single-particle scalar density can be modified with the consideration of the relativistic corrections originated from the picture-change error between the Schrödinger and Dirac pictures, i.e., $$\begin{aligned}
\rho_{s,\textrm{exact}}(r)=\rho_{s,\textrm{non}}(r)+ \Delta \rho_s(r).\end{aligned}$$ It is important to point out that the difference between $\rho_{v,\textrm{non}}$ and $\rho_{s,\textrm{non}}$ appears from the $1/{M^\ast}^2$ order, and the relativistic correction $\Delta \rho_s$ also appears from the same order. Therefore, the relativistic corrections will also play a crucial role in the difference between the vector and scalar densities.

For the systems with the spherical symmetry, the single-particle scalar density reads $$\begin{aligned}
\label{rhos2}
4\pi r^2\rho_s(r)
=\,&
\left.\left( \begin{array}{cc}
\varphi^\dag(r')& 0
\end{array}
\right)
\boldsymbol{\Delta}_{\rm rFW}(r',r)\,\tilde{\gamma}\,
\left( \begin{array}{c}
\varphi(r')\\ 0
\end{array}
\right )\right|_{r'=r}\nonumber\\
=\,&
\varphi^\dag(r)\left[
1-\left(-\overleftarrow{\frac{\textrm{d}^2}{\textrm{d}r^2}}+\frac{\kappa(\kappa+1)}{r^2}\right)\frac{1}{8{M^\ast}^2}
-\frac{1}{8{M^\ast}^2}\left(-\overrightarrow{\frac{\textrm{d}^2}{\textrm{d}r^2}} +\frac{\kappa(\kappa+1)}{r^2}\right)
-\left(\overleftarrow{\frac{\textrm{d}}{\textrm{d}r}}+\frac{\kappa}{r}\right)\frac{1}{4{M^\ast}^2}
\left(\overrightarrow{\frac{\textrm{d}}{\textrm{d}r}}+\frac{\kappa}{r}\right)\right.\nonumber\\
&\qquad\qquad\left.-\overleftarrow{\frac{\textrm{d}}{\textrm{d}r}}\frac{S'}{8{M^\ast}^3} +\frac{V'}{2{M^\ast}^3}\frac{\kappa}{r} -\frac{S''+2V''}{8{M^\ast}^3}
-\frac{S'}{8{M^\ast}^3}\overrightarrow{\frac{\textrm{d}}{\textrm{d}r}}\right]\varphi(r).\end{aligned}$$ In this expression, there are also contributions from the off-diagonal parts of $\boldsymbol{\Delta}_{\rm rFW}(r,r)$ and $\tilde{\gamma}$.

Results and Discussion {#sec:III}
======================

In order to make the comparisons and discussions with the corresponding results obtained by the SRG methods in Ref. [@Guo2019Phys.Rev.C99.054324], we use the Woods-Saxon forms for the scalar and vector potentials, $\Sigma(r)=\Sigma_0f(a_0,r_0,r)$ and $\Delta(r)=\Delta_0f(a_0,r_0,r)$, with $$f(a_0,r_0,r)=\frac{1}{1+e^{\frac{r-r_0}{a_0}}},$$ where the parameters are the same as those in Ref. [@Guo2019Phys.Rev.C99.054324], i.e., $\Sigma_0=-66.0$ MeV, $\Delta_0 = 650.0$ MeV, $r_0= 7.0$ fm, and 