How can the Cauchy integral and Fourier integral produce the same result?

  • Thread starter nrqed
  • Start date
  • Tags
    Qft
In summary, the conversation discusses questions about causality and measurements in quantum field theory (QFT). The concept of causality in QFT is different from classical physics and involves considering the possibility of particles propagating over spacelike intervals. The connection between QFT and non-relativistic quantum mechanics (NRQM) is also discussed, with a question about the classical limit of QED. The issue of causality in QFT is further explored, with a suggestion that allowing for negative energies in the propagator may restore locality. However, the overall understanding of these concepts in QFT remains unclear and there is a lack of clear explanations in literature.
  • #316
Hans de Vries said:
Yes,

Note that the poles of the Klein Gordon propagator are of this simple form 1(z-a)


Regards, Hans

Yes, but as I pointed out earlier, doing a Fourier transform you have a factor exp(...) around, so the whole integrand is not of the simple form.
 
Physics news on Phys.org
  • #317
Micha said:
Yes, but as I pointed out earlier, doing a Fourier transform you have a factor exp(...) around, so the whole integrand is not of the simple form.

entirely correct.
 
  • #318
To go back to the original question:Why does the Cauchy integral produce the same result as the Fourier integral ?
I'm using the latter. Both arguments are exactly the same, only the paths differ.[tex] \mbox{ $\frac{1}{2\pi}$} \int^{\infty}_{-\infty}dE\ e^{iEt} \bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\} ~~~~ = ~~~~ \mbox{ $\frac{1}{2\pi}$} \oint dE\ e^{iEt}\bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\}[/tex]

[tex]~\Rightarrow ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mbox{$\frac{i sgn(t)}{2}$}\ e^{-i\omega_p t}~~ + ~~~~~~\mbox{ $\frac{i}{2}$}\ e^{-i\omega_p t}
~~~~~~~~~~~~~~ = ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~ie^{-i\omega_p t}~~~~~~~~~~+~~~~~~~~~~~~0[/tex]


Fourier integral


Both Hilbert partners provide half of the result at (t>0) while they cancel
at (t<0). This automatically provides the result [itex]i\theta(t)\exp(-i\omega_p t)[/itex]

Cachy integral

The pole provides the whole result. The contour has to be "hand-picked" to
include the pole at (t>0) and exclude the pole at (t<0) to give [itex]i\theta(t)\exp(-i\omega_p t)[/itex]
The Hilbert partner (the delta) has zero contribution. In fact it is generally
not even considered in the calculation at all.

But the Hilbert partner has to be considered. It shows up trivially from the
Fourier transform back:[tex] i\mbox{\Huge $\mathcal{F}$}\ \bigg\{\ \theta(t)\ e^{-i\omega_p t}\ \bigg\} ~~~~=~~~~\frac{i}{2\pi} \mbox{\Huge $\mathcal{F}$}\bigg\{\ \theta(t)\ \bigg\}\ *\ \mbox{\Huge $\mathcal{F}$}\bigg\{\ e^{-i\omega_p t} \bigg\}~~~~=[/tex]


[tex]\frac{i}{2\pi}\bigg\{\ 2\pi\mbox{ $\frac{1}{2}$}\left(\frac{1}{i\pi E}+\delta(E)\right) \bigg\}\ *\ \bigg\{\ 2\pi\ \delta(E-\omega_p) \bigg\}\ ~~~~=~~~~\left(~\frac{1}{E}~~+~~i\pi\ \delta(E)~\right)~*~\delta(E-\omega_p)~~~~=[/tex]

[tex]\frac{1}{E-\omega_p}~~+~~i\pi\ \delta(E-\omega_p)[/tex]The convolution is just a shift by [itex]\omega_p[/itex] applied on the Fourier transform of the
Heaviside step-function [itex]\theta(t)[/itex] given here:

http://en.wikipedia.org/wiki/Fourier_transform#Distributions (formula 310, with [itex]f=2\pi E[/itex])

Regards, Hans
 
Last edited:

Similar threads

  • Quantum Physics
Replies
1
Views
1K
Replies
2
Views
1K
Replies
1
Views
707
Replies
31
Views
2K
Replies
1
Views
839
  • Quantum Physics
Replies
5
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
10
Views
916
Back
Top