How does the transformation of 4-derivative into a 4-momentum actually happen in a derivative coupling? Consider a derivative coupling with $$\mathcal{L}_{int} = \lambda \phi_1 (\partial_\mu \phi_2) (\partial_\mu \phi_3),\tag{7.101}$$ and a scalar field
$$ \phi(x) = \int \frac{d^4p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (a_p e^{-ip_\mu x^\mu} + a_p^\dagger e^{ip_\mu x^\mu}). \tag{7.102}$$
Using Feynman rules, the textbooks make the following transformation
$$ \partial_\mu \rightarrow -i p_\mu $$ 
(incoming particle into a vertex)
$$ \partial_\mu \rightarrow i p_\mu $$ 
(outgoing particle).
I have 2 questions: 


*

*Can anybody help me to understand the derivation of this transformation?

*I would like to understand the impact on results of Feynman rules when I change the Lagrangian in this way (or even others), but I was not able to get it.
References:


*

*M.D. Schwartz, Quantum Field Theory and the Standard Model, 2014; Page: 99.

 A: Let's first look at the case of an incoming particle.
An incoming particle state is something like this
$$ |I\rangle = a_p^\dagger |0\rangle.$$
When we act with $\phi(x)$ on this, the part with a destruction operator, $\phi^+(x) = \int \frac{d^4p'}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}a_{p'}e^{-ip'_\mu x^\mu}$, is the term that will annihilate the incoming particle, and hence be the part relevant for the incoming particle Feynman rule. 
If we take the derivative of this one we will just pull down a factor including momentum from the exponential
$$ \partial_\mu \phi^+(x)a_p^\dagger |0\rangle = -i p_\mu\phi^+(x)a_p^\dagger |0\rangle.$$
For an outgoing particle we have the state
$$ \langle F| = \langle 0 | a_{p},$$
in this case it is the part of $\phi(x)$ with a creation operator, $\phi^-(x) = \int \frac{d^4p'}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}a^\dagger_{p'}e^{ip'_\mu x^\mu}$,  that will annihilate the final particle, and we get
$$ \langle 0 | a_{p}\partial_\mu \phi^-(x) = \langle 0 | a_{p}i p_\mu\phi^-(x).$$
A: First of all, check the indices in your interaction term, the term is not Lorentz Invariant!. The substitution $i\partial_\mu \rightarrow p_\mu$ becomes clear using the Fourier transform. Imagine you have the term
\begin{equation}
\partial_\mu \phi(x).
\end{equation}
In order to derive the "Feynman rule", you have to move to momentum space using the Fourier transform. Define the Fourier transform of $\phi(x)$ i.e $\hat{\phi}(p)$ as
\begin{equation}
\phi(x)=\int \dfrac{d^4p}{(2\pi)^4}\hat{\phi}(p)e^{-ip\cdot x}.
\end{equation}
Then note that,
\begin{equation}
\partial_\mu\phi(x)=\int \dfrac{d^4p}{(2\pi)^4}\hat{\phi}(p)\partial_\mu e^{-ip\cdot x}=\int \dfrac{d^4p}{(2\pi)^4}(-ip_\mu)\hat{\phi}(p) e^{-ip\cdot x}.
\end{equation}
The correspondence is clear. In the case you have more fields proceed as follows
\begin{equation}
\lambda \phi_1(\partial_\mu\phi_2)(\partial^\mu \phi_3
)=\lambda\int \dfrac{d^4p_1}{(2\pi)^4}\hat{\phi}(p_1)e^{-ip_1\cdot x}
\int \dfrac{d^4p_2}{(2\pi)^4}\hat{\phi}(p_2)\partial_\mu e^{-ip_2\cdot x}
\int \dfrac{d^4p_3}{(2\pi)^4}\hat{\phi}(p_3)\partial^\mu e^{-ip_3\cdot x}=
\end{equation}
\begin{equation}
=\lambda\int\dfrac{d^4p_1}{(2\pi)^4}\int \dfrac{d^4p_2}{(2\pi)^4}
\int \dfrac{d^4p_3}{(2\pi)^4}(-ip_{2\mu})(-ip_3^\mu)
\hat{\phi}(p_1)
\hat{\phi}(p_2)
\hat{\phi}(p_3)e^{-i(p_1+p_2+p_3)\cdot x}.
\end{equation}
Remember that the action $S$ has an integration over space $d^4x$ so that you get a factor $(2\pi)^4\delta(p_1+p_2+p_3)$, then you can use the delta function to integrate over one momentum (I choose $p_1$).
\begin{equation}
\int d^4 x \mathcal{L}_{int}
=\lambda\int\dfrac{d^4p_2}{(2\pi)^4}
\int \dfrac{d^4p_3}{(2\pi)^4}(-p_{2}\cdot p_3)
\hat{\phi}(-p_2-p_3)
\hat{\phi}(p_2)
\hat{\phi}(p_3).
\end{equation}
So that we get the correspondence
\begin{equation}
\lambda \phi_1(\partial_\mu\phi_2)(\partial^\mu \phi_3
)\longrightarrow -\lambda(p_2\cdot p_3).
\end{equation}
Finally, this vertex has three incoming lines with three momentums $p_1$ (fixed), $p_2$ and $p_3$, so that you have to sum over permutations $\sigma\in(1,2,3)$. Then the associated Feynman rule is
\begin{equation}
\sum_{\sigma\in(1,2,3)}-\lambda(p_2\cdot p_3)\longrightarrow -2\lambda(p_1\cdot p_2+p_2\cdot p_3+p_3\cdot p_1).
\end{equation}
