the Dirac equation  is shown with the black solid line, while the result by Eq.  is shown with the olive dashed line. The results by Eq.  with the corrections up to the $1/{M^\ast}^2$ and $1/{M^\ast}^3$ orders are shown with the blue dotted and red dash-dotted lines, respectively.[]{data-label="fig6"}](Fig6.eps){width="45.00000%"}

For the discussions on the single-particle scalar density, let us first focus on the difference between the single-particle density and the single-particle scalar density. In the Dirac picture, such a difference comes from the small component of Dirac spinor, i.e., $\rho_v(r) - \rho_s(r) = 2F^2(r)/(4\pi r^2)$ in the present case, which is in general tiny. This tiny difference is shown with the black solid line in Fig. \[fig6\], by taking the $2d_{5/2}$ state as an example. The peak of the black solid line is only around $0.10 \times 10^{-3}$ fm$^{-3}$ at $r=3.75$ fm.

For comparison, the corresponding result by Eq.  without any relativistic (picture-change-error correction) correction is shown with the olive dashed line in Fig. \[fig6\]. In contrast, the results calculated by Eq.  with the relativistic corrections up to the $1/{M^\ast}^2$ and $1/{M^\ast}^3$ orders are shown with the blue dotted and red dash-dotted lines, respectively.

![(Color online) Difference between the single-particle scalar density 