When discussing causality in Chapter 2 of Peskin & Schroeder a couple of equations giving the asymptotic behaviour of the propagator for a scalar field appear:

$$ \text{If} \,\, x^0-y^0=t, \, \, \mathbf{x-y} = 0 \Rightarrow D (x-y) = \frac{1}{4 \pi^2} \int_{m}^{\infty} dE \sqrt{E^2-m^2} e^{-iEt} \underset{t \to \infty}{\sim} e^{-imt} $$ $$ \text{If} \,\, x^0-y^0=0, \, \, \mathbf{x-y} = \mathbf{r} \Rightarrow D (x-y) = \frac{1}{4 \pi^2 r} \int_{m}^{\infty} d\rho \frac{\rho\, e^{-\rho r}}{\sqrt{\rho^2 - m^2}} \underset{r \to \infty}{\sim} e^{-mr} $$

I can't see how you derive these asymptotic behaviours (I have no problem deriving the integral exact expressions, but then I get stuck). All I could do was to rewrite the first integral as follows:

$$ D (x-y) = \frac{1}{4 \pi^2} \int_{m}^{\infty} dE \sqrt{E^2-m^2} e^{-iEt} = \frac{m}{4 \pi^2 i t} K_1(imt) $$

using this article on modified Bessel functions of the second kind. But checking with Mathematica, this vanishes for $t \to \infty$. For the second integral I don't have any clue, so any help would be more than welcome!

Extra (but related) question:

In the first discussion of the chapter something similar appears

$$ U(t) = \frac{1}{2 \pi^2 | \mathbf{x - x_0} | } \int_{0}^{\infty}dp\,p\, \sin (p | \mathbf{x - x_0} | ) e^{-it \sqrt{p^2 + m^2}} $$

Is this also obtained through a similar procedure?


1 Answer 1


The idea here is to use a stationary phase / saddle point approximation, that needs to be slightly adapted.

For example, if you rewrite the integrand of the first integral in the form $$ e^{-it\left(E+\frac{i}{2t}\ln\left(E^2-m^2\right)\right)}, $$ you will find two stationary points $E_\pm=\pm m +\mathcal{O}(t^{-1})$. Only $+m$ can contribute to the integral due to the range of integration, and thus you find that the dominant contribution to the integral is proportional to $e^{-i mt}$.

You can proceed in the same way for the other integrals.

  • $\begingroup$ Ok, that was helpful! I think I will have to look for a little bit more information concerning the stationary phase method because I have never heard of it... However, there is still something that seems odd to me, how is it that if the integral behaves asymptotically as $e^{-imt}$ for large $t$ the limit when $t \to \infty$ is 0? (I'm assuming my previous calulation involving Bessel functions was right, and that the limit of that expression is indeed zero as Mathematica claims). $\endgroup$
    – Alex V.
    Jan 20, 2016 at 17:41
  • 2
    $\begingroup$ @AlexV.: Note that my argument just developped enough to find the leading exponential term. There will also be prefactors, dependent on $t$, which will imply the "damping" of the $e^{-i mt}$ oscillations, that you can obtain if you compute the gaussian correction to the saddle-point expansion. You can get the correct form of the prefactor here : en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms $\endgroup$
    – Adam
    Jan 21, 2016 at 10:07

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