C081 derivation: beta = J optimizes R_t

Physical argument

The collective coupling $\hbar J S_{x,e}^2$ shifts the spin-manifold energies (an AC-Stark-like shift) so that drive (C), nominally resonant, is effectively detuned. The free parameter $\beta$ in $\hbar\beta(\ket{\uparrow}\bra{\uparrow}\otimes \mathbf1+\mathbf1\otimes\ket{\uparrow}\bra{\uparrow})$ detunes the qubit and can be used to put all drives back on resonance. The paper states $\beta=J$ achieves this and is numerically optimal.

Numerical scan (run.py)

At the paper optimum $(\Omega_C,\kappa,\gamma)=(1.95J,2.58J,0.29\pi)$: - coarse scan $\beta/J\in[0,3]$: argmax at $\beta/J=1.000$, $R_t=0.11424\,J$; - fine scan $\beta/J\in[0.7,1.3]$: argmax at $\beta/J=1.0000$; - $R_t(\beta=0)=0.060\,J < R_t(\beta=J)=0.114\,J$ (the on-resonance choice nearly doubles the gap); - two other operating points (e.g. $\Omega_C=6J$) also peak at $\beta/J=1.000$.

The steady state at $\beta=J$ is the singlet with fidelity $1.000000$.

Conclusion

$\beta=J$ is numerically confirmed to maximise the Liouvillian gap $R_t$, with a clear physical interpretation as AC-Stark-shift compensation. Result rests on the reimplemented continuous model (no paper code), so capped at partial with limitation paper_text_only_reimplementation.