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The propagator is nonzero outside of the light cone.

  1. Dec 7, 2013 #1
    I saw an interesting comment in my notes about the [itex]E = \pm \sqrt{\vec{p}^2 + m^2}[/itex] eingenvalues plane wave solution to the free particle Klein Gordon equation. The propagator

    [itex]U = \langle \vec{x} \mid e^{-i H t} \mid \vec{x_0} \rangle = \langle \vec{x} \mid e^{-i t \sqrt{\vec{p}^2 + m^2} } \mid \vec{x_0} \rangle \propto[/itex]
    [itex]\propto e^{-m \sqrt{ (\vec{x} - \vec{x_0})^2 - t^2}} \neq 0 \quad \text{also for} \quad (\vec{x} - \vec{x_0})^2 > c^2 t^2 [/itex]

    that is outside the light cone. About this there's even a whole section on Wikipedia's propagator page, but it lacks the math. My questions are:

    1. how do you get to the above result, after the proportionality symbol?
    2. The imaginary exponent becomes real, how does that happen? Also like this it seems that inside the light cone the exponent is imaginary again, what does that mean?
    Thanks.
     
  2. jcsd
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