- #1
Romanopoulos Stelios
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How the authors came to the conclusion (eq. 4.25) that
$$ U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)} $$
$$ U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)} $$
You may not see this answer, but as an individual memo, I answer your question.
Yeah, I know (ゝω・)
Note that this thread is five years old!
The problem will occur if ##H## is dependent on time. We must perform derivative ##H## in the component ##\exp\left[-iH(t-t_0)\right]##You can plug the expression into equation 4.24 to verify it's correct. The left side is (taking into account the non-commutativity of ##H_0## and ##H##):
$$i\frac{\partial}{\partial t}U(t,t')=i\left[e^{iH_0(t-t_0)}(iH_0)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}\right]=\underbrace{e^{iH_0(t-t_0)}(H-H_0)e^{-iH_0(t-t_0)}}_{H_I(t)}\underbrace{e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}}_{U(t,t')}$$
this is nothing but the right hand side of 4.25. In the second line I've used ##1=e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)}##, ##H_{int}=H-H_0## and the definition of the interaction picture Hamiltonian.