all PT-SRG equations of Evangelista 2014 written in details

Notes/PerturbativeAnalysis.tex

@@ -147,19 +147,18 @@
 
 \maketitle
 
-
+%=================================================================%
-%%%%%%%%%%%%%%%%%%%%%%
 \section{Introduction}
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 
 The aim of this document is two-fold.
 First, we want to re-derive the perturbative analysis of the similarity renormalisation group (SRG) formalism applied to the non-relativistic electronic Hamiltonian.
 In a second time, we want to apply the same formalism to the unfolded GW Hamiltonian.
 Before jumping into these analysis, we do a brief presentation of the SRG formalism.
 
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 \section{The similarity renormalisation group}
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 
 The similarity renormalization group aims at continuously transforming an Hamiltonian to a diagonal form, or more often to a block-diagonal form.
 Therefore, the transformed Hamiltonian
@@ -200,9 +199,9 @@ This generator has the advantage of defining a true renormalisation scheme, \ie
 One of the flaws of this generator is that it generates a 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.
 
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 \section{The electronic Hamiltonian}
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 
 In this part, we derive the perturbative expression for the SRG applied to the non-relativistic electronic Hamiltonian
 \begin{equation}
@@ -219,7 +218,7 @@ In this case, we want to decouple the reference determinant from every singly an
 Hence, we define the off-diagonal Hamiltonian as
 \begin{equation}
   \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}v(s)_{ij}^{ab}\no{ab}{ij}.
 \end{equation}
 Note that each coefficients depend on $s$.
 
@@ -239,6 +238,11 @@ Now, we want to compute the terms at each order of the following development
   \bH(s) = \bH^{(0)}(s) + \bH^{(1)}(s) + \bH^{(2)}(s) + \bH^{(3)}(s) + \dots
 \end{equation}
 by integrating Eq.~\eqref{eq:flowEquation}.
+
+%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Zeroth order Hamiltonian}
+%%%%%%%%%%%%%%%%%%%%%%
+
 First, we start by showing that the zeroth order Hamiltonian is independant of $s$ and therefore equal to $\bH^{(0)}(0)$
 \begin{align}
   \bH(\delta s) &= \bH(0) + \delta s \dv{\bH(s)}{s}\bigg|_{s=0} + O(\delta s^2) \\
@@ -252,6 +256,10 @@ Which gives us the following equality
 \end{equation}
 meaning that the zeroth order Hamiltonian is independant of $s$.
 
+%%%%%%%%%%%%%%%%%%%%%%
+\subsection{First order Hamiltonian}
+%%%%%%%%%%%%%%%%%%%%%%
+
 The right-hand side of \eqref{eq:flowEquation} has no zeroth order, so we want to compute its first order term.
 To do so, we first need to compute the first order term of $\boldsymbol{\eta}(s)$.
 We have seen that $\bH^\text{od}(0)$ has no zeroth order contribution so we have
@@ -262,31 +270,104 @@ According to the appendix the one-body part of $\boldsymbol{\eta}^{(1)}(s) $ has
 In addition, the term $\left[A_{1}, B_{2}\right]_{p}^{q}$ involves the coefficients $A_i^a = f_i^a(0) = 0$.
 So finally we only have one term
 \begin{align}
-  \eta_a^{i,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_1^{\text{od},(1)}(s)} \\
+  \eta_a^{i,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_1^{\text{od},(1)}(s)}_a^i  \\
                     &= \sum_r \left( f_a^r(0) f_r^{i,(1)}(s)  - f_r^i(0) f_a^{r,(1)}(s) \right) \\
                     &= (\epsilon_a - \epsilon_i)f_a^{i,(1)}(s)
 \end{align}
 Now turning to the two-body part of $\boldsymbol{\eta}^{(1)}(s) $ and once again two terms are zero because the two-body part of $\bH^{\text{d},(0)}(0)$ is equal to zero.
 So we have 
 \begin{align}
-  \eta_{ab}^{ij,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_2^{\text{od},(1)}(s)} \\
+  \eta_{ab}^{ij,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_2^{\text{od},(1)}(s)}_{ab}^{ij}  \\
-                    &= \sum_t [P(ab) f_a^t(0) v_{tb}^{ij,(1)}(s) - P(ij) f_t^i(0) v_{ab}^{tj,(1)}(s) ] \\
+                    &= \sum_t [P(ab) f_a^t(0) v_{tb}^{ij,(1)}(s) - P(ij) f_t^i(0) v_{ab}^{tj,(1)}(s) ] \notag \\
-                        &= \sum_t [ P(ab) \epsilon_a \delta_{at}v_{tb}^{ij,(1)}(s) - P(ij) \epsilon_i \delta_{it} v_{ab}^{tj,(1)}(s) ] \\
+                        &= \sum_t [ P(ab) \epsilon_a \delta_{at}v_{tb}^{ij,(1)}(s) - P(ij) \epsilon_i \delta_{it} v_{ab}^{tj,(1)}(s) ] \notag \\
-                        &= P(ab) \epsilon_a v_{ab}^{ij,(1)}(s) - P(ij) \epsilon_i  v_{ab}^{ij,(1)}(s) \\
+                        &= P(ab) \epsilon_a v_{ab}^{ij,(1)}(s) - P(ij) \epsilon_i  v_{ab}^{ij,(1)}(s)  \notag\\
-                        &= \left( \epsilon_a + \epsilon_b - \epsilon_i  - \epsilon_j \right) v_{ab}^{ij,(1)}
+                        &= \left( \epsilon_a + \epsilon_b - \epsilon_i  - \epsilon_j \right) v_{ab}^{ij,(1)} \notag
 \end{align}
 
-WIP...
+We can now compute the first order contribution to Eq.~\eqref{eq:flowEquation}. We have seen that $\eta$ has no zeroth order contribution so
+\begin{equation}
+  \dv{\bH^{(1)}(s)}{s} = \comm{\boldsymbol{\eta}^{(1)}(s)}{\bH^{(0)}(s)}
+\end{equation}
+We start with the scalar contribution, \ie the PT1 energy
+\begin{equation}
+  \dv{E_0^{(1)}(s)}{s} = \mel{\phi}{\comm{\boldsymbol{\eta}_1^{(1)}(s)}{\bH_1^{(0)}(s)}}{\phi} + \mel{\phi}{\comm{\boldsymbol{\eta}_2^{(1)}(s)}{\bH_2^{(0)}(s)}}{\phi}
+\end{equation}
+where the second term is equal to zero.
+Thus we have
+\begin{align}
+  \label{eq:diffEqScalPT1}
+  \dv{E_0^{(1)}(s)}{s} &= \sum_{ip}\eta_i^{p,(1)}f_p^i(0) - \eta_p^{i,(1)}f_i^p(0) \\
+                       &= \sum_i \epsilon_i(\eta_i^{i,(1)} - \eta_i^{i,(1)}) \notag \\
+                       &= 0 \notag
+\end{align}
+Using the exact same reasoning as above we can show that there is only of the four terms of the one-body part of the commutator that is non-zero.
+\begin{align}
+  \dv{f_a^{i,(1)}(s)}{s} &= \comm{\boldsymbol{\eta}_1^{(1)}(s)}{\bH_1^{(0)}(s)}_a^i \\
+                         &= \sum_r \eta_a^{r,(1)}(s)f_r^i(0) - \eta_r^{i,(1)}f_a^r(0)  \notag \\
+                         &= (\epsilon_i - \epsilon_a) \eta_a^{i,(1)}(s) \notag \\
+  &= -(\epsilon_i - \epsilon_a) ^2f_a^{i,(1)}(s)  \notag
+\end{align}
+The derivation for the two-body part to first order is once again similar
+\begin{align}
+  \dv{v_{ab}^{ij,(1)}(s)}{s} &= \comm{\boldsymbol{\eta}_2^{(1)}(s)}{\bH_1^{(0)}(s)}_{ab}^{ij} \\
+                             &= -(\epsilon_i + \epsilon_j - \epsilon_a - \epsilon_b) ^2v_{ab}^{ij,(1)}(s)  \notag
+\end{align}
+These three differential equations can be integrated to obtain the analytical form of the Hamiltonian coefficients up to first order of perturbation theory.
+\begin{align}
+  E_0^{(1)}(s) &= E_0^{(1)}(0) = 0 \\
+  f_a^{i,(1)}(s) &= f_a^{i,(1)}(0) e^{-s (\Delta_a^i )^2} = 0 \\
+  v_{ab}^{ij,(1)}(s) &= v_{ab}^{ij,(1)}(0) e^{-s (\Delta_{ab}^{ij})^2} = \aeri{ij}{ab} e^{-s (\Delta_{ab}^{ij})^2}
+\end{align}
 
 %%%%%%%%%%%%%%%%%%%%%%
-\section{The unfolded GW Hamiltonian}
+\subsection{Second order Hamiltonian}
 %%%%%%%%%%%%%%%%%%%%%%
 
+To compute the second order contribution to the Hamiltonian coefficients, we first need to compute the second order contribution to $\boldsymbol{\eta}(s)$.
+\begin{equation}
+  \boldsymbol{\eta}^{(2)}(s) = \comm{\bH^{\text{d},(0)}(0)}{\bH^{\text{od},(2)}(s)} + \comm{\bH^{\text{d},(1)}(s)}{\bH^{\text{od},(1)}(s)}
+\end{equation}
+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.
+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).
+So we have
+\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 \\
+  &= (\epsilon_a - \epsilon_i)f_a^{i,(2)}(s) + \dots
+\end{align}
+Need to continue this derivation but this not needed for EPT2.
+
+
+Now turning to the differential equations, we start by computing the scalar part of Eq.~\eqref{eq:flowEquation}, \ie the differential equation for the second order energy.
+\begin{align}
+  \dv{E_0^{(2)}(s)}{s} & = \mel{\phi}{\comm{\boldsymbol{\eta}_1^{(2)}(s)}{\bH_1^{(0)}(s)}}{\phi} + \mel{\phi}{\comm{\boldsymbol{\eta}_2^{(2)}(s)}{\bH_2^{(0)}(s)}}{\phi} \\
+  &+ \mel{\phi}{\comm{\boldsymbol{\eta}_1^{(1)}(s)}{\bH_1^{(1)}(s)}}{\phi} + \mel{\phi}{\comm{\boldsymbol{\eta}_2^{(1)}(s)}{\bH_2^{(1)}(s)}}{\phi}
+\end{align}
+The two first terms are equal to zero for the same reason as the PT1 scalar differential equation (see Eq.~\eqref{eq:diffEqScalPT1}).
+In addition, the one-body hamiltonian has no first order contribution so 
+\begin{align}
+  \dv{E_0^{(2)}(s)}{s} &=  \mel{\phi}{\comm{\boldsymbol{\eta}_2^{(1)}(s)}{\bH_2^{(1)}(s)}}{\phi} \\
+                       &= \frac{1}{4} \sum_{i j} \sum_{a b}\left(\eta_{i j}^{ab,(1)} H_{ab}^{i j,(1)} - H_{i j}^{ab,(1)} \eta_{ab}^{i j,(1)}\right) \\
+                       &=\frac{1}{4} \sum_{i j} \sum_{a b}\left(\eta_{i j}^{ab,(1)}  - \eta_{ab}^{i j,(1)}\right) v_{ab}^{ij,(1)} \\
+                       &= \frac{1}{4} \sum_{i j} \sum_{a b}\left(\Delta_{ab}^{ij}v_{ij}^{ab,(1)} - (-\Delta_{ab}^{ij} v_{ab}^{ij,(1)}) \right) \aeri{ij}{ab} e^{-s (\Delta_{ab}^{ij})^2} \\
+                       &= \frac{1}{2} \sum_{i j} \sum_{a b} \Delta_{ab}^{ij} (v_{ab}^{ij,(1)})^2 \\
+                       &= \frac{1}{2} \sum_{i j} \sum_{a b} \Delta_{ab}^{ij} \aeri{ij}{ab}^2 e^{-2s (\Delta_{ab}^{ij})^2}
+\end{align}
+After integration, using the initial condition $E_0^{(2)}(0)=0$, we obtain
+\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)
+\end{equation}
+
+%=================================================================%
+\section{The unfolded GW Hamiltonian}
+%=================================================================%
+
 \appendix
 
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 \section{Matrix elements of $C=[A, B]_{1,2}$}
-%%%%%%%%%%%%%%%%%%%%%%
+%=================================================================%
 
 An operator $A$ containing at most two-body terms may be written in normal ordered form with respect to the reference $\Phi$ as