Total divergence term and corresponding Feynman Diagram A total divergence term added to the Lagrangian doesn’t affect the action because the integral of a total divergence vanishes. But if one attempts to derive the Feynman rules from the Lagrangian with the total divergence term and from that without this term, will not the former give rise to extra diagrams compared to the latter? Isn’t that disturbing?
 A: I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like
\begin{equation}
ig \sum_i p_i \ ,
\end{equation}
where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum conservation. So there is no new interaction and thus the theories are the same.
Edit:
Consider $\phi^4$ theory. A total derivative term would be 
\begin{equation}
\delta \mathcal{L} = gd\phi^3 \ ,
\end{equation}
where $g$ is a dimensionless coupling and $[d_x]=[\phi]=\text{energy}$. Each of this fields $\phi$ now live at spacetime point $x_i$ and has momentum $p_i$, where $i\in \{1,2,3\}$.
The total derivative now gives three terms with the same structure $\sim\phi^2\partial_i\phi$. In Fourier space, the derivative becomes a multiplication with the corresponding momentum and we thus have
\begin{align}
d_i(\phi_1\phi_2\phi_3)&=\phi_1\phi_2\partial_i\phi_3 + \phi_1\phi_3\partial_i\phi_2+\phi_2\phi_3\partial_i\phi_1 \\
&\sim p_3\phi_1\phi_2\phi_3 + p_2\phi_1\phi_3\phi_2+p_1\phi_2\phi_3\phi_1 \\
&=\phi_1\phi_2\phi_3 (p_1+p_2+p_3) \,
\end{align}
which leads then to the vertex rule $iV= ig(p_1+p_2+p_3)$. And this is zero, because the momentum has to be conserved at the vertex.
