diagH_after_exc Interface

public interface diagH_after_exc

Evaluate the energy of a new determinant $D_j$ quickly.

The calculation of a diagonal term of the hamiltonian, scales quadratically with the number of particles $\mathcal{O}(N_{e}^2)$. Often we start from a determinant $D_i$, where we know the diagonal term, and excite to a new determinant $D_j$. Under this circumstance we can calculate the diagonal element of $D_j$ in $\mathcal{O}(N_{e})$ time.

In the following we will derive the necessary equations. We assume the notations and conventions of the “purple book” (Helgaker et al). The diagonal term for a determinant is given as $$\begin{equation*} \langle D_i | \hat{H} | D_i \rangle = \sum_{I \in D_i} h_{II} + \frac{1}{2} \sum_{I \in D_i} \sum_{J \in D_i} (g_{IIJJ} - g_{IJJI}) \end{equation*}$$

We want to calculate $\langle D_j | \hat{H} | D_j \rangle - \langle D_i | \hat{H} | D_i \rangle$. Which we do by separately calculating the difference for the one- and two-electron term.

We can rewrite the one-electron term as: $$\begin{equation*} \sum_{I \in D_i} h_{II} = \sum_{I \in D_i \cap D_j} h_{II} + \sum_{I \in D_i \setminus D_j} h_{II} \end{equation*}$$ Which gives $$\begin{equation*} \sum_{J \in D_j} h_{JJ} - \sum_{I \in D_i} h_{II} = \sum_{J \in D_j \setminus D_i} h_{JJ} - \sum_{I \in D_i \setminus D_j} h_{II} = \sum_{J \in \texttt{tgt}} h_{JJ} - \sum_{I \in \texttt{src}} h_{II} \end{equation*}$$

For the two electron term we define $$\begin{equation*} \gamma_{IJ} = (g_{IIJJ} - g_{IJJI}) \end{equation*}$$ and note the two properties $$\begin{equation*} \gamma_{IJ} = \gamma_{JI} \end{equation*}$$ $$\begin{equation*} \gamma_{II} = 0 \end{equation*}$$

We write $$\begin{equation*} \frac{1}{2} \sum_{I \in D_i} \sum_{J \in D_i} \gamma_{IJ} = \frac{1}{2} \left[ \sum_{I \in D_i \cap D_j} \sum_{J \in D_i \cap D_j} \gamma_{IJ} + \sum_{I \in D_i \cap D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} + \sum_{I \in D_i \setminus D_j} \sum_{J \in D_i \cap D_j} \gamma_{IJ} + \sum_{I \in D_i \setminus D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \right] \end{equation*}$$ $$\begin{equation*} = \frac{1}{2} \left[ \sum_{I \in D_i \cap D_j} \sum_{J \in D_i \cap D_j} \gamma_{IJ} + 2 \Big( \sum_{I \in D_i \cap D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \Big) + \sum_{I \in D_i \setminus D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \right] \end{equation*}$$ In the last equality we used $\gamma_{IJ} = \gamma_{JI}$.

For the difference we get: $$\begin{equation*} \frac{1}{2} \sum_{I \in D_j} \sum_{J \in D_j} \gamma_{IJ} - \frac{1}{2} \sum_{I \in D_i} \sum_{J \in D_i} \gamma_{IJ} = \frac{1}{2} \left[ 2 \Big( \sum_{I \in D_i \cap D_j} \sum_{J \in D_j \setminus D_i} \gamma_{IJ} \Big) + \sum_{I \in D_j \setminus D_i} \sum_{J \in D_j \setminus D_i} \gamma_{IJ} - 2 \Big( \sum_{I \in D_i \cap D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \Big) - \sum_{I \in D_i \setminus D_j} \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \right] \end{equation*}$$ $$\begin{equation*} = \sum_{I \in D_i \cap D_j} \Big( \sum_{J \in D_j \setminus D_i} \gamma_{IJ} - \sum_{J \in D_i \setminus D_j} \gamma_{IJ} \Big) + \sum_{I \in D_j \setminus D_i} \sum_{J \in D_j \setminus D_i, I < J} \gamma_{IJ} - \sum_{I \in D_i \setminus D_j} \sum_{J \in D_i \setminus D_j, I < J} \gamma_{IJ} \end{equation*}$$ $$\begin{equation*} = \sum_{I \in D_i \cap D_j} \Big( \sum_{J \in \texttt{tgt}} \gamma_{IJ} - \sum_{J \in \texttt{src}} \gamma_{IJ} \Big) + \sum_{I \in \texttt{tgt}} \sum_{J \in \texttt{tgt}, I < J} \gamma_{IJ} - \sum_{I \in \texttt{src}} \sum_{J \in \texttt{src}, I < J} \gamma_{IJ} \end{equation*}$$

In total we obtain $$\begin{equation*} \langle D_j | \hat{H} | D_j \rangle - \langle D_i | \hat{H} | D_i \rangle = \sum_{J \in \texttt{tgt}} h_{JJ} - \sum_{I \in \texttt{src}} h_{II} + \sum_{I \in D_i \cap D_j} \Big( \sum_{J \in \texttt{tgt}} \gamma_{IJ} - \sum_{J \in \texttt{src}} \gamma_{IJ} \Big) + \sum_{I \in \texttt{tgt}} \sum_{J \in \texttt{tgt}, I < J} \gamma_{IJ} - \sum_{I \in \texttt{src}} \sum_{J \in \texttt{src}, I < J} \gamma_{IJ} \end{equation*}$$

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