# Polarization Propagator Derivation of TDHF/RPA

Following immediately where we left off last time, when ${A}$ and ${B}$ are one-electron number conserving, e.g. creation/annihilation operators are in pairs, we have

which allow us to define the general polarization propagator. If the elements did not conserve the number of electrons, we could use the same formalism to get the electron propagator, which can describe the attachment/detachment of electrons from a system, thus describing processes such as ionization. In general, it is sufficient to consider the typical elements ${A \rightarrow a_r^{\dagger}a_s}$ and ${B \rightarrow a_t^{\dagger}a_u}$. If spin-orbitals of reference are ${\psi_i, \psi_j}$,…, then a particle-hole basis includes only excitation and de-excitation operators ${a_a^{\dagger}a_i}$ and ${a_i^{\dagger}a_a}$, respectively. Thus the basis and its dual:
$\mathbf{n} = (\mathbf{e} \quad \mathbf{d}), \qquad \mathbf{n}^{\dagger} = (\mathbf{e} \quad \mathbf{d})^{\dagger} \ \ \ \ \$
$e_{ia} = a_a^{\dagger}a_i = E_{ia}, \qquad d_{ai} = a_i^{\dagger}a_a = E_{ia}^{\dagger} \ \ \ \ \ (29)$

This gives, for our resolvent,
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As an example, we give the elements of the first block:
$E^{\dagger}_{ia}(\hbar\omega\hat{1} - \hat{H})E_{jb} = \hbar\omega E^{\dagger}_{ia} E_{jb} - E^{\dagger}_{ia} \left[H,E_{jb}\right] \ \ \ \ \ (31)$

with scalar products
$E^{\dagger}_{ia} E_{jb} = \langle \psi_0 \vert [E^{\dagger}_{ia},E_{jb}]\vert \psi_0 \rangle, \qquad E^{\dagger}_{ia} \left[H,E_{jb}\right] = \langle \psi_0 \vert [E^{\dagger}_{ia},[H,E_{jb}]]\vert \psi_0 \rangle \ \ \ \ \ (32)$

Evaluating these products gives
$E^{\dagger}_{ia} E_{jb} = \langle \psi_0 \vert [E^{\dagger}_{ia},E_{jb}]\vert \psi_0 \rangle = \langle \psi_0 \vert E^{\dagger}_{ia}E_{jb}\vert \psi_0 \rangle - \langle \psi_0 \vert E_{jb}E^{\dagger}_{ia}\vert \psi_0 \rangle = \langle \psi_0 \vert E^{\dagger}_{ia}E_{jb}\vert \psi_0 \rangle = \delta_{ia,jb} \ \ \ \ \ (33)$

and

$E^{\dagger}_{ia} \left[H,E_{jb}\right] = \langle \psi_0 \vert [E^{\dagger}_{ia},[H,E_{jb}]]\vert \psi_0 \rangle \ \ \ \ \$
$E^{\dagger}_{ia} \left[H,E_{jb}\right] = \langle \psi_0 \vert E^{\dagger}_{ia} H E_{jb}\vert \psi_0 \rangle - \langle \psi_0 \vert E^{\dagger}_{ia}E_{jb} H\vert \psi_0 \rangle - \langle \psi_0 \vert H E_{jb} E^{\dagger}_{ia}\vert \psi_0 \rangle + \langle \psi_0 \vert E_{jb} H E^{\dagger}_{ia} \vert \psi_0 \rangle \ \ \ \ \$
$E^{\dagger}_{ia} \left[H,E_{jb}\right] = \langle \psi^a_i \vert H \vert \psi_j^b \rangle - \delta_{ia}\delta_{jb}\langle \psi_0 \vert H \vert \psi_0 \rangle \ \ \ \ \$
$E^{\dagger}_{ia} \left[H,E_{jb}\right] = E_0\delta_{ij}\delta_{ab} + F_{ab}\delta_{ij} - F_{ij}\delta_{ab} + \langle aj \vert \vert ib \rangle - \delta_{ia}\delta{jb}E_0 \ \ \ \ \$
$E^{\dagger}_{ia} \left[H,E_{jb}\right] = (\epsilon_a - \epsilon_i)\delta_{ia}\delta_{ab} + \langle aj \vert \vert ib \rangle = \mathbf{A} \ \ \ \ \$
Which is precisely the same elements as we have derived earlier for TDHF. Completing the rest, we have:
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With ${\mathbf{B} = \langle ab \vert \vert ij \rangle}$. Now, we saw from the definition of the resolvent that ${R(\omega) \rightarrow \infty}$ at ${\omega \rightarrow \omega_{0n}}$, which are the poles of ${R(\omega)}$. Therefore, ${R(\omega)^{-1} \rightarrow 0}$ and
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which are the RPA/TDHF equations. The eigencolumns determine the linear combinations of excitation and de-excitation operators that produce corresponding approximate excited states when working on the reference state ${\vert \psi_0 \rangle}$.