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I have been doing some reading on Wilsonian renormalization and also Effective Field Theories.

It's my understanding, and I could be wrong, that part of the process is to continually rescale the kinetic term to be in canonical form as you lower the momentum cutoff (though I've also seen that you could choose any quadratic term, so I suppose kinetic is just convention).

Why is that? Do you need to rescale the fields in general and choosing quadratic is the traditional way? Is there something special about a quadratic term in Wilson renormalization?

References to rescaling:

Qmechanic
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Cory
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    The RG procedure involves 2 steps, coarse graining and rescaling. If you don't rescale the recursion equations don't have fixed points. The choice of rescaling the kinetic term to keep it fixed is because the free theory is expected to be a fixed point of the transformation. If you knew how fields scale near a nontrivial fixed point, you could choose the rescalings differently (e.g., if you wanted to set up perturbation theory around the non-trivial fp). This is what field renormalization factors are for: modifying the scaling so you can find nontrivial fixed points. – bbrink Mar 09 '23 at 13:26

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As far as I know, the idea is just that generally, you can just redefine the action with a rescaled field without changing the physics. As such this scaling is just a redundancy in your description and is fixed by canonical scaling of the kinetic term. My guess why you use the kinetic term to do so is just that this term has no coupling constant such that you can differentiate between a scaling of the field and a change of the coupling constant.

Thomas Tappeiner
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OP is discussing the Wilsonian effective action (WEA), cf. e.g. my Phys.SE answer here. A path integral contains a freedom to perform field redefinitions and scale fields. Normalization of a non-zero action term is a useful tool in order to have a well-defined non-ambiguous running of the coupling constants. The kinetic term is often picked to be canonically normalized because it is usually present and represents the free theory.

Qmechanic
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