# What is the physical meaning of the Fourier transformed Coulomb potential $1/q^2$?

$V(r)=\frac{1}{r}$ 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 $V(q)=\frac{1}{q^2}$?

• "$V(q)$ must be quantum mechanics in nature" No. Just no. The scattering problem between point-particles can be treated in a fully classical manner using exactly the same formalism. Aug 24, 2018 at 17:24
• The form of $V(q)$ is just Poisson's equation in momentum space. Aug 24, 2018 at 19:12
• $|q_1-q_2|$ does not refer to the momentum difference of two interacting electrons with plane-wave wave function at $q_1$ and $q_2$. In fact with interactions, plane waves are not eigenfunctions (recall hydrogen atom). Instead, if you consider a initial state of one plane-wave electron with momentum $q_1$, it will be scattered into a different plane-wave state with $q_2$ by Coulomb interaction. What is the probability of this process? Well it is $1/|q_1-q_2|^2$. You can do the same thing for the other electron as long as you make sure momentum is conserved. Aug 24, 2018 at 20:55
• yes, reduce to a single electron problem, this is just the Rutherford scattering $| \frac{1}{(q_1-q_1')^2 } |^2 = \frac{1}{(\Delta q)^4 }$, it is coincidentally true, both in classical and quantum case.
– Jian
Aug 24, 2018 at 21:46

Sometimes it happens that in a pursuit of ultimate understanding we overcomplicate simple concepts.

What you should focus on is that:

1. A Fourier transform is a mathematical procedure that allows to expand a given function (continuous, differentiable etc.) into a complete set of orthonormal functions, usually utilizing a plane-wave kernel: $$F'(x) = \int dq \; F(q) e^{i q \cdot x}$$ It does not need to be more complicated than this.

2. Why would one use FT-functions instead of original ones? In many cases it makes math simpler. A good example is Maxwell's equations. In the position basis, they have a complicated form of first-order vector differential equations, while in momentum space they are just linear equations. $$\nabla \cdot \pmb E(\pmb r,t) = \rho(\pmb r,t) \;\; \rightarrow \;\; i \pmb k \cdot \pmb E' = \rho$$ where prime denotes a Fourier-transformed function. You can work out the rest yourself, if curious.

3. Sometimes scientists assign physical meaning to the FT'd functions, but it happens rarely, up to my knowledge (signal theory, information processing etc.). Most of the time, speaking of FT'd values, they actually mean the observables in any space one is comfortable with, because one can always back-transform (if nothing is seriously messed up).

My humble opinion in this case: you are thinking about that too much. An FT of the potential is just an FT of the potential. Since the mapping between V(r) and FT'd V'(q) is one-to-one and onto (in your case and in many other physical ones), you can think of V'(q) as a mathematical and physical equivalent of V(r).

• ObXKCD Aug 24, 2018 at 22:30
• @PM 2Ring Yeah, this is an old and good one X) Aug 24, 2018 at 23:22
• I disagree, I think this is a great question. Just because you can't think of a physical meaning doesn't mean there is one, or that it is useless to try to find one. As an experimentalist, thinking "physically" is indispensable - even when the theorists keep assuring me it's "just a mathematical tool". Physics uses math to describe the natural world. Any mathematics used in physics should have a physical meaning attached. Aug 28, 2018 at 16:28
• @SabrinaChoice Yes, that is why you are a great experimentalist=) Aug 28, 2018 at 19:29

I think this is a good question, so I'll attempt to answer even though I don't feel very confident about my ideas. Here goes:

The potential $$V(r)$$ in position space is pretty intuitive. As you (the particle) move about in space, you 'see' a potential which varies as a function of position.

The potential $$V(q)$$ is in momentum space. So maybe this means that as you (the particle) change (have a different) momentum you 'see' a potential which depends on your momentum. For example, if your momentum is large enough, you don't see a binding potential. You will 'fly' over it. But if your momentum isn't large enough, you do see a binding potential, and you become bound.