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Gradient with respect to a function involving Schrodinger equation

Physics Asked by Junk Warrior on December 3, 2020

For a general $N$-level closed quantum system, the control objective $J$ associated with an observable $O$ at the final time $T/2$ can be defined by
begin{equation}
J=text{Tr}[U(frac{T}{2},-frac{T}{2})rho_0U^dagger(frac{T}{2},-frac{T}{2})O],
end{equation}

where $rho_0$ is the density operator at the initial time $-frac{T}{2}$. The evolution operator $U$ of the quantum system interacting with control fields $mathcal{E}(s,t)$ can be described by the time-dependent Schrodinger equation $partial_tU(t,-frac{T}{2})=-i[H_0-mumathcal{E}(t)]U(t,-frac{T}{2})$ subject to the initial condition $U(-frac{T}{2},-frac{T}{2})equivmathbb{I}$, where $H_0$ is the field-free Hamiltonian, and $mu$ the dipole moment operator. The gradient of $J$ with respect to $mathcal{E}(s,t)$ can thus be written as
begin{equation}
frac{delta J}{deltamathcal{E}(s,t)}=-2text{Im}(text{Tr}{[rho_0,O(frac{T}{2})]mu(t)}),
end{equation}

with $mu(t)=U^dagger(t,-frac{T}{2})mu U(t,-frac{T}{2})$ and $O(T/2)=U^dagger(frac{T}{2},-frac{T}{2})OU(frac{T}{2},-frac{T}{2})$.

I think the above quation is wrong. The reason is that the unit on the left hand side is not the same as the unit on the right hand side. Can anyone tell me how to derive the above quation?

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