Evolution operator always in $SU(n)$?

Question

Think about the evolution operator $U$ in Quantum Mechanics for finite dimensional systems. Then this operator satisfies an equation

$$U'(t) = -iHU(t)..$$

Here, I assume that $H$ is time-independent(!).

Now, I was wondering: In physics you often see that $U^*U= \operatorname{id}$ holds, but can we even say that $U(t) \in SU(n)$ for any $t$?

Answer 1

If $A=\exp(B)$ then $\det(A)=\exp(\mathrm{tr}(B))$ and so $U(t)\in SU(n)$ if and only if the trace of $H$ is zeo.