Quantum Field Theory

  • Thread starter touqra
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  • #1
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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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  • #2
mjsd
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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.
 
  • #3
287
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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.
Why would the arc or circular bit dies away as the variable goes infinity?
 
  • #4
mjsd
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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.
 
  • #5
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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?
 
  • #6
mjsd
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when I said ML-estimate I mean the following:
Suppose C is a piecewise smooth curve. If [tex]h(z)[/tex] is continuous function on C then
[tex]\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq
\int_{C}|h(z)|\, |dz|}.[/tex]
and if C has length L and [tex]|h(z)|\leq M[/tex] on C then
[tex]\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq ML}[/tex]
 

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