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

[Micha]

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.

 

[Hans de Vries]

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.

 

[Hans de Vries]

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.$$\mbox{ \frac{1}{2\pi}} \int^{\infty}_{-\infty}dE\ e^{iEt} \bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\} ~~~~ = ~~~~ \mbox{ \frac{1}{2\pi}} \oint dE\ e^{iEt}\bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\}$$

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


Fourier integral

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

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 $i\theta(t)\exp(-i\omega_p t)$
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:$$i\mbox{\Huge \mathcal{F}}\ \bigg\{\ \theta(t)\ e^{-i\omega_p t}\ \bigg\} ~~~~=~~~~\frac{i}{2\pi} \mbox{\Huge \mathcal{F}}\bigg\{\ \theta(t)\ \bigg\}\ *\ \mbox{\Huge \mathcal{F}}\bigg\{\ e^{-i\omega_p t} \bigg\}~~~~=$$


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

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

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

Regards, Hans