Peskin and Schroeder derivation of Klein-Gordon propagator

Click For Summary
SUMMARY

The discussion centers on the derivation of the Klein-Gordon propagator as presented in "An Introduction to Quantum Field Theory" by Peskin and Schroeder, specifically addressing the transformation of variables in equation (2.54). The change from "ip(x-y)" to "-ip(x-y)" is clarified through a dummy variable substitution, where the integral transformation leads to a consistent exponential form. This transformation elucidates why p^0 equals -E_p, reinforcing the mathematical consistency in the derivation process.

PREREQUISITES
  • Understanding of quantum field theory principles
  • Familiarity with the Klein-Gordon equation
  • Knowledge of integral transformations in physics
  • Proficiency in complex exponentials and their properties
NEXT STEPS
  • Study the derivation of the Klein-Gordon propagator in detail
  • Explore integral transformations in quantum mechanics
  • Learn about the implications of negative energy solutions in quantum field theory
  • Investigate the role of dummy variables in mathematical physics
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on quantum field theory and the mathematical foundations of particle physics.

chern
Messages
10
Reaction score
0
In page 30 of book "An introduction to quantum field theory" by Peskin and Schroeder in the derivation of Klein-Gordon propagator, why p^0=-E_p in the second step in equation (2.54). and why change "ip(x-y)" to "-ip(x-y)"? I thought a lot time, but get no idea. Thank you for your giving me an explanation.
 
Last edited:
Physics news on Phys.org
chern said:
In page 30 of book "An introduction to quantum field theory" by Peskin and Schroeder in the derivation of Klein-Gordon propagator, why p^0=-E_p in the second step in equation (2.54). and why change "ip(x-y)" to "-ip(x-y)"?
It took me a long time to figure that out too, when I first studied P+S.

First, look at this 1D integral:
$$
\int_{-\infty}^{+\infty} dp \; e^{-ipx} ~.
$$ If you perform a change of dummy variable ##p \to p' = -p##, what do you get?

So in the 2nd step of (2.54), they're just converting the ##e^{ip\cdot(x-y)}## of the 2nd term in the last line on the previous page 29, so that both exponentials are the same, i.e., ##e^{-ip\cdot(x-y)}##. (The latter explains "why p^0=-E_p").
 
  • Like
Likes   Reactions: 1 person
Thank you!
 

Similar threads

Replies
18
Views
3K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K