# Feynman rule for deriative interaction: an example

Consider a theory for a finite number of real scalar fields $\phi _i$ with interaction terms of the form $$-\lambda _{ijk}\phi _i\partial _\mu \phi _j\partial ^\mu \phi _k,$$ with the sum over $i,j,k$ being implicit. Without loss of generality, assume that $\lambda _{ijk}$ is symmetric in $j$ and $k$.

Consider the thee-point interaction vertex between three of these fields of type $i$, $j$, and $k$ with momenta respectively $p_1$, $p_2$, and $p_3$. I just want to check that I have the Feynman rule for this vertex correct (so I can proceed on with the rest of my computation without being unsure if my Feynman rule is even correct). I believe the Feynman rule associated to this vertex should be $$-2\mathrm{i}\, (p_1\cdot p_2\lambda _{kij}+p_1\cdot p_3\lambda _{jik}+p_2\cdot p_3\lambda _{ijk}).$$

Is this correct?

• Looks good, derivatives result in factors of momentum. Dec 4, 2013 at 8:47
• The structure is correct, this may be derived from the interaction term (written in momentum space) $L_I= -\int dp_1 dp_2 dp_3~\phi_i(p_1)\phi_j(p_2)\phi_k(p_3)~\lambda_{ijk}~p_2.p_3~\delta(p_1+p_2+p_3)$. By using symmetries of the term $\phi_i(p_1)\phi_j(p_2)\phi_k(p_3)$, for instance $i \leftrightarrow j, p_1 \leftrightarrow p_2$, and symmetries of $\lambda$ ($\lambda_{ijk} = \lambda_{ikj}$), one obtains your structure. I trust you for the global factor... Dec 4, 2013 at 10:52

Usually such terms alter the Feynman rules in a subtle way through the measure in the functional integral. A non-linear sigma model in d=2 is a standard example. You have to add terms to the action containing $\delta^d(0)$ so as to cancel non-renormalizable loop diagrams with $k^{-2}$ in the propagator and two $k$'s in the numerator. If you omit these terms the theory loses symmetries. You can sometimes get away without these terms if you use dimensional regularization, as that sets all power law divergences to zero by fiat.