Understanding the Klein-Gordon Propagator and its Satisfying Equation

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Homework Help Overview

The discussion revolves around the Klein-Gordon propagator and its relationship to the equation involving the d'Alembertian operator and the Dirac delta function. Participants are examining the mathematical properties and implications of the propagator in the context of quantum field theory.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the manipulation of the propagator expression and questioning the cancellation of terms in the integral. There is a focus on the role of the i epsilon term and its implications for contour integration.

Discussion Status

The discussion is ongoing, with participants raising questions about specific steps in the derivation and the validity of assumptions regarding the i epsilon term. Some guidance has been offered regarding the interpretation of the integral, but no consensus has been reached on the cancellation process.

Contextual Notes

There are indications of uncertainty regarding the treatment of the i epsilon term and its significance in the limit as epsilon approaches zero. The original poster's approach and the subsequent responses highlight differing interpretations of the mathematical expressions involved.

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Homework Statement


Homework Equations


Show that the KG propagator
[tex]G_F (x) = \int \frac{d^4p}{(2\pi)^4} e^{-ip.x} \frac{1}{p^2-m^2+i\epsilon}[/tex]
satsify
[tex](\square + m^2) G_F (x) = -\delta(x)[/tex]

The Attempt at a Solution


I get
[tex](\square + m^2) G_F (x) = - \int \frac{d^4p}{(2\pi)^4} (p^2-m^2) e^{-ip.x} \frac{1}{p^2-m^2+i\epsilon}[/tex]
but where do I go from there?
 
Last edited:
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Cancel the numerate and denominator p^2-m^2.
The i epsilon is just a direction how to take the contour, and is negligible here.
The remaining integral is \delta^4.
 
how does (p^2-m^2)/(p^2-m^+i*epsilon) cancel?

I would try to do the limit of epsilon -> 0+
 
Last edited:
malawi_glenn said:
how does (p^2-m^2)/(p^2-m^+i*epsilon) cancel?

I would try to do the limit of epsilon -> 0+
That's what
"The i epsilon is just a direction how to take the contour, and is negligible here."
means.
 

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