following discussions.

In Ref. [@Guo2019Chin.Phys.C43.114105], the relation between the conventional SRG method and FW transformation has been investigated. Similarly, by comparing Eqs. (\[neq1\]), (\[neq2\]), (\[neq3\]) and (\[neq4\]) with the corresponding results obtained in the non-relativistic expansion by the reconstituted SRG method [@Guo2019Phys.Rev.C99.054324], one finds the differences as

\[eq:diff\] $$\begin{aligned}
\mathcal{H}_{\rm rFW,0}-\mathcal{H}_{\rm rSRG,0}=\,&0\\
\mathcal{H}_{\rm rFW,1}-\mathcal{H}_{\rm rSRG,1}=\,&0\\
\mathcal{H}_{\rm rFW,2}-\mathcal{H}_{\rm rSRG,2}=\,&0\\
\mathcal{H}_{\rm rFW,3}-\mathcal{H}_{\rm rSRG,3}=\,&\frac{\Delta'\Sigma'}{16{M^\ast}^3}.\label{neq5}\end{aligned}$$

It turns out that the differences shown in Eq.  come from an additional unitary transformation, after the Dirac Hamiltonian is decoupled into the upper and lower parts. Let $$\begin{aligned}
\Xi=\,&-\frac{i\beta}{64}\left(\frac{1}{{M^\ast}^3}[\tilde{O}^2,\tilde{\varepsilon}-\beta{M^\ast}]+[\tilde{O}^2,\tilde{\varepsilon}-\beta{M^\ast}]
\frac{1}{{M^\ast}^3}\right)\nonumber\\
&+\cdots\end{aligned}$$ It is a Hermitian and diagonal operator, i.e., $\Xi^\dag = \Xi$ and $\beta\Xi = \Xi\beta$. Acting an additional unitary transformation on $\mathcal{H}_{\rm rFW}$, it reads $$\begin{aligned}
e^{i\Xi}\mathcal{H}_{\rm rFW}e^{-i\Xi}=\mathcal{H}_{\rm rFW}+i[\Xi,\mathcal{H}_{\rm rFW}]+\cdots\end{aligned}$$

Keeping all the terms up to the $1/{M^\ast}^3$ order, the results are $$\begin{aligned}
\label{neq6}
\mathcal{H}_{\rm rFW}+\frac{\beta}{64}\Big[\frac{1}{{M^\ast}^3}[\tilde{O}^2,\tilde{\varepsilon}-\beta{M^\ast}]
+[\tilde{O}^2,\tilde{\varepsilon}-\beta{M^\ast}]\frac{1}{{M^\ast}^3},\beta M^\ast+\tilde{\varepsilon}\Big].\end{aligned}$$ As a step further, it is not complicated to obtain the upper-left component of Eq.  as $$\begin{aligned}
\mathcal{H}_{\rm rFW,0}+\mathcal{H}_{\rm rFW,1}+\mathcal{H}_{\rm rFW,2}+\mathcal{H}_{\rm rFW,3}-\frac{\Delta'\Sigma'}{16{M^\ast}^3},\end{aligned}$$ which is nothing but $\mathcal{H}_{\rm rSRG}$.

Since $e^{-i\Xi}$ is a unitary operator acting on the already decoupled Hamiltonian, it does not affect the non-relativistic expansion of the Dirac equation, in the sense that the single-particle spectrum obtained 