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Green's function & Klein Gordon

  1. Dec 4, 2012 #1

    I was wondering what the use in the Green's function for the Klein-Gordon equation was, I have listed it below:

    [tex]\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{ip\cdot(x-x')}[/tex]

    We find this gives an infinite result when the Klein gordon equation is applied to it and if x=x', what does this mean, i.e.

    [tex](\frac{\partial^2 }{\partial t^2}-\nabla^2+m^2)\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{ip\cdot(x-x')}=i\delta^{4})(x-x')[/tex]

    Does this mean the particle described by the green's function in the first equation can only be found in one place when x=x'?

    Also if we set x=x', the Green's function above turns to

    [tex]\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}[/tex]

    Which is alike to the 2 point function first order correction of form

    [tex]-\frac{i\lambda}{2}\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}[/tex]

    What is the relationship here? What use are the Green's function to the above equation? p is the four vector :


  2. jcsd
  3. Dec 4, 2012 #2
    I'm now aware that the first equation in my previous post is also a position space propagator for the Klein-Gordon equation; would anyone be able to explain to me why for this diagram


    The correction to the 2 point function for this first order can be given as an integral over the 3 position space propagators associated with (x',x'), (x,x') and (x',y)?

    Along with the -iλ/2 from the vertex (/symmetry factor = 1/2)

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