Bare Charges of Identical Particles: Is It Possible?

Click For Summary
SUMMARY

The discussion centers on the relationship between bare charges and physical charges of identical particles under electromagnetism. It concludes that while bare charges are fundamental, gauge invariance ensures that they renormalize identically, as dictated by the Ward identity. This identity leads to the conclusion that for two particles with the same bare charge, their physical charges must also be identical. The equations referenced from Peskin/Schroeder illustrate this relationship clearly.

PREREQUISITES
  • Understanding of gauge invariance in quantum field theory
  • Familiarity with the Ward identity
  • Knowledge of renormalization in particle physics
  • Basic concepts from quantum electrodynamics (QED)
NEXT STEPS
  • Study the implications of gauge invariance in quantum field theories
  • Explore the derivation and applications of the Ward identity
  • Investigate the process of renormalization in QED
  • Read Peskin and Schroeder's "An Introduction to Quantum Field Theory" for deeper insights
USEFUL FOR

The discussion is beneficial for theoretical physicists, quantum field theorists, and students studying particle physics, particularly those interested in the nuances of charge renormalization and gauge theories.

Riposte
Messages
15
Reaction score
0
For two particles which have identical physical charges (say under electromagnetism), are the bare charges necessarily the same? Since the physical charge is related to the bare charge by photon and particle renormalization factors, I don't see how this could be the case in general.

In some sense, the bare charges are fundamental and therefore having two distinct charges renormalize to be identical strikes me as fine-tuned. Any thoughts?
 
Physics news on Phys.org
This fine-tuning is avoided thanks to gauge-invariance. A consequence of gauge-invariance is the Ward identity, which tells us that the charges will renormalize in the same way so as to preserve equality.
 
Thanks, I think I get it now. In the terminology of Peskin/Schroeder we have:

eZ1 = e0 Z2 Z3^(1/2),
e'Z1' = e0 Z2' Z3^(1/2),

for two particles with the same bare charge. The Ward identity forces Z1 = Z2 and Z1' = Z2', and therefore the two must have the same physical charge as well
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
999
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K