Solving the CCD equations
One way to solve these equations, is to write equation
(3) as a series of iterative nonlinear algebraic equations
$$
\begin{align}
t_{ij}^{ab}{}^{(n+1)} = \frac{1}{\epsilon^{ab}_{ij}} \bigg(\langle
ab \vert \hat{v} \vert ij \rangle & \nonumber
\\ +\frac{1}{2}\sum_{cd} \langle ab \vert \hat{v} \vert cd \rangle
t_{ij}^{cd}{}^{(n)}+\frac{1}{2}\sum_{kl} \langle kl \vert \hat{v}
\vert ij \rangle t_{kl}^{ab}{}^{(n)}+\hat{P}(ij\vert ab)\sum_{kc}
\langle kb \vert \hat{v} \vert cj \rangle t_{ik}^{ac}{}^{(n)} &
\nonumber \\ +\frac{1}{4}\sum_{klcd} \langle kl \vert \hat{v} \vert
cd \rangle
t_{ij}^{cd}{}^{(n)}t_{kl}^{ab}{}^{(n)}+\hat{P}(ij)\sum_{klcd}
\langle kl \vert \hat{v} \vert cd \rangle
t_{ik}^{ac}{}^{(n)}t_{jl}^{bd}{}^{(n)}& \nonumber
\\ -\frac{1}{2}\hat{P}(ij)\sum_{klcd} \langle kl \vert \hat{v} \vert
cd \rangle
t_{ik}^{dc}{}^{(n)}t_{lj}^{ab}{}^{(n)}-\frac{1}{2}\hat{P}(ab)\sum_{klcd}
\langle kl \vert \hat{v} \vert cd \rangle
t_{lk}^{ac}{}^{(n)}t_{ij}^{db}{}^{(n)} \bigg),&
\tag{11}
\end{align}
$$
for all \( i < j \) and all \( a < b \), where \( \epsilon^{ab}_{ij} =\left(\epsilon_i+\epsilon_j-\epsilon_a-\epsilon_b\right) \), and
\( t_{ij}^{ab}{}^{(n)} \) is the \( t \) amplitude for the nth iteration of
the series. This way, given some starting guess
\( t_{ij}^{ab}{}^{(0)} \), we can generate subsequent \( t \) amplitudes
that converges to some value.