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

  1. Sep 3, 2016 #1
    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. jcsd
  3. Sep 5, 2016 #2
    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.
     
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