Fourier transform of a scatering potential The Fermi golden rules states
$$ \Gamma(\vec{k},\vec{k}') = \frac{2\pi}{\hbar} \left| \left \langle \vec{k}|V|\vec{k}' \right \rangle \right|^2 \delta(E(\vec{k})-E(\vec{k}')) \, .$$
Many places (for example - Simon, Solid State, p. 141) state the scattering potential's matrix element is the Fourier transform of potential's position representation, i.e.
$$ \left \langle \vec{k}|V|\vec{k}' \right \rangle = \intop dr \frac{e^{-i\vec{k}'\cdot\vec{r}}}{\sqrt{V^3}} V(\vec{r}) \frac{e^{i\vec{k}\cdot\vec{r}}}{\sqrt{V^3}} =\frac{1}{V^3} \intop dr e^{-i(\vec{k}'-\vec{k}) \cdot \vec{r}} V(\vec{r}) \, .$$
I don't understand how this formula was derived. It seems like there was a double use of the completeness relation, but without assuming that $V$
 is diagonal in the position representation I don't understand where does it go, but this seems like a strong assumption.
 A: Let's dispense with all the vector arrows to simplify the notation.
\begin{align}
\left \langle k | V | k' \right \rangle
=& \int dx' \int dx \langle k \underbrace{| x' \rangle \langle x'|}_\text{identity} V \underbrace{| x \rangle \langle x |}_\text{identity} k' \rangle \, .
\end{align}
Now the question is what to do with $\langle x' | V | x \rangle$.
As already pointed out in the original post, we could make progress if $V$ were diagonal in the position representation.
If $V$ represented a two-particle interaction, then of course it would depend on more than one position, i.e. it would depend on the two positions of the interacting particles.
However, by construction in the problem, $V$ represents a scattering potential that is assumed to be fixed, i.e. induced by some external fields that we are not treating as having their own internal degrees of freedom.
In that case, the potential energy of a particle moving under the influence of $V$ depends only on the position of the particle, so in other words, $V$ is diagonal in the position representation.
Therefore, continuing the computation, we have
\begin{align}
=& \int dx' \int dx \, e^{i k x'} V(x) \delta(x - x') e^{-i k' x} \\
=& \int dx \, e^{i k x} \, V(x) \, e^{-i k' x} \\
=& \int dx \,e^{i (k - k') x} \, V(x) \, ,
\end{align}
which is what we were trying to show.
Treating the potential $V$ as an externally supplied, fixed function of space that acts on one particle at a time, is always an approximation.
In real life, the only real potential energies are interactions between particles or fields.
For example, we might have electrons interacting with photons, or in other words we might have the electromagnetic field interacting with the electron field.
A full quantum field theory calculation treats both fields as dynamic quantities.
There's an interaction energy between the two fields (or between the photon and the electron) that depends on two positions.
However, if the electromagnetic field is strong enough to not change much during its interactions with the electrons, we can treat the electromagnetic field as static, in which case the interaction energy only depends on the position of the electron, and the static shape of the electromagnetic field is captured in the form of $V(x_\text{electron})$.
