 Solution

The terminal term $\mathcal{A}$ is the objective reward. With the Hilbert-Schmidt inner product $⟨⟨A|\rho ⟩⟩=\mathrm{Tr}(A^{†}\rho )$, a Hermitian target observable gives its final expected value. The vector ${\rho }_v$ is the density operator represented as a vector in operator space; it retains the same physical content as the density matrix, not a wavefunction of the system.

The term $\mathcal{D}$ imposes the dynamical equation by a time-dependent costate $A_v$. It is a Lagrange-multiplier operator in the same operator space, propagated backward from a terminal condition. It is not required to be a positive, trace-one density matrix. DefineThen the enforced state equation is ${\dot{\rho }}_v=K[f]{\rho }_v$. The uncontrolled coherent part is ${\mathcal{L}}_0$, the dissipative contribution is ${\mathcal{L}}_D$, and each $f_m{\mathcal{L}}_m$ describes a control coupling. For Quantum Hamiltonian control, ${\mathcal{L}}_m\rho =[H_m,\rho ]$.

The term $\mathcal{C}$ is a quadratic field-energy or fluence penalty. For the positive scale $p_0$ and positive weights ${\lambda }_m$, larger ${\lambda }_m$ make a given control amplitude more costly. Thus maximizing $J$ trades terminal performance against control fluence while enforcing the dynamics. The multiplier term vanishes on any dynamically feasible trajectory. The weights do not represent decoherence rates; they specify control-resource preferences.