Unitary conserving Schrödinger equation

Question

If we work in the spaces of $N\times N$ matrices we can write the propagator Schrödinger equation as: $$ \dot{U}(t)=-iH(t)U(t) $$ For some time-dependent Hermitian Hamiltonian $H(t)$. If we assume that this Hamiltonian is bounded in the supremum norm and we can prove existence of a solution for $U$ by considering the integral equation $$ U(t)=U(0)-i\int_{0}^tH(s)U(s)ds$$. And showing that the the operator $$ (TU)(t):=U(0)-i\int_{0}^tH(s)U(s)ds$$ is a contraction in the following norm $$ \|U\|=\sup_{t\in[0,T]}(\|U(t)\|_Fe^{-\lambda t})$$ for small enough $\lambda$. Now I am stuck showing that this also leads to a unitary solution for $U(t)$ as long as $U(0)$ is unitary. Any help would be welcome! I am trying to circumvent the use of the Dyson series, unless it is somehow possible to couple the solution for U(t) found by the contraction to the Dyson series

Answer 1

You dont need to use any advanced techniques here - just use that your matrix $H(t)$ is hermitian for all times. Furthermore, you have $$ (-iH(t)U(t))^{\dagger}=+iU^{\dagger}(t)H(t) $$ Then we can use the Schroedinger equation to deduce that $$ \frac{d}{dt}U^{\dagger}(t)U(t)=(\frac{d}{dt}U^{\dagger}(t))U(t)+U^{\dagger}(t)\frac{d}{dt}U(t)=(+iU^{\dagger}(t)H(t))U(t)+U^{\dagger}(t)(-iH(t)U(t))=\pm i U^{\dagger}(t)H(t)U(t)=0 $$ You can do the same calculation for $\frac{d}{dt}UU^{\dagger}$. The you can use that $U(0)U^{\dagger}(0)=1$ and conclude that your matrix is unitary for all times, since the product $U(t)U^{\dagger}(t)$ is constant in time.