A contour integral frequenctly encountered

In summary: I am not sure if I am understanding the problem correctly.In summary, the student is having trouble understanding how to calculate the inner product of two wave functions in momentum space using the Lippmann-Schwinger equation.
  • #1
hneder
3
0

Homework Statement



Need some help here on a frequently encountered integral in Green's function formalism.

Homework Equations



I have an integral/summation as a product of a retarded and advanced Green's functions, looks simply like
[itex]\sum_{p^{\prime}}\frac{1}{p^{2}-{p^{\prime}}^{2}-i\epsilon}\frac{1}{p^{2}-{p^{\prime}}^{2}+i\epsilon}[/itex]
where I have omitted the mass [itex]m[/itex] to make the notation simple. [itex]\epsilon[/itex] is a positive infinitesimal.

The Attempt at a Solution


I can convert the summation to an integral over momentum [itex]p[/itex], this is standard. Then I follow the complex analysis and identify 4 poles and further apply the residue theorem, what I get is an expression proportional to [itex]\frac{1}{\epsilon}[/itex]. This means it is divergent.

Can anyone point out to me what have I missed in this calculation? What mistakes did I make? Many many thanks
 
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  • #2
hneder said:

Homework Statement



Need some help here on a frequently encountered integral in Green's function formalism.

Homework Equations



I have an integral/summation as a product of a retarded and advanced Green's functions, looks simply like
[itex]\sum_{p^{\prime}}\frac{1}{p^{2}-{p^{\prime}}^{2}-i\epsilon}\frac{1}{p^{2}-{p^{\prime}}^{2}+i\epsilon}[/itex]
where I have omitted the mass [itex]m[/itex] to make the notation simple. [itex]\epsilon[/itex] is a positive infinitesimal.

The Attempt at a Solution


I can convert the summation to an integral over momentum [itex]p[/itex], this is standard. Then I follow the complex analysis and identify 4 poles and further apply the residue theorem, what I get is an expression proportional to [itex]\frac{1}{\epsilon}[/itex]. This means it is divergent.

Can anyone point out to me what have I missed in this calculation? What mistakes did I make? Many many thanks

Hi.

Can you provide more details on your calculation? I am assuming you are doing the four-dimensional integral over [itex] d^4 p [/itex]. Then did you do the [itex] p_0 [/itex] integral using contour integration and got the [itex] 1/ \epsilon [/itex] term before doing the remaining 3-dimensional integral?
 
  • #3
Thanks for the reply. This is simply a product of two Green's functions, one retarded and one advanced. No, it is a regular 3D integral (summation) and momentum [itex]p[/itex] and [itex]p^{\prime}[/itex] here are both 3D vectors. Somehow I know there is a contour integral to be done. But it is not clear to me how should I do it.

What I did was the following. If I perform the [itex]p[/itex] integration using residue theorem, as I mentioned in the post, I end up with four poles [itex]p\pm i\epsilon[/itex], [itex]-p\pm i\epsilon[/itex] and the final result [itex]\sim \frac{1}{\epsilon}[/itex]. The same result comes out if I convert the [itex]p[/itex] integration into one with energy using [itex]E=p^{2}/{2m}[/itex].
 
Last edited:
  • #4
It doesn't make physical sense if the vectors [itex]p[/itex] and [itex]p'[/itex] are Euclidean.

Anyway, as a purely mathematical exercise, it's an interesting problem, because it demonstrates the problem of Pinch singularities that sometimes occur in real-time many-body QFT if one is not careful enough, because, as you seem to have realized, the expressions do not make sense in the weak limit [itex]\epsilon \rightarrow 0^+[/itex]. Fortunately, one can prove rigorously that such pinch singularities are absent when the correct Schwinger-Keldysh Contour techniques are applied. The details are a bit tedious, and one has to use some care to correctly make sense of the distributions involved in the calculation. See my lecture notes on relativistic many-body theory:

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

(the long Sect. 2.2).
 
  • #5
Thanks, vanhees71, for your explanation. Your notes in the link are a bit too advanced for a junior student. Hopefully I will get there later. Is there any chance to explain in a simplified language? To be more specific, my problem arises from calculating, for example, the inner product of two scattered wave functions built up with Lippmann-Schwinger equation in momentum space (see, for example, in the Sakurai book), where you immediately encounter the product of two propagators described above.
 

Related to A contour integral frequenctly encountered

1. What is a contour integral and how is it used in science?

A contour integral is a mathematical concept used in complex analysis to calculate the integral of a function along a path or contour in the complex plane. In science, it is commonly used to solve problems in electromagnetics, fluid dynamics, and quantum mechanics.

2. What is the difference between a contour integral and a regular integral?

A regular integral calculates the area under a curve in the real number line, while a contour integral calculates the integral along a path in the complex plane. This allows for the evaluation of functions that are not defined on the real number line.

3. How do you choose the contour for a contour integral?

The contour is chosen based on the function being integrated and the desired properties of the integral. It is often chosen to avoid singularities, simplify the integral, or to ensure convergence.

4. Can a contour integral be applied to functions with multiple variables?

Yes, a contour integral can be extended to functions with multiple variables through the use of multi-dimensional contour integration. This is commonly used in fields such as mathematical physics and engineering.

5. What are some real-world applications of contour integrals?

Contour integrals have many real-world applications, including calculating electric and magnetic fields in electromagnetics, determining fluid flow patterns in fluid dynamics, and solving problems in quantum mechanics such as the Schrödinger equation.

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