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16
Notes/PerturbativeAnalysis.tex
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@ -178,7 +178,7 @@ Therefore, the transformed Hamiltonian
depends on a flow parameter $s$. depends on a flow parameter $s$.
The resulting Hamiltonian possess up to $N$-body operators with $N$ the number of particle. The resulting Hamiltonian possess up to $N$-body operators with $N$ the number of particle.
\begin{equation} \begin{equation}
\bH(s) = E_0(s) + \bF(s) + \bV(s) + \bW(s) + \dots \bH(s) = E_0(s) + \bF(s) + \titou{\bV(s)} + \bW(s) + \dots
\end{equation} \end{equation}
In the following, we will truncate every contribution superior to two-body operators. In the following, we will truncate every contribution superior to two-body operators.
We can easily derive an evolution equation for this Hamiltonian by taking the derivative of $\bH(s)$. This gives We can easily derive an evolution equation for this Hamiltonian by taking the derivative of $\bH(s)$. This gives
@ -206,7 +206,7 @@ In this work, we will use Wegner's canonical generator which is defined as
\boldsymbol{\eta}^\text{W}(s) = \comm{\bH^\text{d}(s)}{\bH(s)} = \comm{\bH^\text{d}(s)}{\bH^\text{od}(s)}. \boldsymbol{\eta}^\text{W}(s) = \comm{\bH^\text{d}(s)}{\bH(s)} = \comm{\bH^\text{d}(s)}{\bH^\text{od}(s)}.
\end{equation} \end{equation}
This generator has the advantage of defining a true renormalisation scheme, \ie the coupling coefficients with the highest energy determinants are removed first. This generator has the advantage of defining a true renormalisation scheme, \ie the coupling coefficients with the highest energy determinants are removed first.
One of the flaws of this generator is that it generates a stiff set of ODE which is difficult to solve numerically. One of the flaws of this generator is that it generates a \titou{stiff} set of ODE which is difficult to solve numerically.
However, here we consider analytical perturbative expressions so we will not be affected by this problem. However, here we consider analytical perturbative expressions so we will not be affected by this problem.
%=================================================================% %=================================================================%
@ -228,7 +228,7 @@ In this case, we want to decouple the reference determinant from every singly an
Hence, we define the off-diagonal Hamiltonian as Hence, we define the off-diagonal Hamiltonian as
\begin{equation} \begin{equation}
\label{eq:hamiltonianOffDiagonal} \label{eq:hamiltonianOffDiagonal}
\hH^{\text{od}}(s) = \sum_{ia} f_i^a(s)\no{a}{i} + \frac{1}{4} \sum_{ijab}v(s)_{ij}^{ab}\no{ab}{ij}. \hH^{\text{od}}(s) = \sum_{ia} f_i^a(s)\no{a}{i} + \frac{1}{4} \sum_{ijab}\titou{v(s)_{ij}^{ab}}\no{ab}{ij}.
\end{equation} \end{equation}
Note that each coefficients depend on $s$. Note that each coefficients depend on $s$.
@ -237,7 +237,7 @@ The perturbative parameter $\la$ is such that
\bH(0) = E_0(0) + F(0) + \la V(0) \bH(0) = E_0(0) + F(0) + \la V(0)
\end{equation} \end{equation}
In addition, we know the following initial conditions. In addition, we know the following initial conditions.
We use the HF basis set of the reference such that $F^{\text{od}}(0) = 0$ and $F^{\mathrm{d}}(0)=\delta_{pq}\epsilon_p$ We use the HF basis set of the reference such that $F^{\text{od}}(0) = 0$ and \titou{$F^{\mathrm{d}}(0)=\delta_{pq}\epsilon_p$}
Therefore, we have Therefore, we have
\begin{align} \begin{align}
\bH^\text{d}(0)&=E_0(0) + F^{\mathrm{d}}(0) + \la V^{\mathrm{d}}(0) & \bH^\text{od}(0)&= \la V^{\mathrm{od}}(0) \bH^\text{d}(0)&=E_0(0) + F^{\mathrm{d}}(0) + \la V^{\mathrm{d}}(0) & \bH^\text{od}(0)&= \la V^{\mathrm{od}}(0)
@ -340,7 +340,7 @@ To compute the second order contribution to the Hamiltonian coefficients, we fir
The expressions for the first commutator are computed analogously to the one of the previous subsection. The expressions for the first commutator are computed analogously to the one of the previous subsection.
We focus on deriving expressions for the second term. We focus on deriving expressions for the second term.
The one-body part of $\bH^{\text{od},(1)}(s)$ is equal to zero so two of the four terms contributing to the one-body part of $\comm{\bH^{\text{d},(1)}(s)}{\bH^{\text{od},(1)}(s)}$ are zero. The one-body part of $\bH^{\text{od},(1)}(s)$ is equal to zero so two of the four terms contributing to the one-body part of $\comm{\bH^{\text{d},(1)}(s)}{\bH^{\text{od},(1)}(s)}$ are zero.
In addition, the term $\comm{A_1}{B_2}$ is equal to zero as well because the coefficients $A_{1,i}^a$ are zero (see expression in Appendix). In addition, the term $\comm{A_1}{B_2}$ is equal to zero as well because the coefficients \titou{$A_{1,i}^a$} are zero (see expression in Appendix).
So we have So we have
\begin{align} \begin{align}
\eta_a^{i,(2)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_1^{\text{od},(2)}(s)}_a^i + \comm{\bH_2^{\text{d},(1)}(s)}{\bH_2^{\text{od},(1)}(s)}_a^i \\ \eta_a^{i,(2)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_1^{\text{od},(2)}(s)}_a^i + \comm{\bH_2^{\text{d},(1)}(s)}{\bH_2^{\text{od},(1)}(s)}_a^i \\
@ -366,7 +366,7 @@ In addition, the one-body hamiltonian has no first order contribution so
\end{align} \end{align}
After integration, using the initial condition $E_0^{(2)}(0)=0$, we obtain After integration, using the initial condition $E_0^{(2)}(0)=0$, we obtain
\begin{equation} \begin{equation}
E_0^{(2)}(s) = \frac{1}{4} \sum_{i j} \sum_{a b} \frac{\Delta_{ab}^{ij}}{\aeri{ij}{ab}}\left(1-e^{-2s (\Delta_{ab}^{ij})^2}\right) E_0^{(2)}(s) = \frac{1}{4} \sum_{i j} \sum_{a b} \titou{\frac{\Delta_{ab}^{ij}}{\aeri{ij}{ab}}}\left(1-e^{-2s (\Delta_{ab}^{ij})^2}\right)
\end{equation} \end{equation}
%=================================================================% %=================================================================%
@ -417,7 +417,7 @@ which gives the following conditions
\subsection{Zeroth order Hamiltonian} \subsection{Zeroth order Hamiltonian}
%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%
The zero-th order commutator of the Wegner generator therefore gives The zeroth-order commutator of the Wegner generator therefore gives
\begin{equation} \begin{equation}
\bEta{0} = \comm{\bHd{0}}{\bHod{0}} = \bO \bEta{0} = \comm{\bHd{0}}{\bHod{0}} = \bO
\end{equation} \end{equation}
@ -456,7 +456,7 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
\dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(1),\dagger}\bF^{(0)} - \bV{}{(1),\dagger}(\bF^{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(1),\dagger} \\ \dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(1),\dagger}\bF^{(0)} - \bV{}{(1),\dagger}(\bF^{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(1),\dagger} \\
\dv{\bC{}{(1)}}{s} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(1)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(1)} - \bC{\text{od}}{(1)}(\bC{\text{d}}{(0)})^2 \dv{\bC{}{(1)}}{s} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(1)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(1)} - \bC{\text{od}}{(1)}(\bC{\text{d}}{(0)})^2
\end{align} \end{align}
The two last equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$. The last two equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$.
\subsubsection*{Diagonal $\bC{}{(0)}$} \subsubsection*{Diagonal $\bC{}{(0)}$}
In the following, upper case indices correspond to the 2h1p and 2p1h sectors while lower case indices correspond to the 1h and 1p sectors. Also the $\Delta\eps_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors. In the following, upper case indices correspond to the 2h1p and 2p1h sectors while lower case indices correspond to the 1h and 1p sectors. Also the $\Delta\eps_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors.