Minimal Subsitution from Lorentz Invariance

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SUMMARY

The discussion centers on the coupling of photons to the Dirac field for electrons, specifically through the Dirac equation (i\not{\partial -m })\psi=0. The key point is the distinction between Lorentz invariance and gauge invariance. While the transformation of the space-time measure is expressed as ∂^μ → ∂^μ + ieA^μ (D^μ), it is gauge invariance that ensures the invariance of this change, as demonstrated by the transformation of the electromagnetic gauge Aμ and the wave function ψ.

PREREQUISITES
  • Understanding of the Dirac equation and its implications in quantum mechanics.
  • Familiarity with Lorentz invariance and its role in relativistic physics.
  • Knowledge of gauge invariance and electromagnetic gauge transformations.
  • Basic concepts of quantum field theory, particularly in relation to particle interactions.
NEXT STEPS
  • Study gauge invariance in quantum electrodynamics (QED).
  • Explore the implications of the Dirac equation in particle physics.
  • Learn about electromagnetic gauge transformations and their mathematical formulations.
  • Investigate the role of covariant derivatives in quantum field theories.
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The discussion is beneficial for theoretical physicists, quantum field theorists, and students studying advanced concepts in particle physics and gauge theories.

Sekonda
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Hello,

My question is on coupling the photons to our Dirac field for electrons, we have the Dirac equation:

(i\not{\partial -m })\psi=0

By Lorentz invariance we can change our space-time measure by:

\partial ^\mu \rightarrow \partial ^\mu+ieA^\mu\equiv D^\mu

Though I cannot see why Lorentz invariance implies that this change is invariant?

Sorry if I haven't explained my issue well, any help appreciated!

SK
 
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It's not Lorentz invariance that's involved here, it's gauge invariance. Under an electromagnetic gauge transformation, Aμ → Aμ + λ and ψ → ψ exp(-ieλ(x)). So ∂μψ → (∂μψ - ieψλ) exp(-ieλ(x)), which by itself is not covariant, but the combination

Dμψ = (∂μ + ieAμ)ψ → (∂μψ + ieAμ - ieψλ + ieψλ) exp(-ieλ(x)) = (Dμψ) exp(-ieλ(x)) is.
 

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