derived in Ref. [@Guo2019Phys.Rev.C99.054324] can help us investigate the results in Fig. \[fig1\] more deeply here. It can be seen that except the terms that have already included so far with the replacement of $M$ by $M^\ast$, the three biggest contributions in the $1/M^4$ order are $-\frac{3S'}{4M^4}p^2\frac{\textrm{d}}{\textrm{d}r}$, $-\frac{9S''}{16M^4}p^2$, and $\frac{3\Delta'}{16M^4}\frac{\kappa}{r}p^2$ shown in Table. VIII in Ref. [@Guo2019Phys.Rev.C99.054324]. It is found that the sum of the contributions from these three terms are positive (negative) for the spin-up states with $\kappa<0$ (the spin-down states with $\kappa>0$), and the absolute value becomes bigger with the increase of $l$ (or $\kappa$). Consequently, the large $l$ (or $\kappa$) states show the biggest discrepancy here, and the $\kappa<0$ ($\kappa>0$) states show negative (positive) discrepancy from the corresponding exact values here. A possible way to improve the results further is to derive the $1/{M^\ast}^4$ order in the reconstituted FW transformation, and pay attention to the terms proportional to $\frac{1}{{M^\ast}^4}p^2\frac{\textrm{d}}{\textrm{d}r}$, $\frac{1}{{M^\ast}^4}p^2$, and $\frac{1}{{M^\ast}^4}\frac{\kappa}{r}p^2$.

By summing up all the single-particle energies of the occupied states for $126$ neutrons in $^{208}$Pb, the exact value is $-3045.501$ MeV, while the result obtained by the $1/{M^\ast}^3$-order reconstituted FW transformation is $-3055.767$ MeV. The corresponding relative discrepancy is only about $0.34\%$.

Single-particle and 