# What does it mean for a potential not to have a Fourier transform?

Consider an isotropic potential $$\phi(r)$$ corresponding to the classical force of interaction between two point-particles. The Fourier transform of this potential $$\Phi(k) = \int \exp(-i\vec k\cdot\vec r) \phi(r) d^3r = \frac{4\pi}{k}\int_0^\infty \sin(kr)r\phi(r) dr$$ exists so long as

• $$\phi(r)$$ decays faster than $$1/r$$ as $$r\to\infty$$
• $$\phi(r)$$ diverges no faster than $$1/r^3$$ as $$r\to0$$ (if at all)

There is of course the borderline case of the Coulomb potential, for which $$\Phi(k)\propto 1/k^2$$. I take these properties from considering inverse-power potentials $$\phi(r) = \alpha r^{-n}$$ (with $$n>0$$), for which \begin{align} \Phi(k) &= \frac{4\pi\alpha}{k}\int_0^\infty\frac{\sin(kr)}{r^{n-1}}dr \\ &= \frac{4\pi\alpha}{k^{3-n}}\int_0^\infty\sin(x)x^{1-n}dx \\ &= \frac{4\pi\alpha}{k^{3-n}}\left[\Gamma(2-n)\sin\left(\frac{n\pi}{2}\right)\right] \end{align} which is finite as long as $$1.

But there are lots of models that use potentials that violate this condition, e.g., Maxwell molecules or the Lennard-Jones fluid. I'm curious if there's some physical insight to be had from this. If a potential does not have a Fourier transform, then that seems to imply that the interaction between two particles has no equivalent description in terms of plane-wave scattering. But that sounds contradictory to me, because it's perfectly routine to study simple liquids as a superposition of density fluctuations in $$k$$-space.

• Lennard-Jones and the like all decay as $r^6$ or with higher powers, but only at $r\to\infty$. Close to the origin, all these potentials are strongly modified and are, typically, finite there (they are $\mathcal O(1)$ instead of $\mathcal O(r^{-6})$). – AccidentalFourierTransform Oct 11 '18 at 22:49
• I have in mind the classic 6-12 potential. However, the exp-6 potential you mention still is not regular at the origin when correctly parameterized. – Endulum Oct 12 '18 at 2:11
• Classically, you could "cut off" the center of the potential and as a result end up with correct answers for all scatterings below a certain energy, as set by the turning point potential at the place where you cut off the pole. I suspect that you could do something similar in wave mechanics to obtain an approximate potential that had a Fourier transform, and was correct for scatterings up to some energy. Let me know if you try that. ;) – Display Name Oct 12 '18 at 2:37
• Oh, that's a cute idea, but I'm not sure it would pan out. Introducing a cutoff (or any really any kind of short-ranged attenuation) $\phi(r)$ would affect the transform over all $k$-space, not just the high-$k$ behavior. – Endulum Oct 15 '18 at 22:01