Commuting with time evolution operator implies commuting with Hamiltonian

Question

Consider a quantum system (finite dimensional) has overall Hamiltonian:

$H_t = H_0 + w(t)H_c$

with $H_0, H_c$ constant in time and traceless and $w(t)$ a, not too badly behaved, function of time.

It is true that if an operator commutes with the overall Hamiltonian (at all times) then it commutes with the time evolution operator, i.e. $[A, H_t]=0$ for all time implies $[A, U_t] = 0$.

Question: Is it also true that $[A, U_t] = 0$ implies $[A,H_0]=0$ and $[A,H_c]=0$ separately? If so, how can this be proven?

Answer 1

Assuming $A$ itself is time-independent. If $[A, U_t]=0$ for all $t$, then it can be proven that $[A, H]=0$: We have

$H=i \partial_t U_t\cdot U_t^\dagger$

Because $[A, U_t]=0$, it follows that $[A, \partial_t U_t]=0$ since we assume $\partial_t A=0$.

Now $[A, H]=[A, H_0]+w(t)[A, H_c]=0$. Again we take derivative with respect to $t$, we get $w'(t)[A, H_c]=0$. So we conclude that as long as $w'(t)\neq 0$ for some $t$, $[A, H_c]=[A, H_0]=0$.