Quantum Field Theory: Evaluating Integrals on Page 27

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SUMMARY

The discussion focuses on the evaluation of integrals in Quantum Field Theory as presented in Peskin & Schroeder's work, specifically on page 27. Participants analyze the use of contour integration techniques, including key-hole contours and the application of Cauchy's theorem. The conversation highlights the importance of understanding the behavior of integrands as the variable approaches infinity, referencing Jordan's lemma and the ML estimate to support their conclusions. The integral's evaluation is confirmed by the behavior of the semicircular path, which tends to zero as R approaches infinity.

PREREQUISITES
  • Understanding of contour integration techniques in complex analysis
  • Familiarity with Cauchy's theorem and its applications
  • Knowledge of Jordan's lemma and its implications for integrals
  • Basic concepts of branch cuts in complex functions
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  • Study the application of Cauchy's theorem in complex analysis
  • Learn about Jordan's lemma and its use in evaluating integrals
  • Explore the concept of branch cuts and their significance in complex integrals
  • Investigate the ML estimate and its applications in contour integration
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I don't understand how Peskin & Schroeder can evaluate the integral on page 27 by having the real axis wrapping around branch cuts just like that. The picture of the contours are on page 28.
 
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I think what they have done is simply completed a loop (like a key-hole contour) but the arc/circular bit dies away as your variable go to infinity so effectively the flat/horizontal bit is same as the two vertical bits (by Cauchy theorem... as no poles inside loop)

describing the loop: first bit is the original bit the flat/horizontal (-R,+R) bit with R eventually taken to infinity, then to complete the loop you need to add a 1/4 of an arc going from +R to +iR, then comes down to avoid the branch cut, go around the pole and goes up again before arch back from +iR to -R.
 
mjsd said:
I think what they have done is simply completed a loop (like a key-hole contour) but the arc/circular bit dies away as your variable go to infinity so effectively the flat/horizontal bit is same as the two vertical bits (by Cauchy theorem... as no poles inside loop)

describing the loop: first bit is the original bit the flat/horizontal (-R,+R) bit with R eventually taken to infinity, then to complete the loop you need to add a 1/4 of an arc going from +R to +iR, then comes down to avoid the branch cut, go around the pole and goes up again before arch back from +iR to -R.

Why would the arc or circular bit dies away as the variable goes infinity?
 
I haven't check this particular example and see if it does goes away... but it usually does and that's why we close the contour in the first place...by the way, I did say "I think"...perhaps you can check that... to prove that you need to look at your integrand and see what happen when R becomes large (ie. when the integration variable expressed in polar form becomes large). Sometimes Jordon's lemma or ML-estimate maybe used to help.
 
mjsd said:
I haven't check this particular example and see if it does goes away... but it usually does and that's why we close the contour in the first place...by the way, I did say "I think"...perhaps you can check that... to prove that you need to look at your integrand and see what happen when R becomes large (ie. when the integration variable expressed in polar form becomes large). Sometimes Jordon's lemma or ML-estimate maybe used to help.

I looked up on Jordan's lemma, and yeah the integrand of the semicircular path (excluding the real axis) tends to zero as R goes infinity.
OOOooo contour integrals are so interesting !
Thank you!

Is ML estimate maximum likelihood estimate? How can ML estimate be used here since it is about probability?
 
when I said ML-estimate I mean the following:
Suppose C is a piecewise smooth curve. If h(z) is continuous function on C then
\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq <br /> \int_{C}|h(z)|\, |dz|}.
and if C has length L and |h(z)|\leq M on C then
\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq ML}
 

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