# Differentiation of unitary operator U(t,t') in Peskin and Sc

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1. Sep 3, 2016

### Romanopoulos Stelios

How the authors came to the conclusion (eq. 4.25) that
$$U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)}$$

2. Sep 5, 2016

### kontejnjer

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.