RedX
Feb4-11, 06:52 PM
I've seen written the following manipulation of the KG Feynman propagator:
\frac{1}{p^2-m^2+i\epsilon}=\frac{1}{p^2+i\epsilon} \frac{1}{(1-\frac{m^2}{p^2+i\epsilon})}= \frac{1}{p^2+i\epsilon} (1+\frac{m^2}{p^2+i\epsilon}+\left(\frac{m^2}{p^2+ i\epsilon}\right)^2+...)
I don't think this can be valid unless |p^2+i\epsilon|>m^2 .
However if you are going to integrate this expression to get the inverse Fourier transform:
\int d^4p \frac{e^{ip(x-y)}}{p^2-m^2+i\epsilon}
then this integrals picks out the poles which forces p^2=m^2-2iE_p\epsilon .
However, since the epsilon term comes in with a minus sign, it could potentialy cancel out the +iepsilon term in |p^2+i\epsilon|>m^2 .
So is the binomial expansion valid?
\frac{1}{p^2-m^2+i\epsilon}=\frac{1}{p^2+i\epsilon} \frac{1}{(1-\frac{m^2}{p^2+i\epsilon})}= \frac{1}{p^2+i\epsilon} (1+\frac{m^2}{p^2+i\epsilon}+\left(\frac{m^2}{p^2+ i\epsilon}\right)^2+...)
I don't think this can be valid unless |p^2+i\epsilon|>m^2 .
However if you are going to integrate this expression to get the inverse Fourier transform:
\int d^4p \frac{e^{ip(x-y)}}{p^2-m^2+i\epsilon}
then this integrals picks out the poles which forces p^2=m^2-2iE_p\epsilon .
However, since the epsilon term comes in with a minus sign, it could potentialy cancel out the +iepsilon term in |p^2+i\epsilon|>m^2 .
So is the binomial expansion valid?