In my recent research concerning the unitary quantum control, I have met a problem which can be described as follows:

Given two Hermitian matrices $A,B$ on a $d$-dimensional Hilbert space $\mathcal H_d$, where $A$ is positive semi-definite and $B$ has a vanishing trace. Let $H$ be a fixed Hermitian matrix on $\mathcal H_d$. Assume that we know all eigenvalues of $A,B$ and $H$.

How can we identify the solution of the following equation in a real argument $t$:

$$

\mathrm{Tr}(e^{itH}A e^{-itH}B) = 0.

$$

Here we want to identify which $t$ can have the above equality hold. I have never met this kinds of equations which is beyond a scope of my knowledge.

My analysis is: for two fixed quantum states $\rho,\sigma$ on $\mathcal H_d$,let $H$ be a fixed Hamiltonian. The evolustion of the closed system with its energy $H$ is described by a unitary transformation $U_t = e^{itH}$ because of the quantum mechanical theory.

Set $f(t) := \mathrm{Tr}(U_t \rho U^\dagger_t \sigma)$. Then:

$$

df(t)/dt = i\mathrm{Tr}(U_t \rho U^\dagger_t [H,\sigma]).

$$

I want to get the extremal points of $f(t)$, thus it is obtaind that

$$

\mathrm{Tr}(U_t \rho U^\dagger_t [H,\sigma]) = 0.

$$

This condition is just the equation mentioned above.