Vector Boson Propagator and gauge

Click For Summary
SUMMARY

The discussion focuses on the differences between two vector boson propagators: \(\frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}\) and \(\frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}\). The second propagator is associated with a gauge transformation that eliminates Goldstone modes, which are not present when dealing with massive vector bosons. The conversation highlights the importance of gauge invariance in quantum field theory (QFT) and suggests consulting Srednicki's textbook for a deeper understanding of these concepts.

PREREQUISITES
  • Understanding of quantum field theory (QFT)
  • Familiarity with gauge transformations
  • Knowledge of vector bosons and their properties
  • Basic comprehension of Feynman diagrams
NEXT STEPS
  • Study gauge invariance in quantum field theory
  • Learn about Goldstone bosons and their role in particle physics
  • Explore the implications of different gauge choices in QFT
  • Read "Quantum Field Theory" by Mark Srednicki for detailed insights
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on quantum field theory, gauge theories, and particle physics. This discussion is beneficial for anyone seeking to understand the implications of gauge choices on vector boson propagators.

ChrisVer
Science Advisor
Messages
3,372
Reaction score
465
Well I'm trying to understand the difference between these propagators:

\frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}

and

\frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}

My professor told me that they are different gauges, and the from the second you rule out the goldstone modes. Can someone explain it to me a little better or refer me to some textbook? I don't also get the meaning of goldstone modes- if you have a massive vector boson then you don't have goldstone modes [they become the longitudial dofs]

Also I am having one more question. If the first doesn't kill out the goldstone bosons, then can someone -after arriving at the end result- kill them? Maybe by a gauge transformation? and what gauge transformation?
 
Last edited:
Physics news on Phys.org
ChrisVer said:
Well I'm trying to understand the difference between these propagators:

\frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}

and

\frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}

My professor told me that they are different gauges, and the from the second you rule out the goldstone modes. Can someone explain it to me a little better or refer me to some textbook? I don't also get the meaning of goldstone modes- if you have a massive vector boson then you don't have goldstone modes [they become the longitudial dofs]

Also I am having one more question. If the first doesn't kill out the goldstone bosons, then can someone -after arriving at the end result- kill them? Maybe by a gauge transformation? and what gauge transformation?

I am not sure what your prof meant by "killing off the Goldstone modes". The two expressions correspond to different gauges. If you put a parameter ##\xi## in front of the ##k^\mu k^\nu## term, upon summing up any gauge invariant combination of Feynman diagrams (in particular the sum of the diagrams of a given loop order), the ##\xi## dependent terms will cancel out, showing that adding that term does not change anything to the final result. I guess it is possible to think of these extra terms as being associate to the Goldstone modes (like some pieces of the ##g^{\mu \nu}## piece in QED correspond to the longitudinal modes) .
 
You need to look at a QFT text for details; try Srednicki.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
6K