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Contour Deformation, Pinches, and Causal Prescriptions

A contour may be moved only through a region in which the complete integrand is holomorphic. For an infinite contour, that statement also requires control of the arcs at infinity; for a parameter-dependent integral, it requires enough uniformity to pass limits through the integral. A crossed pole changes the answer by a residue. A branch cut requires a specified continuation. A pinch is more severe: singularities approaching from opposite sides can remove every admissible path between the old and proposed contours.

Required background. Laurent Series, Poles, and Residues supplies the residue picked up by a crossed pole; Branches, Sheets, Analytic Continuation, and Monodromy supplies the cut and continuation data needed when the obstruction is not isolated.

This page turns those observations into a reusable method. Its QFT-facing calculation tests the mathematical legality of a one-variable Lorentzian-to-Euclidean energy rotation. State selection, Euclidean correlators, and reflection positivity belong to the Foundations treatment.

Finite contour deformations are homotopies

Section titled “Finite contour deformations are homotopies”

Let ff be holomorphic on a domain DCD\subset\mathbb C. Suppose γ0\gamma_0 and γ1\gamma_1 are oriented piecewise smooth paths in DD with the same endpoints and are homotopic in DD while those endpoints are held fixed. Then

γ0f(z)dz=γ1f(z)dz.\int_{\gamma_0}f(z)\,\mathrm dz = \int_{\gamma_1}f(z)\,\mathrm dz.

One useful formulation asks for a continuous family

H:[0,1]×[0,1]D,γu(t)=H(u,t),H:[0,1]\times[0,1]\longrightarrow D, \qquad \gamma_u(t)=H(u,t),

whose endpoint values are independent of uu. The entire swept image must remain in DD. Checking only the two final contours is not enough: a singularity may lie between them.

For closed contours, homology gives the corresponding statement. If two cycles have the same winding number about every point outside DD, their integrals of every holomorphic ff on DD agree. This is often the cleanest language when a contour has several components; see Conway 1978, Chapter IV, §§ 4–6.

Let ff be meromorphic on a domain UU, and let

C=γ1γ0C=\gamma_1-\gamma_0

be a cycle in UU whose trace avoids the poles of ff. Assume that CC is null-homologous in UU: equivalently, Ind(C,w)=0\operatorname{Ind}(C,w)=0 for every wUw\notin U. The residue theorem then gives

γ1f(z)dzγ0f(z)dz=2πiaInd(C,a)Resz=af(z).\int_{\gamma_1}f(z)\,\mathrm dz - \int_{\gamma_0}f(z)\,\mathrm dz = 2\pi i \sum_a \operatorname{Ind}(C,a) \operatorname{Res}_{z=a}f(z).

The sum runs over the poles aUa\in U with nonzero winding number. The ambient-domain hypothesis prevents an unmentioned hole from being filled. The winding number fixes the sign; a verbal instruction such as “move the contour past the pole” does not.

A branch cut is different. On a chosen sheet it is part of the excluded set, so a deformation cannot simply pass through it. One must specify analytic continuation around the branch point or replace the cut by its two banks and use their boundary-value difference. The dispersion-integral page implements the second option.

Endpoints also matter. If an endpoint moves, the two open contours no longer form a closed chain without connector paths. Their integrals differ by the connector contributions as well as by any enclosed residues.

Infinite contours are limits, not pictures

Section titled “Infinite contours are limits, not pictures”

An integral over an infinite contour should first be truncated. Let ARA_R be a connecting arc at radius of order RR. The estimation lemma gives

ARf(z)dzlength(AR)supzARf(z).\left| \int_{A_R}f(z)\,\mathrm dz \right| \leq \operatorname{length}(A_R) \sup_{z\in A_R}|f(z)|.

For a quarter-circle, length(AR)=π2R\operatorname{length}(A_R)=\frac{\pi}{2}R. Thus a uniform estimate

f(z)=O(R1δ),δ>0,f(z)=O(R^{-1-\delta}), \qquad \delta>0,

on that arc makes its contribution O(Rδ)O(R^{-\delta}). Pointwise decay along the real axis does not establish this bound in the intervening quadrant.

There are four separate limiting questions:

  1. Do the truncated integrals converge?
  2. Do all connecting arcs vanish with their orientations tracked?
  3. Is the bound uniform in any external parameter or remaining integration variable?
  4. May a regulator or boundary-value limit be taken after the deformation?

For an iterated loop integral, legality at each fixed value of the other variables need not be enough. A dominating integrable bound or another theorem justifying interchange of deformation and integration is still needed. Ultraviolet regularization and contour deformation are therefore logically distinct steps.

Use the site’s Fourier convention

f~(p)=ddxe+ipxf(x),f(x)=ddp(2π)deipxf~(p),\widetilde f(p) = \int\mathrm d^d x\, e^{+ip\cdot x}f(x), \qquad f(x) = \int\frac{\mathrm d^d p}{(2\pi)^d}\, e^{-ip\cdot x}\widetilde f(p),

and the metric (+)(+---). For fixed spatial momentum, set

Ep=p2+m2>0.E_{\mathbf p} = \sqrt{\mathbf p^2+m^2}>0.

The following scalar boundary-value kernels have different pole placements. The notation records the limit from positive ϵ\epsilon; it is not an ordinary algebraic infinitesimal.

PrescriptionRepresentative kernel denominatorPoles in the p0p^0 plane
Feynman(p0)2Ep2+i0(p^0)^2-E_{\mathbf p}^2+i0+Ep+E_{\mathbf p} below, Ep-E_{\mathbf p} above
Retarded(p0+i0)2Ep2(p^0+i0)^2-E_{\mathbf p}^2both below
Advanced(p0i0)2Ep2(p^0-i0)^2-E_{\mathbf p}^2both above

For example, at finite ϵ>0\epsilon>0 the Feynman roots are

p+0=+Epiϵ2Ep+O ⁣(ϵ2Ep3),p0=Ep+iϵ2Ep+O ⁣(ϵ2Ep3).\begin{aligned} p^0_+ &= +E_{\mathbf p} -\frac{i\epsilon}{2E_{\mathbf p}} +O\!\left(\frac{\epsilon^2}{E_{\mathbf p}^3}\right),\\ p^0_- &= -E_{\mathbf p} +\frac{i\epsilon}{2E_{\mathbf p}} +O\!\left(\frac{\epsilon^2}{E_{\mathbf p}^3}\right). \end{aligned}

Multiplying ϵ\epsilon by a positive smooth factor does not change the resulting boundary value. This is why the same placement is often written formally as

1(p0Ep+i0)(p0+Epi0).\frac{1}{ (p^0-E_{\mathbf p}+i0) (p^0+E_{\mathbf p}-i0) }.

With the inverse transform eip0te^{-ip^0t}, closing below for t>0t>0 and above for t<0t<0 shows why placing both poles below gives

GR(t)=0(t<0),G_R(t)=0 \qquad (t<0),

while placing both poles above gives

GA(t)=0(t>0).G_A(t)=0 \qquad (t>0).

The Feynman placement instead selects opposite half-planes for positive- and negative-energy poles. These prescriptions are not interchangeable decorations on the same integral. Their inverse-Fourier support properties are derived in Tong 2006–2007, §§ 2.7.1–2.7.2.

Controlled Wick rotation at fixed spatial momentum

Section titled “Controlled Wick rotation at fixed spatial momentum”

Consider the convergent energy integral

I(E)=limϵ0+dk02πi(k0)2E2+iϵ,E>0.I(E) = \lim_{\epsilon\to0^+} \int_{-\infty}^{\infty} \frac{\mathrm dk^0}{2\pi}\, \frac{i}{ (k^0)^2-E^2+i\epsilon }, \qquad E>0.

Keep ϵ\epsilon finite during the deformation. The positive-energy pole lies in the fourth quadrant and the negative-energy pole lies in the second. Rotating the whole oriented real line counterclockwise through an angle π/2\pi/2 sweeps the first and third quadrants, so it crosses neither pole. On the two quarter-circle arcs the integrand is O(R2)O(R^{-2}), and the arc length is O(R)O(R); both arc contributions therefore vanish.

The rotated contour runs upward from i-i\infty to +i+i\infty. Set

k0=ikE0,dk0=idkE0.k^0=ik_E^0, \qquad \mathrm dk^0=i\,\mathrm dk_E^0.

Then

I(E)=limϵ0+dkE02π1(kE0)2+E2iϵ.I(E) = \lim_{\epsilon\to0^+} \int_{-\infty}^{\infty} \frac{\mathrm dk_E^0}{2\pi}\, \frac{1}{ (k_E^0)^2+E^2-i\epsilon }.

For every ϵ>0\epsilon>0,

1(kE0)2+E2iϵ1(kE0)2+E2,\left| \frac{1}{ (k_E^0)^2+E^2-i\epsilon } \right| \leq \frac{1}{ (k_E^0)^2+E^2 },

and the right-hand side is integrable. Dominated convergence therefore justifies the boundary-value limit:

I(E)=dkE02π1(kE0)2+E2=12E.\begin{aligned} I(E) &= \int_{-\infty}^{\infty} \frac{\mathrm dk_E^0}{2\pi}\, \frac{1}{ (k_E^0)^2+E^2 } = \frac{1}{2E}. \end{aligned}

The signs follow from both the Jacobian and the denominator:

dk0i(k0)2E2+i0dkE01(kE0)2+E2.\mathrm dk^0\, \frac{i}{(k^0)^2-E^2+i0} \longmapsto \mathrm dk_E^0\, \frac{1}{(k_E^0)^2+E^2}.

In four dimensions the same coordinate substitution gives

k0=ikE0,d4k=id4kE,k2=kE2k^0=ik_E^0, \qquad \mathrm d^4k=i\,\mathrm d^4k_E, \qquad k^2=-k_E^2

under the (+)(+---) convention. These algebraic replacements are consequences of a legal contour deformation, not a proof that one exists. A full loop integral additionally needs ultraviolet control and a deformation that remains valid uniformly in the spatial momentum and all other loop variables.

Only after the contour is on the imaginary axis, where the Euclidean denominator is nonzero, was ϵ0+\epsilon\to0^+ taken. Reversing that order would put poles on the original contour and erase the information that made the rotation unambiguous. For the same pole geometry, orientation, and Jacobian check, compare Schwartz 2014, Appendix B.2, pp. 823–825 and Weinberg 1995, § 11.2, pp. 475–476.

Consider the local model

J(a,ϵ)=dx(xa+iϵ)(x+aiϵ),a>0,ϵ>0.J(a,\epsilon) = \int_{-\infty}^{\infty} \frac{\mathrm dx}{ (x-a+i\epsilon) (x+a-i\epsilon) }, \qquad a>0,\quad\epsilon>0.

Its poles are

x=aiϵ,x=a+iϵ.x=a-i\epsilon, \qquad x=-a+i\epsilon.

One approaches the real contour from below and the other from above. For a>0a>0 a path can still thread between their real parts. As a0+a\to0^+ and ϵ0+\epsilon\to0^+, that corridor collapses at the origin. The contour is pinched.

Closing in the upper half-plane makes the singular behavior explicit:

J(a,ϵ)=πia+iϵ.J(a,\epsilon) = \frac{\pi i}{-a+i\epsilon}.

The divergence as (a,ϵ)(0,0)(a,\epsilon)\to(0,0) is the analytic warning that the original parameter dependence need not continue through the pinch. Detouring both poles to the same side would change the boundary-value problem rather than repair the given one.

This local picture has several generalizations.

  • A pole can collide with an endpoint instead of with a second pole.
  • Branch points or cut endpoints can trap a contour.
  • In several integration variables, singular hypersurfaces can obstruct a deformation even when no single one-variable slice reveals a uniform problem.
  • For Feynman integrals, Landau analysis supplies systematic conditions for candidate pinches. Determining the physical sheet and the actual singularity still requires the prescribed contour and further analysis; see Zwicky 2016, § 4.2.

The QFT classification of Landau singularities belongs to Singularities, Cuts, and Integrand Reconstruction. Here the reusable conclusion is simpler: a legal deformation is a homotopy through nonsingular contours, and a pinch is the loss of that homotopy.

For a proposed rotation or contour move:

  1. Retain finite regulators. Write iϵi\epsilon rather than erasing it at the start.
  2. Locate every singularity. Include poles, branch points, chosen cuts, endpoints, and singularities introduced by regulators or numerators.
  3. Specify the oriented homotopy. State which region each contour segment sweeps.
  4. Test crossings and pinches. If a pole is crossed, compute the oriented residue term. If the corridor collapses, stop: the proposed deformation is not available.
  5. Bound every added arc. Give an estimate uniform in parameters that remain to be integrated or limited.
  6. Track Jacobians and metric signs. A mnemonic such as k0ikE0k^0\mapsto ik_E^0 is incomplete by itself.
  7. Take limits in the justified order. Record when RR\to\infty, ϵ0+\epsilon\to0^+, regulator removal, and external-parameter continuation occur.

This protocol distinguishes three outcomes: equality by deformation, equality plus explicit residue or discontinuity terms, and failure because no admissible deformation exists.

Treating a drawn empty quadrant as a proof. The full integrand may have additional poles, cuts, or parameter-dependent singularities. List them and bound the arcs before deforming.

Setting i0=0i0=0 before moving the contour. The prescription determines which side of the contour contains each pole. Removing it first turns a well-defined boundary value into an ambiguous singular integral.

Checking decay only on the original axis. Arc estimates require control throughout the swept sector. Pointwise real-axis decay does not supply that control.

Calling every obstruction a crossed pole. A crossed isolated pole can be accounted for by a residue. A pinch means that no contour with the required endpoints can pass between the approaching singularities.

Interpreting Wick rotation as automatic physical equivalence. This page establishes a contour criterion for a controlled integral. State reconstruction, operator ordering, reflection positivity, and interacting correlator domains require the Foundations analysis.

  1. For ϵ>0\epsilon>0, locate the two poles of

    1(z0)2E2+iϵ\frac{1}{(z^0)^2-E^2+i\epsilon}

    to first order in ϵ\epsilon. Which quadrants are swept by the rotation z0=eiθxz^0=e^{i\theta}x, 0θπ/20\leq\theta\leq\pi/2?

    Check

    Since

    E2iϵ=Eiϵ2E+O ⁣(ϵ2E3),\sqrt{E^2-i\epsilon} = E-\frac{i\epsilon}{2E} +O\!\left(\frac{\epsilon^2}{E^3}\right),

    the poles are

    z+0=Eiϵ2E+O ⁣(ϵ2E3),z0=E+iϵ2E+O ⁣(ϵ2E3).z^0_+ = E-\frac{i\epsilon}{2E} +O\!\left(\frac{\epsilon^2}{E^3}\right), \qquad z^0_- = -E+\frac{i\epsilon}{2E} +O\!\left(\frac{\epsilon^2}{E^3}\right).

    They lie in the fourth and second quadrants. The positive real ray sweeps the first quadrant; the negative real ray sweeps the third. Neither pole is crossed.

  2. Evaluate I(E)I(E) by closing the original contour in the upper half-plane and compare with the Euclidean integral.

    Check

    The upper pole approaches E+i0-E+i0. Its residue in the integrand i/((k0)2E2+i0)i/((k^0)^2-E^2+i0) is i/(2E)-i/(2E). Including the counterclockwise contour factor and the measure 1/(2π)1/(2\pi) gives

    I(E)=i(i2E)=12E.I(E) = i\left(-\frac{i}{2E}\right) = \frac{1}{2E}.

    The rotated integral is

    dkE02π1(kE0)2+E2=12E,\int_{-\infty}^{\infty} \frac{\mathrm dk_E^0}{2\pi} \frac{1}{(k_E^0)^2+E^2} = \frac{1}{2E},

    so the orientation and Jacobian agree.

  3. Compute the pinch model J(a,ϵ)J(a,\epsilon) by residues and explain why its limit cannot be made finite by an innocuous contour shift.

    Check

    The upper pole is a+iϵ-a+i\epsilon, with residue

    12a+2iϵ.\frac{1}{-2a+2i\epsilon}.

    Hence

    J(a,ϵ)=2πi12a+2iϵ=πia+iϵ.J(a,\epsilon) = 2\pi i \frac{1}{-2a+2i\epsilon} = \frac{\pi i}{-a+i\epsilon}.

    The poles approach the same contour point from opposite sides. Any shift that puts both on one side must cross one of them and changes the prescribed integral by its residue; it is not a homotopy in the nonsingular domain.

  • John B. Conway, Functions of One Complex Variable I, 2nd ed., Chapter IV §§4–6 and Chapter V §2, Springer, 1978. Book record. Chapter IV is the structural source for Cauchy’s theorem, homotopy invariance, winding numbers, and contour estimates; Chapter V supplies the residue theorem.
  • Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Appendix B.2, pp. 823–825, Cambridge University Press, 2014. Book record. This supplies the explicit Feynman-pole geometry, Jacobian, metric signs, and the warning that additional poles can invalidate a Wick rotation.
  • David Tong, Quantum Field Theory, §§2.7.1–2.7.2, Cambridge Part III lecture notes, University of Cambridge, 2006–2007. This is the source for the inverse-Fourier contour closures and the Feynman, retarded, and advanced support conventions.
  • Steven Weinberg, The Quantum Theory of Fields, Vol. I, § 11.2, pp. 475–476, Cambridge University Press, 1995. Book record. This independently fixes the counterclockwise real-to-imaginary rotation, the upward orientation, and the need to regulate divergent loop integrals.
  • Roman Zwicky, “A Brief Introduction to Dispersion Relations and Analyticity”, §4.2, 2016. This is the QFT-facing source for Landau conditions, endpoint singularities, and pinch singularities.