Feynman rules out of the Lagrangian

Question

Accordingly to chapter 10, section 10.6 Feynman Rules of 'Introduction to Elementary Particles' by David Griffiths, there is a way to extract the vertex and propagators just by inspection of the Lagrangian:

1) Propagators: take the Euler-Lagrange equations of the free fields and the inverse of the operators in momentum space that act on the fields and multiplied by $i$ are the propagators for each one.

2) Vertex: take interaction Lagrangian and multiply it by $i$. Make use of the prescription $i\partial_\mu \rightarrow k_\mu$ and rub out the fields. The remaining is the vertex.

My questions are:

a) If I want to compute the vertex for the interaction Lagrangian ${\cal L}_{int} = g\varphi\partial_\mu \varphi \partial^\mu\varphi$ (all scalar fields), by rule 2) I get ${\rm vertex} = -igk_1k_2$. Nevertheless, apparently the solution is $-2ig(k_1k_2 + k_1k_3 + k_2k_3)$. Where did these extra factors come out?

b) If I had a similar interaction as in a) but changing one field for a new one $\chi$ (scalar too), so ${\cal L}_{int} = g\chi \partial_\mu \varphi \partial^\mu\varphi$, would the solution be $-2igk_1k_2$ with $k_i$ the momenta of $\varphi$ fields?

c) This book grants you that for QCD, with ${\cal L}_{int}^{3\ fields} = g\{[\partial^\mu A^\nu - \partial^\nu A^\mu]·(A_\mu \times A_\nu) + (A^\mu \times A^\nu)·[\partial_\mu A_\nu - \partial_\nu A_\mu]\}$, you can obtain the 3 fields vertex known as ${\rm vertex} = -gf^{\alpha \beta \gamma}[g_{\mu \nu}(k_1 - k_2)_\lambda + g_{\nu \lambda}(k_2 - k_3)_\mu + g_{\lambda \mu}(k_3 - k_1)_\nu]$. I've tried to get it from rule 2) but I wasn't able to.

Answer 1

I don't get how to apply these rules, when the interaction contains derivatives, either. If you are unsure you can go the long way. If you do not know how, here is how one could do it:

Try to evaluate (e.g. for the $g\phi \partial_\mu \phi \partial^{\mu} \phi$ interaction) the $\mathcal{O}(g)$ greens functions $G(x_1,x_2,x_3)$ of 3 external scalar particles, e.g. by using wicks theorem, and see what the vertex rule is. In position space you get 6 terms of the form: $-ig\int d^4x D(x_1-x) (\partial_\mu D)(x_2-x) (\partial^\mu D)(x_3-x)$. Then consider the momentum space greensfunction $\tilde{G}(k_1,k_2,k_3) = \int d^4x_1 \int d^4x_2 \int d^4x_3 G(x_1,x_2,x_3) e^{-ik_1x_1}e^{-ik_2x_2}e^{-ik_3x_3}$ (usual convention is all momenta ingoing, i.e. $e^{-ikx}$). This gives you the $g k_2 k_3$ term. When you do this for all the other contractions you get $3$ non equal contributions and $2$ each of the total $6$ are equal (therefore the factor $2$).

When you have different scalar fields it works almost the same, you just only consider contractions between same types of scalar fields.

The QCD case is also the same, however all $6$ are non equal, because they you have an additional space time and colour index. Therefore you get the $6$ terms.