Schrodinger equation $iψ'(t) = H(t)ψ(t),\,\,\,\,\,\,\,\,\,\, ψ(t_0) = ψ_0,$ (Quantum Mechanics). A quantum mechanical system which
can only attain finitely many states is described by a complex-valued vector
$ψ(t) ∈ \mathbb{C}^n$. The square of the absolute values of the components $|ψ_j(t)|^2$ is
interpreted as the probability of finding the system in the $j’th$ state at time
$t$. Since there are only $n$ possible states, these probabilities must add up to one, that is, $ψ(t)$ must be normalized, $|ψ(t)| = 1$. The time evolution of the
system is governed by the Schrodinger equation
$iψ'(t) = H(t)ψ(t),\quad ψ(t_0) = ψ_0,$
where $H(t)$, is a self-adjoint matrix, that is, $H(t)^∗ = H(t)$. (Here $A^∗$ is the
adjoint (complex conjugate of the transposed) matrix.) The matrix $H(t)$ is
called the Hamiltonian and describes the interaction. Show that the solution
is given by
$ψ(t) = U(t, t_0)ψ_0, \quad U(t_0, t_0) = \mathbb{I},$
where $U(t, t_0)$ is unitary, that is, $U(t, t_0)^{−1} = U(t, t_0)^∗$ .
(Hint: Given $A(t)$ and B(t) matrices, holds: $\dfrac{d(A(t).B(t))}{dt}=A'(t)B(t)+A(t)B'(t)$ and $\dfrac{d(A(t)^{-1})}{dt}=-A(t)^{-1}A'(t)A(t)^{-1}$).
Conclude that $ψ(t)$ remains normalized for all $t$ if $ψ_0$ is.
Each observable (quantity you can measure) corresponds to a self-adjoint
matrix, say $L_0$. The expectation value for a measurement of $L_0$ if the system
is in the state $ψ(t)$ is given by
$\langle ψ(t), L_0ψ(t) \rangle,$
where $ \langle ϕ, ψ \rangle = ϕ^∗ · ψ$ is the scalar product in $\mathbb{C}^n$. Show that
$\dfrac{d\langle ψ(t), L_0ψ(t) \rangle}{dt}= i \langle ψ(t), [H(t), L_0]ψ(t) \rangle$
where $[H, L] = HL − LH$ is the commutator.
In fact I do not know how to start solving this problem. Can someone help me? 
The biggest problem is that I do not know anything about the behavior of this array $H(t)$.
 A: Suppose that $\psi(t) = U(t, t_0) \psi(t_0)$ for some matrix $U$ (which you don't know is unitary just yet). Plug this into Schrodinger's equation
$$
i\frac{{\rm d} U(t,t_0)}{{\rm d}t} = H(t) U(t, t_0) \tag{1}
$$
Take the adjoint ($A^\dagger$ means take the transpose of $A$ and then complex conjugate each element)
$$
-i\frac{{\rm d} U^\dagger(t,t_0)}{{\rm d}t} = U(t, t_0)^\dagger H(t) \tag{2}
$$
Since I used the fact that $H^{\dagger}(t) = H(t)$. Now consider
\begin{eqnarray}
\frac{{\rm d} }{{\rm d}t}\left(U^\dagger(t,t_0)U(t,t_0)\right) &=& U^\dagger(t,t_0) \frac{{\rm d} U(t,t_0)}{{\rm d}t} + \frac{{\rm d} U^\dagger(t,t_0)}{{\rm d}t}U(t,t_0) \\
&\stackrel{(1),(2)}=& -iU^{\dagger}(t,t_0)H(t) U(t,t_0)+iU^{\dagger}(t,t_0)H(t) U(t,t_0) \\
&=& 0 \tag{3}
\end{eqnarray}
That means that $U^\dagger(t,t_0) U(t,t_0)$ is constant at all times, but you have the initial condition $U^\dagger(t_0,t_0) U(t_0,t_0) = 1$, which implies that 
$$
U^\dagger(t,t_0) U(t,t_0) = 1 \tag{4}
$$
This shows that $U(t,t_0)$ is unitary. 
For the second part, consider the norm of the vector $\psi(t)$
\begin{eqnarray}
|\psi(t)|^2 &=& \psi(t)^\dagger\psi(t) = \left(U(t,t_0) \psi_0\right)^\dagger \left(U(t,t_0) \psi_0\right) \\
&=&\psi(t_0)^\dagger \left(U(t,t_0)^\dagger U(t,t_0) \right) \psi(t_0) \\
&\stackrel{(4)}{=}&  \psi(t_0)^\dagger 1 \psi(t_0)\\
&=& \psi(t_0)^\dagger \psi(t_0) \\
&=& |\psi(t_0)|^2 \tag{5}
\end{eqnarray}
The norm is preserved under unitary evolution
