 Solution

Take the driving and feedback strengths to be real, so the feedback Quantum Hamiltonian is Hermitian. The lowering operator satisfies ${\sigma }^2=({\sigma }^{†}{)}^2=0$. Using the supplied convention for ${\sigma }_y$,The additional Quantum Hamiltonian term is consequentlyWith no drift Quantum Hamiltonian, the total Quantum Hamiltonian is just $\alpha {\sigma }_y$. The feedback-modified Lindblad operator isSubstitution into the measurement-feedback equation givesThus 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.