QFT: Gauge Invariance, Ghosts, Symmetry & Lorentz Invariance

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SUMMARY

The discussion centers on the complexities of gauge invariance in quantum field theory, particularly in relation to ghosts and gauge-fixing terms. It establishes that the unitary gauge maintains gauge invariance, while the background field gauge requires the boson field to equal the background field to uphold this principle. The conversation also explores the relationship between Lorentz invariance and CPT symmetry, concluding that while Lorentz breaking does not necessarily imply a failure of CPT, any theory that fails CPT likely breaks Lorentz invariance. The necessity of hermiticity in interactions and the implications of bounded Hamiltonians are also emphasized.

PREREQUISITES
  • Understanding of quantum field theory concepts
  • Familiarity with gauge invariance and gauge-fixing terms
  • Knowledge of Lorentz invariance and CPT symmetry
  • Basic principles of hermiticity in quantum mechanics
NEXT STEPS
  • Research the implications of ghosts in quantum field theories
  • Study the background field gauge and its applications
  • Explore the relationship between CPT symmetry and Lorentz invariance
  • Investigate non-hermitian Hamiltonians in quantum mechanics
USEFUL FOR

The discussion is beneficial for theoretical physicists, quantum field theorists, and advanced students seeking to deepen their understanding of gauge theories, Lorentz invariance, and the implications of symmetry in quantum mechanics.

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When quantizing boson fields, ghosts and gauge-fixing terms seem to break gauge invariance. The unitary gauge (where there are no ghosts or gauge-fixing terms) respects gauge invariance however. So which is correct - is the Standard Model a gauge theory or not?

Sometimes I hear people speak about breaking Lorentz invariance. Does anyone have any idea what they mean? I think it has to do with CPT symmetry. There is a close relationship between CPT symmetry and Lorentz symmetry - one practically implies the other. So if CPT is broken, then I think that's where they are saying Lorentz symmetry is broken. Does this sound like pseudo-science, because Lorentz symmetry seems to be sacred?
 
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Any local quantum field theory that respects lorentz invariance and that has a bounded hamiltonian automatically respects CPT. Does the converse hold true? Certainly failure of CPT implies that you don't have a field theory and any such theory probably breaks lorentz invariance so that much I think is correct.

But does lorentz breaking imply a corresponding failure of CPT? I don't think so.

As for ghosts. We require that ghosts cancel at the end of calculation in order to have a consistent theory. When this fails to happen, it implies the theory is unstable and breaks unitarity at some point (probabilities don't add up to 1).
 
Haelfix said:
Any local quantum field theory that respects lorentz invariance and that has a bounded hamiltonian automatically respects CPT.
Let's say we have an interaction phi^\dagger phi^2. This respects Lorentz invariance. Does this respect CPT?
 
nrqed said:
Let's say we have an interaction phi^\dagger phi^2. This respects Lorentz invariance. Does this respect CPT?

The condition is Lorentz invariance and hermicity, so if you add its hermitian conjugate than it should. The bounded Hamiltonian I think is needed for thermodynamical reasons, but not quantum reasons.

Anyways, there is a particular gauge called the background field gauge that respects gauge invariance, but in order for that to happen, you have to set the boson field equal to the background field, and that is confusing to me as to why you can do that, so I thought I'd start by asking some simple questions about gauge invariance in general.
 
Thats a good catch, you definitely need hermiticity in the interaction hamiltonian in there (although that probably follows from other conditions if we shift the axioms around a bit, eg that the field theory is unitarity, that its built up from free fields as well as being bounded from below).

I think I read something about the possibility of using a nonhermitian hamiltonian once, but it was a little ackward and I didn't understand it.
 

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