3
$\begingroup$

In Peskin and Schroeder Introduction to Quantum Field Theory book, above equation 6.64 on pg. 200, it was said that "to gain better understanding, of the divergence, let us evaluate the coefficient of the log in the limit of $-q^2\rightarrow\infty$." Why does he want to take $-q^2$ to infinity rather than $0$ to extract the divergence? I thought the form factor $F_1(q^2)$ contains an infrared divergence at $q^2=0$.

$\endgroup$
0
$\begingroup$

The photon which generates the divergence has momentum $p-k$ which can be seen at the beginning of section 6.3 in P&S. The fact that you take $-q^{2}\rightarrow\infty$ doesn't affect the possibility of a soft photon appearing. When $-q^{2}\ll1$ you can only have soft photons while in the case of $-q^{2}\rightarrow\infty$ you can get several soft photon and/or hard photons. This is due to the conservation of 4-momentum.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.