5
$\begingroup$

In the book N. D. Birrell and P. C. W. Davies, "Quantum Fields in Curved Space" at pp. 52-53 the four dimensional (positive frequency) Wightman Green function is expressed in position space as $$ D^{+}(x,x') = -\frac{1}{4\pi[(t-t'-i\epsilon)^2 -|\vec{x}-\vec{x}'|^2}.\tag{3.59}$$ It is said that this result comes from solving the momentum space integral $$\frac{1}{(2\pi)^4}\int d^4p\frac{1}{p_0^2 - |\vec{p}|^2}e^{-ip_0(t-t')+i\vec{p} ({\vec{x}-\vec{x}')}}$$ with the specific contour reported in figure 3 at p. 22 (a circle arount the positive pole). Anyway, I cannot find the above result. Instead, I find something very similar to the retarded Green function, that depends only from $\frac{1}{4\pi|\vec{x}-\vec{x}'|}$ and a $\delta$-function which fixes the time. I think that I'm getting wrong on the evaluation of something involving the $i\epsilon$ prescription used to shift the poles. Can anybody help me getting through this?

$\endgroup$
3
  • 2
    $\begingroup$ I guess it's the right-inverse of the d'Alembert operator in 3+1 dimensions. It's similar to $\frac{1}{4\pi|\vec{x}-\vec{x}'|}$ which is the inverse of the Laplacian in three dimensions. $\endgroup$
    – Valac
    Jul 6, 2022 at 15:07
  • 1
    $\begingroup$ But the Wightman function is solution of the KG equation, so of the homogeneous and not of the impulsive equation. Cannot apply the Fourier transform method directly. Also, if the expression as a momentum space expansion is the one given above, in principle I should be able to find (3.59) by solving the integral... $\endgroup$ Jul 6, 2022 at 15:14
  • 1
    $\begingroup$ Addendum: it is argued, always in Birrel-Davies, that the above function can be found by a contour integration of a function which is the same of the one giving the Feynman propagator (the trick to go from a 3D p-integral to a 4D p-integral is known) $\frac{e^{ip(x-x')}}{\omega_p^2 -|p|^2-i\epsilon}$. However, solving with the contour reported in figure 3 at p. 22 does not bring the above result (3.59). $\endgroup$ Jul 8, 2022 at 12:07

1 Answer 1

6
+50
$\begingroup$

The $\epsilon$ in (3.59) does not shift the poles but rather regulates the integral over 3-momentum.

More concretely, we start with (note the extra $i$ in fig. 3) $$ iD^+(t,\vec x) = \frac{1}{(2\pi)^4}\int_{\gamma^+} dp_0\int \ d^3\vec p \,\frac{e^{-ip_0t+i\vec p\cdot \vec x}}{p_0^2-|\vec p|^2}, $$ where $\gamma^+$ is a contour that encircle only the pole in the right half-plane. By the residue theorem, we find $$ iD^+(t,\vec x) =\frac{i}{(2\pi)^3} \int \ d^3\vec p\, \frac{e^{-i|\vec p|t+i\vec p\cdot \vec x}}{2|\vec p|}. $$ The $\epsilon$ now enters the game to regulate the above integral as $t \rightarrow t-i\epsilon$ such that it is convergent. The rest of the details are relatively straightforward $$ \begin{align*} iD^+(t,\vec x) &=\frac{i}{(2\pi)^2} \int_{-1}^1d(\cos\theta)\int _0^\infty d|\vec p|\, |\vec p|^2 \frac{e^{-|\vec p|(\epsilon+it)+i|\vec p| |\vec x| \cos \theta}}{2|\vec p|}\\ &=\frac{i}{8\pi^2}\frac{1}{i|\vec x|}\int _0^\infty d|\vec p|\,e^{-|\vec p|(\epsilon+it)}(e^{i|\vec p||\vec x|}-e^{-i|\vec p||\vec x|})\\ &=\frac{i}{8\pi^2}\frac{1}{i|\vec x|}\left(\frac{1}{\epsilon+it-i|\vec x|}-\frac{1}{\epsilon+it+i|\vec x|}\right)\\ &=\frac{i}{8\pi^2}\frac{1}{i|\vec x|}\frac{2i|\vec x|}{(\epsilon+it)^2+|\vec x|^2} = -\frac{i}{4\pi^2}\frac{1}{(t-i\epsilon)^2-|\vec x|^2} \end{align*} $$ The regulator can be "removed" using the Sokhotski–Plemelj theorem as explained in (3.60).

$\endgroup$
1
  • $\begingroup$ Dude I know that I should use answers only to enrich the conversation, but a special thanks deserves an answer on its own. Allow me to give you these 50 points. Thank you. $\endgroup$ Jul 22, 2022 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.