Explicit form of scalar propagator

  • Thread starter parton
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
1,948
201
Make a change of variables k'=ik or k'=-ik as needed
 
  • #3
83
1
Make a change of variables k'=ik or k'=-ik as needed

Thank you for your hint, but somehow I don't see how that could help.
If I make the substitution [itex] k' = i k [/itex] I find:

[tex]
\begin{align}
\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] =
\\
& = -i \int_{0}^{i \infty} \mathrm{d}k \left[ \mathrm{e}^{k(x_{0}+r)} - \mathrm{e}^{-k(x_{0}+r)} + \mathrm{e}^{-k(x_{0}-r)} - \mathrm{e}^{k(x_{0}-r)} \right]
\\
& = -i \left[ \dfrac{1}{x_{0} + r} \left( \mathrm{e}^{k(x_{0}+r)} + \mathrm{e}^{-k(x_{0}+r)} \right) - \dfrac{1}{x_{0} - r} \left( \mathrm{e}^{k(x_{0}-r)} + \mathrm{e}^{-k(x_{0}-r)} \right) \right] \Bigg|_{0}^{i \infty}
\\
& = -2 i \left[ \dfrac{1}{x_{0} + r} \mathrm{cosh}(k (x_{0} + r)) - \dfrac{1}{x_{0} - r} \mathrm{cosh}(k(x_{0}-r)) \right] \Bigg|_{0}^{i \infty}
\end{align}
[/tex]

And this will diverge for [itex] k \to i \infty [/itex].

Do you have an idea to resolve this problem?
 
  • #4
178
18
Multiply integrand by [itex]\exp(-\alpha k)[/itex], integrate it and then take limit of [itex]\alpha[/itex] as zero.
[itex]\alpha [/itex] is real and positive.
 
  • #5
83
1
Multiply integrand by [itex]\exp(-\alpha k)[/itex], integrate it and then take limit of [itex]\alpha[/itex] as zero.
[itex]\alpha [/itex] is real and positive.

OK, thanks a lot :smile:. This seems to work.

But what I still not undestand is: Why can the result [itex] - \dfrac{1}{4 \pi^{2} x^{2}} [/itex] be undestood as Cauchy's principal value?
 

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