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almost finsihed deriving analytical expression for third order coupling elements (not really elegant tbh...)

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Antoine Marie 2022-11-23 17:01:03 +01:00
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Notes/Notes.tex
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@ -705,6 +705,31 @@ Therefore the differential equations can be simplified to
\end{align}
and can be solved by integration of the right side using the previous analytic formula for Hamiltonian elements.
\begin{align}
\text{Part B}: &(\dv{\bW^{(3)}}{s})_{pR} = - \left[\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)\bW^{(1)}\right]_{pR} \notag \\
&= - \sum_q\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)_{pq}W^{(1)}_{qR} \notag \\
&= - \sum_q(\epsilon_p+\epsilon_q)F_{pq}^{(2)}W^{(1)}_{qR} \notag \\
&= - \sum_{qS} (\epsilon_p+\epsilon_q) \frac{\Delta_{pS}+\Delta_{qS}}{\Delta_{pS}^2+\Delta_{qS}^2}W^{(1)}_{pS}(0)W^{(1)}_{qS}(0) W^{(1)}_{qR}(0) \notag \\
&\times (1 - e^{-(\Delta_{pS}^2+\Delta_{qS}^2)s})e^{-\Delta_{qR}^2s} \notag \\
\text{Part B}: & (\bW^{(3)}(s))_{pR} = - \sum_{qS} (\epsilon_p+\epsilon_q) \frac{\Delta_{pS}+\Delta_{qS}}{\Delta_{pS}^2+\Delta_{qS}^2}W^{(1)}_{pS}(0)W^{(1)}_{qS}(0) W^{(1)}_{qR}(0) \notag \\
&\times \left(-\frac{e^{-\Delta_{qR}^2s} }{\Delta_{qR}^2} + \frac{e^{-(\Delta_{pS}^2+\Delta_{qS}^2+\Delta_{qR}^2)s}}{\Delta_{pS}^2+\Delta_{qS}^2+\Delta_{qR}^2}\right)
\end{align}
\begin{align}
\text{Part C}: &(\dv{\bW^{(3)}}{s})_{pR} = - \left[\bW^{(1)}\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right)\right]_{pR} \notag \\
&= - \sum_S W^{(1)}_{pS} \left( \bD^{(0)}\bD^{(2)} + \bD^{(2)}\bD^{(0)} \right)_{SR}\notag \\
&= - \sum_SW^{(1)}_{pS}(D_R+D_S)D_{SR}^{(2)}\notag \\
&= - \sum_{Sq}W^{(1)}_{pS}(0)e^{-\Delta_{pS}^2s} (D_R+D_S)\frac{-\Delta_{qS}-\Delta_{qR}}{\Delta_{qS}^2+\Delta_{qR}^2} \notag \\
&\times W^{(1)}_{qS}(0)W^{(1)}_{qR}(0)(1-e^{-(\Delta_{qS}^2+\Delta_{qR}^2)s})) \notag \\
\text{Part C}: & (\bW^{(3)}(s))_{pR} = + \sum_{Sq} (D_R+D_S) \frac{\Delta_{qS}+\Delta_{qR}}{\Delta_{qS}^2+\Delta_{qR}^2} W^{(1)}_{pS}(0)W^{(1)}_{qS}(0) W^{(1)}_{qR}(0) \notag \\
&\times \left(-\frac{e^{-\Delta_{pS}^2s} }{\Delta_{pS}^2} + \frac{e^{-(\Delta_{pS}^2+\Delta_{qS}^2+\Delta_{qR}^2)s}}{\Delta_{pS}^2+\Delta_{qS}^2+\Delta_{qR}^2}\right) \notag
\end{align}
\begin{align}
\text{Part A}: &(\dv{\bW^{(3)}}{s})_{pR} = (2 \bF{}{(0)}\bW^{(1)}\bD^{(2)} + 2 \bF{}{(2)}\bW^{(1)}\bD^{(0)})_{pR} \notag \\
&= 2\sum_{qS} \epsilon_p\delta_{pq} W_{qS}^{(1)}D_{SR}^{(2)} + F_{pq}^{(2)}W_{qS}^{(1)}D_S\delta_{SR}
\end{align}
%///////////////////////////%
\subsubsection{Forth order Hamiltonian elements}
%///////////////////////////%