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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?

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  • $\begingroup$ 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. $\endgroup$ – Ron Maimon Sep 14 '11 at 19:52
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I really appreciate the physical explanations made in other answers, but I want to add that Fourier transform of the Coulomb potential makes mathematical sense, too. This answer is meant to clarify on what sense the standard calculation is valid mathematically.

Firstly, and maybe more importantly, I want to emphasize that

The Fourier transform of f is not simply just $\int{f(x)e^{-ikx}dx}$.

For an $L^1$ function (a function which is norm integrable), this is always the case but Coulomb potential is definitely not in $L^1$. So the Fourier transform of it, if it ever exists, is not expected to be the integral above.

So here comes the second question: can Fourier transformation be defined on functons other than $L^1$?

The answer is "yes", and there are many Fourier transformations. Here are two examples.

  1. Fourier transformation on $L^2$ functions (i.e., square integrable functions).

It turns out that the Fourier transform behaves more nicely on $L^2$ than on $L^1$, thanks to the Plancherel's theorem. However, as we mentioned above, if an $L^2$ function is not in $L^1$, then the above integral may not exist and Fourier transform is not given by that integral, either. (However, it has a simple characterization theorem, saying that in this case the Fourier transform is given by the principle-value integration of the above integral.)

  1. Fourier transform of distributions (generalized functions)

It is in this sense that the Forier transform of Coulomb potential holds. The Coulomb potential, although not an $L^1$ or $L^2$ function, is a distribution. So we need to use the definition of the Fourier transform to distributions in this case. Indeed, one can check the definition and directly calculate the Fourier transform of it. However, the physicists' calculation illustrates another point.

Fourier transformation on distributions (however it is defined) is continuous (under a certain topology on the distribution space, but let's not be too specific about it).

Remeber that if f is continuous, then $x_\epsilon\rightarrow x$ implies that $f(x_\epsilon)\rightarrow f(x)$.

Now $\frac{1}{r}e^{-\mu r}\rightarrow\frac{1}{r}$ when $\mu\rightarrow 0$ (again, under the "certain topology" mentioned above), and therefore continuity implies

$$\operatorname{Fourier}\left\{\frac{1}{r}e^{-\mu r}\right\}\rightarrow \operatorname{Fourier}\left\{\frac{1}{r}\right\}.$$

However, $\frac{1}{r}e^{-\mu r}$ is in $L^1$ and therefore its Fourier transformation can be computed using the integral $\int{f(x)e^{-ikx}dx}$.

Therefore, those physicists' computations make perfect mathematical sense, but it's on Fourier transform of distributions, which is much more general than that on $L^1$ functions.

Wish this answer can build people's confidence that the Coulomb potential Fourier transformation problem is not only physically reasonable but also mathematically justifiable.

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  • $\begingroup$ When you introduce $\exp\left(-\mu r \right)$ - is that not just a Laplace transform? $\endgroup$ – NewDogOldTricks Apr 25 '17 at 5:33
  • $\begingroup$ Hi. What matters is that you create a sequence of distributions which converges to the distribution you are concerned with and in which each element belongs to $L^1$, so that its Fourier transform is expressible as an integral. The exponential factor $\exp{(-\mu r)}$ is just one example of such a sequence, which happens to be widely used by physicists (mainly because it's easy to deal with). $\endgroup$ – Zheng Liu May 1 '17 at 12:02
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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: $$\int {\frac{1}{{\left( {2\pi } \right)^3 }}\frac{{e^{i\vec k \cdot \vec r} }}{{k^2 }}d^3 \vec k} = \frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{\sin \left( {kr} \right)}}{k}} dk = $$ $$\frac{1}{{\left( {2\pi } \right)^2 }}\frac{1}{r}\int\limits_{ - \infty }^\infty {\frac{{\sin t}}{t}dt = \frac{1}{{4\pi r}}} $$

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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.

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    $\begingroup$ 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. $\endgroup$ – genneth Mar 24 '11 at 20:15
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    $\begingroup$ 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. $\endgroup$ – Vladimir Kalitvianski Mar 25 '11 at 16:43
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If the final result is unique and independent of the regularization method, then it is a mathematically correct answer.

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  • $\begingroup$ Is it possible to prove that the result is unique and independent of regularization in this case? $\endgroup$ – jinawee Mar 23 '16 at 14:55
  • $\begingroup$ Yes, it is. Nobody arrived at a different result so far ;-) $\endgroup$ – Vladimir Kalitvianski Mar 24 '16 at 13:02
  • $\begingroup$ That nobody arrived at a different result is no indication of uniqueness. For more than a year and a half I had the following (unrelated to Coulombic potential) question at Math.SE without an answer, until I finally figured out a way to break uniqueness assumption: Does such divergent integral assume the same values for any regularization? Although it's not decades, it's still an example against your comment. $\endgroup$ – Ruslan Mar 7 '17 at 9:52
  • $\begingroup$ @Ruslan: I meant exactly that - independence of regularization. Otherwise it would be discussed somewhere without fail, believe me. $\endgroup$ – Vladimir Kalitvianski Mar 7 '17 at 9:57
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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.

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This is not really a complete answer, but more of an interesting observation which might help to understand the regularized result better. First observe that \begin{equation} \int_V d^3 r \frac{e^{i\mathbf q \cdot \mathbf r}}{r} = 2\pi \int_0^R dr \int_{-1}^1 du \, r e^{iqru} = \frac{4\pi}{q} \int_0^R dr \, \sin qr = \frac{4 \pi}{q^2} \left( 1 - \cos qR \right). \end{equation} In the limit $R \rightarrow \infty$ the integral diverges since it fluctuates between $0$ and $8\pi/q^2$ indefinitely. The regularized Fourier transform is equal to the average value of these extremes. This is similar to the divergent series $1-1+1-1+\cdots$ which can be regularized to $1/2$.

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  • $\begingroup$ Is taking the average a legit trick? Because wouldn't by analogy $1-1+1-1 \dots=0$? $\endgroup$ – jinawee Mar 23 '16 at 22:56
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    $\begingroup$ @jinawee It's not a 'trick'. It's a way to give meaning to a series that diverges by redefining the concept of a diverging series. See here, for example. $\endgroup$ – Praan Mar 24 '16 at 21:00

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