Question

Let us compare a system of $ N $-independent electrons with pseudo-independent electrons described by the mean-field Hartree-Fock model. Fill the table below. Use Koopmans theorem when appropriate. Define any notations you use, i.e.: \[ \begin{array}{r} h_{i} \equiv h\left(x_{i}\right)=-\frac{1}{2} \nabla_{i}^{2}-\sum_{A} \frac{Z_{A}}{\left|R_{a}-r_{i}\right|} \\ v_{i j} \equiv v\left(x_{i}, x_{j}\right)=\frac{1}{\left|r_{i}-r_{j}\right|} \\ \Phi_{k}=\mid \phi_{1} \phi_{2} \ldots \not \not_{K} \ldots \phi_{N}> \\ \Phi_{i}^{a}=\mid \phi_{1} \phi_{2} \ldots\left\{\not X\left\langle\phi_{a}\right\} \ldots \phi_{N}>\right. \end{array} \] \begin{tabular}{||l|l|l||} \hline & Independent electrons & Pseudo-independent electrons \\ \hline Hamiltonian & $ H=\sum_{i=1}^{N} \hat{h}_{i} $ & $ H=\sum_{i=1}^{N} \hat{h}_{i}+\frac{1}{2} \sum_{i j=1}^{N} v_{i j} $ \\ \hline Wfn $ \left|\Phi_{0}\right\rangle= $ & $ \left|\phi_{1} \phi_{2} \ldots \phi_{N}\right\rangle $ & $ \left|\phi_{1} \phi_{2} \ldots \phi_{N}\right\rangle $ \\ \hline$ \phi_{i} $ are solutions of: & & \\ \hline Total energy is: & & \\ \hline \begin{tabular}{l} Energy required to remove \\ electron from $ \phi_{k} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy for attaching \\ electron to $ \phi_{a} \notin \Phi_{0} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy difference between \\ $ \Phi_{k} $ and $ \Phi_{j} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy difference between \\ $ \Phi_{0} $ and $ \Phi_{i}^{a} $ \end{tabular} & & \\ \hline \end{tabular}

          Let us compare a system of \( N \)-independent electrons with pseudo-independent electrons described by the mean-field Hartree-Fock model. Fill the table below. Use Koopmans theorem when appropriate. Define any notations you use, i.e.:
\[
\begin{array}{r}
h_{i} \equiv h\left(x_{i}\right)=-\frac{1}{2} \nabla_{i}^{2}-\sum_{A} \frac{Z_{A}}{\left|R_{a}-r_{i}\right|} \\
v_{i j} \equiv v\left(x_{i}, x_{j}\right)=\frac{1}{\left|r_{i}-r_{j}\right|} \\
\Phi_{k}=\mid \phi_{1} \phi_{2} \ldots \not \not_{K} \ldots \phi_{N}> \\
\Phi_{i}^{a}=\mid \phi_{1} \phi_{2} \ldots\left\{\not X\left\langle\phi_{a}\right\} \ldots \phi_{N}>\right.
\end{array}
\]
\begin{tabular}{||l|l|l||}
\hline & Independent electrons & Pseudo-independent electrons \\
\hline Hamiltonian & \( H=\sum_{i=1}^{N} \hat{h}_{i} \) & \( H=\sum_{i=1}^{N} \hat{h}_{i}+\frac{1}{2} \sum_{i j=1}^{N} v_{i j} \) \\
\hline Wfn \( \left|\Phi_{0}\right\rangle= \) & \( \left|\phi_{1} \phi_{2} \ldots \phi_{N}\right\rangle \) & \( \left|\phi_{1} \phi_{2} \ldots \phi_{N}\right\rangle \) \\
\hline\( \phi_{i} \) are solutions of: & & \\
\hline Total energy is: & & \\
\hline \begin{tabular}{l} 
Energy required to remove \\
electron from \( \phi_{k} \)
\end{tabular} & & \\
\hline \begin{tabular}{l} 
Energy for attaching \\
electron to \( \phi_{a} \notin \Phi_{0} \)
\end{tabular} & & \\
\hline \begin{tabular}{l} 
Energy difference between \\
\( \Phi_{k} \) and \( \Phi_{j} \)
\end{tabular} & & \\
\hline \begin{tabular}{l} 
Energy difference between \\
\( \Phi_{0} \) and \( \Phi_{i}^{a} \)
\end{tabular} & & \\
\hline
\end{tabular}

$Let us compare a system of $ N $-independent electrons with pseudo-independent electrons described by the mean-field Hartree-Fock model. Fill the table below. Use Koopmans theorem when appropriate. Define any notations you use, i.e.: \[ \begin{array}{r} h{i} \equiv h\left(x{i}\right)=-\frac{1}{2} \nabla{i}^{2}-\sum{A} \frac{Z{A}}{\left|R{a}-r{i}\right|} \\ v{i j} \equiv v\left(x{i}, x{j}\right)=\frac{1}{\left|r{i}-r{j}\right|} \\ \Phi{k}=\mid \phi{1} \phi{2} \ldots \not \not{K} \ldots \phi{N}> \\ \Phi{i}^{a}=\mid \phi{1} \phi{2} \ldots\left\{\not X\left\langle\phi{a}\right\} \ldots \phi{N}>\right. \end{array} \] \begin{tabular}{||l|l|l||} \hline & Independent electrons & Pseudo-independent electrons \\ \hline Hamiltonian & $ H=\sum{i=1}^{N} \hat{h}{i} $ & $ H=\sum{i=1}^{N} \hat{h}{i}+\frac{1}{2} \sum{i j=1}^{N} v{i j} $ \\ \hline Wfn $ \left|\Phi{0}\right\rangle= $ & $ \left|\phi{1} \phi{2} \ldots \phi{N}\right\rangle $ & $ \left|\phi{1} \phi{2} \ldots \phi{N}\right\rangle $ \\ \hline$ \phi{i} $ are solutions of: & & \\ \hline Total energy is: & & \\ \hline \begin{tabular}{l} Energy required to remove \\ electron from $ \phi{k} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy for attaching \\ electron to $ \phi{a} \notin \Phi{0} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy difference between \\ $ \Phi{k} $ and $ \Phi{j} $ \end{tabular} & & \\ \hline \begin{tabular}{l} Energy difference between \\ $ \Phi{0} $ and $ \Phi{i}^{a} $ \end{tabular} & & \\ \hline \end{tabular}$

Added by Keerthy P.

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