The Kepler Problem (Part 7)

Azimuth 2025-07-29

I’ve explained a cool way to treat bound states of the hydrogen atom as wavefunctions on a sphere in 4-dimensional space. But so far I’ve been neglecting the electron’s spin. Now let’s throw that in too!

This will wind up leading us in some surprising directions. So far I’ve just been reviewing known ideas, but now we’re getting into my new paper:

• Second quantization for the Kepler problem.

It starts out being quite routine: to include spin, we just tensor our previous Hilbert space $L^2(S^3)$ with a copy of $\mathbb{C}^2$ describing the electron’s spin. The resulting space

$L^2(S^3) \otimes \mathbb{C}^2$

is the Hilbert space of bound states of a spinor-valued version of the Schrödinger equation for the hydrogen atom.

Beware: this is a simplification of a more careful treatment of hydrogen using the Dirac equation: it neglects all spin-dependent terms in Hamiltonian, like spin-orbit interactions. These spin-dependent terms give corrections that go to zero in the limit where the speed of light approaches infinity. So what we’re doing now is giving a nonrelativistic treatment of the hydrogen atom, but taking into account the fact that the electron is a spin-½ particle.

Things get fun now. The Hilbert space $L^2(S^3) \otimes \mathbb{C}^2$ becomes a unitary representation of $\text{SU}(2)$ in three important ways. The first two come from the actions of $\text{SU}(2)$ on $L^2(S^3)$ by left and right translation, which I explained in Part 5. The third comes from the natural action of $\text{SU}(2)$ on $\mathbb{C}^2.$ All three of these actions of $\text{SU}(2)$ on $L^2(S^3) \otimes \mathbb{C}^2$ commute with each other. We thus get a unitary representation of $\text{SU}(2) \times \text{SU}(2) \times \text{SU}(2)$ on $L^2(S^3) \otimes \mathbb{C}^2.$

It is useful to spell this out at the Lie algebra level. In Part 5, I introduced self-adjoint operators $A_j$ and $B_j$ on $L^2(S^3)$ : the self-adjoint generators of the left and right translation actions of $\text{SU}(2),$ respectively. Now we’ll tensor these operators with the identity on $\mathbb{C}^2$ and get operators on $L^2(S^3) \otimes \mathbb{C}^2,$ which by abuse of notation we’ll denote with the same names: $A_j$ and $B_j.$ But we’ll also introduce spin angular momentum operators

$S_j = 1 \otimes \frac{1}{2} \sigma_j$

on $L^2(S^3) \otimes \mathbb{C}^2.$ These operators obey the following commutation relations:

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

Once we have 3 commuting actions of $\text{SU}(2)$ on a Hilbert space we can get more by mixing and matching them. I won’t go overboard and describe all 2³ = 8 of them, but I’ll mention some that we need for physics. First we can define orbital angular momentum operators

$L_j = A_j + B_j$

These obey

$\begin{array}{ccl} [L_j, L_k] &=& i\epsilon_{jk\ell} L_\ell \\ [2pt] [S_j, S_k] &=& i \epsilon_{jk\ell} S_\ell \\ [2pt] [L_j, S_k] &=& 0 \end{array}$

Physically speaking, the $L_j$ generate an action of $\text{SU}(2)$ that rotates the position of the electron in space while not changing its spin state, just as the $S_j$ rotate the electron’s spin state while not changing its position.

Adding the spin and orbital angular momentum, we get total angular momentum operators

$J_j = L_j + S_j$

which obey

$[J_j, J_k] = i \epsilon_{jk\ell} J_\ell$

These generate an action of $\text{SU}(2)$ that rotates the electron’s wavefunction along with its spin state!

Finally, we define a Hamiltonian for our new hydrogen atom with spin:

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

This is just the Hamiltonian $H_0$ for the simplified hydrogen atom neglecting spin that we studied in Part 5, tensored with the identity operator on $\mathbb{C}^2.$ Thus it has the same spectrum, but the multiplicity of each eigenvalue has doubled. This Hamiltonian $H$ commutes with all the operators $A_j, B_j, S_j,$ and thus also $L_j$ and $J_j.$

Now we can reuse our work from Part 5 and decompose our new Hilbert space into eigenspaces of the Hamiltonian $H,$ labeled by $n = 1, 2, 3, \dots$ , and the orbital angular momentum operator $J^2,$ labeled by $\ell = 0 , \dots, n-1.$ We get this:

$\displaystyle{ L^2(S^3) \otimes \mathbb{C}^2 \cong \bigoplus_{n = 1}^\infty \bigoplus_{\ell = 0}^{n-1} V_\ell \otimes \mathbb{C}^2 }$

where $V_\ell$ is the spin- $\ell$ representation of the $\text{SU}(2)$ that rotates the electron’s position but not its spin.

In Part 5 we saw a basis $|n, \ell, m \rangle$ of $L^2(S^3).$ If we tensor that with the standard basis of $\mathbb{C}^2,$ we get an orthonormal basis $|n , \ell, m, s \rangle$ of $L^2(S^3) \otimes \mathbb{C}^2$ where:

• the principal quantum number $n$ ranges over positive integers;

• the azimuthal quantum number $\ell$ ranges from $0$ to $n-1$ in integer steps;

• the magnetic quantum number $m$ ranges from $-\ell$ to $\ell$ in integer steps;

• the spin quantum number $s$ is $+\frac{1}{2}$ or $-\frac{1}{2}.$

The calculations we did in Part 5 now imply that

$\begin{array}{ccl} A^2 |n, \ell, m, s \rangle &=& B^2 |n, \ell, m, s \rangle \; = \; \frac{1}{4}( n^2 - 1) |n, \ell, m, s\rangle \\ [8pt] H |n, \ell, m, s \rangle &=& \displaystyle{ - \frac{1}{2n^2}\, |n, \ell, m, s \rangle } \\ [12pt] L^2 |n, \ell, m, s\rangle &=& \ell(\ell + 1) |n , \ell, m, s \rangle \\ [3pt] L_3 |n , \ell, m, s \rangle &=& m |n , \ell, m, s \rangle \\ [3pt] S^2 |n , \ell, m, s \rangle &=& \frac{3}{4} |n , \ell, m, s \rangle \\ [3pt] S_3 |n , \ell, m, s \rangle &=& s |n , \ell, m, s \rangle \end{array}$

Combining this with the textbook treatment of the hydrogen atom, it follows that $L^2(S^3) \otimes \mathbb{C}^2$ is indeed unitarily equivalent to the subspace of $L^2(\mathbb{R}^3) \otimes \mathbb{C}^2$ consisting of bound states of the spinor-valued Schrödinger equation

$i \frac{\partial \psi}{\partial t} = -\frac{1}{2} \nabla^2 \psi - \frac{1}{r} \psi$

with the operators $H, L_j$ and $S_j$ having their usual definitions:

$\begin{array}{ccl} H &=& -\frac{1}{2} \nabla^2 - \frac{1}{r} \\ [12pt] L_j &=& -i\epsilon_{jk\ell} x_k \frac{\partial}{\partial x_\ell} \\ [10pt] S_j &=& \frac{1}{2} \sigma_j \end{array}$

In short, the Hamiltonian $H$ on $L^2(S^3) \otimes \mathbb{C}^2$ is unitarily equivalent to the Hamiltonian on bound states of the hydrogen atom defined in the usual way! We’ve turned hydrogen into a festival of commuting $\text{SU}(2)$ actions.

Next we’ll do something a bit wild, and new.