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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 $L_n$?

Qmechanic
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Alex
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1 Answers1

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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 \begin{equation}\mathbf{R}=\frac{1}{2m}\left(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p}\right)-e^2\frac{\mathbf{r}}{r},\end{equation}

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

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

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

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

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

\begin{equation}\left[L_i,R_j\right]=\epsilon_{ijk}R_k.\end{equation}

From equations above, we can construct two abstract angular momenta

\begin{equation}\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..\end{equation}

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

\begin{equation}\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}\end{equation}

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

Thus, we can see that \begin{equation}\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}\end{equation}

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 $\oplus^{2j}_{i=0}H_i=\oplus^{\nu-1}_{i=0}H_i$.

Ladmon Draxngfüskiii
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