D_E(Q^2)=\int_0^\infty ds\,{\rho(s)\over Q^2+s} ={Z\over Q^2+m^2}+\int_{s_{\rm th}}^\infty ds\,{\rho_c(s)\over Q^2+s}.$$

In Euclidean coordinate space, a state of invariant mass $\sqrt{s}$ contributes at large separation $R$ roughly as

$$e^{-\sqrt{s}\,R}$$

up to powers of $R$. The lightest state dominates. If $Z\neq0$ and $m^2<s_{\rm th}$, the one-particle pole gives

$$G_E(R)\sim Z\,C(R)e^{-mR},$$

where $C(R)$ is a dimension-dependent power-law prefactor. The continuum is more suppressed, beginning at

$$e^{-\sqrt{s_{\rm th}}R}.$$

Thus large Euclidean distance measures the lightest state that the operator can create.

References and further reading

• Mark Srednicki, Quantum Field Theory, sections 13–15, for the Källén–Lehmann representation and one-loop spectral/dispersion examples.
• Steven Weinberg, The Quantum Theory of Fields, vol. I, sections 10.7–10.8, for spectral representations, causality, and dispersion relations.
• Sidney Coleman, Lectures on Quantum Field Theory, chapters 13–15, for Green functions, LSZ, and the spectral representation in the renormalization discussion.
• A. Zee, Quantum Field Theory in a Nutshell, chapter III.8, for a compact discussion of imaginary parts, cuts, and dispersion relations.