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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 $[A, U_t] = 0$.

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.

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?

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 $[A, U_t] = 0$.

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

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 $[A, U_t] = 0$.

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.