Here is some material on this question. For related classical issues, see

In quantum mechanics, a particle moving in an inverse cube force law has a Hamiltonian like this:

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

The first term describes the kinetic energy, while the second describes the potential energy. I'm setting $ \hbar = 1$ and $ 2m = 1$ to remove some clutter that doesn't really affect the key issues.

To see how strange this Hamiltonian is, let me compare an easier case. If $ p \lt 2,$ the Hamiltonian

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

is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}),$ which is the space of compactly supported smooth functions on 3d Euclidean space minus the origin. What this means is that first of all, $ H$ is defined on this domain: it maps functions in this domain to functions in $ L^2(\mathbb{R}^3)$. But more importantly, it means we can uniquely extend $ H$ from this domain to a self-adjoint operator on some larger domain. In quantum physics, we want our Hamiltonians to be self-adjoint. So, this fact is good.

Proving this fact is fairly hard! It uses something called the Kato–Lax–Milgram–Nelson theorem together with this beautiful inequality:

$$ \displaystyle{ \int_{\mathbb{R}^3} \frac{1}{4r^2} |\psi(x)|^2 \,d^3 x \le \int_{\mathbb{R}^3} |\nabla \psi(x)|^2 \,d^3 x } $$

for any $ \psi\in C_0^\infty(\mathbb{R}^3).$

If you think hard, you can see this inequality is actually a fact about the quantum mechanics of the inverse cube law! It says that if $ c \ge -1/4,$ the energy of a quantum particle in the potential $ c r^{-2}$ is bounded below. And in a sense, this inequality is optimal: if $ c \lt -1/4$, the energy is *not* bounded below. This is the quantum version of how a classical particle can spiral in to its doom in an attractive inverse cube law, if it doesn't have enough angular momentum. But it's subtly and mysteriously different.

You may wonder how this inequality is used to prove good things about potentials that are 'less singular' than the $ c r^{-2}$ potential: that is, potentials $ c r^{-p}$ with $ p \lt 2.$ For that, you have to use some tricks that I don't want to explain here. I also don't want to prove this inequality, or explain why its optimal! You can find most of this in some old course notes of mine:

See especially section 15.

But it's pretty easy to see how this inequality implies things about the expected energy of a quantum particle in the potential $ c r^{-2}$. So let's do that.

In this potential, the expected energy of a state $ \psi$ is:

$$ \displaystyle{ \langle \psi, H \psi \rangle =
\int_{\mathbb{R}^3} \overline\psi(x)\, (-\nabla^2 + c r^{-2})\psi(x) \, d^3 x }$$

Doing an integration by parts, this gives:

$$ \displaystyle{ \langle \psi, H \psi \rangle = \int_{\mathbb{R}^3} |\nabla \psi(x)|^2 + cr^{-2} |\psi(x)|^2 \,d^3 x } $$

The inequality I showed you says precisely that when $ c = -1/4,$ this is greater than or equal to zero. So, the expected energy is actually *nonnegative* in this case! And making $ c$ greater than $ -1/4$ only makes the expected energy bigger.

Note that in classical mechanics, the energy of a particle in this potential ceases to be bounded below as soon as $ c \lt 0.$ Quantum mechanics is different because of the uncertainty principle! To get a lot of negative potential energy, the particle's wavefunction must be squished near the origin, but that gives it kinetic energy.

It turns out that the Hamiltonian for a quantum particle in an inverse cube force law has exquisitely subtle and tricky behavior. Many people have written about it, running into 'paradoxes' when they weren't careful enough. Only rather recently have things been straightened out.

For starters, the Hamiltonian for this kind of particle

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

has different behaviors depending on $ c.$ Obviously the force is attractive when $c \gt 0$ and repulsive when $ c \lt 0,$ but that's not the only thing that matters! Here's a summary:

$ c \ge 3/4.$ In this case $ H$ is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}).$ So, it admits a unique self-adjoint extension and there's no ambiguity about this case.

$ c \lt 3/4.$ In this case $ H$ is *not* essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}).$ In fact, it admits more than one self-adjoint extension! This means that we need *extra input from physics* to choose the Hamiltonian in this case. It turns out that we need to say what happens when the particle hits the singularity at $ r = 0.$

$ c \ge -1/4.$ In this case the expected energy $ \langle \psi, H \psi \rangle $ is bounded below for $ \psi \in C_0^\infty(\mathbb{R}^3 - \{0\}).$ It turns out that whenever we have a Hamiltonian that is bounded below, even if there is not a *unique* self-adjoint extension, there exists a canonical 'best choice' of self-adjoint extension, called the Friedrichs extension. I explain this in my course notes.

$ c \lt -1/4.$ In this case the expected energy is not bounded below, so we don't have the Friedrichs extension to help us choose which self-adjoint extension is 'best'.

To go all the way down this rabbit hole, I recommend these two papers:

The first is good for a broad overview of problems associated to singular potentials such as the inverse cube force law; there is attention to mathematical rigor the focus is on physical insight. The second is good if you want—as I wanted—to really get to the bottom of the inverse cube force law in quantum mechanics. Both have lots of references.

Also, both point out a crucial fact I haven't mentioned yet: in quantum mechanics the inverse cube force law is special because, *naively, at least* it has a kind of symmetry under rescaling! You can see this from the formula

$$ H = -\nabla^2 + cr^{-2} $$

by noting that both the Laplacian and $ r^{-2}$ have units of length^{-2}. So, they both transform in the same way under rescaling: if you take any smooth function $ \psi$, apply $ H$ and then expand the result by a factor of $ k,$, you get $ k^2$ times what you get if you do those operations in the other order.

In particular, this means that if you have a smooth eigenfunction of $ H$ with eigenvalue $ \lambda,$ you will also have one with eigenfunction $ k^2 \lambda$ for any $ k \gt 0.$ And if your original eigenfunction was normalizable, so will be the new one!

With some calculation you can show that when $ c \le -1/4,$ the Hamiltonian $ H$ has a smooth normalizable eigenfunction with a negative eigenvalue. In fact it's spherically symmetric, so finding it is not so terribly hard. But this instantly implies that $ H$ has smooth normalizable eigenfunctions with *any* negative eigenvalue.

This implies various things, some terrifying. First of all, it means that $ H$ is not bounded below, at least not on the space of smooth normalizable functions. A similar but more delicate scaling argument shows that it's also not bounded below on $ C_0^\infty(\mathbb{R}^3 - \{0\}),$ as I claimed earlier.

This is scary but not terrifying: it simply means that when $ c \le -1/4,$ the potential is too strongly negative for the Hamiltonian to be bounded below.

The terrifying part is this: we're getting uncountably many normalizable eigenfunctions, all with different eigenvalues, one for each choice of $ k.$ A self-adjoint operator on a countable-dimensional Hilbert space like $ L^2(\mathbb{R}^3)$ can't have uncountably many normalizable eigenvectors with different eigenvalues, since then they'd all be orthogonal to each other, and that's too many orthogonal vectors to fit in a Hilbert space of countable dimension!

This sounds like a paradox, but it's not. These functions are not all orthogonal, and they're not all eigenfunctions of a self-adjoint operator. You see, the operator $ H$ is not self-adjoint on the domain we've chosen, the space of all smooth functions in $ L^2(\mathbb{R}^3).$ We can carefully choose a domain to get a self-adjoint operator... but it turns out there are many ways to do it.

Intriguingly, in most cases this choice breaks the naive dilation symmetry. So, we're getting what physicists call an 'anomaly': a symmetry of a classical system that fails to give a symmetry of the corresponding quantum system.

Of course, if you've made it this far, you probably want to understand what the different choices of Hamiltonian for a particle in an inverse cube force law *actually mean, physically.* The idea seems to be that they say how the particle changes phase when it hits the singularity at $ r = 0$ and bounces back out.

(Why does it bounce back out? Well, if it didn't, time evolution would not be unitary, so it would not be described by a self-adjoint Hamiltonian! We could try to describe the physics of a quantum particle that *does not* come back out when it hits the singularity, and I believe people have tried, but this requires a different set of mathematical tools.)

For a detailed analysis of this, it seems one should take Schrödinger's equation and do a separation of variables into the angular part and the radial part:

$$ \psi(r,\theta,\phi) = \Psi(r) \Phi(\theta,\phi) $$

For each choice of $ \ell = 0,1,2,\dots$ one gets a space of spherical harmonics that one can use for the angular part $ \Phi.$ The interesting part is the radial part, $ \Psi.$ Here it is helpful to make a change of variables

$$ u(r) = \Psi(r)/r $$

At least naively, Schrödinger's equation for the particle in the $ cr^{-2}$ potential then becomes

$$ \displaystyle{ \frac{d}{dt} u = -iH u }$$

where

$$ \displaystyle{ H = -\frac{d^2}{dr^2} + \frac{c + \ell(\ell+1)}{r^2} }$$

Beware: here I'm using $H$ for a new Hamiltonian, one that describes only the *radial part* of the dynamics of a quantum particle in an inverse cube force. As in classical mechanics, the centrifugal force and the inverse cube force join forces in an 'effective potential'

$$ \displaystyle{ U(r) = kr^{-2} } $$

where

$$ k = c + \ell(\ell+1)$$

So, we have reduced the problem to that of a particle on the open half-line $ (0,\infty)$ moving in the potential $ kr^{-2}.$ The Hamiltonian for this problem:

$$ \displaystyle{ H = -\frac{d^2}{dr^2} + \frac{k}{r^2} }$$

is called the **Calogero Hamiltonian**. Needless to say, it has fascinating and somewhat scary properties, since to make it into a bona fide self-adjoint operator, we must make some choice about what happens when the particle hits $ r = 0.$ The formula above does not really specify the Hamiltonian.

This is more or less where Gitman, Tyutin and Voronov *begin* their analysis, after a long and pleasant review of the problem. They describe all the possible choices of self-adjoint operator that are allowed. The answer depends on the values of $ k,$ but very crudely, the choice says something like how the phase of your particle changes when it bounces off the singularity. Most choices break the dilation invariance of the problem. But intriguingly, some choices retain invariance under a *discrete subgroup* of dilations!

Finally, some remarks on the same problem in 4 dimensions. It's a bit odd to study the inverse cube force law in *3-dimensonal* space, since Newtonian gravity and the electrostatic force would actually obey an inverse cube law in *4-dimensional* space. For the classical 2-body problem it doesn't matter much whether you're in 3d or 4d space, since the motion stays on the plane. But for quantum 2-body problem it makes more of a difference!

Just for the record, let me say how the quantum 2-body problem works in 4 dimensions. As before, we can work in the center of mass frame and consider this Hamiltonian:

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

And as before, the behavior of this Hamiltonian depends on $ c.$ Here's the story this time:

$ c \ge 0.$ In this case $ H$ is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^4 - \{0\}).$ So, it admits a unique self-adjoint extension and there's no ambiguity about this case.

$ c \lt 0.$ In this case $ H$ is not essentially self-adjoint on $ C_0^\infty(\mathbb{R}^4 - \{0\}).$

$ c \ge -1.$ In this case the expected energy $ \langle \psi, H \psi \rangle $ is bounded below for $ \psi \in C_0^\infty(\mathbb{R}^3 - \{0\}).$ So, there is exists a canonical 'best choice' of self-adjoint extension, called the Friedrichs extension.

$ c \lt -1.$ In this case the expected energy is not bounded below, so we don't have the Friedrichs extension to help us choose which self-adjoint extension is 'best'.

I've been assured these are correct by Barry Simon, and a lot of this material will appear in Section 7.4 of his book:

- Barry Simon,
*A Comprehensive Course in Analysis, Part 4: Operator Theory*, American Mathematical Society, Providence, RI, 2015.

See also:

- Barry Simon, Essential self-adjointness of Schrödinger operators with singular potentials,
*Arch. Rational Mech. Analysis* **52** (1973), 44-48.