So I am aware of a thread at Propagator of a scalar in position space but it does not answer my question, which is more about poles in position space.

Starting from

$$D_F(x_1-x_2) = \int \frac{d^4 k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}e^{i k\cdot x}$$

I have been able to show that

$$D_F(x_1-x_2) = \frac{-i}{16\pi^2}\int_{0}^{\infty}\frac{ds}{s^2}\exp\left[-i\frac{X^2}{4s}\right]\exp\left[-i(m^2-i\epsilon)s\right]$$

which by change of variable can be written as

$$ D_F(x_1-x_2) = \frac{-i}{16\pi^2}[i(m^2-i\epsilon)]\int_0^\infty\frac{dt}{t^2}\exp\left[-t-\frac{[-(m^2-i\epsilon)X^2]}{4t}\right].$$

Using the integral representation of $K_1(z)$ (the modified Bessel function of the second kind) I can see that

$$D_F(x_1-x_2) = \frac{(m^2-i\epsilon)}{16\pi^2}\frac{4}{\sqrt{-(m^2-i\epsilon)X^2}}K_1(\sqrt{-(m^2-i\epsilon)X^2}).$$

But I know that the correct answer is

$$D_F(x_1-x_2) = -\frac{i}{4\pi^2}\frac{1}{\sqrt{-X^2 + i\epsilon}}K_1(im\sqrt{-X^2 + i\epsilon}).$$

What bothers me is how $\sqrt{-(m^2-i\epsilon)X^2}$ is equal to $\sqrt{im(-X^2 + i\epsilon)}$, because according to me

$$\sqrt{-(m^2-i\epsilon)X^2)} = \sqrt{-m^2(1-i\epsilon)X^2} = im\sqrt{X^2 - i\epsilon}$$

What's the error here?

  • $\begingroup$ So my confusion in the last step stems from the book 'Analytic Tools for Feynman Integrals' by Vladimir A. Smirnov (Springer 2012). The equation just above equation (2.17) gives the "correct answer" which has a manifestly different form. $\endgroup$ Mar 3, 2015 at 13:32
  • $\begingroup$ Have you solved this problem? I have a similar one $\endgroup$ Apr 29, 2015 at 7:38
  • $\begingroup$ My solution appears in my post. $\endgroup$ Apr 29, 2015 at 10:10
  • $\begingroup$ I refer to the book you mentioned and get to the step before changing variable. Can you teach me how you change $s$ to $t$? Thanks! $\endgroup$ May 2, 2015 at 11:49
  • $\begingroup$ I took $t = i X^24s$ from the looks of it... $\endgroup$ May 2, 2015 at 11:51


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.