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In QED, when the Dirac Lagrangian is found to be not invariant under a local phase transformation,

$\psi$ $\longrightarrow$ $\psi'$ = $e^{i\theta(x)} \psi$

one tries to force it to get the desired invariance by introducing the covariant derivative. Is there a compelling argument to this? Does it have to be invariant locally because it is globally?

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  • $\begingroup$ This isn't an answer, but local invariance implies global invariance, not the other way around. $\endgroup$
    – Eletie
    Commented Feb 17, 2022 at 17:23
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    $\begingroup$ Does this answer your question? Why do we require local gauge invariance. See links on this question. $\endgroup$
    – Eletie
    Commented Feb 17, 2022 at 17:25

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There is only an A-field introduced if the field is locally gauged (in QED). The Lagrangian is trivially invariant under a global transformation. If you promote the globality to locality, there naturally appears a mediating field to compensate for the local gauges, so the Lagrangian stays invariant. Which is exactly what you want. There is no fundamental need for it to hold but interactions between particles exìst. Are interactions a fundamental need?

In a world in which interaction would be absent, the local gauge symmetry wouldn't hold. It introduces interaction, the charge being the generator of the transformation (the exponent in the gauge function).

In other words, it's the very fact that there interactions that forces us to introduce local gauge invariance and not the symmetry causing a force mediating field. In a non-interacting world, the principle doesn't give interaction! Non-interacting matter fields would not be invariant under local gauges and there would be no need to add symmetrizing fields. By the simple fact they didn't exist.

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  • $\begingroup$ Hi Felicia, thank you very much for your inspiring answer! So it is basically introduces to match the world we see, right? $\endgroup$ Commented Feb 18, 2022 at 1:43

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