1
$\begingroup$

We have an operator A given by:

$A = \int_0^{2π} dФ \int_0^\infty \frac{d|\mathbf p|}{\left(2π\right)^3} |\mathbf p|^2 \int_{-1}^1 d\left(cos ϴ\right)e^{i|\mathbf p||\mathbf x|cos ϴ}e^{-iE_\mathbf p t}$

$=\frac{1}{\left(2π\right)^2 i|\mathbf x|} \int_0^\infty d|\mathbf p||\mathbf p| \left(e^{i|\mathbf p||\mathbf x|} - e^{-i|\mathbf p||\mathbf x|}\right) e^{-iE_\mathbf p t}$

$=\frac{-i}{\left(2π\right)^2 |\mathbf x|} \int_0^\infty d|\mathbf p||\mathbf p|e^{i|\mathbf p||\mathbf x|} - e^{-it{\sqrt {|\mathbf p|^2 + m^2}}}$

$=\frac{-i}{\left(2π\right)^2 |\mathbf x|} \int_m^\infty d\left(iz\right)ize^{-z|\mathbf x|} \left(e^{t{\sqrt {z^2 - m^2}}}-e^{-t{\sqrt {z^2 - m^2}}}\right)$

$=\frac{i}{\left(2π\right)^2 |\mathbf x|} e^{-m|\mathbf x|} \int_m^\infty dzze^{-(z-m)|\mathbf x|} sinh\left(t{\sqrt {z^2 - m^2}}\right)$

where $\mathbf p$ is the complex momentum vector.

what I'd like to know is, how did the $e^{i|\mathbf p||\mathbf x|}$ of the third to the last line has become $e^{-m|\mathbf x|} \int_{m}^\infty dzz e^{-(z-m)|\mathbf x|}$ in the last line, and $e^{-it {\sqrt {|\mathbf p|^2 + m^2}}}$ of the same, third to the last line, has become $sinh(t{\sqrt {z^2 - m^2 }})$ in the last line. Because I didn't understand the evolution of the complex integration part, I also did not understand how we draw the contour according to its evaluation. I'd also appreciate if you could involve the explanation for the contour as well.

The contour drawn according to the complex integral is in the below link:

Contour drawn by the operator A's complex integration

This calculation is the last step of the proof of single-particle Quantum Mechanics being not compatible with General Relativity as the result should turn out zero but its non-zero, but there was no need for the earlier steps, so I only included the part I'd like to know the answer of.There is no mistake in the question, at least no mistake in my part, as I've copied the page of the book correctly and exactly. For those who might own the book I've taken this part from; Quantum Field Theory for the gifted amateur by Tom Lancaster and Stephen J. Blundell, page75, equation 8.18 and page 76, eqt. 8.19 with Figure 8.3

Thank you and Kind Regards

$\endgroup$
  • $\begingroup$ Note: I have asked this question under a different topic but it was put on hold, I have corrected the question as much as I could but it wasn't taken back from being on hold position(I believe the picture still had typos, and that was the reason), so I had to re-write it. I'm not trying to reincarnate an already sold problem, I'm only re-writing the incorrectly formed question of the past corrected and understandable. $\endgroup$ – bergdi Aug 3 '18 at 17:58
  • $\begingroup$ Related post by OP: math.stackexchange.com/q/2864667/11127 $\endgroup$ – Qmechanic Sep 6 '18 at 14:53
  • $\begingroup$ @Qmechanic this link is my question, the one I've been talking about in the comment right above your answer, that hasn't been answered as well but it's understandable, there were atleast 2 typo's that made comprehending the question difficult, and since it was an uploaded picture instead of text-based equation written w/ equation-editor, I left it like that, after editing not that successfully, a couple of times $\endgroup$ – bergdi Sep 23 '18 at 12:21

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.