# Is the $i\varepsilon$ prescription in the Feynman propagator just as "outrageous" as $1+2+3+... = -1/12$?

When the calculation of the Feynman propagator is introduced in QFT (or QM), the inclusion of the $$i\varepsilon$$ term is often described as a minor technical detail that is there just to "make the integral converge":

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

I feel however that this hides quite a remarkable result. We can see the essence of it we can just consider the simple integral:

$$\int_{-\infty}^{+\infty} \frac{1}{x^2-a^2}dx = \frac{i\pi}{a}$$

which holds for complex $$a$$ with $$\text{Im}(a)>0$$. This is easy to derive using Cauchy's integral formula or standard calculus methods. Now set $$a=m+i\varepsilon$$ with real $$m,\varepsilon$$ and take the limit $$\varepsilon \to 0$$:

$$\lim_{\varepsilon \to 0} \int_{-\infty}^{+\infty} \frac{1}{x^2-(m+i\varepsilon)^2}dx = \frac{i\pi}{m}.$$

The remarkable thing here is that a divergent integral of a real function $$\int_{-\infty}^{+\infty} \frac{1}{x^2-m^2} dx$$ evaluates to a finite, purely imaginary number!

This looks analogous to the way the divergent sum of positive integers $$1+2+3+...$$ is assigned the negative value $$-\frac{1}{12}$$, also by analytic continuation. This sum of course is very famous, but the Feynman propagator, being one of the most basic physical quantities, is used much more in practical calculations.

Is it fair to say that these results should be considered as equally "outrageous" ? did the latter become more famous just because the apparent absurdity of it can be appreciated by a wider audience ?

EDIT

To clarify, the question is not about how we choose the sign of $$i\varepsilon$$, which resolves the ambiguity of the real integral (in the example above, taking the limit from the negative direction will lead to a different answer). It is about assigning any value to the divergent integral.

In the path integral formulation, the propagator arises from calculating the vacuum expectation value

$$\langle \phi(x)\phi(y) \rangle = \frac{\int D\phi \phi(x)\phi(y) e^{iS[\phi]}}{\int D\phi e^{iS[\phi]}}$$

which is done by treating the path integral as a (finite) multivariate gaussian integral and using the gaussian integral formula

$$\langle x_i x_j \rangle = \frac{\int dx_1 ... dx_n x_i x_j \exp(\frac{i}{2}x^T A x)}{\int dx_1 ... dx_n\exp(\frac{i}{2}x^T A x)} = i(A^{-1})_{ij}$$

where the "matrix" $$A$$ is the differential operator $$\partial^2 - m^2$$. The problem is that this matrix has zero eigenvalues ($$p^2 = m^2$$, $$p^2$$ being the eigenvalues of $$\partial^2$$), and is therefore singular, so $$A^{-1}$$ doesn't exists. It is somewhat interesting to notice that this singularity can appear already in the discretized version of the integral, and is actually only resolved in the continuum limit (there is no equivalent trick that can invert singular matrices that I know of, or is there?). Anyway the point is that the path integral doesn't exists, unless we treat it as a limit of an analytic continuation using the $$i\varepsilon$$ trick.

I acknowledge that this might seem like an opinion-based question, so the more concrete answer I am looking for is: are there any concrete mathematical criteria that applies only to one of those examples of assigning values to divergent quantities, that would allow us to say that they are fundamentally different from each other ?

• Are you aware of the use of principal value in physics? see for example math.stackexchange.com/q/2450848 Apr 19, 2023 at 12:15
• It is not at all outrageous. The ie-prescription defines the propagator in the momentum space as a particular distribution. This ensures that you get the time-ordered propagator in position space. It is simply a way of writing down a particular solution of the free EOM with a source term. It may not be explained properly in some (most?) expositions, but there is absolutely zero trickery of the $-1/12$ kind involved. Apr 19, 2023 at 12:18
• @PeterKravchuk Of course, within the appropriate mathematical framework, both results are perfectly fine. But both can be viewed as assigning a value by analytic continuation to an otherwise divergent quantity. So why does one involves "trickery" while the other is straightforward ? Apr 19, 2023 at 12:36
• @Mauricio the point of this question is that it's not the principal value. I'd indeed agree that a principal value would not be outrageous at all (though here it also diverges), indeed it's quite intuitive. But the trick here where we get an imaginary value after integrating a real-valued function does indeed look suspicious. Apr 19, 2023 at 12:43
• This is an off-topic opinion-based question. Apr 19, 2023 at 15:35

## 1 Answer

No, you misunderstand what the prescription is supposed to do.

What the integral without $$i \epsilon$$ needs is not a way to make the integral converge." What it needs is an auxiliary specification of extra properties needed to figure out if the singularities are supposed to contribute or not. That is, the integral as directly written without a specification is ambiguous, and we would not know if it is supposed to be the causal one, the anti-causal one, or Feynman or anti-Feynman propagators. They all have the same basic integral.

What Feynman realised is that, since for the various scenarios (really, positive or negative evaluation time $$t$$) the integration over the real number line can be closed in the complex plane by a gigantic semi-circular contour, we can get a cheap but visually obvious way to include the prescription chosen by adding these $$i \epsilon$$ whenever we write these integrals. It is much cheaper than having to write down in words everywhere that these integrals appear, and definitely much cheaper than people getting confused about them all the time.

That is, we have an annoying auxiliary condition to have to state at every place we write these integrals for them to make sense, and instead of writing long words every time, we instead use a proven fact of complex analysis to omit all those words and possibility of confusion. There is thus nothing controversial. It is nowhere near as scary as renormalisation.