Derivative of unitary time evolution operator

Question

Consider the unitary time evolution operator

$$U(t) = \exp\left(\frac{-iHt}{\hbar}\right)$$ and its hermitian conjugate:

$$U(t)^{\dagger} = \exp\left(\frac{iHt}{\hbar}\right)$$

The derivatives of these operators are as follows:

$$\frac{\partial U(t)}{\partial t} = \frac{-i}{\hbar}H(t)U(t)$$

and

$$\frac{\partial U(t)^{\dagger}}{\partial t} = \frac{i}{\hbar}U(t)^{\dagger} H(t)$$

My question is why are the $U(t)$ and $H(t)$ in the derivatives in the order that they are. In other words why are the derivatives of the operator not the following:

$$\frac{\partial U(t)}{\partial t} = \frac{-i}{\hbar}H(t)U(t)$$

and

$$\frac{\partial U(t)^{\dagger}}{\partial t} = \frac{i}{\hbar} H(t)U(t)^{\dagger} $$

Answer 1

$$[H,U(t)]=0\Rightarrow \text{Doesn't matter :)}$$

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Note that

$$\text{In general } \ \ U(t)\not=\exp\left(-\frac{iHt}{\hbar}\right)$$ You have used time dependent Hamiltonian and above is not valid for such case.

Answer 2

We can do it either by first taking the derivative and then taking the conjugate transpose of it.

$(\partial U/\partial t)^\dagger$ = $(-{\frac\iota \hbar} \space U(t) \space H(t))^\dagger$ and we know $(AB)^\dagger$ = $ B^\dagger \space A^\dagger$

So we get,

$\frac {∂U(t)^†} {∂t}=\frac iℏU(t)^†H(t)$

so we get the expression. We can also find it buy expanding the exponential term and then taking the conjugate transpose of the expanded terms. Either way it will come the same. But as pointed out H(t) commutes with U(t) so the order does not matter if H(t) is taken to be time independent