Consider the interaction Lagrangians given by
\begin{align*} \mathcal{L}_1 = A_\mu \ \pi^{0} \ \partial^{\mu} \pi^{0}\\ \mathcal{L}_2 = B^{\mu\nu} \ \partial_\mu \pi^{0} \partial_{\nu} \pi^{0}\\ \mathcal{L}_3 = C_{\nu} \ \ \partial_\mu \pi^{0} \partial^\mu \pi^0 \partial^{\nu} \pi^{0} \end{align*}
My question concerns finding the vertex rules for the amputated diagrams corresponding to these interactions (I don't care about $A,B,C$ I just added them for Lorentz invariance).
Notation: I will use $\stackrel{FT}{=}$ to mean "corresponds to the momentum space interaction", and $\sim$ to mean "corresponds to the vertex rule".
Suppose that all the $\pi^0$ are all exiting the vertex and that $A,B,C$ are incident. We use the sign convention where the Feynman rule for outgoing momenta is $-ip_\mu$.
Then naively, we obtain (again, I don't care about $A,B,C$)
\begin{align*} \mathcal{L}_1 &\stackrel{FT}{=} (-ip^\pi_\mu) A^\mu \pi^0 \pi^0 \ \sim \ A^\mu (-ip^\pi_\mu) \\ \mathcal{L}_2 &\stackrel{FT}{=} (-ip^\pi_{1 \mu})(-ip^\pi_{2\nu}) B^{\mu\nu} \pi^0 \pi^0 \ \sim\ B^{\mu\nu}(-ip^\pi_{1 \mu})(-ip^\pi_{2\nu}) \\ \mathcal{L}_3 &\stackrel{FT}{=} (-ip^\pi_{1\mu})(-ip^\pi_{2\mu})(-ip^\pi_{3\nu}) C^\nu \pi^0 \pi^0 \pi^0 \ \sim \ C^\nu(-ip^\pi_{1\mu})(-ip^\pi_{2\mu})(-ip^\pi_{3\nu}) \end{align*}
However, since the $\pi^0$ are identical the vertex rules should be symmetric under their exchange so shouldn't momenta be symmetrized? That is, shouldn't we have that
\begin{align*} \mathcal{L}_1 &\sim A^\mu (-i)(p^\pi_{1\mu} + p^\pi_{2\mu}) \\ \mathcal{L}_2 &\sim B^{\mu\nu}(-i)^2\bigg[p^\pi_{1 \mu}p^\pi_{2\nu} + p^\pi_{2 \mu} p^\pi_{1\nu}\bigg]\\ \mathcal{L}_3 & \sim C^\nu (-i)^3\bigg[ p^\pi_{1\mu}p^\pi_{2\mu} +p^\pi_{2\mu}p^\pi_{3\mu} +p^\pi_{1\mu}p^\pi_{3\mu} \bigg]\bigg[ p^\pi_{1\nu} + p^\pi_{2\nu} + p^\pi_{3\nu} \bigg] \end{align*}
I'm about to get involved in a long, nasty computation so I just wanted to make sure.
Additionally, do I need to add any additional symmetry factors when performing the symmetrization of momenta? My intuition is telling me that I should add a factor $1/n!$ to $\mathcal{L}_n$ but I'm not sure.
