# Derivative of unitary time evolution operator

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

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

• $H$ commutes with $U(t)$ so the order does not matter. Commented Apr 7, 2021 at 18:52
• @PraharMitra $H$ commutes with $U(t)$ only when the Hamilonian is time-independent (which to be fair, is the assumption in the question). However, when you have a time-dependent $H(t)$ the time-evolution operator is a time-ordered exponential and $H(t)$ no longer commutes with $U(t)$ (fundamentally this is because in general $H(t)$ does not commute with itself at different times). Commented Apr 7, 2021 at 18:58
• @QuantumEyedea - I didn't say anything about the general case. It is clear from the formulas in the question that the Hamiltonian in question is time-independent. If it was, $U(t)$ wouldn't take the form shown and we would need a time-ordering operator in there. Once you have the time-ordering operator, we could again commute the Hamiltonian past that at the cost of changing the times at which it is evaluated. In this case, the order would matter and the RHS would either be $H(t) U(t,t')$ or $U(t,t') H(t')$ and they are both equivalent. Commented Apr 7, 2021 at 19:00

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

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.

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