$V(r)=\frac{1}{r}$ means for any two electrons at position $r_1$ and $r_2$, the electric potential is given by $\frac{1}{|r_1-r_2|}$
The Fourier transform of $\frac{1}{r}$ is $\frac{1}{q^2}$.
How can we interpret $V(q)=\frac{1}{q^2}$ physically?
This is what I'm thinking about:
(1) $V(q)=\frac{1}{|q_1-q_2|^2}$ does not describe the potential between two electrons with wave vector $q_1$ and $q_2$
If we have two electrons with wave vector $q_1$ and $q_2$ (in othe words wavefunctions are $e^{iq_1r}$ and $e^{iq_2r}$), the probability distributions are uniform in space, then the potential $V(q)$ should be constant for all combination of $q_1$ and $q_2$.
(2) $V(q)$ must be quantum mechanics in nature ?
It is hard to explain in classical physics, but if we think about quantum Hilbert basis, $$ \langle r_1',r_2'|\hat{V}|r_1,r_2\rangle =\frac{1}{|r_1-r_2|} \delta_{r_1,r_1'} \delta_{r_2,r_2'}$$ $$ \langle q_1',q_2'|\hat{V}|q_1,q_2\rangle =\frac{1}{|q_1-q_1'|^2} \delta_{q_1+q_2,q_1'+q_2'} $$ $\hat{V}$ is two particles' operator. $\frac{1}{r}$ or $\frac{1}{q^2}$ are just the representation in certain basis.
Therefore $V(q)$ is an off-diagonal term, so it must be something quantum.
(3) what does the energy value $V(q)$ corresponds to?
$V(q)$ has the dimensional of energy, my naive understanding is, it corresponds to the energy you need to "push" $|q_1,q_2 \rangle$ to a new state $|q_1-q,q_2+q\rangle$.
In classical picture, both $|q_1,q_2 \rangle$ and $|q_1-q,q_2+q\rangle$ configuration, the electrons are uniformly distributed, no work need to be done to change the Coulomb potential.
In quantum mechanics, let's take the analogy of two level system $$|\uparrow\rangle :=|q_1,q_2 \rangle $$ $$|\downarrow\rangle := |q_1-q,q_2+q\rangle $$
Then the off-diagonial term $\Delta=\langle \uparrow |\hat{V}|\downarrow \rangle =\frac{1}{q^2} $ is more like tunnelling rate ( hopping rate ~ frequency ~ energy).
So, "push" means tunnelling. The energy value $V(q)$ is tunnelling rate.
People are talking about $\frac{1}{q^2}$ divergence, I don't understand it. Does it mean two electrons exchange their momentum very often, and the rate is divergent for small momentum ($q\rightarrow 0$) exchanges?
(4) $V(q)$ as a Laplace transformation in classical E&M
Contradict to (2), we do have $V(q)$ in purely classical systems.
This often solve the problem of screened plasma potential, classical Drude model of metal, etc..
$$ \nabla^2 \phi(r)=\delta(r) $$ The Laplace transformation gives : $$ q^2 \phi(q)=1 $$
And $q$ does not mean the momentum of electron wavefunction $e^{iqr}$ (there is no such concept of wavefunction in classical physics ). $q$ means the oscillation of charge density:
given charge density wave $\rho(r)=\cos(qr)$, we have potential wave $V(r) = \frac{1}{q^2}\cos(qr)$
Classical: a point charge can be represented by a superposition of cosine charge waves from all wave-vector. $$ \rho(x)=\delta(x-r)=\sum_q \cos(q(x-r)) $$
Quantum: a point charge state can be represented by a superposition of momentum q eigen-waves from all wave-vector. $$ |r\rangle =\sum_q e^{iqr}|q\rangle $$
I couldn't find a natural way to connect quantum and classical. Firstly, classical density $\cos(qr)$ changes it signs, while quantum $e^{iqr}$ is uniform in density; secondly, $\sum_q$ is coherent superpositions in the quantum case. This might be a bad analogy.
What is your interpretation of $V(q)=\frac{1}{q^2}$?