fix absolute value of coupling constant

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Valentin Boettcher 2023-06-20 11:23:01 -04:00
parent 383da7c139
commit 5e66bfafa0

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@ -482,13 +482,13 @@ constant. With these considerations in mind we can simplify
\cref{eq:32} to \cref{eq:32} to
\begin{equation} \begin{equation}
\label{eq:64} \label{eq:64}
η_{m}=κ\frac{πn_{B}}{c}_{σ=\pm,β,β'}U_{βm}U^\ast_{β'm}δ_{\sgn(β),σ} δ_{\sgn(β'),σ} η_{m}=\abs{κ}\frac{πn_{B}}{c}_{σ=\pm,β,β'}U_{βm}U^\ast_{β'm}δ_{\sgn(β),σ} δ_{\sgn(β'),σ}
\end{equation} \end{equation}
and and
\begin{gather} \begin{gather}
\label{eq:34} \label{eq:34}
\dot{d}_{γ} = \dot{d}_{γ} =
-\iu\bqty{\pqty{ω_{γ}-\iu \tilde{η}_{γ}}d_{γ} + \sqrt{κ} ∑_{σ=\pm} -\iu\bqty{\pqty{ω_{γ}-\iu \tilde{η}_{γ}}d_{γ} + \sqrt{κ^\ast} ∑_{σ=\pm}
U^{σ}_{γ}(t) \frac{b_{\inputf}(t)}{\sqrt{ω_{0}}}}\\ U^{σ}_{γ}(t) \frac{b_{\inputf}(t)}{\sqrt{ω_{0}}}}\\
U^{σ}_{γ}(t) = ∑_{m,β} δ_{\sgn({β}),σ}U^\ast_{βm}\pqty{O^{-1}(t)}_{γm} \eu^{\iu ω_{m}^{0}t}= ∑_{m,β} δ_{\sgn({β}),σ}U^\ast_{βm}\pqty{O^{-1}}_{γm}\eu^{\iu_{m}^{0}_{m})t}. U^{σ}_{γ}(t) = ∑_{m,β} δ_{\sgn({β}),σ}U^\ast_{βm}\pqty{O^{-1}(t)}_{γm} \eu^{\iu ω_{m}^{0}t}= ∑_{m,β} δ_{\sgn({β}),σ}U^\ast_{βm}\pqty{O^{-1}}_{γm}\eu^{\iu_{m}^{0}_{m})t}.
\end{gather} \end{gather}
@ -506,7 +506,7 @@ We can now proceed to integrate \cref{eq:34} to obtain
with with
\begin{equation} \begin{equation}
\label{eq:37} \label{eq:37}
χ_{γ}(t) = κ \eu^{-\pqty{\iu ω_{γ} + \tilde{η}_{γ}}t}. χ_{γ}(t) = \abs{κ} \eu^{-\pqty{\iu ω_{γ} + \tilde{η}_{γ}}t}.
\end{equation} \end{equation}
When constructing the total output field, we have to remember how the When constructing the total output field, we have to remember how the