the other hand, it is even more important to point out its deviation from the exact density around the node of the wave function, which is shown in Panel (b) of Fig. \[fig2\] in a smaller scale. Because of the existence of the small component of Dirac spinor, i.e., $F(r)$ in Eq. , the single-particle density in the Dirac picture, $\rho_v(r) = [G^2(r)+F^2(r)]/(4\pi r^2)$, will never become exactly zero at any finite radius $r$. However, the corresponding single-particle density in the Schrödinger picture (Eq. ) becomes exactly zero at the nodes of the single-particle wave function $\varphi(r)$. This is one of the explicit reasons why the relativistic corrections to the densities are indispensable. In other words, the picture-change error belongs to the systematic error, instead of the matter of truncation or numerical accuracy.

By taking into account the relativistic corrections, the results obtained by Eq.  are almost identical to the exact density in the whole region, including the node region as shown in Panel (b) of Fig. \[fig2\]. Indeed, in Eq. , the terms such as $$\varphi^\dag(r) \overleftarrow{\frac{\textrm{d}}{\textrm{d}r}} \frac{1}{4{M^\ast}^2} \overrightarrow{\frac{\textrm{d}}
{\textrm{d}r}} \varphi(r)$$ prevent the single-particle density from becoming exactly zero at any finite radius $r$.

![(Color online) Difference between the 