# Eigenspaces of the hydrogen atom as representations of $SO(3)$

When we computing the discrete spectrum of the hamiltonian of the hydrogen atom

$$H=\Big(-\frac{\hbar^2}{2m} \Delta - \frac{e^2}{r} \large),$$

by some explicit computation we get that eigenspace $$L_n=\{\psi \in \mathbb{H} | H \psi = E_n \psi\}$$ with principal quantum number $$n$$ is a direct sum of irreducible representations of $$SO(3)$$ $$L_n = D_0 \oplus D_1 \oplus \ldots D_{n-1}.$$ The index $$l$$ of irreducible representation $$D_l$$ is the azimuthal quantum number.

Suppose I'm not interested in formulas for eigenfunctions, I only want to understand decomposition of the eigenspace $$L_n$$ into irreducible representations of $$SO(3)$$, is there a short way of doing it without computing explicitly basis of $$L_n$$?

• Related: physics.stackexchange.com/q/116244/2451 , physics.stackexchange.com/q/89654/2451 and links therein. – Qmechanic Sep 8 at 19:46
• Indeed, Pauli found this spectrum in 1926 before Schroedinger used his eponymous equation; He utilizing the SO(4) ~ SO(3) × SO(3) symmetry of the problem, cf Valent 2003. Group theory trumps all. – Cosmas Zachos Sep 8 at 20:03
• WP. Pauli barely "beat" Schroedinger by submitting 10 days earlier, of no significance. However, it might be of significance that Pauli's straightforward group theory argument is much simpler than the complicated TISE that Schroedinger relied on Weyl's help to solve! – Cosmas Zachos Sep 8 at 20:26
• Motl’s blog if the details are not evident. – Cosmas Zachos Sep 8 at 22:20

First, hypothise $$E=-\frac{e^2}{2a_{0}}\frac{1}{\nu^2}$$ where $$\nu$$ is an unknown parameter. This is plausible since $$-\frac{e^2}{2a_{0}}$$ are simply constants that give the correct units. Then, we introduce the Runge-Lenz vector $$$$\mathbf{R}=\frac{1}{2m}\left(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p}\right)-e^2\frac{\mathbf{r}}{r},$$$$

to reveal the hidden symmetries of this system, where $$\mathbf{L}$$ is the angular momentum operator. And we can prove that

$$$$\label{eq:1}\left[R_i,R_j\right]=-i\hbar\frac{2H}{me^2}\varepsilon_{ijk}L_k.$$$$ By using the familiar identity $$$$\label{eq:6}\mathbf{L}\times\mathbf{u}+\mathbf{u}\times\mathbf{L}=2i\hbar\mathbf{u},$$$$ it also indicates that $$$$\label{eq:2} \mathbf{R}\times\mathbf{R}=i\hbar\left(-\frac{2H}{me^4}\mathbf{L}\right).$$$$

We can also explicitly calculate the value of $$\mathbf{R}^2$$, which reads

$$$$\label{eq:3}\mathbf{R}^2=1+\frac{2}{me^2}H\left(\mathbf{L}^2+\hbar^2\right).$$$$

Since $$\mathbf{R}$$ is only consist of the sum of vector under rotations, (the cross product of vector under rotations is also a vector under rotation, such as $$\mathbf{p}\times\mathbf{L}$$), $$\mathbf{R}$$ itself is a vector under rotation.

Thus

$$$$\left[L_i,R_j\right]=\epsilon_{ijk}R_k.$$$$

From equations above, we can construct two abstract angular momenta

\left\{\begin{aligned} \mathbf{J}_{-}=\frac{1}{2}\left(\mathbf{L}-\hbar\nu\mathbf{R}\right)\\ \mathbf{J}_{+}=\frac{1}{2}\left(\mathbf{L}+\hbar\nu\mathbf{R}\right) \end{aligned}\right..

We can also prove that $$\mathbf{J}_{+}$$ and $$\mathbf{J}_{-}$$ satisfy the properties of angular momenta

\label{eq:4}\begin{aligned} &\mathbf{J}_{\pm}\times\mathbf{J}_{\pm}=i \hbar \mathbf{J}_{\pm}\\ &\left[(\mathbf{J}_{\pm})_{i},(\mathbf{J}_{\pm})_{j}\right]=0\\ &\mathbf{J}_{+}^2=\mathbf{J}_{-}^2=\hbar^2 j(j+1), j\in\frac{\mathbb{Z}}{2} \end{aligned}

Since $$$$\label{eq:5} \mathbf{J}_{+}^2\mathbf{L}^2+\hbar^2\nu^2\mathbf{R}^2=\hbar^2(\nu^2+1)$$$$

Thus, we can see that \label{eq:7}\begin{aligned} &\nu=2j+1\in\frac{\mathbb{Z}}{2}\\ &E=-\frac{me^4}{2\hbar}\frac{1}{(2j+1)^2}. \end{aligned}

Since $$\mathbf{L}=\mathbf{J}_1+\mathbf{J}_2$$, the eigenspace for every j is $$H_j\otimes H_j$$ (the direct product of two j multiplets) which equals $$\oplus^{2j}_{i=0}H_i=\oplus^{\nu-1}_{i=0}H_i$$.