Regularization is mandatory. What about renormalization?

We need to regularize in order to declare with confidence that infinities drop out from measurable quantities, e.g. in the form of a cutoff scale. In general, the amplitudes in QFT depend on the regulator, e.g., the cutoff scale $$\Lambda$$: $$\Gamma = \Gamma(\mu, \Lambda)$$ where $$\mu$$ is the scale at which we evaluate the amplitude.

However, the difference between a given amplitude at two different energy scales is actually independent of $$\Lambda$$: $$\Gamma(\mu, \Lambda) - \Gamma(\mu', \Lambda) \propto \ln(\mu/\mu')\, .$$ In this sense, the regulator drops out from measurable quantities and with it the dangerous UV infinites.

Therefore, I was wondering if renormalization is really mandatory or just a convenient method to make calculations simple?

• Everything here seems backwards and confused. Regularization is not mandatory (e.g., some theories are naturally finite; moreover, if you consider fields to be distributions rather than functions, no infinities appear at any point, cf. Glaser-Epstein). Similarly, renormalisation has nothing to do with divergences, and it is not a "method"; rather, it is just the statement that non-linearities (either classical or QM; either finite or divergent) modify the parameters that appear in the linear theory, cf. physics.stackexchange.com/a/300734/84967 – AccidentalFourierTransform Jul 8 '19 at 13:58

Regularization is a way to rewrite the divergent integrals in QFT in a nicer form so that you can point out the divergent piece easily. Now, renormalization techniques at first relate model parameters to real measurable quantities and then the scale dependence of those parameters. For example, fine structure constant at low energies is 1/137 but at sufficiently high energies (TeV)is 1/127. To see these we have to study renormalisation group equation which describes how measurable quantities "run" with scale $$\mu$$.