I'm doing a question in Mark Srednicki's Quantum Field Theory. (Question 10.5) Which says that, when one changes a free field $\phi$ to $\phi + \lambda\phi^2$, the Lagrangian density would include an interacting term. However, if one still calculates the meson-meson (i.e. $\phi\phi->\phi\phi$) scattering, one would still get a zero scattering amplitude.

From the solution it seems that one can expand the interacting lagrangian into two terms, namely: $$L_3=\lambda(-2\phi（\delta\phi）^2-m^2\phi^3)$$ and $$L_4=\lambda^2(-2\phi^2（\delta\phi）^2-{1\over2}m^2\phi^4)$$

From hereon, the solution converts the two lagrangians into two kinds of Feynman diagrams, with 3-vertices and 4-vertices respectively. The "three-point" vertex factor is $$(-2i\lambda)[(k_1^2+m^2)+(k_2^2+m^2)+(k_3^2+m^2)]$$ And the "four-point" vertex factor is $$(-4i\lambda^2)[(k_1^2+m^2)+(k_2^2+m^2)+(k_3^2+m^2)+(k_4^2+m^2)]+4i\lambda^2m^2$$

This is where I'm actually lost...As a beginner in QFT, I'm actually still a little confused on how exactly one goes from an expression like $L_3$, to $<f|S|i>$ (which, according to Dyson's formula, would be a complicated integral like $e^{i\int{d^4xL_3}}$, which in principle can be simplified into propagators...) and then to the corresponding "three line" or "four line" vertices in graph, and finally to the actual graph for scattering like $\phi\phi->\phi\phi$ (How are the line numbers determined...? And what would be the scattering amplitude, would it be $<0|S|k_1k_2k_3>$?)

and a last question..would there be some direct way to calculate the differential cross section $d\sigma \over dcos\theta$ from the expression of the scattering amplitude $<f|S|i>$ (which we can acquire by counting the lines in the Feynman's diagram)? (I know that this might be a big question...but could someone please briefly explain the logic behind the transition from amplitude to cross section? The part on textbook is a bit overwhelming in calculation and I cannot seem to grasp the principle train of thought...)

Thank you very much!


Your Answer


Browse other questions tagged or ask your own question.