A What is the solution for two electrons on a sphere?

[bob012345]

The excited states for the $unperturbed$ Hamiltonian are constructed from basis one-electron states;
$$\psi=N|\ell,m\rangle_i={{Y_\ell^m (\theta_i, \varphi_i )}\over R}~~~ i={1,2}$$

Where there are twelve possible states for the lowest possible transition from $\Psi(\ell_1,\ell_2)$ going from $\Psi(0,0) \rightarrow \Psi(0,1)$.

Written in $|ket\rangle$ notation;
$$
\left.\begin{cases}
{\frac{1}{\sqrt{2}}}(|0,0\rangle_1|1,1\rangle_2+|0,0\rangle_2|1,1\rangle_1)|singlet\rangle ~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,0\rangle_2+|0,0\rangle_2|1,0\rangle_1)|singlet\rangle~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,\bar1\rangle_2+|0,0\rangle_2|1,\bar1\rangle_1)|singlet\rangle~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,1\rangle_2-|0,0\rangle_2|1,1\rangle_1)|triplet\rangle~~ortho\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,0\rangle_2-|0,0\rangle_2|1,0\rangle_1)|triplet\rangle~~ortho\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,\bar1\rangle_2-|0,0\rangle_2|1,\bar1\rangle_1)|triplet\rangle~~ortho\\
\end{cases}\right\}
$$

Where $[S,m_s\rangle$ equals;
$$
\left.\begin{cases}
|1,1\rangle & =\;|\uparrow\uparrow\rangle\\
|1,0\rangle & =\;\frac{1}{\sqrt{2}}|\uparrow\downarrow + \downarrow\uparrow\rangle\\
|1,-1\rangle & =\;|\downarrow\downarrow\rangle
\end{cases}\right\}\quad S=1\quad\mathrm{(triplet)}
$$
And;
$$
\left.\begin{cases}
|0,0\rangle & =\;\frac{1}{\sqrt{2}}|\uparrow\downarrow - \downarrow\uparrow\rangle
\end{cases}\right\}\quad S=0\quad\mathrm{(singlet)}
$$
The energy of these twelve degenerate lowest excited states is given by;

$$ E_{total} = {\hbar^2 \over 2 m R^2} [ \ell_1 \left (\ell_1+1\right) + \ell_2 \left (\ell_2+1\right)]= {\hbar^2 \over m R^2}~~~ \ell_{1,2}=0,1$$

The next step is adding the Coulomb interaction for the ground and then the lowest excited states.