- #1
parton
- 83
- 1
Hi!
I have encountered a little problem. I want to show
that the explicit form of the Feynman propagator for massless scalar fields is given by:
[tex]
\begin{align}
G_F(x) & = - \lim_{\epsilon \to +0} \int \dfrac{\mathrm{d}^{4}k}{(2 \pi)^{4}} \dfrac{1}{k^{2} + i \epsilon} \mathrm{e}^{- i k x}
\\
& = - \lim_{\epsilon \to +0} \dfrac{1}{4 \pi^{2}} \dfrac{1}{x^{2} -i \epsilon}
\end{align}
[/tex]
And I would like to do that directly, i.e., without starting from the massive case and considering the limit [itex] m \to 0 [/itex].
I found a script where one can find a derivation of that result,
http://mo.pa.msu.edu/phy853/lectures/lectures.pdf
on pages 58-59.
There, the integration is split up into the imaginary and real part.
But for the imaginary part, the author finds:
[tex]
\begin{align}
G_F,i(x) & = \dfrac{1}{4 \pi^{2} r} \int_{0}^{\infty} \mathrm{d}k \cos(k x_{0}) \sin(k r)
\\
& = \dfrac{1}{16 \pi^{2} r i} \int_{0}^{\infty} \mathrm{d}k \left[ \mathrm{e}^{ik(x_{0}+r)} - \mathrm{e}^{-ik(x_{0}+r)} + \mathrm{e}^{-ik(x_{0}-r)} - \mathrm{e}^{ik(x_{0}-r)} \right]
\\
& = - \frac{1}{8 \pi^{2} r} \left[ \dfrac{1}{x_{0} + r} - \dfrac{1}{x_{0} - r} \right]
\\
& = - \frac{1}{4 \pi^{2} x^{2}}
\end{align}
[/tex]
Here, I don't understand how the integration is performed. I think if you integrate every exponential term it will diverge at [itex]\infty[/itex], but somehow the author ends up with a finite term.
Does anyone understand what is going on here?
Furthermore, I think we should end up with Cauchy's principal value and not just 1/x^2, right?
Maybe, someone has a more "elegant" way of deriving the massless propagator?
I have encountered a little problem. I want to show
that the explicit form of the Feynman propagator for massless scalar fields is given by:
[tex]
\begin{align}
G_F(x) & = - \lim_{\epsilon \to +0} \int \dfrac{\mathrm{d}^{4}k}{(2 \pi)^{4}} \dfrac{1}{k^{2} + i \epsilon} \mathrm{e}^{- i k x}
\\
& = - \lim_{\epsilon \to +0} \dfrac{1}{4 \pi^{2}} \dfrac{1}{x^{2} -i \epsilon}
\end{align}
[/tex]
And I would like to do that directly, i.e., without starting from the massive case and considering the limit [itex] m \to 0 [/itex].
I found a script where one can find a derivation of that result,
http://mo.pa.msu.edu/phy853/lectures/lectures.pdf
on pages 58-59.
There, the integration is split up into the imaginary and real part.
But for the imaginary part, the author finds:
[tex]
\begin{align}
G_F,i(x) & = \dfrac{1}{4 \pi^{2} r} \int_{0}^{\infty} \mathrm{d}k \cos(k x_{0}) \sin(k r)
\\
& = \dfrac{1}{16 \pi^{2} r i} \int_{0}^{\infty} \mathrm{d}k \left[ \mathrm{e}^{ik(x_{0}+r)} - \mathrm{e}^{-ik(x_{0}+r)} + \mathrm{e}^{-ik(x_{0}-r)} - \mathrm{e}^{ik(x_{0}-r)} \right]
\\
& = - \frac{1}{8 \pi^{2} r} \left[ \dfrac{1}{x_{0} + r} - \dfrac{1}{x_{0} - r} \right]
\\
& = - \frac{1}{4 \pi^{2} x^{2}}
\end{align}
[/tex]
Here, I don't understand how the integration is performed. I think if you integrate every exponential term it will diverge at [itex]\infty[/itex], but somehow the author ends up with a finite term.
Does anyone understand what is going on here?
Furthermore, I think we should end up with Cauchy's principal value and not just 1/x^2, right?
Maybe, someone has a more "elegant" way of deriving the massless propagator?