Problem 3.29 from Gerald Teschl ODE.
Let the Schrödinger Equation, $$i\psi'(t)=H(t)\psi(t),\ \psi(t_{0})=\psi_{0},$$ where $H(t)$, is a self-adjoint matrix, that is, $H(t)^{*}=H(t) $ . Show that the solution is given by $$\psi(t)=U(t,t_{0})\psi_{0},\ U(t_{0},t_{0})=I$$, with $U(t,t_{0})$ is unitary, that is, $U(t,t_{0})^{-1}=U(t,t_{0})^{*}$.
Compute that $$\frac{d}{dt}ψ^*(t)ψ(t)=0$$ by the product rule and ODE and symmetries, and thus $$ψ_0^*U(t,t_0)^*U(t,t_0)ψ_0=ψ_0^*ψ_0$$ from where the claim follows by polarization.