Feynman rules for interactions with derivatives: How exactly do the momentum factors appear? I know how to treat Feynman interactions without derivatives by Wick contraction. But now, take for example $$\mathcal{L}_{int}=\lambda \phi (\partial_{\mu}\phi)(\partial^{\mu}\phi).$$
Now many books write that in momentum space the derivatives turn into momenta. While I can imagine this happening, I don't really know how to write this down explicitly. At what point do I consider the Fourier transform of the field? Am I still using Wick contractions, but now with the field depending on the momenta? I have not found a source doing this explicitly.
 A: For your case, starting with this interaction term, let us substitute the expansion of $\phi$ in Fourier modes:
$$
\phi = \sum_k \phi_k e^{i kx}
$$
The action of derivative produces a factor of $ik$. Then, in the action you sum(integrate) over all $x$ :
$$
\sum_x \sum_{k_1, k_2, k_3} (ik_{2 \mu}) (ik^{3 \mu}) \lambda \phi_{k_1} \phi_{k_2} \phi_{k_3} e^{i (k_1 + k_2 + k_3) x} = 
 \sum_{k_1, k_2, k_3} (ik_{2 \mu}) (ik^{3 \mu}) \lambda \phi_{k_1} \phi_{k_2} \phi_{k_3} \delta (k_1 + k_2 + k_3) 
$$
Where in the last expression we have employed the well-known integral for exponent. The change from derivatives to momenta is simply results of change from positional basis, to momentum basis and has nothing to do with Wick theorem.
A: Firstly, from the Wick theorem, the position-space Feynman rules for a Feynman diagram is just the multiplication of Feynman propagators or, the two point correlation function in the free theory, i.e.
$$D_F(x,y)=\int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m^2+i\epsilon} e^{ip(x-y)}.$$
It is easy to verify the Wick theorem is still applicable, i.e., the vacuum matrix element of a time-ordered product of free fields is given by the sum of all possible full contractions, so what we need is to calculate the new contractions:
$\langle 0|T\{\phi(x)\partial_{\mu}\phi(y)\}|0\rangle$
and
$\langle 0|T\{\partial_{\mu}\phi(x)\partial_{\nu}\phi(y)\}|0\rangle$.
After similar calculation for the ordinary Feynman propagator, we can arrived at
$$\langle 0|T\{\phi(x)\partial_{\mu}\phi(y)\}|0\rangle
=\int \frac{d^4p}{(2\pi)^4} \frac{p_{\mu}}{p^2-m^2+i\epsilon} e^{ip(x-y)},$$
it needs to be noted that there is a minus sign between $\langle 0|\phi(x)\partial_{\mu}\phi(y)|0\rangle$ and $\langle 0|\partial_{\mu}\phi(y)\phi(x)|0\rangle$.
We just find that the derivative coupling Feynman rules is to modify the "propagator", but after LSZ projection, the denominator of the "new propagator" is canceled by the on-shell operator ($\partial^2 +m^2$), so the momentum in the nominator is left.
Remember that to derive the momentum-space Feynman rules, we need to handle the integral over several positions and momentums. Note that the integrals of the internal points lead to conservation of momentum in each vertex, while the integrals of the external points lead to the equality for momentums of the initial and final states. The integrals of the position just consume the delta function $\delta^4(\sum p)$ for conservation of momentum. So the Feynman rules for the derivative coupling is proportional to $p_{\mu}$ effectively.
Moreover, we need to identify the sign before $p_{\mu}$. Unlike the ordinary Feynman propagator $D_F(x,y)$ which is symmetric for the two position coordinates, i.e., $D_F(x,y)=D_F(y,x)$, there will be a additional minus sign after exchange coordinates for the "new propagator":
$$\langle 0|T\{\phi(y)\partial_{\mu}\phi(x)\}|0\rangle
=\int \frac{d^4p}{(2\pi)^4} \frac{- p_{\mu}}{p^2-m^2+i\epsilon} e^{ip(x-y)}.$$
Finally, let us count the number of $i$ to get the correct sign before $p_{\mu}$. There is an $i$ from LSZ projector $i\int d^4 x_i e^{\pm ip_i x_i}(\partial^2 +m^2)$ (minus for initial states and plus for final states), and a $-1$ for the $\partial^2+m^2$ after cancel the denominator of "new propagator", so we are left with $\mp ip_{\mu}$ for the vertex effectively, with $-p_\mu$ for the incoming momentum, $+p_{\mu}$ for the outgoing momentum.
From the explicit deduction above, we finally arrive at the Feynman rules for the derivative coupling.
