The Kepler Problem (Part 6)

Azimuth 2025-07-26

In Part 4 we saw that the classical Kepler problem—the problem of a single classical particle in an inverse square force—has symmetry under the group of rotations of 4-dimensional space $\text{SO}(4).$ Since the Lie algebra of this group is

$\mathfrak{so}(4) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$

we must have conserved quantities

$\vec{A} = (A_1, A_2, A_3)$

and

$\vec{B} = (B_1, B_2, B_3)$

corresponding to these two copies of $\mathfrak{su}(2).$ The physical meaning of these quantities is a bit obscure until we form linear combinations

$\begin{array}{ccl} \vec{L} &=& \vec{A} + \vec{B} \\ \vec{M} &=& \vec{A} - \vec{B} \end{array}$

Then $\vec{L}$ is the angular momentum of the particle, while $\vec{M}$ is a subtler conserved quantity: it’s the eccentricity vector of the particle divided by $\sqrt{-2 E},$ where the energy $E$ is negative for bound states (that is, elliptical orbits)

The advantage of working with $\vec{A}$ and $\vec{B}$ is that these quantities have very nice Poisson brackets:

$\begin{array}{ccl} \{ A_j, A_k\} &=& \epsilon_{jk\ell} A_\ell \\ \{ B_j, B_k\} &=& \epsilon_{jk\ell} B_\ell \\ \{ A_j, B_k\} &=& 0 \end{array}$

This says they generate two commuting $\mathfrak{su}(2)$ symmetries. Moreover since

$\vec{L} \cdot \vec{M} = 0$

the vectors $\vec{A}$ and $\vec{B}$ have equal lengths, since

$A^2 = B^2$

Even better, the energy of the particle, in suitable units, turns out to be

$\displaystyle{ E \;=\; - \frac{1}{8A^2} \;=\;- \frac{1}{8B^2} }$

(All this is only good for bound states, where the energy is negative.)

In Part 5 we turned to the quantum Kepler problem—switching from the solar system to the hydrogen atom, as it were. Things become very nice. The bound states for the hydrogen atom, neglecting the electron’s spin, can be thought of as functions

$\phi \colon S^3 \to \mathbb{C}$

Indeed the Hilbert space of these bound states is $L^2(S^3).$ The conserved quantities $A_1,A_2,A_3$ and $B_1, B_2, B_3$ become self-adjoint operators on this Hilbert space, obeying commutation relations that perfectly mimic the Poisson brackets we had in the classical case:

$\begin{array}{ccl} [A_j, A_k] &=& i\epsilon_{jk\ell} A_\ell \\ [2pt] [B_j, B_k] &=& i\epsilon_{jk\ell} B_\ell \\ [2pt] [A_j, B_k] &=& 0 \end{array}$

Indeed there’s a beautiful geometrical picture: we can think of $S^3$ as being the group $\text{SU}(2).$ Like any group, this group acts on itself by left and right multiplication—speaking geometrically, we say left and right ‘translation’. So we get an action of $\text{SU}(2) \times \text{SU}(2)$ on $S^3.$ This gives a representation of $\text{SU}(2) \times \text{SU}(2)$ on the Hilbert space $L^2(S^3).$ And the self-adjoint operators $A_i$ and $B_i$ generate these two $\text{SU}(2)$ representations!

All this is as close to ideal perfection as you could ever want. But then comes a twist. I sketched out why the quantum version of the Hamiltonian is

$\displaystyle{ H \;=\; - \frac{1}{8(A^2 + \frac{1}{4})} \;=\;- \frac{1}{8(B^2 + \frac{1}{4})} }$

Why does the quantum version have a $\frac{1}{4}$ in it that’s not present in the classical case? From a pragmatic viewpoint, we need this to match the spectrum obtained in the usual approach to quantizing the Kepler problem, which agrees with experiment. We also need some term like this to avoid dividing by zero, since $A^2 = B^2$ has zero as an eigenvalue, coming from the constant functions on the sphere. But there is yet another explanation: it arises from the Duflo isomorphism.

The Poincaré–Birkoff–Witt theorem gives for any Lie algebra $\mathfrak{g}$ a natural linear map from the polynomial algebra $S(\mathfrak{g})$ to the universal enveloping algebra $U(\mathfrak{g}).$ This map is an isomorphism of vector spaces, but clearly not of algebras: $S(\mathfrak{g})$ is commutative while $U(\mathfrak{g})$ is not. This map is compatible with the natural representation of $\mathfrak{g}$ on these spaces, so it restricts to a vector space isomorphism

$F \colon S(\mathfrak{g})^\mathfrak{g} \to U(\mathfrak{g})^\mathfrak{g}$

where the superscript indicates the subspace annihilated by the action of $\mathfrak{g}.$

What does this mean to the everyday physicist?

You should think of $S(\mathfrak{g})^\mathfrak{g}$ as the algebra of all classical observables coming from $\mathfrak{g}$ that have vanishing Poisson brackets with all other observables of this form. In other words, they’re the observables that are invariant under the Lie group $G,$ which acts on $S(\mathfrak{g}).$

Similarly, $U(\mathfrak{g})^\mathfrak{g}$ is the algebra of all quantum observables coming from $\mathfrak{g}$ that are invariant under $G.$

In either the classical or quantum case, physicists call these invariant observables Casimir elements.

Now here’s where things get interesting and tricky. $F$ is not an algebra homomorphism! But in 1977, Duflo proved that in some cases we can compose $F$ with a map

$G \colon S(\mathfrak{g})^\mathfrak{g} \to S(\mathfrak{g})^\mathfrak{g}$

to get an algebra isomorphism

$F \circ G \colon S(\mathfrak{g})^\mathfrak{g} \to U(\mathfrak{g})^\mathfrak{g}$

Later Kontsevich showed that this works for all finite-dimensional Lie algebras. So, we have a perfect prescription for taking any classical Casimir element and finding its quantum analogue. But it’s not the one you’d first guess.

Calaque and Rossi have written a clear pedagogical account, and I recommend starting there if you want to learn more:

• Damien Calaque and Carlo A. Rossi, Lectures on Duflo Isomorphisms in Lie Algebra and Complex Geometry, EMS Series of Lectures in Mathematics, 2011.

Let’s see how it’s relevant to the Kepler problem. In the case $\mathfrak{g} = \mathfrak{su}(2),$ the algebra $S(\mathfrak{g})^\mathfrak{g}$ is generated by the one element

$J_1^2 + J_2^2 + J_3^2$

where $J_j$ are the usual basis for $\mathfrak{su}(2).$

Applying $G$ to this element we obtain

$\tilde{J}^2 = J_1^2 + J_2^2 + J_3^2 + \frac{1}{4}$

So yes, we get that $\frac{1}{4}$ showing up!

Examining the details of the map $G$ as explained in the above references, one can see that the value $\frac{1}{4}$ arises from the fact that

$\displaystyle{ \frac{\sinh (x/2)}{(x/2)} = 1 + \frac{x^2}{24} + \cdots }$

together with the fact that each of the 3 elements $J_j^2$ has trace $2$ in the adjoint representation, giving a correction of

$3 \times 2 \times \tfrac{1}{24} = \frac{1}{4}$

This is pretty exciting to me. Why? Because the function

$\displaystyle{ \frac{\sinh (x/2)}{(x/2)} = \frac{e^{x/2} - e^{-x/2}}{x} }$

is connected to a topological invariant called the Â genus, and the formula

$\displaystyle{ \frac{\sinh (x/2)}{(x/2)} = 1 + \frac{x^2}{24} + \cdots }$

is closely connected to the special role of the number 24 in conformal field theory, the theory of modular forms, topology and other subjects! Thus, the $\frac{1}{4}$ in the hydrogen atom Hamiltonian is part of a larger story. A few details can be found in this paper:

• Luigi Rosa and Patrizia Vitale, On the ⋆-product quantization and the Duflo map in three dimensions, Mod. Phys. Lett. A 27 (2012), 1250207.

But there’s a lot not discussed there, and a lot I don’t understand about this yet.

In general, quantization is full of little ‘fudge factors’ or ‘corrections’ that wind up being related to rather deep mathematics. Too deep to get into here! But for starters, we can think of the number 1/24 as coming from how the function

$e^{x/2} - e^{-x/2}$

is close to $x,$ but not quite equal, when $x$ is small:

$\displaystyle{ e^{x/2} - e^{-x/2} = x \left( 1 + \frac{x^2}{24} + \cdots \right) }$

But why should we care about this? One reason is that it says how the difference operator

$\displaystyle{ \frac{f(t + \epsilon/2) - f(t - \epsilon/2)}{\epsilon} = \left(\frac{e^{\epsilon D/2} - e^{-\epsilon D/2}}{\epsilon} f\right)(t) }$

reduces to the derivative operator

$\displaystyle{ D = \frac{d}{dt} }$

as $\epsilon \to 0.$ Namely:

$\displaystyle{ \frac{e^{\epsilon D/2} - e^{-\epsilon D/2}}{\epsilon} = D\left(1 + \frac{\epsilon^2 D^2}{24} + \cdots \right) }$

I dig into this in a bit more detail here:

• John Baez, Bernoulli numbers and the harmonic oscillator, The n-Category Café, 16 August 2024.

But the whole subject deserves a lot more exploration.