How Does Spherical Polar Coordinates Simplify the Coulomb Potential in K-Space?

[ognik]

Hi - I've got myself mixed up here, please see what I am missing below ...

Show $ \int e^{ik \cdot (r - r')} \frac{d^3 k}{(2 \pi)^3 k^2} = \frac{1}{4 \pi}|r-r'| $

Let R = r-r', then $k \cdot (r - r') = kR cos \theta$

Next I would translate into spherical polar coords, using $\int d^3 k = \int_{0}^{\infty} r^2 \,dr \int_{0}^{2 \pi} \,d\phi \int_{0}^{\pi}sin \theta \,d\theta $

I am not sure I can just do that in 'k-space'? I know that $k^2 = \frac{\omega}{m}$ ... ? If my coordinate transform above is wrong (because of being in k space) please point me at the correct transform?

PS: I really would appreciate some help with this one, it shows one of those gaps in my background that I need to plug quickly - I write the exam next week ...

 

[jvicens]

Hi there,

It seems like you are on the right track with your approach. However, there are a few things that can be clarified to help you understand this integral better.

Firstly, the integral you are trying to solve is known as the Coulomb potential in k-space. This is a well-known result in physics and it represents the potential energy between two charged particles in Fourier space.

Now, to address your specific questions:

1. Can you just translate into spherical polar coordinates in k-space?

Yes, you can. In fact, this is the standard approach to solving this integral. The reason for this is that spherical polar coordinates are more suitable for dealing with the angular dependence of the Coulomb potential.

2. Is your coordinate transform incorrect because you are in k-space?

No, your coordinate transform is correct. In fact, in k-space, we use the same coordinate system as in real space. The only difference is that the coordinates are now in terms of wave vectors (k) instead of position vectors (r).

3. How does $k^2 = \frac{\omega}{m}$ fit into this integral?

In this case, $\omega$ represents the frequency of the wave and m represents the mass of the particle. This relation is known as the dispersion relation and it relates the frequency of the wave to its wave vector. However, in this integral, we are not dealing with a specific wave or particle, so this relation is not directly applicable here.

To solve this integral, you can use the substitution $kRcos\theta = x$ and then use the identity $\int_{0}^{2\pi} e^{ikRcos\theta} d\phi = 2\pi J_0(kR)$, where $J_0$ is the Bessel function of the first kind. This will simplify the integral to $\int_{0}^{\infty} \frac{2\pi}{k} J_0(kR) dk$, which can be solved using the integral representation of $J_0$.

I hope this helps you understand the integral better. Good luck with your exam!