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I ask myself if the demand of local gauge invariance - say $U(1)$ invariance in free Dirac theory -$$L_D=\bar{\Psi}(i\gamma^\mu\partial_\mu-m)\Psi$$ is enough to define the full Maxwell-Dirac-Lagrangian UNIQUELY? I read that adding the Maxwell term $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $ demands for other requirements like 1) simplicity, 2) Lorentz invariance and 3) renormalizability. Similarly you need those other requirements for the interaction term $L_{int}=-j_\mu A^\mu$. While simplicity is clearly a metaphysical requirement Lorentz invariance is a genuinely physical one. But I am not sure about the role of renormalizability here?

What is the deeper physical meaning of it and why is the demand for renormalizability physically like the demand for Lorentz invariance?

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When you do naive calculations based on your Lagrangian (perturbation theory and thus Feynman diagrams, setting e.g. the parameter $m$ to the mass of the particle) you will at some point run into infinite results.

This is due to the fact, that $m$ in your Lagrangian (and other coupling constants) is not physical. The physical mass could for example be defined as the pole of the propagator, which will change with the order in perturbation theory.

Now these infinities can be erased by carefully defining what the physical parameters are and it then turns out, that the parameters of the Lagrangian (e.g. $m$) cancel the infinities of your naive calculation.

At least this is so in a renormalizable theory. In a non-renormalizable theory it can happen, that at some point in your perturbation theory you would get infinities you could not get rid of, without introducing new terms in the Lagrangian.

However, their are certainly people who believe that renormalizability is not a fundamental requirement for a theory which we only use in perturbation theory and which we know to break down at some high energy. In that sense I would not say that you can compare the demand for renormalizability with the demand for Lorentz invariance in its physical inevitability. But it produces "nicer" theories, which could also be argued to be simpler.

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  • $\begingroup$ But isn't it that the understanding of renormalizability has advanced in the last decades and went beyond what you just described? I know that it was originally invented just to remove infinities from perturbative calculations but hasn't it gained further physical meaning through considerations about effective field theories, energy scales und cutoffs? Is it that we can demand renormalizability say for QED because we expect it to be an effective field theory and these has to be renormalizable? $\endgroup$
    – Jasper Do
    Sep 30, 2020 at 7:15
  • $\begingroup$ Sure, renormalization is a large topic for itself. Effective theories are generally non-renormalizable, i.e. their Lagrangian includes an infinite number of interactions. You can, however, argue that QED is indeed an effective theory, where the non-renormalizable interactions are just suppressed by some large energy-factor. Due to the suppression we cannot measure the effects of these terms. In Weinberg's Vol. 1 Sec. 12.3, where he argues this and that renormalizability is not a requirement for a fundamental QFT by itself. $\endgroup$
    – drfk
    Sep 30, 2020 at 16:06
  • $\begingroup$ How to arrive at an effective theory starting with some Hamiltonian can be seen in Wilson's approach to renormalization. This is used in Statistical Mechanics/Field Theory and is not motivated by infinities in the theory but by coarse-graining procedures, i.e. integrating out degrees of freedom below some length scale. So renormalization is not just some strange method to get rid of infinities. For instance, changing the cut-off in a cut-off regularization means "integrating out" microscopic effects (e.g. the "real" underlying theory) leaving us with a theory whose effects we can measure. $\endgroup$
    – drfk
    Sep 30, 2020 at 16:58

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