= Solution
The single-electron isotropic <quantum harmonic oscillator> has energies
$$
\epsilon_{n_xn_yn_z}=\hbar\omega_0\left(n_x+n_y+n_z+\frac32\right).
$$
Its lowest spatial orbital, denoted $s$, is unique, with energy $3\hbar\omega_0/2$. Including <spin one-half> gives two spin orbitals. Two identical <Electrons> can occupy both only with opposite spin in the unique <singlet state>. Thus the two-electron <ground state> has energy $3\hbar\omega_0$ and multiplicity one.
The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hbar\omega_0/2$, giving six spin orbitals. A first-excited two-electron <Slater determinant> occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. Hence
$$
\boxed{E_{\rm ground}=3\hbar\omega_0,\quad d_{\rm ground}=1;
\qquad E_{\rm first}=4\hbar\omega_0,\quad d_{\rm first}=2\times6=12.}
$$
No further factor of two is inserted for interchanging the electrons, since the <fermionic exchange symmetry> is already built into a <Slater determinant>.
The spatial <Hamiltonian operator> is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total <orbital angular momentum>. It is also spin independent, so $[H_0,S_k]=0$ for each total <spin> component. Therefore
$$
\boxed{[H_0,L^2]=[H_0,S^2]=0,}
$$
expressing <conservation of angular momentum> and conservation of the total spin magnitude for this model. The <Pauli exclusion principle> restricts which states occur but does not break either symmetry.
To change the first-shell basis, take normalized real Cartesian orbitals $\phi_x,\phi_y,\phi_z$. Using <spherical coordinates>, $z=r\cos\theta$ and $x\pm iy=r\sin\theta e^{\pm i\varphi}$. Comparison with the <spherical harmonics> gives the <Cartesian and spherical bases of the first oscillator shell>:
$$
\boxed{\phi_{+1}=-\frac{\phi_x+i\phi_y}{\sqrt2},\qquad
\phi_0=\phi_z,\qquad
\phi_{-1}=\frac{\phi_x-i\phi_y}{\sqrt2}.}
$$
Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta,\varphi)$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a <unitary> change of basis. The ground spatial orbital has $l=0$.
For two ground-shell electrons, <addition of angular momentum> gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinations
$$
\Phi_m^\pm(1,2)=\frac{s(1)\phi_m(2)\pm\phi_m(1)s(2)}{\sqrt2}.
$$
The <exchange symmetry of two identical spin-one-half fermions> permits $\Phi_m^+$ with the <singlet state> $\chi_{00}$ and $\Phi_m^-$ with each <triplet state> $\chi_{1M}$. In <term symbol> notation the manifolds are therefore
$$
\boxed{\begin{array}{c|c|c|c}
\text{energy}&(L,S)&\text{term}&\text{multiplicity}\\ \hline
3\hbar\omega_0&(0,0)&{}^1S&1\\
4\hbar\omega_0&(1,0)&{}^1P&3\\
4\hbar\omega_0&(1,1)&{}^3P&9
\end{array}}
$$
The count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the <two-electron states in the first two shells of an isotropic oscillator>.
For the specified <spin-orbit interaction>, $\mathbf S$ and $\mathbf L$ are the total two-electron operators. Set $\mathbf J=\mathbf L+\mathbf S$. The <spin-dot-product eigenvalue> identity gives
$$
\Delta E_{LSJ}=\frac{\lambda\hbar^2}{2}[J(J+1)-L(L+1)-S(S+1)].
$$
For the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The <spin-orbit splitting of the first excited two-electron oscillator level> is consequently
$$
\boxed{\begin{array}{c|c|c}
\text{level}&\Delta E&\text{multiplicity}\\ \hline
{}^3P_0&-2\lambda\hbar^2&1\\
{}^3P_1&-\lambda\hbar^2&3\\
{}^1P_1&0&3\\
{}^3P_2&+\lambda\hbar^2&5
\end{array}}
$$
Add each shift to $4\hbar\omega_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda|\hbar^2\ll\hbar\omega_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2019-q5-oscillator-multiplets.png]
{title=Spin-orbit splitting of the twelve-state first-excited two-electron oscillator manifold}
{description=Original plot of shifts from the common unperturbed energy $4\hbar\omega_0$. Labels give the <term symbol> and number of magnetic substates. Both signs of the coupling are shown; the singlet branch remains unshifted.}
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