A non-practical way of solving the eigenvalue problem
To see this, we look at the contributions arising from
$$
\langle \Phi_H^P | = \langle \Phi_0|
$$
in Eq. (1), that is we multiply with \( \langle \Phi_0 | \)
from the left in
$$
(\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0.
$$