Calculating Z0 Decay Matrix Element: Finding the Error

  • Context: Graduate 
  • Thread starter Thread starter marlon1
  • Start date Start date
  • Tags Tags
    Decay
Click For Summary
SUMMARY

The discussion focuses on calculating the Z0 decay matrix element into a fermion-antifermion pair using the propagator, vertex element, and Fermi's golden rule. The user identifies a discrepancy in their calculations, noting that while the propagator contributes a term of 1/Mz^4, the vertex factor introduces a term of Mz^2, leading to confusion regarding the expected result of Mz^2. A suggestion is made to review the Feynman diagram to clarify the role of the propagator in this tree-level decay process.

PREREQUISITES
  • Understanding of Fermi's golden rule
  • Familiarity with Feynman diagrams
  • Knowledge of particle physics, specifically Z0 boson decay
  • Proficiency in matrix element calculations
NEXT STEPS
  • Review the principles of Fermi's golden rule in particle decay
  • Study the construction and interpretation of Feynman diagrams
  • Learn about the role of propagators in quantum field theory
  • Investigate the specific decay processes of the Z0 boson
USEFUL FOR

Particle physicists, students studying quantum field theory, and researchers involved in decay matrix element calculations will benefit from this discussion.

marlon1
Messages
5
Reaction score
0
I want to calculate the matrix element of Z0 decay into a fermion, antofermion pair given the propagator, vertex element and fermi's golden rule.

But somewthing is going wrong.

First the propagator has a term 1/Mz^2 so
the MatrixElement^2 will go like 1/Mz^4.

but the vertex factor has a term Mz^2 so the matrix element will finally go
1/Mz^2

But according to the answer the matrix element must go like Mz^2 so I'm wrong a factor Mz^4

take a look here;

https://www.physicsforums.com/attachment.php?attachmentid=8956&d=1169839397"
 
Last edited by a moderator:
Physics news on Phys.org
why do you have a propagtor when it is a tree-level decay... look at your Feymann diagram and you should see no propagtor.. unless I have misunderstood your interaction.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K