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Why does the Schwinger parameter correspond to proper length?

  1. Dec 21, 2013 #1
    I have just learned from nice article


    that the propagator of a massive particle can be rewritten as an integral over the so-called Schwinger parameter t as

    \frac{1}{p^2 + m^2} = \int\limits_0^\infty dt \exp(-t(p^2 + m^2))

    In addition, in the blog article it is said that this Schwinger parameter p can be interpreted as the proper length of the propagator. I dont see this, so can somebody give a derivation/further explanation?
  2. jcsd
  3. Dec 21, 2013 #2
  4. Dec 21, 2013 #3


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