# Contour Integral from Peskin & Schroeder Intro to QFT

#### rolltider

1. I'm having some trouble with some of the contour integrals covered in Chapter 2 of Peskin & Schroeder's Intro to QTF. I'm not so much as looking for answers to the integral (in fact, the answers are given in the textbook), but I was hoping someone could point me to some resources to use to become more familiar and confident with integrals like this as so far my understanding of it is very "hand-wavy."

2. Relevant equations: In section 2.4 of P&S, the propagator for the Klein Gordon field is derived and found to have the form: $$D(x-y) = \frac{-i}{2(2\pi)^2 r} \int_{-\infty}^{\infty}{dp \frac{p e^{ipr}}{\sqrt{p^2+m^2}}}$$

Now, I can readily see how this is a contour integral along the real p-axis with (isolated?) singular points at $$p=\pm im$$

But, I'm not familiar with the contour deformation done in the book: 3. My attempt: I'm not very clear on how pushing the contour up like that doesn't change the value of the integral, or why this is easier to do in practice than to do the integral along the real line. I suppose it makes sense that you'd want to wrap around one of the poles so that you can use the residue theorem to show that the part of the contour that wraps around the pole goes to zero, since the Residue at p=+/- im is zero (or I may be completely wrong about that) leaving only the nonzero part of the contour along the branch cut, but this only makes sense to me in a "hand-wavy" sort of way.

I did take complex analysis a long time ago as an undergrad, but either my professor wasn't that great or I just have a terrible memory because I don't remember doing anything like this. From what I recall, we used the book "Complex Variables and Applications" by Churchill but that book doesn't seem to prove entirely useful when trying to review integrals like this.

Any point in the right direction would be much appreciated!

Related Advanced Physics Homework News on Phys.org

#### RUber

Homework Helper
To do this, you are essentially saying that:
*edited*
$\int_{-\infty}^\infty f(p) dp = \int_{-\infty}^\infty f(p) dp + \lim_{r\to \infty} \int_{0}^\pi f(re^{i\theta})ire^{i\theta}d\theta +\lim_{r\to \infty} \int_{\pi}^0 f(re^{i\theta}) ire^{i\theta}d\theta$
Your function appears to go to zero as r goes to infinity, so the last integral can be set to zero, leaving you with a contour around the entire upper half plane.
From there, you know that an counter-clockwise closed contour integral is equal to zero if it contains no singularities, so you break the contour into one counterclockwise contour with no singularities and one clockwise contour around the area with singularities.

#### rolltider

So, please correct me if I'm wrong, but it seems to me that by writing: $$\int_{-\infty}^{\infty}{f(p)dp} = \int_{-\infty}^{\infty}{f(p)dp} + \lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(re^{i\theta})ire^{i\theta}d\theta} + \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(ire^{i\theta})ire^{i\theta}d\theta}$$

what you're doing is closing the contour from -infinity to infinity on the real p-axis with a semicircle over the upper-half on the plane and then subtracting the same circle so that the value of the overall integral stays the same? And then by rewriting the last term:

$$-\lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(ire^{ir\theta})ire^{ire\theta}d\theta} = \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(ire^{ir\theta})ire^{ire\theta}d\theta}$$

You claim that it goes to zero as r goes to infinity. Now, why would the last term go to zero as r goes to infinity and not the second term, too? Aren't they the same terms save for a sign? That is, unless I'm completely misunderstanding your explanation.

#### rolltider

To do this, you are essentially saying that:
*edited*
$\int_{-\infty}^\infty f(p) dp = \int_{-\infty}^\infty f(p) dp + \lim_{r\to \infty} \int_{0}^\pi f(re^{i\theta})ire^{i\theta}d\theta +\lim_{r\to \infty} \int_{\pi}^0 f(re^{i\theta}) ire^{i\theta}d\theta$
Your function appears to go to zero as r goes to infinity, so the last integral can be set to zero, leaving you with a contour around the entire upper half plane.
From there, you know that an counter-clockwise closed contour integral is equal to zero if it contains no singularities, so you break the contour into one counterclockwise contour with no singularities and one clockwise contour around the area with singularities.
Okay, I think I get it now after thinking about it some more:

So the integral $$\int_{-\infty}^{\infty}{f(p) dp}$$ can be closed above the p-axis by adding a term (and subtracting off the same term to make sure the integral remains unchanged): $$\int_{-\infty}^{\infty}{f(p) dp} = \int_{-\infty}^{\infty}{f(p) dp} + \lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(re^{i\theta})ire^{i\theta} d\theta} + \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(re^{i\theta})ire^{i\theta} d\theta}$$

But since the integrand of both the new terms go to zero as r goes to infinity, you can set the third term to zero and keep the second term at no cost. Now, since I have a closed contour, I can break down the following contour like I drew here: And since the first contour in the image does not enclose any singularities, that term is zero leaving only the second term, which is precisely the contour deformation done in Peskin & Schroeder.

Is that correct?

#### RUber

Homework Helper
That is how I understand it.

#### TSny

Homework Helper
Gold Member
Equivalently, but perhaps a bit shorter, is to consider just your middle figure. The integral around the entire contour is zero. Integrals along the arcs at infinity are zero. So, the integral along the real axis must be the negative of the integral along the "dipping" portion of the contour.

#### Attachments

• 12.3 KB Views: 310

#### rolltider

Equivalently, but perhaps a bit shorter, is to consider just your middle figure. The integral around the entire contour is zero. Integrals along the arcs at infinity are zero. So, the integral along the real axis must be the negative of the integral along the "dipping" portion of the contour.
That makes a lot of sense, too. So basically by flipping the direction of the "dipping" portion of the contour you make the two equivalent. Makes much more sense now, thanks!

#### DerHutmacher

Hello all, I have a additional question to this topic ( I don't regard is at off-topic which is why I didn't open a new thread. If I'm mistaken, please correct me). In general I understand the argument made above, but I do not see, why one can use the residue theorem, since by closing the contour across the line $$[im,\infty)$$ one crosses the branch cut and therefore the integrand is not continuous along the path and so is not analytical.

Best regards,
Lev

#### RUber

Homework Helper
Good point, Lev.
The integral that you want to compute is the one around the branch cut. I think you would agree that the larger contour with the notch cut out encloses a region where the function is analytic, and therefore would have an integral equal to zero.
Instead of using Cauchy's thm to simplify the contour, we could more generally use the concept of path independence. Define a domain D such that the function is analytic everywhere on D. Then, through path independence, any other path between the points should have the same value.
In this case, our path is between $-\infty$ and $+\infty$. Using the same rationale, we can justify setting some pieces of the contour integral to zero in the limit near infinite radius.

#### DerHutmacher

Good point, Lev.
The integral that you want to compute is the one around the branch cut. I think you would agree that the larger contour with the notch cut out encloses a region where the function is analytic, and therefore would have an integral equal to zero.
Instead of using Cauchy's thm to simplify the contour, we could more generally use the concept of path independence. Define a domain D such that the function is analytic everywhere on D. Then, through path independence, any other path between the points should have the same value.
In this case, our path is between $-\infty$ and $+\infty$. Using the same rationale, we can justify setting some pieces of the contour integral to zero in the limit near infinite radius.
Thank you for the quick answer but I still do not quite understand what we are doing to finally evaluate the integral? After we showed that our former integral is equal to the contour integral that is attached in the file, I would still not know what I should do, to get any value from the integral, since I can not use the residue formula. In the book they say: "To evaluate the integral we push the contour up to wrap around the upper branch cut. defining $$\rho = ip$$, we obtain $$\frac{1}{4 \pi^2 e}\int_{m}^{\infty} d \rho \frac{\rho e^{- \rho r}}{\sqrt{\rho^2-m^2}} \tilde{\tiny{r \to \infty}} e^{-mr}$$

For me, it seems that the attempt in this posts aims to use the residue theorem but I do not necessarily see that this is, what they are doing in the book. I hope I made my confusion clear.

Best,
Lev

#### Attachments

• 12.3 KB Views: 28

#### DerHutmacher

Hi all, I finally understood what is happening. So no residue is taken, but after the deformation of the contour, for every point of the path left to the branch cut, one subtracts the value of the horizontal point right to the branch cut. Because of the branch cut, this value is the same, if we take the limit of the path distance to the branch cut, but only with a minus sign, which gives us a total of 2 times the integral along the Interval $$[im,\infty)$$, which gives the integral in my last comment. Then I guess that they argue that the biggest part of the integral comes from near the singularity for a big r, which makes the last "conclusion".

"Contour Integral from Peskin & Schroeder Intro to QFT"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving