# Equation of Motion (EOM) derivation of RPA

The Equation of Motion derivations of excited state and response properties are elegant, and (in my opinion) very direct. Here I’ll give an example of how it can be used to derive TDHF/RPA. We’ve already done it two other ways, and the agreement between them is important if we wish to extend these methods further.

Given an exact ground state, we can say that

Define operator ${Q_n^{\dagger}}$ and ${Q_n}$:
$\vert n \rangle = Q_n^{\dagger} \vert 0 \rangle, \qquad Q_n\vert 0 \rangle = 0 \implies Q_n^{\dagger} = \vert n \rangle \langle 0 \vert \ \ \ \ \$

These operators generate excited states from the ground state (not excited determinants, as in the case of post-HF correlation methods). So it is clear that, when acting on an exact ground state:

Multiply on left by arbitrary state of form ${\langle 0 \vert \delta Q}$, giving
$\langle 0 \vert [\delta Q, [H, Q_n^{\dagger}]]\vert 0 \rangle = \hbar\omega_{0n} \langle 0 \vert [\delta Q, Q_n^{\dagger}] \vert 0 \rangle \ \ \ \ \$

Where we have made use of the fact that ${\langle 0 \vert Q_n^{\dagger} = \langle 0 \vert HQ_n^{\dagger} = 0}$. Note that is we express ${Q}$ by particle-hole operators ${a_p^{\dagger}a_q}$, ${a_p^{\dagger}a_q^{\dagger}a_r a_s}$ with coefficients ${C_{pq}}$ and ${C_{pqrs}}$, then ${\delta Q}$ is given by ${\frac{\partial Q}{\partial C} \delta C}$ for arbitrary variations ${\delta C}$. These are in principle exact, since ${\delta Q \vert 0 \rangle}$ exhausts the whole Hilbert space, such that the above equation corresponds to the full Schrödinger equation. Tamm-Dancoff (or Configuration Interaction Singles) can be obtained by approximating ${\vert 0 \rangle \rightarrow \vert \hbox{HF} \rangle}$ and the operator ${Q_n^{\dagger} = \sum\limits_{ia} C_{ia} a_a^{\dagger}a_i}$, restricting ourselves to 1p-1h excitations. Thus ${\delta Q \vert 0 \rangle = \sum_{ia} a_a^{\dagger}a_i \vert \hbox{HF} \rangle \delta C_{ai}}$, (${\delta C_{ai}}$ cancels), and
$\sum\limits_{bj} \langle \hbox{HF} \vert [a_i^{\dagger}a_a, [H, a_b^{\dagger}a_j]]\vert \hbox{HF} \rangle C_{jb} = \hbar\omega \langle \hbox{HF} \vert [a_i^{\dagger}a_a, a_a^{\dagger}a_i] \vert \hbox{HF} \rangle C_{ia} \ \ \ \ \$

These are the CIS equations. Put another way:
$\sum\limits_{bj} \left\{(\epsilon_a - \epsilon_i)\delta_{ab}\delta_{ij} + \langle aj \vert \vert ib \rangle \right\} C_{bj} = E^{CIS}C_{ai} \ \ \ \ \$

Similarly, for RPA/TDHF, if we consider a ground state containing 2p-2h correlations, we can not only create a p-h pair, but also destroy one. Thus (choosing the minus sign for convenience):
$Q_n^{\dagger} = \sum\limits_{ia} X_{ia} a_a^{\dagger}a_i - \sum\limits_{ia} Y_{ia} a_i^{\dagger}a_a, \qquad \hbox{and} Q_n \vert RPA \rangle = 0 \ \ \ \ \$

So instead of the basis of only single excitations, and therefore one matrix ${C_{ia}}$, we work in a basis of single excitations and single de-excitations, and have two matrices ${X_{ia}}$ and ${Y_{ia}}$. We also have two kinds of variations ${\delta Q \vert 0 \rangle}$, namely ${a_a^{\dagger}a_i \vert 0 \rangle}$ and ${a_i^{\dagger}a_a \vert 0 \rangle}$. This gives us two sets of equations:
$\langle \hbox{RPA} \vert [a_i^{\dagger}a_a, [H,Q_n^{\dagger}]]\vert \hbox{RPA} \rangle = \hbar \omega \langle \hbox{RPA} \vert [a_i^{\dagger}a_a, Q_n^{\dagger}] \vert \hbox{RPA} \rangle \notag \ \ \ \ \$

These contain only expectation values of our four Fermion operators, which cannot be calculated since we still do not know ${\vert \hbox{RPA} \rangle}$. Thus we assume ${\vert \hbox{RPA} \rangle \rightarrow \vert \hbox{HF} \rangle}$. This gives
$\langle \hbox{RPA} \vert [a_i^{\dagger}a_a, a_b^{\dagger}a_j] \vert \hbox{RPA} \rangle = \langle \hbox{HF} \vert [a_i^{\dagger}a_a, a_b^{\dagger}a_j] \vert \hbox{HF} \rangle = \delta_{ij}\delta_{ab} \ \ \ \ \$

The probability of finding states ${a_a^{\dagger}a_i \vert 0 \rangle}$ and ${a_i^{\dagger}a_a \vert 0 \rangle}$ in excited state ${\vert n\rangle}$, that is, the p-h and h-p matrix elements of transition density matrix ${\rho^{(1)}}$ are:
$\rho^{(1)}_{ai} = \langle 0 \vert a_i^{\dagger}a_a \vert n \rangle \simeq \langle \hbox{HF} \vert [a_i^{\dagger}a_a, Q_n^{\dagger}] \vert \hbox{HF} \rangle = X_{ia} \ \ \ \ \$

Thus altogether
$% $

with
$A_{ia,jb} = \langle \hbox{HF} \vert [a_i^{\dagger}a_a, [H,a_b^{\dagger}a_j]]\vert \hbox{HF} \rangle \ \ \ \ \$

and
$B_{ia,jb} = - \langle \hbox{HF} \vert [a_i^{\dagger}a_a, [H,a_j^{\dagger}a_b]]\vert \hbox{HF} \rangle \ \ \ \ \$

which are the TDHF/RPA equations.