Propagator using cauchy integral

In summary: But because the integral includes the pole inside the contour, it would be nonzero. However, because the singularity is outside of the contour, this would give you a larger value than the step function would give you. Consider the case when \tau>0. It would give you one. But because the integral includes the pole inside the contour, it would be nonzero. However, because the singularity is outside of the contour, this would give you a larger value than the step function would give you.
  • #1
Gary Weiss
2
0
hi
I don't understand this bit about the derivation of propagator expressions.
Bjorken and Drell describe the step function as:
[tex]\theta(\tau)=lim_{\epsilon \to 0}\frac{-1}{2\pi i}\oint_{-\infty}^{\infty}\frac{d\omega e^{-i\omega r}}{\omega + i \epsilon } [/tex]

the singularity is at [tex] -i \omega \epslion [/tex]

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

the singularity is at [tex] -i \omega \epslion [/tex]

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 [tex]\tau >0[/tex] and [tex]\tau <0 [/tex] separately.
The choice the upper semicircle contour or the lower semicircle contour depends on the sign of [tex]\tau[/tex], 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.
 
  • #3
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.
 
  • #4
Gary Weiss said:
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.
 

Related to Propagator using cauchy integral

1. What is the Cauchy integral?

The Cauchy integral is a mathematical tool used in complex analysis to calculate the value of a function at a point inside a region using the values of the function on the boundary of the region.

2. What is a propagator in physics?

In physics, a propagator is a mathematical tool used to describe the evolution of a physical system over time. It takes into account the initial state of the system and the forces acting upon it to determine its future state.

3. How is the Cauchy integral used in propagators?

The Cauchy integral is used in propagators to calculate the probability amplitude for a particle to move from one point to another in a given time frame. It takes into account the initial and final states of the particle and the forces acting upon it.

4. What is the relationship between the Cauchy integral and Green's function?

The Cauchy integral and Green's function are both mathematical tools used in complex analysis and physics, respectively. They are closely related, as the Green's function can be obtained from the Cauchy integral formula by taking the limit as the radius of the integration path approaches infinity.

5. What are some applications of the propagator using Cauchy integral?

The propagator using Cauchy integral has various applications in physics, such as calculating the probability amplitude for a particle to move from one point to another, as well as in quantum mechanics for calculating the evolution of a quantum system over time. It also has applications in engineering for solving boundary value problems and in signal processing for analyzing time series data.

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