Consider the perturbation treatment of He $1s2s$ excited state configuration. The quantum mechanical Hamiltonian for the He-atom is:
\begin{aligned} \hat{H} = \sum_{i=1}^2 h_i + \frac 1 {r_{1 2}} \\ h_i = -\tfrac 12 \nabla_i^2 - \frac 1 r_i \end{aligned}
In the following perturbation treatment, the unperturbed Hamiltonian $\hat{H}^0$ is $\sum_{i=1}^2 h_i$, while the perturbation Hamiltonian $\hat{H}'$ includes the inter-electron repulsion $r_{1 2}^{-1}$.
The four unperturbed degenerate wave functions $\psi^{(0)}$ are shown below as Slater determinants:
$$ \begin{aligned} &\psi_1^{(0)} = \left|\begin{array}{} 1s(1) \overline{2s}(1) \\ 1s(2) \overline{2s}(2) \end{array}\right| \equiv \left| 1 \overline{2} \right\rangle & \psi_2^{(0)} = \left| \overline{1} 2 \right\rangle \\ &\psi_3^{(0)} = \left| 1 2 \right\rangle &\psi_4^{(0)} = \left| \overline{1} \overline{2} \right\rangle \end{aligned} $$
The correct zeroth-order wave functions $\phi^{(0)}$ for the perturbation Hamiltonian $\hat{H}'$ are linear combinations of unperturbed eigenfunctions such that $\phi^{(0)} = \sum_i c_i \psi_i^{(0)}$. Solving for $\phi^{(0)}$ using the linear variational method is equivalent to diagonalizing the perturbation Hamiltonian $\hat{H}'$ in a basis of $\psi^{(0)}$. The eigenvectors and eigenvalues of the $\mathbf{H}'$ matrix, whose elements are $\left\langle \psi_m^{(0)} \middle| \hat{H}' \middle| \psi_i^{(0)} \right\rangle$, are the correct zeroth-order wave functions and the corresponding first-order corrections to the energy, respectively.
Because $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$ (the Condon-Slater rules), $\psi_3^{(0)} = \left| 1 2 \right\rangle$ is already the eigenvector of the perturbation Hamiltonian $\hat{H}'$ with the eigenvalue $E_3^{(1)} = \left\langle 1 2 \mid \mid 1 2 \right\rangle = \left\langle 1 2 \mid 1 2 \right\rangle - \left\langle 1 2 \mid 2 1 \right\rangle$. Could you suggest whether it is correct to go on to state the eigen-equation of $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ ?