# Time evolution operator in terms of Hamiltonian

dEdt

## Homework Statement

"Show that if the Hamiltonian depends on time and $[H(t_1),H(t_2)]=0$, the time development operator is given by
$$U(t)=\mathrm{exp}\left[-\frac{i}{\hbar}\int_0^t H(t')dt'\right]."$$

## Homework Equations

$i\hbar\frac{d}{dt}U=HU$
$U(dt)=I-\frac{i}{\hbar}H(t)dt$

## The Attempt at a Solution

The first thing I tried was to rearrange the first of the relevant equations:
$$\left(\frac{d}{dt}U\right)U^{-1}=-\frac{i}{\hbar}H(t).$$
I can then integrate both sides; if the LHS could turn into an expression like $\ln{U}$ I'd be done, but that didn't work out. Any hints?

Aimless
What is the derivative of $\ln U$?

dEdt
If U was an ordinary function I would say $U^{-1}\frac{d}{dt}U$, but considering that U is an operator function, I'm not sure. I don't know how $\ln{U}$ is even defined, much less how to apply chain rule to it.

Staff Emeritus
If U was an ordinary function I would say $U^{-1}\frac{d}{dt}U$, but considering that U is an operator function, I'm not sure. I don't know how $\ln{U}$ is even defined, much less how to apply chain rule to it.
Like any function of an operator, $\ln{U}$ is defined in terms of it's Taylor series. More specifically in this case, it's the inverse series of the exponential function. It exists iff $U^{-1}$ exists.
$$U^{-1} \partial_t U = -\frac{i}{\hbar}U^{-1}HU \;.$$