by the reconstituted FW transformation is identical to that obtained by the reconstituted SRG method. At the same time, the above conclusions provide the foundation of the reconstituted SRG method proposed in Ref. [@Guo2019Phys.Rev.C99.054324].

Single-particle density {#sec:IIB}
-----------------------

The relativistic single-particle wave-function $|\psi\rangle$ satisfies $$\begin{aligned}
H|\psi\rangle=E|\psi\rangle,\end{aligned}$$ and the non-relativistic counterpart $|\varphi\rangle$ in the FW transformation satisfies $$\begin{aligned}
\mathcal{H}_{\rm FW}|\varphi\rangle=E|\varphi\rangle\end{aligned}$$ with $$\begin{aligned}
\mathcal{H}_{\rm FW}=e^{i{\Lambda_{\rm FW}}}He^{-i{\Lambda_{\rm FW}}}\end{aligned}$$ and $$\begin{aligned}
|\varphi\rangle=e^{i{\Lambda_{\rm FW}}}|\psi\rangle.\end{aligned}$$

Since only the unitary transformations have been applied in the FW transformation, it seems naturally that the single-particle density $\rho_v(\bm{r})=\psi^\dag(\bm{r})\psi(\bm{r})$ will be calculated as $$\begin{aligned}
\label{rhov1}
\rho_{v, \textrm{non}}(r)
=\frac{1}{4\pi r^2}\varphi^\dag(r)\varphi(r).\end{aligned}$$ However, by checking the details, it can be found that $$\begin{aligned}
\rho_{v,\textrm{exact}}(r)=\rho_{v,\textrm{non}}(r)+\Delta\rho_v(r),\end{aligned}$$ where $\Delta\rho_v$ actually comes from the so-called picture-change error [@Douglas1974Ann.Phys.N.Y.82.89--155; @Hess1986Phys.Rev.A33.3742--3748], i.e., the difference between the densities calculated in the Schrödinger picture and the ones obtained in the Dirac picture.

In order to investigate the correction to the single-particle density explicitly, we start from its strict definition, which reads $$\begin{aligned}
\rho_v(\bm{r})=\,&\langle\psi|\hat{c}^\dag(\bm{r})\hat{c}(\bm{r})|\psi\rangle\nonumber\\
=\,&\int\, \textrm{d}\bm{r}'\psi^\dag(\bm{r}')\delta(\bm{r}'-\bm{r})\psi(\bm{r}').\end{aligned}$$ Here $\hat{c}^\dag(\bm{r})$ and $\hat{c}(\bm{r})$ are the field operators (i.e., the particle creation and annihilation operators with the index of spatial coordinate). Therefore, the single-particle density reads $$\begin{aligned}
\rho_v(\bm{r})=\,&\langle\varphi|e^{i{\Lambda_{\rm FW}}}\hat{c}^\dag(\bm{r})\hat{c}(\bm{r})e^{-i{\Lambda_{\rm FW}}}|\varphi\rangle\nonumber\\
=\,&\int\,\textrm{d}\bm{r}'\varphi^\dag(\bm{r}')e^{i{\Lambda_{\rm FW}}(\bm{r}')}\delta(\bm{r}'-\bm{r})e^{-i{\Lambda_{\rm FW}}(\bm{r}')}\varphi(\bm{r}').\end{aligned}$$ Hereafter, the delta 