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I would say that $D_\mu '\psi' = U(D_\mu \psi)$ is a requirement, rather than an assumption: When $U$ is a global transformation (i.e., when $\lambda$ is independent of $x$), we have trivially $\partial_\mu \psi' = U(\partial_\mu \psi)$, and so $(\partial_\mu \psi)^\dagger (\partial_\mu \psi) = (\partial_\mu \psi')^\dagger (\partial_\mu \psi')$. We define ...


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The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...



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