Time-ordered products derivation in "QFT and the SM" by Schwartz

  • A
  • Thread starter Hill
  • Start date
  • Tags
    Derivation
  • #1
Hill
Homework Helper
Gold Member
565
449
TL;DR Summary
Replacing a field eigenstate by the field operator
This question is not crucial, but I'd like to understand better the equation (14.35) in this derivation:

1710605797266.png

1710605837206.png

Here ##\Phi## is an eigenvalue of ##\hat \phi##, i.e., ##\hat \phi (\vec x ) |\Phi \rangle = \Phi (\vec x) |\Phi \rangle##.

First, I think that there is a typo in (14.35): the Hamiltonian should be evaluated at time ##t_{j+1}## rather than ##t_n##. Is it right?

But the question is, why the exponential is included in (14.35)? Wouldn't it be correct just to write, $$\int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \Phi_j (\vec x_j) \langle \Phi_j| = \hat \phi (x_j) \int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \langle \Phi_j|$$?
 
Last edited:
Physics news on Phys.org
  • #2
Your formula without the exponential is correct as well. A formula with an exponential is studied because that's what one needs in (14.34).
 
  • Like
Likes pines-demon
  • #3
Demystifier said:
Your formula without the exponential is correct as well. A formula with an exponential is studied because that's what one needs in (14.34).
Thank you for the clarification.
 

Similar threads

Replies
2
Views
345
Replies
5
Views
401
  • Quantum Physics
Replies
13
Views
1K
Replies
1
Views
663
  • Quantum Physics
Replies
4
Views
2K
  • Quantum Physics
Replies
4
Views
939
  • Quantum Physics
Replies
1
Views
1K
  • Quantum Physics
Replies
12
Views
2K
Replies
15
Views
1K
  • Quantum Physics
Replies
4
Views
1K
Back
Top