Time evolution operator in terms of Hamiltonian

dEdt
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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?
 
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What is the derivative of \ln U?
 
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.
 
Try expanding the expression you have for U(t) using a Taylor series and show that it satisfies the Schrödinger equation.
 
dEdt said:
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.

So, working along the lines of your attempted solution, you would get something like:
U^{-1} \partial_t U = -\frac{i}{\hbar}U^{-1}HU \;.
 

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