Question about the Klein Gordon Propagator

[maverick280857]

Hi,

I'm teaching myself QFT. I don't understand some of the calculations described on pages 29 and 30 of Peskin and Schroeder's book on QFT.

...we can write $[\phi(x),\phi(y)] = \langle 0|[\phi(x),\phi(y)]|0\rangle$. This can be rewritten as a 4 D integral as follows, assuming for now that $x^{0} > y^{0}$:

$$\langle 0|[\phi(x),\phi(y)]|0\rangle = \int \frac{d^{3}p}{(2\pi)^3}\left\{\frac{1}{2E_p}e^{-ip\cdot(x-y)}|_{p^0 = E_{p}} + \frac{1}{-2E_p}}e^{-ip\cdot(x-y)}|_{p^0 = -E_{p}}\right\}$$

The next step is where I have a problem

$$=_{x^0 > y^0} \int \frac{d^{3}p}{(2\pi)^3}\int \frac{dp^0}{2\pi i}\frac{-1}{p^2-m^2}e^{-ip\cdot(x-y)}$$

I think I'm making a mathematical mistake somewhere, but I haven't been able to figure it out yet.

Making a digression, the propagator can be written in momentum space as

$$\Delta_{F}(p) = \frac{1}{2E_{p}}\left[\frac{1}{p^0 - (E_p - i\epsilon)}-\frac{1}{p^0 + (E_p - i\epsilon)}\right]$$

(Feynman prescription)

Plugging this back into the expression for the Green's function $G(x-x') = \int \frac{d^4 p}{(2\pi)^4}e^{-ip\cdot(x-x')}\frac{1}{(p^0)^2 - E_{p}^2}$ we get

$$\Delta_{F}(x-x') = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E_{p}}\int \frac{dp^{0}}{2\pi} e^{ip\cdot(x-x')}\left[\frac{1}{p^0 - (E_p - i\epsilon)}-\frac{1}{p^0 + (E_p - i\epsilon)}\right]$$

This doesn't give me what I want, but I think I am close to it as there is a -i and two theta functions that enter into the final expression if this is integrated.

Specifically,

1. I don't understand how the extra factor of {-i} in the expression in the quote box above (from Peskin & Schroeder) comes in.

2. What is the motivation for the choice of the contour shown on page 30?

3. What is the significance of $x^0 > y^0$.

Thanks!

 

[tirrel]

didn't really follow your calcuation, but it seems to me that you are trying to abtain formula 2.60.. and this one contains theta functions...

1. which extra facor?

2. the motivation is having the last line of 2.54 equal to the precedent one... only with that contour you get the two poles that you need (you need two residues!),,,

3. the significance is that the last line of 2.54 and the line before are equal only if x_0>y_0 (otherwise, as explained the last line is zero)

 

[Entropia]

Dear student,

I am happy to see that you are teaching yourself QFT and tackling challenging topics such as the Klein Gordon propagator. Let me address your questions one by one.

1. The extra factor of {-i} in the expression from Peskin & Schroeder comes from the fact that we are using the Feynman prescription for the propagator. This prescription introduces a small imaginary part to the energy of the particle, which helps to avoid any singularities in the integral. This is a common technique in quantum field theory, and it is known as "i\epsilon prescription". You can think of it as a mathematical trick to make sure that our calculations are well-behaved.

2. The choice of the contour on page 30 is motivated by the fact that we want to perform the integral over the energy component of the momentum. This can be done by using the residue theorem from complex analysis, which requires us to choose a closed contour in the complex plane. The contour chosen in the book is a commonly used one, known as the "Feynman contour". It allows us to perform the integral over the energy component by encircling the poles in the complex plane.

3. The significance of x^0 > y^0 is related to the choice of the contour. In order to use the residue theorem, we need to make sure that the contour does not cross any singularities. In this case, the singularities are located at p^0 = \pm E_p. By choosing x^0 > y^0, we are avoiding crossing the poles in the complex plane, and thus ensuring that our calculation is well-defined.

I hope this helps clarify your doubts. Keep up the good work in teaching yourself QFT, and don't hesitate to reach out if you have any further questions.

Best regards,