= Solution
Take the driving and feedback strengths to be real, so the feedback <Quantum Hamiltonian> is Hermitian. The lowering operator satisfies $\sigma^2=(\sigma^\dagger)^2=0$. Using the supplied convention for $\sigma_y$,
$$
\sigma^\dagger\sigma_y=i\sigma^\dagger\sigma,\qquad
\sigma_y\sigma=-i\sigma^\dagger\sigma.
$$
The additional <Quantum Hamiltonian> term is consequently
$$
\tfrac12(M^\dagger F+FM)
=\tfrac\lambda2(\sigma^\dagger\sigma_y+\sigma_y\sigma)=0.
$$
With no drift <Quantum Hamiltonian>, the total <Quantum Hamiltonian> is just $\alpha\sigma_y$. The feedback-modified <Lindblad operator> is
$$
K=\sigma-i\lambda\sigma_y=(1+\lambda)\sigma-\lambda\sigma^\dagger
=\begin{pmatrix}0&1+\lambda\\-\lambda&0\end{pmatrix}.
$$
Substitution into the measurement-feedback equation gives
$$
\boxed{\dot\rho=-i[\alpha\sigma_y,\rho]+\mathcal D[K]\rho.}
$$
Thus the feedback affects the <Lindblad dissipator> even though its extra coherent <Quantum Hamiltonian> contribution cancels. This is a concrete mechanism of <Markovian reservoir engineering>, rather than merely a classical control force added to an unchanged dissipative equation.
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