\rangle 
\nonumber \\ 
            &=& \sum_{ph}  
[ X_{ph}^{\nu , {J}^{\pi}           } \!\mathcal{A}_{ph}^{{JM} \dagger}  
- Y_{ph}^{\nu , {J}^{\pi}           } \!(-1)^{{J+M}} \!\mathcal{A}_{ph}^{{J-M}}] | 0 \rangle , 
\label{E:nu} \end{aligned}$$ where $| 0 \rangle $ is the ground state, for which $$Q_{\nu ,{J}^{\pi}{M}}| 0 \rangle = 0,$$ and the sum runs over $ph$ pairs which can couple to total spin and parity ${J}^{\pi}$. The renormalized operators are written as linear combinations of $ph$ operators, $$\mathcal{A}_{ph}^{{JM} \dagger} = 
\sum_{n_p'n_h'} N_{n_pn_h,n_p'n_h'}^{\ell_pj_pt_p;\ell_hj_ht_h} 
[a_{n_p'\ell_pj_pt_p}^{\dagger} a_{n_h'\ell_hj_ht_h}]^{{JM}} 
, 
\label{E:Aop}$$ where the angular-momentum coupled operators are given by $$[a_{p}^{\dagger} a_{h}]^{{JM}} = \sum_{m_pm_h} 
\langle j_p m_p j_h m_h | {JM}\rangle 
(-1)^{j_h+m_h} 
a_{pm_p}^{\dagger} a_{h-m_h} .$$ The choice of the single particle basis and the requirement that $$\langle 0 | 
[ \mathcal{A}_{p'h'}^{{JM} }, 
  \mathcal{A}_{ph}^{{JM} \dagger}  ] 
| 0 \rangle = \delta_{pp'}\delta_{hh'} 
 ,$$ following from the fermionic character of the operators, dictate that the matrix $N$ be given by [@CGP1998] $$\begin{aligned}
N_{n_pn_h,n_p'n_h'}^{\ell_pj_pt_p;\ell_hj_ht_h} 
&=& 
\delta_{n_pn_p'}  
\delta_{n_hn_h'} 
(\rho_h - \rho_p)^{-1/2} 
\nonumber \\ 
&\equiv&  
\delta_{n_pn_p'}  
\delta_{n_hn_h'} 
D_{ph}^{-1/2}  .  
\label{E:Nmat} \end{aligned}$$ Here $$\rho_{\alpha} = \langle 0 | a^{\dagger}_{\alpha m_{\alpha}} a_{\alpha m_{\alpha}} 
| 0 \rangle  
$$ denotes the occupation probability of the single-particle state $\alpha$, which is independent of $m_{\alpha}$. It can be shown that, under the above circumstances, the amplitudes 