Time Ordering Operator: Integrals & Step Function

  • Context: Graduate 
  • Thread starter Thread starter aaaa202
  • Start date Start date
  • Tags Tags
    Operator Time
Click For Summary
SUMMARY

The discussion centers on the application of the Heaviside step function, denoted as θ(t1-t2), in the context of rewriting integrals related to time-ordering operators in quantum field theory (QFT). The Heaviside function is defined such that θ(t1-t2) equals 1 when t1 is greater than t2 and 0 otherwise. The integration domains are extended from t1 or t2 to t, necessitating the multiplication by the step function to ensure that integrals over [t1,t] or [t2,t] yield zero. This relationship is crucial for understanding the time-evolution operator's behavior with a time-dependent Hamiltonian.

PREREQUISITES
  • Understanding of Heaviside step function in mathematical analysis
  • Familiarity with quantum field theory concepts, particularly time-ordering operators
  • Knowledge of integral calculus and its applications in physics
  • Basic understanding of Hamiltonian mechanics in quantum systems
NEXT STEPS
  • Study the properties and applications of the Heaviside step function in physics
  • Explore time-ordering operators in quantum field theory
  • Learn about the time evolution operator in the interaction picture
  • Review integral calculus techniques used in quantum mechanics
USEFUL FOR

Researchers and students in quantum field theory, physicists working with time-dependent Hamiltonians, and anyone interested in the mathematical foundations of quantum mechanics.

aaaa202
Messages
1,144
Reaction score
2
I really asked this question in another thread but it seems the original respondent gave up explaining me. My question is about the rewriting of the integrals from first to second line on the attached picture. The θ denotes the heaviside step function such that:
θ(t1-t2) = {1 t1>t2 , 0 t1<t2}
I think the idea is to extend the integration domain from respectively t1 or t2 to t but doing so we have to multiply by the step function in such a way that the integration over [t1,t] or [t2,t] yields zero. But how is this exactly related to when t1>t2 or t2>t1?
 

Attachments

  • integral.png
    integral.png
    18.3 KB · Views: 658
Physics news on Phys.org
Have a look at p. 16 of my QFT manuscript:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

I hope that helps to explain the appearance of the time-ordering operator in the solution of the time-evolution operator with a time-dependent Hamiltonian (or interaction part in the interaction picture).
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K