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In the theoretical framework of covariant density functional theory [43−45, 51−56], the Dirac Hamiltonian for nucleons reads
$ H = { \alpha} \cdot { p} + \beta (M+S) + V, $
(1) where
$ \alpha $ and$ \beta $ are the Dirac matrices, M is the mass of the nucleon, and S and V are the scalar and vector potentials, respectively. In this section, we recall the technique of SRG [47, 48] and show the results up to the$ 1/M^4 $ order [46, 50].The basic idea of SRG is that the Dirac Hamiltonian in Eq. (1) is transformed by a unitary operator
$ U(l) $ as$ H(l) = U(l)HU^{†}(l),\quad H(0) = H, $
(2) with a flow parameter l. While
$ U(l) $ is not known explicitly, an anti-Hermitian generator$ \eta(l) $ related to$ U(l) $ as$ \eta(l) = \frac{ {\rm d}U(l)}{ {\rm d}l} U^{†}(l) $ is chosen, and the Hamiltonian$ H(l) $ evolves with the differential flow equation$ \frac{ {\rm d}H(l)}{ {\rm d}l} = [\eta(l),H(l)]. $
(3) One of the convenient and appropriate choices of
$ \eta(l) $ reads [48]$ \eta(l) = [\beta M,H(l)], $
(4) which transforms the Dirac Hamiltonian into a diagonal form.
For solving Eq. (3), the Hamiltonian is first divided into two parts:
$ H(l) = \varepsilon(l)+o(l), $
(5) according to the commutation and anti-commutation relations with respect to
$ \beta $ , i.e.,$ [\varepsilon, \beta] = 0 $ and$ \{o, \beta\} = 0 $ . As a result,$ \frac{ {\rm d}\varepsilon(l)}{ {\rm d}l} = 4M\beta o^2(l), \tag{6a} $
(6a) $ \frac{ {\rm d}o(l)}{ {\rm d}l}= 2M\beta[o(l),\varepsilon(l)], \tag{6b} $
(6b) with the initial conditions
$ \varepsilon(0) = \beta(M+S)+V,\qquad o(0) = { \alpha} \cdot { p}. $
(7) Equations (6a) and (6b) can then be solved with the expansion into the series of
$ {1}/{M^i} $ [48], which gives$ \frac{\varepsilon(\lambda)}{M}= \sum\limits_{i = 0}^\infty\frac{\varepsilon_i(\lambda)}{M^i},\tag{8a} $
(8a) $ \frac{o(\lambda)}{M}= \sum\limits_{j = 1}^\infty\frac{o_j(\lambda)}{M^j},\tag{8b}$
(8b) while the flow parameter is replaced by a dimensionless parameter
$ \lambda = lM^2 $ . The solutions are obtained as [46]$ \varepsilon_n(\lambda) = \varepsilon_n(0) +4\beta\int_0^\lambda\sum \limits_{k = 1}^{n - 1}o_k(\lambda')o_{n-k}(\lambda')\, {\rm d}\lambda',\tag{9a}$
(9a) $ o_n(\lambda) = o_n(0){\rm e}^{-4\lambda} +2\beta {\rm e}^{-4\lambda}\int_0^\lambda\sum \limits_{k = 1}^{n - 1}[{\rm e}^{4\lambda'}o_k(\lambda'),\varepsilon_{n-k}(\lambda')] \, {\rm d}\lambda', \tag{9b}$
(9b) with the initial conditions,
$ \begin{array}{l} \varepsilon_0(0) = \beta,\quad \varepsilon_1(0) = \beta S+V,\quad \varepsilon_n(0) = 0\quad \mbox{if}\quad n\geqslant 2,\\ o_1(0) = { \alpha} \cdot { p},\quad o_n(0) = 0\quad \mbox{if}\quad n\geqslant 2. \end{array} $
(10) Therefore, at the end of the flow
$ \lambda \rightarrow \infty $ , all the off-diagonal parts vanish, i.e.,$ o_n(\infty) = 0 $ for all n. The diagonalized Dirac operator up to the$ 1/M^4 $ order is not difficult to obtain as [50]$ \begin{split} {\cal H}_{\rm SRG} =& \varepsilon(\infty) = M\varepsilon_0(\infty)+\varepsilon_1(\infty)+\frac{\varepsilon_2(\infty)}{M}+\frac{\varepsilon_3(\infty)}{M^2} +\frac{\varepsilon_4(\infty)}{M^3}+\frac{\varepsilon_5(\infty)}{M^4}+\cdots\\ =&M\varepsilon_0(0)+\varepsilon_1(0)+\frac{1}{2M}\beta o_1^2(0) +\frac{1}{8M^2}[[o_1(0),\varepsilon_1(0)],o_1(0)] +\frac{1}{32M^3}\beta\Big( {-4o_1^4(0)+[[o_1(0),\varepsilon_1(0)],\varepsilon_1(0)]o_1(0)} \\ & {+o_1(0)[[o_1(0),\varepsilon_1(0)],\varepsilon_1(0)] -2[o_1(0),\varepsilon_1(0)][o_1(0),\varepsilon_1(0)]} \Big) +\frac{1}{128M^4}\Big( {-9[[o_1(0),\varepsilon_1(0)],o_1^3(0)] +3[o_1(0),\varepsilon_1(0)]^2\varepsilon_1(0)} \\ &+3\varepsilon_1(0)[o_1(0),\varepsilon_1(0)]^2 -6[o_1(0),\varepsilon_1(0)]\varepsilon_1(0)[o_1(0),\varepsilon_1(0)]\\ & {+3[o_1(0)[o_1(0),\varepsilon_1(0)]o_1(0),o_1(0) +[[[[o_1(0),\varepsilon_1(0)],\varepsilon_1(0)],\varepsilon_1(0)],o_1(0)]} \Big) +\cdots \end{split} $
(11) -
In this section, we perform the exact non-relativistic expansion of the single-nucleon Dirac equation up to the
$ 1/M^4 $ order by using the FW transformation for a general case, in particular for the scalar potential.According to the FW transformation in the presence of external fields [2, 4, 8], the corresponding operators are defined as
$ O = { \alpha} \cdot { p},\qquad \varepsilon = \beta S+V,\qquad \Lambda = -\frac{i}{2M}\beta O, $
(12) where the operators O and
$ \varepsilon $ satisfy$ O \beta = -\beta O $ and$ \varepsilon \beta = \beta \varepsilon $ , respectively. The Hamiltonian (1) reads$ \begin{align} H = \beta M+O+\varepsilon. \end{align} $
(13) Based on the above definitions, the Dirac Hamiltonian is transformed by a unitary operation into [8]
$ \begin{split} H' =&{\rm e}^{i\Lambda}H{\rm e}^{-i\Lambda} =H+i[\Lambda,H] +\frac{i^2}{2!}[\Lambda,[\Lambda,H]]\\& +\cdots +\frac{i^n}{n!}[\underbrace{\Lambda,[\Lambda,\cdots,[\Lambda}_n,H]\cdots]] +\cdots \end{split} $
(14) where
$\begin{split} \frac{i^n}{n!}[\underbrace{\Lambda,[\Lambda,\cdots,[\Lambda}_n,H]\cdots]] =& (-1)^{\textstyle\frac{n(n-1)}{2}}\frac{\beta^n}{n!M^n}(O^n\beta M+O^{n+1} \\& +\frac{1}{2^n}[\underbrace{O,[O,\cdots,[O}_n,\varepsilon]\cdots]]).\end{split}$
(15) Keeping all the terms up to the
$ 1/M^n $ order, the unitary transformed Hamiltonian reads$ \begin{split} H'_{1/M^n} =& \beta M+\varepsilon+\sum \limits_{k = 1}^{n + 1} (-1)^{\textstyle\frac{(k-1)(k-2)}{2}}\frac{k-1}{k!M^{k-1}}\beta^{k-1}O^{k}\\ &+\sum \limits_{k = 0}^n (-1)^{\textstyle\frac{k(k-1)}{2}}\frac{\beta^k}{2^kk!M^k}[\underbrace{O,[O,\cdots,[O}_k,\varepsilon]\cdots]]. \end{split} $
(16) For example, the corresponding result up to the
$ 1/M^4 $ order is$ \begin{split} H'_{1/M^4} =& \beta M +\varepsilon +\frac{1}{2M}\beta O^2 -\frac{1}{3M^2}O^3 -\frac{1}{8M^3}\beta O^4\\& +\frac{1}{30M^4}O^5 +\frac{1}{2M}\beta[O,\varepsilon] -\frac{1}{8M^2}[O,[O,\varepsilon]] \\& -\frac{1}{48M^3}\beta[O,[O,[O,\varepsilon]]] +\frac{1}{384M^4}[O,[O,[O,[O,\varepsilon]]]]. \end{split} $
(17) In the unitary transformed Hamiltonian, e.g., Eq. (17), the off-diagonal parts are not zero but raised by one order from O to
$ \dfrac{1}{2M}\beta[O,\varepsilon] $ (plus higher-order terms). In order to make the off-diagonal parts vanish, more precisely speaking to make them higher than a given order, one should repeat the FW transformation until the accuracy is achieved [8]. For that purpose, the operators O and$ \varepsilon $ can be redefined according to Eq. (17), and one has$\begin{split} \varepsilon' =& \varepsilon +\frac{1}{2M}\beta O^2 -\frac{1}{8M^2}[O,[O,\varepsilon]] -\frac{1}{8M^3}\beta O^4 \\&+\frac{1}{384M^4}[O,[O,[O,[O,\varepsilon]]]], \end{split} \tag{18a} $
(18a) $\begin{split} O' =&\frac{1}{2M}\beta[O,\varepsilon] -\frac{1}{3M^2}O^3 -\frac{1}{48M^3}\beta[O,[O,[O,\varepsilon]]] \\& +\frac{1}{30M^4}O^5, \end{split}\tag{18b}$
(18b) $ \Lambda' =-\frac{i}{2M}\beta O'. \tag{18c}$
(18c) The corresponding FW transformation reads
$ H'' = {\rm e}^{i\Lambda'} H' {\rm e}^{-i\Lambda'}. $
(19) It can be seen that the off-diagonal parts in
$ H'' $ are of the order of$ 1/M^2 $ . Therefore, one should repeat this procedure to$ H''''' $ to make the off-diagonal parts of the order of$ 1/M^5 $ . Of course, to get the resultant diagonal parts in$ H''''' $ up to the$ 1/M^4 $ order, a recursion technique makes the calculation less complicated than it looks.As a result, the non-relativistic expansion with the FW transformation up to the
$ 1/M^4 $ order reads$ \begin{split} {\cal H}_{\rm FW} =&\beta M +\varepsilon +\frac{1}{2M}\beta O^2 -\frac{1}{8M^2}[O,[O,\varepsilon]]-\frac{1}{8M^3}\beta O^4 \\&-\frac{1}{8M^3}\beta[O,\varepsilon][O,\varepsilon] +\frac{1}{384M^4}[O,[O,[O,[O,\varepsilon]]]] \\ &+\frac{1}{12M^4}[O^3,[O,\varepsilon]] +\frac{1}{32M^4}[O,\varepsilon][O,\varepsilon]\varepsilon\\& -\frac{1}{16M^4}[O,\varepsilon]\varepsilon[O,\varepsilon] +\frac{1}{32M^4}\varepsilon[O,\varepsilon][O,\varepsilon]. \end{split} $
(20) -
For the systems with spherical symmetry, i.e., for spherical nuclei, the corresponding radial single-nucleon Dirac equation reads [43−45]
$ \left( \begin{array}{*{20}{c}} \Sigma(r)+M & -\dfrac{ {\rm d}}{ {\rm d}r}+\dfrac{\kappa}{r}\\ \dfrac{ {\rm d}}{ {\rm d}r}+\dfrac{\kappa}{r} & \Delta(r)-M \end{array} \right ) \left( \begin{array}{*{20}{c}} G(r) \\ F(r) \end{array} \right ) = E \left( \begin{array}{*{20}{c}} G(r) \\ F(r) \end{array} \right ), $
(21) where
$ \kappa $ is a good quantum number defined as$ \kappa = \mp\; (j+{1}/{2}) $ for$ j = l\pm{1}/{2} $ , and$ \Sigma(r) = V(r) + S(r) $ and$ \Delta(r) = V(r) - S(r) $ are the sum and the difference of the vector and scalar potentials, respectively. The single-particle energy$ E = \varepsilon +M $ includes the mass of the nucleon. The operators$ \varepsilon $ and O read$ \varepsilon = {\left( \begin{array}{*{20}{c}} \Sigma(r) & 0\\ 0 & \Delta(r) \end{array} \right )},\quad O = {\left( \begin{array}{*{20}{c}} 0 &-\dfrac{ {\rm d}}{ {\rm d}r}+\dfrac{\kappa}{r}\\ \dfrac{ {\rm d}}{ {\rm d}r}+\dfrac{\kappa}{r}& 0 \end{array} \right )}. $
(22) According to Eq. (20), the Dirac Hamiltonian is transformed by the FW transformation as
$ {\left( \begin{array}{*{20}{c}} {\cal H}^{\rm (F)}_{\rm FW} + M & O\bigg(\dfrac{1}{M^5}\bigg) \\ O\bigg(\dfrac{1}{M^5}\bigg) & {\cal H}^{\rm (D)}_{\rm FW} - M \end{array} \right)}, $
(23) where the superscripts (F) and (D) represent the components of the Dirac Hamiltonian acting on single-particle states in the Fermi sea and Dirac sea, respectively. It is seen that the off-diagonal parts are not strictly zero, which is different from the results of SRG, but are of higher order than the required one. Focusing on single-particle states in the Fermi sea which correspond to their counterparts in the non-relativistic framework, the explicit expansion of
$ {\cal H}^{\rm (F)}_{\rm FW} $ up to the$ 1/M^4 $ order is worked out in detail as$ {\cal H}^{\rm (F)}_{0, {\rm FW}} = \Sigma(r), \tag{24a}$
(24a) $ {\cal H}^{\rm (F)}_{1, {\rm FW}} = \frac{1}{2M}p^2, \tag{24b}$
(24b) $ {\cal H}^{\rm (F)}_{2, {\rm FW}} = \frac{1}{8M^2}\left( {-4Sp^2 +4S'\frac{ {\rm d}}{ {\rm d}r} -2\frac{\kappa}{r}\Delta' +\Sigma''} \right),\tag{24c} $
(24c) $\begin{split} {\cal H}^{\rm (F)}_{3, {\rm FW}} =& \frac{1}{8M^3}\left( {-p^4 +4S^2p^2 -8SS'\frac{ {\rm d}}{ {\rm d}r}-2S\Sigma'' }\right.\\&\left.{+4S\Delta'\frac{\kappa}{r} +\Sigma'\Delta'} \right),\end{split}\tag{24d}$
(24d) $ \begin{split} {\cal H}^{\rm (F)}_{4, {\rm FW}} =&\frac{1}{384M^4}\bigg\{144Sp^4 -288S'p^2\frac{ {\rm d}}{ {\rm d}r} +\left[ {72\Delta'\frac{\kappa}{r}}\right.\\&\left.{-24(4\Sigma''+9S'')-192S^3} \right]p^2 +\left[ {72\Delta'\frac{\kappa}{r^2} }\right.\\&\left.{-72\Delta''\frac{\kappa}{r} +24(3S'''+4\Sigma''') +576S^2S'} \right]\frac{ {\rm d}}{ {\rm d}r} \\ &+\left[ {-12(5\Sigma'\!-\!36S')\frac{\kappa(\kappa+1)}{r^3} \!+\!12(5\Sigma''\!+\!12S'')\frac{\kappa(\kappa+1)}{r^2}}\right.\\&\left.{ -72\Delta'\frac{\kappa}{r^3} +72\Delta''\frac{\kappa}{r^2}} {-36\Delta'''\frac{\kappa}{r} -288S^2\Delta'\frac{\kappa}{r} +33\Sigma'''' }\right.\\&\left.{+144S^2\Sigma'' +24S\Sigma'(2\Sigma'-6\Delta')} \right]\bigg\}, \end{split}\tag{24e} $
(24e) where
$ p^2 = -\frac{ {\rm d}^2}{ {\rm d}r^2}+\frac{\kappa(\kappa+1)}{r^2}, $
(25) and
$\begin{split} p^4 =& \frac{ {\rm d}^4}{ {\rm d}r^4} -2\frac{\kappa(\kappa+1)}{r^2}\frac{ {\rm d}^2}{ {\rm d}r^2} +4\frac{\kappa(\kappa+1)}{r^3}\frac{ {\rm d}}{ {\rm d}r}\\& +\frac{\kappa(\kappa+1)(\kappa+3)(\kappa-2)}{r^4}. \end{split}$
(26) In contrast, the Dirac Hamiltonian transformed by the SRG method reads
$ {\left( \begin{array}{*{20}{c}} {\cal H}^{\rm (F)}_{\rm SRG} + M & 0 \\ 0 & {\cal H}^{\rm (D)}_{\rm SRG} - M \end{array} \right)}. $
(27) Its off-diagonal parts strictly vanish up to infinite order. According to Eqs. (11), the explicit expansion of
$ {\cal H}^{\rm (F)}_{\rm SRG} $ up to the$ 1/M^4 $ order reads [50]$ {\cal H}^{\rm (F)}_{0, {\rm SRG}} = \Sigma(r),\tag{28a}$
(28a) $ {\cal H}^{\rm (F)}_{1, {\rm SRG}} = \frac{1}{2M}p^2, \tag{28b} $
(28b) $ {\cal H}^{\rm (F)}_{2, {\rm SRG}} = \frac{1}{8M^2}\left( {-4Sp^2 +4S'\frac{ {\rm d}}{ {\rm d}r}-2\frac{\kappa}{r}\Delta' +\Sigma''} \right), \tag{28c}$
(28c) $\begin{split} {\cal H}^{\rm (F)}_{3, {\rm SRG}} =&\frac{1}{32M^3}\left( {-4p^4 +16S^2p^2-32SS'\frac{ {\rm d}}{ {\rm d}r}-8S\Sigma'' }\right.\\&\left.{+16S\Delta'\frac{\kappa}{r} -2\Sigma'^2 +4\Sigma'\Delta'} \right), \end{split}\tag{28d} $
(28d) $ \begin{split} {\cal H}^{\rm (F)}_{4, {\rm SRG}} =& \frac{1}{128M^4} \bigg\{ 48Sp^4 - 96S'p^2\frac{{\rm d}}{{\rm d}r} \\& +\left[ {24\Delta'\frac{\kappa}{r} - 24(\Sigma''+3S'')}-64S^3 \right] p^2 \\& +\left[ {24\Delta'\frac{\kappa}{r^2} -24\Delta''\frac{\kappa}{r} +24(S'''+\Sigma''')}+192S^2S' \right]\frac{{\rm d}}{ {\rm d}r} \end{split}$
$ \begin{split} \\& +\bigg[ {-12(\Sigma'-12S')\frac{\kappa(\kappa+1)}{r^3}}\\ & {+12(\Sigma''+4S'')\frac{\kappa(\kappa+1)}{r^2} -24\Delta'\frac{\kappa}{r^3} +24\Delta''\frac{\kappa}{r^2}} \\& { {-12\Delta'''\frac{\kappa}{r} -96S^2\Delta'\frac{\kappa}{r} +9\Sigma'''' +48S^2\Sigma''}}\\&{{+24S\Sigma'(\Sigma'-2\Delta')} \bigg]} \bigg\}. \end{split} \tag{28e}$
(28e) -
By comparing Eqs. (11) and (20), it is found that
$ {\cal H}_{0,{\rm SRG}} - {\cal H}_{0,{\rm FW}} =0, \tag{29a}$
(29a) $ {\cal H}_{1,{\rm SRG}} - {\cal H}_{1,{\rm FW}} =0, \tag{29b}$
(29b) $ {\cal H}_{2,{\rm SRG}} - {\cal H}_{2,{\rm FW}} = 0, \tag{29c}$
(29c) $ {\cal H}_{3,{\rm SRG}} - {\cal H}_{3,{\rm FW}} = \frac{\beta}{32M^3}[[O^2,\varepsilon], \varepsilon], \tag{29d}$
(29d) $ \begin{split}{\cal H}_{4,{\rm SRG}} - {\cal H}_{4,{\rm FW}} =&\frac{1}{64M^4}[[O^2,\varepsilon], O^2] \\& +\frac{1}{128M^4}\left[\left([\varepsilon^2,O^2]-2[\varepsilon, O\varepsilon O]\right),\varepsilon\right].\end{split} \tag{29e}$
(29e) This result seems to indicate that the FW transformation and the SRG method agree with each other up to the
$ 1/M^2 $ order, but they lead to different results starting from the$ 1/M^3 $ order. However, considering the infinite mass limit$ M \rightarrow \infty $ , all expressions should be strictly organized order by order, and thus the difference between Eqs. (27) and (23) in the$ 1/M^5 $ order cannot lead to the differences of the results in the$ 1/M^3 $ and$ 1/M^4 $ orders. This puzzle needs to be further investigated.It turns out that the difference shown in Eq. (29) comes from an additional unitary transformation, after the Dirac Hamiltonian is decoupled into the upper and lower parts. Let
$ \Xi = -\frac{i\beta}{32M^3}[O^2, \varepsilon] - \frac{i}{128M^4}\left([\varepsilon^2,O^2]-2[\varepsilon, O\varepsilon O]\right). $
(30) This is a Hermitian and diagonal operator, i.e.,
$ \Xi^{†}= \Xi $ and$ \beta\Xi = \Xi\beta $ . Acting an additional unitary transformation on$ {\cal H}_{\rm FW} $ , it reads$ {\rm e}^{i\Xi}\, {\cal H}_{\rm FW}\, {\rm e}^{-i\Xi} = {\cal H}_{\rm FW} + i [\Xi, {\cal H}_{\rm FW}] + \cdots $
(31) Keeping all terms up to the
$ 1/M^4 $ order, the result is$ \begin{split} {\rm e}^{i\Xi}\, {\cal H}_{\rm FW}\, {\rm e}^{-i\Xi} =&{\cal H}_{\rm FW} + \frac{\beta}{32M^3}[[O^2,\varepsilon], \varepsilon] + \frac{1}{64M^4}[[O^2,\varepsilon], O^2] \\& +\frac{1}{128M^4}\left[\left([\varepsilon^2,O^2]-2[\varepsilon, O\varepsilon O]\right),\varepsilon\right], \end{split} $
(32) which is nothing else but
$ {\cal H}_{\rm SRG} $ .In the spherical case, the explicit expressions for operators
$ \dfrac{\beta}{32M^3}[[O^2,\varepsilon], \varepsilon] $ ,$ \dfrac{1}{64M^4}[[O^2,\varepsilon], O^2] $ , and$ \dfrac{1}{128M^4} $ $\left[\left([\varepsilon^2,O^2]-2[\varepsilon, O\varepsilon O]\right),\varepsilon\right] $ , acting on single-particle states in the Fermi sea, respectively read$ -\frac{1}{16M^3} {\Sigma'}^2, \tag{33a} $
(33a) $ \frac{1}{64M^4} \left[ {4\Sigma'' p^2 - 4\Sigma'''\frac{\rm d}{{\rm d}r} + 4\Sigma'\frac{\kappa(\kappa+1)}{r^3} - 4\Sigma''\frac{\kappa(\kappa+1)}{r^2} - \Sigma'''' } \right], \tag{33b}$
(33b) $ \frac{1}{16M^4} S {\Sigma'}^2. \tag{33c}$
(33c) These equations explain the difference between Eqs. (24d) and (28d), as well as between Eqs. (24h) and (28h).
Since
$ {\rm e}^{-i\Xi} $ is an additional unitary transformation acting on the already decoupled Hamiltonian, it does not affect the non-relativistic expansion of the Dirac equation, and the single-particle spectra obtained by the FW transformation and the SRG method are the same.The unitary transformation
$ {\rm e}^{-i\Xi} $ is equivalent to a rotation in a linear and closed complex space, and in quantum mechanics, the operation or rotation in such a space should be associated with a corresponding generator. However, in the present context, it is nontrivial to write down a general expression for such a generator, or derive a conserved quantity that is asymptotically related to a generator up to an arbitrary order. This is because the number of unitary transformations from the FW results to the SRG ones would in principle be infinite, since the SRG method employs infinite infinitesimal unitary transformations along the flow. Therefore, the physics behind the introduced unitary transformation is open for future studies.
Non-relativistic expansion of single-nucleon Dirac equation: Comparison between Foldy-Wouthuysen transformation andsimilarity renormalization group
- Received Date: 2019-06-23
- Available Online: 2019-11-01
Abstract: By following the Foldy-Wouthuysen (FW) transformation of the Dirac equation, we derive the exact analytic expression up to the 1/M4 order for general cases in the covariant density functional theory. The results are compared with the corresponding ones derived from another novel non-relativistic expansion method, the similarity renormalization group (SRG). Based on this comparison, the origin of the difference between the results obtained with the FW transformation and the SRG method is explored.