I'am not sure about some things about derivative interactions. Lets say I have to following 2 interaction terms:

$$\mathcal{L} = \left(a\phi + b\phi^{2}\right)\partial_{\mu}\phi\partial^{\mu}\phi.$$

Then I can write the vertex factor for the Feynman rules as follows (following Srednicki QFT conventions):

$$V_{a} = \left( -ia \right)\left(-ik^{\mu}\right)\left(ik^{\mu}\right) $$

$$V_{b} = \left( -ib \right)\left(-ik^{\mu}\right)\left(ik^{\mu}\right) $$

with $\left(-ik^{\mu}\right)$ for incoming momenta and a $\left(ik^{\mu}\right)$ for outgoing momenta.Now I wish to calculate the te loop corrections to the propegator which I think are these two:

enter image description here

with derivative fields in the loops. My question is, is the final result for the amplitude for the diagrams always proportional to $k^{2}$? Because in my calculations I have something like this:

$$\Pi_b \propto \frac{-i}{2} \int \frac{\mathrm{d}^{4}l}{\left(2\pi \right)^{4}}\frac{\left(-i\right)}{l^{2} + m^{2} - i\epsilon}\left(-il^{\mu}\right)\left( il^{\mu}\right)\left(-ib\right) = \frac{ib}{2} \int \frac{\mathrm{d}^{4}l}{\left(2\pi \right)^{4}}\frac{\left(l^{2}\right)}{l^{2} + m^{2} - i\epsilon} $$

$$\Pi_a \propto \frac{-i}{2} \int \frac{\mathrm{d}^{4}l}{\left(2\pi \right)^{4}}\frac{\left(-i\right)}{l^{2} + m^{2} - i\epsilon}\frac{\left(-i\right)}{\left(l + k\right)^{2} + m^{2} - i\epsilon}\left\{\left[-il^{\mu}\right]\left[ i\left(l + k\right)^{\mu}\right]\right\}^{2}\left(-ia\right)^{2}\\ \qquad = \frac{-ia^{2}}{2}\int \frac{\mathrm{d}^{4}l}{\left(2\pi \right)^{4}} \frac{1}{l^{2} + m^{2} - i\epsilon}\frac{1}{\left(l + k\right)^{2} + m^{2} - i\epsilon} \left[ l^{2} + \left(k\cdot l\right) \right]^{2} $$

does the above calculation looks right to you, or am I labeling the momenta wrong?

References for derivative couplings:

  1. Page 21 of https://www.maths.tcd.ie/~cblair/notes/list.pdf

  2. Chapter 7.4.2 of qft Schwartz

  • $\begingroup$ Why are you putting only the legs with derivatives inside the loop? There are definitely contributions to the propagator where the $\partial_\mu\phi$ are external legs. $\endgroup$
    – TwoBs
    Feb 22, 2018 at 0:50

1 Answer 1


Looks fine to me (up to the comment by TwoBs above, and up to conventional signs and factors of two). Btw, it would help to label your diagrams with momenta, and to place indices appropriately (i.e. $k_\mu k^\mu$ rather than $k^\mu k^\mu$ unless you do not actually mean to sum).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.