# Quantum Field Theory

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.

mjsd
Homework Helper
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.

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?

mjsd
Homework Helper
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 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?

mjsd
Homework Helper
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
$$\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq \int_{C}|h(z)|\, |dz|}.$$
and if C has length L and $$|h(z)|\leq M$$ on C then
$$\displaystyle{\left|\int_{C} h(z)\, dz\right| \leq ML}$$