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I Dirac vs KG propagation amplitude

  1. Jul 7, 2017 #1
    Hello! Can someone explain to me the physical meaning of ##\bar{\psi}=\psi^\dagger\gamma^0## in the Dirac equation? For example when calculating propagation amplitude I see that what we calculate is ##<0|\psi(x)\bar{\psi(y)}|0>## and not ##<0|\psi(x)\psi(y)|0>## (as we do for KG equation) and I am not sure I understand why. Can someone explain? Thank you!
  2. jcsd
  3. Jul 12, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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