Quantum Mechanics of the Inverse Cube Force Law

Azimuth 2026-05-01

In the last episode of my column in Notices of the American Mathematical Society, we looked at a particle moving in an attractive central force whose strength is proportional to the inverse cube of the distance from the origin. Among other things, we saw that a particle moving in such a force can spiral in to the origin in a finite time. But that was classical mechanics. What about quantum mechanics?

Here things get more tricky. The uncertainty principle tends to prevent the particle from falling in to the origin. But when the attractive force is strong enough, the particle can still fall in. We can make up a theory where the particle shoots back out, but there are choices involved: we need to say how the particle changes phase when it shoots back out. So there is not just a single theory, but many!

Why does the particle come back out? We can make up theories where it does not. In these theories time evolution is usually nonunitary: that is, the probability of finding the particle somewhere or other does not stay equal to $1,$ because the particle simply disappears when it hits the origin. Here we focus on theories where time evolution is unitary and the particle comes back out. Many people have written about these, running into ‘paradoxes’ when they weren’t careful enough. Only rather recently have things been straightened out.

Let us dig into the details. In quantum mechanics, the Hilbert space of states of a particle in $\mathbb{R}^3$ is $L^2(\mathbb{R}^3).$ In a central force whose strength is proportional to $1/r^3,$ such a particle has a Hamiltonian of this form:

$H = -\nabla^2 + c r^{-2}$

The first term describes the particle’s kinetic energy, while the second describes its potential energy: remember, taking the gradient of an inverse square potential gives an inverse cube force. I have set some constants to $1$ to remove irrelevant clutter, but we need the constant $c$ to say how strong the force is. When $c \, < \, 0,$ the force is attractive.

In this game, analysis is paramount. We should interpret $H$ as a densely defined linear operator on $L^2(\mathbb{R}^3).$ For this, we choose a dense linear subspace $D \subset L^2(\mathbb{R}^3)$ and treat $H$ as a linear map from $D$ to $L^2(\mathbb{R}^3).$ Different choices of $D$ correspond to different physical assumptions: for example, assumptions about what happens when the particle falls into the origin.

To get unitary time evolution in quantum mechanics, we need the Hamiltonian to be self-adjoint. But adjoints of densely defined operators are tricky. Let us briefly recall how they work. Given a Hilbert space $\mathcal{H}$ and a linear operator $A$ from a dense linear subspace $D(A) \subseteq \mathcal{H}$ to $\mathcal{H},$ we define $D(A^*)$ to be the set of all $\psi \in \mathcal{H}$ for which there exist $\psi' \in \mathcal{H}$ such that

$\langle \psi' , \phi \rangle = \langle \psi, A \phi \rangle \; \text{ for all } \; \phi \in D(A)$

If such a vector $\psi'$ exists, it is unique, and it depends linearly on $\psi.$ Thus, for $\psi \in D(A^\ast)$ we define $A^\ast \psi$ to be the vector $\psi'$ with the above property. The adjoint of $A$ is then the linear operator $A^\ast \colon D(A^\ast) \to \mathcal{H}.$ We say $A$ is self-adjoint if $A = A^\ast.$ We say that $A$ is essentially self-adjoint if it has a unique extension to a self-adjoint operator. If it does, this extension must be $A^\ast.$

All this raises the question of whether the Hamiltonian $H$ for the inverse cube force law can be made self-adjoint with a suitable choice of domain. It turns out we can always do it, but sometimes in more than one way. There are three regimes:

• $c \ge 3/4.$ In this case we can start with the domain $C_0^\infty(\mathbb{R}^3 - \{0\})$ consisting of smooth functions that are compactly supported on $\mathbb{R}^3$ minus the origin. The operator $H$ is unambiguously defined on this domain, and it is essentially self-adjoint.

• $-1/4 \le c \, < \, 3/4.$ In this case $H$ is still well-defined on the domain $C_0^\infty(\mathbb{R}^3 - \{0\}),$ but it is not essentially self-adjoint. In fact, it admits more than one self-adjoint extension! However, $H$ is bounded below: there is a constant $E_0$ such that

$\langle \psi, H \psi \rangle \ge E_0 \langle \psi, \psi \rangle$

for all $\psi \in C_0^\infty(\mathbb{R}^3 - \{0\}).$ Physically, this means that the particle’s energy is bounded below by $E_0.$ Mathematically, this implies that $H$ has a canonical choice of self-adjoint extension called the ‘Friedrichs extension’, with the smallest possible domain. But there is another canonical choice, the ‘Krein extension’, with the largest possible domain.

• $c \, < \, -1/4.$ In this case $H$ is well-defined on the domain $C_0^\infty(\mathbb{R}^3 - \{0\}),$ and it has more than one self-adjoint extension, but it is not bounded below.

These strange results demand explanation. For example, what is special about $c =-1/4$ ? In classical mechanics, the energy of a particle in the inverse cube force ceases to be bounded below as soon as $c \, < \,0.$ Quantum mechanics is different. To get a lot of negative potential energy, the particle’s wavefunction must be peaked near the origin, but that gives it kinetic energy. The tradeoff is captured by Hardy’s inequality. This says that for any $\psi\in C_0^\infty(\mathbb{R}^3)$ we have

$\langle \psi, (-\nabla^2 - \tfrac{1}{4} r^{-2}) \psi \rangle \ge 0$

This is why $H$ is bounded below when $c \ge -1/4.$

On the other hand, the constant $1/4$ in Hardy’s inequality cannot be improved, so if $c \, < \, 1/4$ we can find $\psi$ with $\langle \psi, H \psi \rangle \, < \, 0.$ Then we can use a remarkable property of the $r^{-2}$ potential to show that $H$ is not bounded below. Namely, $H$ has a kind of symmetry under dilations. You can guess this by noting that both the Laplacian and $r^{-2}$ have units of 1/length². Indeed, if you take any smooth function $\psi,$ dilate it by a factor of $\alpha,$ and then apply $H,$ you get $\alpha^{-2}$ times what you get if you do these operations in the other order. This implies that if

$\langle \psi, H \psi \rangle = E \langle \psi, \psi \rangle$

we can dilate $\psi$ and get a function obeying the same equation with $E$ replaced by $\alpha^{-2} E.$ Thus, as soon as $E$ can be negative, it can be made arbitrarily large and negative by choosing $\alpha$ to be very small. Thus $H$ is not bounded below.

Next, what is special about $c = 3/4$ ? This is more subtle. For any value of $c \in \mathbb{R}$ we can find spherically symmetric solutions of

$( -\nabla^2 + c r^{-2})\psi = i \psi$

on $\mathbb{R}^3 - \{0\}$ that are nonzero and smooth. When $c \, < \, 3/4,$ and only in this case, some of these solutions $\psi$ lie in $L^2(\mathbb{R}^3).$ This dooms the chance of $H$ being essentially self-adjoint, because it implies $H^\ast \psi = i \psi.$ If $H$ were essentially self-adjoint $H^\ast$ would be self-adjoint, and it is easy to see that a self-adjoint operator cannot have $i$ as an eigenvalue.

When $c \, < \, \frac{3}{4}$ the operator $H$ has more than one self-adjoint extension from $C_0^\infty(\mathbb{R}^3 - \{0\})$ to some larger domain. To classify these we can use separation of variables, writing $\nabla^2$ as a sum of a radial part and an angular part, assuming the angular dependence of $\psi$ is given by a spherical harmonic $Y_{\ell m},$ and doing a change of variables $u = \psi/r$ to reduce $H$ to the ordinary differential operator

$\displaystyle{ - \frac{d^2}{d r^2} + \left(c + \ell(\ell+1)\right) \frac{1}{r^2} }$

on the half-line $(0,\infty).$ We can completely classify self-adjoint extensions of this differential operators from $C_0^\infty(0,\infty)$ to larger domains; the answer depends on $c$ and $\ell.$ A choice of self-adjoint extension is a choice of boundary conditions at $r = 0,$ and this says how the phase of an incoming wave changes as it reflects off the origin and bounces back. Finally, we can assemble the results for different spherical harmonics to classify self-adjoint extensions of $H.$

There exist many self-adjoint extensions of $H$ that respect the rotational symmetry of the inverse cube force law, but for $c \, < \, -1/4$ the extension must break the dilation symmetry discussed above. This is what physicists call an ‘anomaly’: a symmetry of a classical system that fails to be a symmetry of the corresponding quantum system. But intriguingly, for $-5/4 \le c \, < \, -1/4$ one can choose a Hamiltonian that is symmetrical under a discrete subgroup of dilations! For $c < -5/4$ one cannot even do this.

To explore this topic thoroughly, I recommend first this:

• S. Gopalakrishnan, Self-Adjointness and the Renormalization of Singular Potentials, B.A. Thesis, Amherst College, 2006.

then this:

• D. M. Gitman, I. V. Tyutin and B. L. Voronov, Self-adjoint extensions and spectral analysis in Calogero problem.

and finally this:

• J. Dereziński and S. Richard, On Schrödinger operators with inverse square potentials on the half-line, Ann. Henri Poincaré 18 (2017), 869–928.

The first is an excellent overview of problems associated to singular potentials, including the inverse cube force. The second delves into self-adjoint extensions of the ordinary differential operators mentioned above, and the third works them out with exquisite thoroughness.