Propagator using cauchy integral

[Gary Weiss]

hi
I don't understand this bit about the derivation of propagator expressions.
Bjorken and Drell describe the step function as:
$$\theta(\tau)=lim_{\epsilon \to 0}\frac{-1}{2\pi i}\oint_{-\infty}^{\infty}\frac{d\omega e^{-i\omega r}}{\omega + i \epsilon }$$

the singularity is at $$-i \omega \epslion$$

I understand that if I evaluate this integral with a path above the real axis
I get zero, and that if I evaluate it below the real axis I get 1.

however I don't understand why it's ok to choose the paths in this way.
What happened at tau<0 that allowed me to integrate only the upper half of
the real/imaginary axis?

thanks!

 

[ismaili]

hi
I don't understand this bit about the derivation of propagator expressions.
Bjorken and Drell describe the step function as:
$$\theta(\tau)=lim_{\epsilon \to 0}\frac{-1}{2\pi i}\oint_{-\infty}^{\infty}\frac{d\omega e^{-i\omega r}}{\omega + i \epsilon }$$

the singularity is at $$-i \omega \epslion$$

I understand that if I evaluate this integral with a path above the real axis
I get zero, and that if I evaluate it below the real axis I get 1.

however I don't understand why it's ok to choose the paths in this way.
What happened at tau<0 that allowed me to integrate only the upper half of
the real/imaginary axis?

thanks!

Consider the case $$\tau >0$$ and $$\tau <0$$ separately.
The choice the upper semicircle contour or the lower semicircle contour depends on the sign of $$\tau$$, i.e. if the semicircle part of the contour contributed, that's not the original contour integral you want to calculate.

Calculate these two cases separately, you will see that's exactly step function.

 

[Gary Weiss]

sorry I don't get it...
why is it that when tau < 0 I can choose the contour above the real axis and ignore the singularity.

 

[ismaili]

sorry I don't get it...
why is it that when tau < 0 I can choose the contour above the real axis and ignore the singularity.

When \tau <0, you contour integral is equivalent to the closed contour integral which goes above the real axis.(because the upper semicircle doesn't contribute to the integral, this is ensured by Jordan's lemma)

Since there is no pole inside the contour, the integral is zero, consistent with step function.

You can consider the case when \tau>0.
It would give you one.