# Homework Help: QFT in a nutshell: Propagators

1. Nov 16, 2007

### noospace

1. The problem statement, all variables and given/known data

I'm trying to show that the general form of the propagator is

$D(x) = - \int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_k t - \vec{k}\cdot\vec{x})}\theta(x^0) + e^{i(\omega_k - \vec{k}\cdot\vec{x})}\theta(-x^0)]$

but my answers always seem to differ by a sign.

2. Relevant equations

$D(x-y) = \int \frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}$

3. The attempt at a solution

If we take $x^0 > 0$ then Zee states that we take the integration over $k^0$ to be in the upper half of the complex plane.

This gives

$D(x) = \int \frac{d\vec{k}dk^0}{(2\pi)^4}\frac{e^{i k\cdot x}}{k^2 - m^2 + i\epsilon}$
$D(x) = \int \frac{d\vec{k}}{(2\pi)^4}2\pi i\,Res\left(\frac{e^{i k\cdot x}}{(k^0)^2 - \vec{k}^2-m^2 + i \epsilon} , -\sqrt{\omega_k^2-i\epsilon} \right)$
$D(x) = -i\int \frac{d\vec{k}dk^0}{(2\pi)^3}\frac{e^{-i(\omega_k + \vec{k}\cdot\vec{x})}}{2\omega_k}$

Zee gets

$D(x) = -i\int \frac{d\vec{k}dk^0}{(2\pi)^3}\frac{e^{i(\omega_k - \vec{k}\cdot\vec{x})}}{2\omega_k}$

2. Nov 17, 2007

### nrqed

there are several mistakes and I don't know if they are simple typos or realmistakes. See my comments below
Do you see why one must do that?
Sounds ok (note that since epsilon is small, you basically substitute $k_0 = \omega_k$ )
the dk_0 should not be there anymore. And a time is missing.
The time is missing.

I am confused. This is not the equation you quote at the top of your message.
can you give the page number?
In any case, since you are integrating over all the values of $k_i$ (positive and negative), you may always switch their sign in the integrand in which case I think your answer agrees with his.

3. Nov 18, 2007

### noospace

Hi nrqed,

Thanks for your reply. The page reference is p. 23 Eq. (23).

We want to show that

$D(x) = - i\int \frac{d^3\vec{k}}{(2\pi)^32\omega_k}[e^{-i(\omega_k t - \vec{k}\cdot\vec{x})}\theta(x^0) + e^{i(\omega_k - \vec{k}\cdot\vec{x})}\theta(-x^0)]$.

Although Zee states the contour prescription for the solutions without justification, I think I at least understand half of the motivation. For the positive energy solutions, the particles are traveling forward in time so can be expected to satisfy the causlity principle. If we interpret the Fourier transform of the Green's function as a Laplace transform, then the theory of inverse laplace transforms prescribes the correct contour (integrate below the pole).

For the negative energy solutions I'm not totally sure, but I believe it is to do with Feynman who said it's better to treat these as postive energy solutions traveling backward in time, so the opposite of the causality principle applies. Please correct me if I have a mis-understanding.

You're right that the dk_0 should be absent, that's a typo. I thought about using the transformation $\vec{k} \to - \vec{k}$, but then it occured to me that this would imply $d\vec{k} \to -d\vec{k}$ also, which is no good. Perhaps I should show you my complete working:

$D(x) = \int \frac{d\vec{k}}{(2\pi)^4}2\pi i\,Res\left(\frac{e^{i k\cdot x}}{(k^0)^2 - \vec{k}^2-m^2 + i \epsilon} , -\sqrt{\omega_k^2-i\epsilon} \right)$

$D(x) = \frac{i}{(2\pi)^3}\int d\vec{k} \lim_{k^0 \to -\sqrt{\omega_k^2-i\epsilon}}\frac{e^{i(k^0x^0-\vec{k}\cdot\vec{x})}(k^0+\sqrt{\omega_k^2-i\epsilon})}{(k^0)^2-\vec{k}^2-m^2+i\epsilon}}$

using the definition of a residue. Then using l'Hopital's rule for limits gives

$D(x) = \frac{i}{(2\pi)^3}\int d\vec{k} \lim_{k^0 \to -\sqrt{\omega_k^2-i\epsilon}}\frac{e^{i(k^0x^0-\vec{k}\cdot\vec{x})}+(k^0+\sqrt{\omega_k^2-i\epsilon})e^{i(k^0x^0-\vec{k}\cdot\vec{x})}ix^0}{2k^0}$

Setting epsilon to zero gives

$D(x) = -i\int \frac{d\vec{k}}{(2\pi)^3} \frac{e^{-i(\omega_k t+\vec{k}\cdot\vec{x})}}{2\omega_k}$.

Last edited: Nov 18, 2007
4. Nov 18, 2007

### noospace

Hang on. It just occured to me that the implication $\vec{k} \to - \vec{k} \implies d\vec{k} \to -d\vec{k}$ might not be valid since we're doing volume integrals here. Perhaps this saves me?

5. Nov 18, 2007

### nrqed

Actually, the choice of contour follows directly from mathematical considerations. You want the contribution from the contour to go to zero so you must choose the contour that will make the exponential being of the form e^(- C) with C C a positive quantity.
You are right that changing the sign of the k would change the overall sign. I am puzzled too, to be honest. (By the way, in the second line just above equation 23 when he gives the result of the integration for x_0 >0 within the paragraph there is a typo there and that typo is mentioned in his errata page). But I get the same result as you.
I am puzzled.

6. Nov 18, 2007

### jdstokes

Yep, changing k does change the integral's sign because the jacobian determinant in this case is -1.

I don't understand your argument about contours. In every exponential I can see there is an imaginary unit so none of them will die out, regardless of sign. Are you talking about the exponentials you get after doing the integration over 3-space? If so, that might make sense.

Another thing: what is theta? Is it possible that he has absorbed the -1 into one of these?

Btw to avoid any confusion, I'm the OP.

7. Nov 18, 2007

### nrqed

What I mean is the exponent is $e^{i k_0 x_0}$. If x_0 is positive, you want k_0 to go to + i infinity in order to dampen the exponential so you must close above the axis.

Theta (x_0) is just defined to be +1 for x_0 >0 so no help from that.

8. Nov 18, 2007

### jdstokes

Okay, I think that makes sense. Thanks. Perhaps this is another error?

Incidentally, I've been trying to answer I.3.1 but keep coming up against integrals I have no idea how to evaluate. Right now I'm trying to show that the propagator decays exponentially for purely spacelike separations. This basically amounts to doing the following integral from p. 27 of Peskin and Schoeder

The problem is, I'm not sure I understand the first step of their derivation!

9. Nov 19, 2007

### Avodyne

Correct, but it's also true in one dimension, because when dk is replaced by -dk, the limits of integration also change to +infinity to -infinity, and swapping these removes the overall minus sign.

More generally, it is the absolute value of the jacobian that multiplies the integrand after a transformation.

Last edited: Nov 19, 2007
10. Mar 22, 2011

### jmlaniel

Is there a conclusion on the sign problem on "k"?

I just tried to do the integral myself and I found exactly Zee's solution (as shown in post #1) except for the first term (I have the wrong sign on $$\vec{k}$$). I don't have the sign problem in the second term.

From what I have read in this thread, the solution is just to change the sign of $$\vec{k}$$ in the first term and everything is OK? This would be in agreement with a comment that I found on Zee's website saying that k's sign are not important since we integrate over all k's (http://www.kitp.ucsb.edu/~zee/QFTedit.html" [Broken], see the comment on p.23 of the 1st edition).

(BTW, I am using the second edition of Zee which has the corrected sign term mentionned before).

Last edited by a moderator: May 5, 2017