- #1

- 69

- 1

[tex]\delta \mathcal{L}_1 = -\dfrac{gm^2}{32\pi \epsilon^2}\phi^2[/tex]

to the Lagrangian, which should give rise to a propagator term of the form:

[tex]-\dfrac{igm^2}{16\pi^2 \epsilon}[/tex]

that will cancel the divergent term in ##\Gamma^{(2)}(p)##. My problem is with the expression above, it's unclear to me how to reach this propagator correction just from the Lagrangian. How can this be achieved?