Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

When trying to find the Fourier transform of the Coulomb potential

$$V(\mathbf{r})=-\frac{e^2}{r}$$

one is faced with the problem that the resulting integral is divergent. Usually, it is then argued to introduce a screening factor $e^{-\mu r}$ and take the limit $\lim_{\mu \to 0}$ at the end of the calculation.

This always seemed somewhat ad hoc to me, and I would like to know what the mathematical justification for this procedure is. I could imagine the answer to be along the lines of: well, the Coulomb potential doesn't have a proper FT, but it has a weak FT defined like this …

Edit: Let me try another formulation:

What would a mathematician (being unaware of any physical meanings) do when asked to find the Fourier transform of the Coulomb potential?

share|improve this question
    
The integral is not divergent at zero, because 1/r is too weak, and it is not divergent at infinity. It is not absolutely convergent, but the oscillations fix that, so long as you take the limit in any reasonable way. You can just do the integral, it requires nothing special. The "mu" trick is just a nice way of cutting off large distances. The answer is not particularly divergent there anyway. –  Ron Maimon Sep 14 '11 at 19:52

5 Answers 5

up vote 8 down vote accepted

It can be justified using distribution theory.

share|improve this answer
4  
I like this answer much better than the "who needs a justification?" answers! It's true that physicists often decide not to worry about whether a particular mathematical operation can be rigorously justified, but the original question was asking for just such a justification, and it's not an obviously unreasonable thing to ask for. –  Ted Bunn Mar 23 '11 at 19:44
1  
Correct answer. But to fill in details: Laplacian of the $r^{-1}$ potential will give delta function. That means after Fourier transform $q^2 V(q) \sim 1$ which is precisely what you want. –  Marek Mar 24 '11 at 1:43
1  
@Ted: The mathematics of limits is thoroughly rigorous in and of itself. The whole business was developed to put calculus on a firm foundation. QGR has just offered a alternative way of phrasing the limit, because what is the first thing you see when you ask for a definition of a distribution: a limit. –  dmckee Mar 24 '11 at 1:55
3  
This is all good, but remember that there are limits (enjoy the pun) to distribution theory. Crudely, we can exchange limits only given certain conditions, and distributions can be understood often to be an effective way to manage certain kinds of limits, even though they are strictly elements in a space of linear maps from a function space, usually (in Physics) into either the reals or the complex numbers. The people commenting above know this stuff well, but it is well to know when and how distributions can be consistently used, precisely because they are used a lot in Physics. –  Peter Morgan Mar 24 '11 at 12:08
2  
(Continuing previous comment.) I'm like you, and like most physicists, in that I generally don't worry about this stuff and hope for the best. And I completely agree that that's turned out to be a very fruitful procedure in physics. But the mathematicians like to dot all the i's and cross all the t's, and I'm actually glad that they do. –  Ted Bunn Mar 24 '11 at 13:55

I'm not sure that a justification is required.

We don't feel the need to justify solutions obtained by making a change of variables, nor those obtained by differentiating under the integral sign, nor those obtained by analytic continuation and application of contour integrals and so on for a host of other tricks of symbol manipulations.

But if you insist...

You can interpret this as the potential due to a massive scaler boson (the Yukawa potential), and then take the limit of vanishing mass.

share|improve this answer

Dear Emerson, the "weak Fourier transform" of the potential, which is simply $4\pi / q^2$, is the "proper physical" value of the Fourier transform.

In physics, it simply makes no sense to say that a Fourier transform "doesn't exist": the Coulomb potential clearly does exist, and when we're calculating something that depends on its Fourier transform, the answer also has to exist because the world has to behave in some way. It's just generally impossible in physics to say that a valid theory produces an "ill-defined answer". We must always struggle to get an actual, well-defined answer.

And quantum electrodynamics, even though it has a superficially divergent Fourier transform of the Coulomb potential, is an excellent theory. Still, we need the Fourier transform to answer many questions. So the only question can be how to find out the right answer - not whether an answer exists. Of course that it does. The physically correct answer is most easily obtained by regulating the potential - as a limit of the Yukawa potential - and by exchanging the limit with the integration in the middle of the calculation, to guarantee that the result is sensible.

This is like giving a mass to the photon - we call it an infrared regulator. Such things are omnipresent in quantum field theory. If an expression doesn't naively make sense, one tries to express it as a limit of a more general expression, and calculate a more general calculation, which is more well-defined, and try to take the required limit at the very end. This is the right physicist's approach to seemingly ill-defined expressions and it always leads us out of the trouble and it always tells us something more meaningful.

A failure - a refusal to calculate the answers - is just not acceptable in physics.

share|improve this answer
4  
That reminds me of a lecturer I had: "this series obvious converges because it's physical". I went through multiple cycles of utter belief and disbelief in a few minutes thinking about that one. –  genneth Mar 24 '11 at 20:15
1  
Once I encountered a problem with divergent matrix elements so the perturbation theory series were useless whereas the problem had clearly physical and mathematical finite solution and their power expansions. I had a very strong temptation to discard divergent parts in the matrix elements. Fortunately, I found the true reason of their divergence and a way to reformulate the problem without this difficulty. This taught me that sometimes divergent results may be correct. –  Vladimir Kalitvianski Mar 25 '11 at 16:43

If the final result is unique and independent of the regularization method, then it is a mathematically correct answer.

share|improve this answer

I think you need this integral $$\int {\frac{1}{{\left( {2\pi } \right)^3 }}\frac{{e^{i\vec k \cdot \vec r} }}{{k^2 + a^2 }}d^3 \vec k} = \frac{{e^{ - ar} }}{{4\pi r}}$$ We suppoes initially that $a \ne 0$. If we use polar coordinates with $\vec k$ in the z-axis we have: $$\int {\frac{1}{{\left( {2\pi } \right)^3 }}\frac{{e^{i\vec k \cdot \vec r} }}{{k^2 + a^2 }}d^3 \vec k} = \frac{1}{{\left( {2\pi } \right)^3 }}\int\limits_0^\infty {\frac{{k^2 }}{{k^2 + a^2 }}dk} \int\limits_0^{2\pi } {d\varphi } \int\limits_{ - \pi /2}^{\pi /2} {\cos \theta e^{ikr\sin \theta } } d\theta = $$ $$ = \frac{1}{{\left( {2\pi } \right)^2 }}\int\limits_0^\infty {\frac{{k^2 }}{{k^2 + a^2 }}dk} \int\limits_{ - 1}^1 {e^{ikrx} dx} = \frac{2}{{\left( {2\pi } \right)^2 }}\int\limits_0^\infty {\frac{{k^2 }}{{k^2 + a^2 }}\frac{{i\sin \left( {kr} \right)}}{{ikr}}dk} $$ $$ = \frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{k\sin \left( {kr} \right)}}{{k^2 + a^2 }}} dk$$ It is the immaginary part of: $$ \frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{ke^{ikr} }}{{k^2 + a^2 }}} dk = \frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}2\pi i\frac{{iae^{ - ar} }}{{2ia}} = \frac{1}{{4\pi r}}e^{ - ar} i$$ so: $$\int {\frac{1}{{\left( {2\pi } \right)^3 }}\frac{{e^{i\vec k \cdot \vec r} }}{{k^2 + a^2 }}d^3 \vec k} = \frac{1}{{4\pi r}}e^{ - ar} $$ The unique singular point inside the circumference is $k=ia$. We now determine the case $a=0$. We have: $$\int {\frac{1}{{\left( {2\pi } \right)^3 }}\frac{{e^{i\vec k \cdot \vec r} }}{{k^2 }}d^3 \vec k} = \frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{\sin \left( {kr} \right)}}{k}} dk = $$ $$\frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{\sin t}}{t}dt = \frac{1}{{4\pi r}}} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.