The number of $J=0$ pairs in $^{44,46,48}$Ti

In the single $j$-shell, the configuration of an even--even Ti isotope consists of 2 protons and $n$ neutrons. The $I=0$ wave function can be written as $$\Psi=\sum_{Jv} D(J,Jv) [(j^2)^J_\pi (j^n)^J_\nu ]^{I=0},$$ where $v$ is the seniority quantum number. There are several states with isospin $T_{\rm min}=|(N-Z)/2|$, but only one with $T_{\rm max}=T_{\rm min}+2$. By demanding that the $T_{\rm max}$ wave function be orthogonal to the $T_{\rm min}$ ones, we obtain the following relation involving a one-particle cfp: $$ D(00)=\frac{n}{2j+1} \sum_J D(J,Jv)(j^{n-1}(jv=1)j|}j^nJ) \sqrt{2J+1} $$ This leads to the following simple expressions for the number of $J=0$ $np$ pairs in these Ti isotopes: \begin{itemize} \item For $T=T_{\rm min}$, \ \ \ \# of pairs $(J_{12}=0)=2|D(00)|^2/n$ \item For $T=T_{\rm max}$, \ \ \# of pairs $(J_{12}=0)=2n|D(00)|^2=\frac{2n(2j+1-n)}{(2j+1)(n+1)}$ \end{itemize} For $^{44}$Ti we have also the result for {\em even} $J_{12}$ $$ \#\ {\rm of}\ nn\ {\rm pairs}\ =\ \#\ {\rm of}\ pp\ {\rm pairs}\ =\ \#\ {\rm of}\ np\ {\rm pairs}\ =\ |D(J_{12},J_{12})|^2 $$