\\ &=& \langle 0 | a_{hm_h}^{\dagger} \delta (E-H_{\text{int}}) a_{hm_h} | 0 \rangle , \end{aligned}$$ where $ | f \rangle $, $E_f$ are the eigenstates and energies of the $(A-1)$-system, $H_{\text{int}} |f\rangle = E_f|f\rangle$. One finds, using again a linearization procedure, $$\varepsilon^{(-)}_{h} = -\sum_{\alpha, J} \rho_{\alpha} 
(1+\delta_{\alpha h}) 
\langle \alpha h ; J | H_{\text{int}} | \alpha h ; J \rangle \equiv - \varepsilon_h 
. 
\label{E:spvareh}$$ Similarly one finds the centroid of the energy corresponding to picking up a particle at the state $p$, $$\varepsilon^{(+)}_{p} =  \sum_{\alpha, J} \rho_{\alpha} 
(1+\delta_{\alpha p}) 
\langle \alpha p ; J | H_{\text{int}} | \alpha p ; J \rangle \equiv  \varepsilon_p 
. 
\label{E:spvarep}$$ Defined in the natural-orbital basis, $\varepsilon_h$ and $\varepsilon_p$ no longer correspond to the eigenvalues of the single-particle (s.p.) Hamiltonian — in contrast to the HF case. A s.p. Hamiltonian $H_{\text{int}} ^{(1)}$ can be defined, in principle, as the one-body term of the nuclear Hamiltonian operator, Eq. (\[E:ham\]), when written in a normal-ordered form [@Row1970; @Hey1990]. It differs from the HF s.p. Hamiltonian only by a similarity transformation and therefore posesses the same eigenvalues. If, however, one [linearizes]{} the two-body part of the normal-ordered total Hamiltonian $H_{\text{int}}$ and adds the thus obtained one-body operator 