# Evaluating propagator without the epsilon trick

Consider the Kleinâ€“Gordon equation and its propagator: $$G(x,y) = \frac{1}{(2\pi)^4}\int d^4 p \frac{e^{-i p.(x-y)}}{p^2 - m^2} \; .$$

I'd like to see a method of evaluating explicit form of $G$ which does not involve avoiding singularities by the $\varepsilon$ trick. Can you provide such a method?

Before answering the question more or less directly, I'd like to point out that this is a good question that provides an object lesson and opens a foray into the topics of singular integral equations, analytic continuation and dispersion relations. Here are some references of these more advanced topics: Muskhelishvili, Singular Integral Equations; Courant & Hilbert, Methods of Mathematical Physics, Vol I, Ch 3; Dispersion Theory in High Energy Physics, Queen & Violini; Eden et.al., The Analytic S-matrix. There is also a condensed discussion of `invariant functions' in Schweber, An Intro to Relativistic QFT Ch13d.

The quick answer is that, for $m^2 \in\mathbb{R}$, there's no "shortcut." One must choose a path around the singularities in the denominator. The appropriate choice is governed by the boundary conditions of the problem at hand. The $+i\epsilon$ "trick" (it's not a "trick") simply encodes the boundary conditions relevant for causal propagation of particles and antiparticles in field theory.

We briefly study the analytic form of $G(x-y;m)$ to demonstrate some of these features.

Note, first, that for real values of $p^2$, the singularity in the denominator of the integrand signals the presence of (a) branch point(s). In fact, [Huang, Quantum Field Theory: From Operators to Path Integrals, p29] the Feynman propagator for the scalar field (your equation) may be explicitly evaluated: \begin{align} G(x-y;m) &= \lim_{\epsilon \to 0} \frac{1}{(2 \pi)^4} \int d^4p \, \frac{e^{-ip\cdot(x-y)}}{p^2 - m^2 + i\epsilon} \nonumber \\ &= \left \{ \begin{matrix} -\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H_1^{(1)}(m \sqrt{s}) & \textrm{ if }\, s \geq 0 \\ -\frac{i m}{ 4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) & \textrm{if }\, s < 0. \end{matrix} \right. \end{align} where $s=(x-y)^2$.

The first-order Hankel function of the first kind $H^{(1)}_1$ has a logarithmic branch point at $x=0$; so does the modified Bessel function of the second kind, $K_1$. (Look at the small $x$ behavior of these functions to see this.)

A branch point indicates that the Cauchy-Riemann conditions have broken down at $x=0$ (or $z=x+iy=0$). And the fact that these singularities are logarithmic is an indication that we have an endpoint singularity [eg. Eden et. al., Ch 2.1]. (To see this, consider $m=0$, then the integrand, $p^{-2}$, has a zero at the lower limit of integration in $dp^2$.)

Coming back to the question of boundary conditions, there is a good discussion in Sakurai, Advanced Quantum Mechanics, Ch4.4 [NB: "East Coast" metric]. You can see that for large values of $s>0$ from the above expression that we have an outgoing wave from the asymptotic form of the Hankel function.

Connecting it back to the original references I cited above, the $+i\epsilon$ form is a version of the Plemelj formula [Muskhelishvili]. And the expression for the propagator is a type of Cauchy integral [Musk.; Eden et.al.]. And this notions lead quickly to the topics I mentioned above -- certainly a rich landscape for research.

• 'it's not a "trick"' Agreed. This is a standard procedure from complex analysis. It has the blessing of the same kinds of OCD mathematicians who balk at other shortcuts physicist sometimes take. – dmckee --- ex-moderator kitten Dec 4 '12 at 22:57
• This "standard procedure" was certainly not covered in any of the multiple analysis courses I've taken. I guess we had a different focus... In any event, there is still some disconnect between mathematicians and physicists, where only the latter tell me integrals can depend on path deformations within the domain of analyticity. – user10851 Dec 5 '12 at 2:00
• @ChrisWhite Both my undergrad and graduate institution had course called Mathematical Methods in Physics and it was covered in them. The people I know whose programs did not have such a course often haven't seen it. If someone asked me for a text recommendation I'd say Arfkin, but that's just 'cause it was the one I used that I liked more. – dmckee --- ex-moderator kitten Dec 5 '12 at 2:03
• @dmckee Thanks. I'll look into that, as well as the other suggestions. The relevant courses I took/audited were taught by a pure mathematician, an applied mathematician, a dynamics mathematician, and a string theorist. I didn't take the one taught by a run-of-the-mill quantum physicist, alas. – user10851 Dec 5 '12 at 2:16
• @ChrisWhite: "only the latter tell me integrals can depend on path deformations within the domain of analyticity" -- then you're talking to someone -- I'm not sure why it's relevant that the "telling" came from someone in a specific profession -- who doesn't understand complex analysis. If you can provide a specific example of this, I'm sure we can clear up the confusion. Sometimes it's difficult to see when a singularity has been 'crossed' etc. – MarkWayne Dec 5 '12 at 3:21

Expanding on dmckee's comment:

The $+i\epsilon$-trick has the blessing of OCD mathematicians because it follows directly from a deep fact about the group of spacetime translations: the group $\{e^{-i\langle P,x\rangle/\hbar}| x \in \mathbb{R}^n\}$ of spacetime translations is the boundary of an analytic semigroup $\{e^{-i\langle P,\xi\rangle/\hbar}| x \in \mathbb{C}^n \mbox{ and } Im(\xi) \leq 0\}$.

Many quantities in field theory are expressed in terms of these translations, and frequently these quantities can be computed more easily by analytically continuing from real "Minkowski" time to imagininary "Euclidean" time, where the delicate cancellation of phases becomes the crude suppression of exponential damping. When you use the $+i\epsilon$-trick, what you're really doing is saying that the particular cancellation of phases you want is the one which respects this analyticity. This is precisely what's happening when you use the $+i\epsilon$-trick to evaluate the Klein-Gordon propagator. You've got an integral which does not converge absolutely, and you're picking out a certain resummation which does. The $+i\epsilon$ is not just a trick here; it's really the definition of the quantity you're after.

• This is very interesting! I didn't know that the group of spacetime translations is the boundary of an analytic semigroup! Where can I read more about this? – QuantumDot Dec 5 '12 at 4:29
• You can look in Streater & Wightman's book. But more seriously: just try it out by hand in the 1d case, when there's only time translations. You'll be able to draw a picture of the whole group on the plane. – user1504 Dec 5 '12 at 13:38

As far as my experience goes, the problem stems from writing the right solution for all reals to the problem: $$(p-m)G(p)=1.$$ which reads: $$G(p)=\text{P.v.} \frac {1}{p-m}+c_0\delta(p-m)$$ where $\text{P.v.}$ stands for principal value. The $\delta(\epsilon-\omega)$ function appears as it is the Kernel of $(\omega-\epsilon)$ and $c_0$ is some constant to be fixed. If we now take the Fourier transform we obtain: $$\int e^{ipt }G(p)=\:\left( i\pi \text{sign}(t)+c_0\right)e^{i m t}$$ $c_0$ must now be fixed according to boundary conditions; for the retarded and advanced Green functions, one has $c_0=\pm i\pi$ and the solution given by the $i \epsilon$ trick is recovered. In my opinion though, it is a rather bad method as it only works when you have poles or first order, as the $\delta$ function can be approached by square integrable functions. If you now are looking for the solution of: $$(p-m)^kG(p)=1.$$ with k an integer, you now have $$G(p)=\text{P.v.} \frac {1}{(p-m)^k}+\sum_{j=0}^kc_j\delta^{(j)}(p-m)$$ with $\delta^{(k)}$ the k-th derivative of the delta function. The Fourier transform reads $$G(t)=\left( i\pi \frac{(i t)^{k-1}}{k-1!}\text{sign}(t)+\sum_{j=0}^kc_j (-it)^j\right)e^{i m t}$$ And again, the $c_j$s are fixed depending on boundary conditions. Yet I don't know any way to recover this solution with the $i\epsilon$ trick.