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I've seen both positive and negative answers to this question, though most part of the community seem to agree it can be said it is an EFT up to the electroweak scale. My question is: What are the main arguments from each side? References are welcome too.

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  • $\begingroup$ Hi Martin, imo this question is obviously important, but I think it may judged as both too broad and too personal opinion based. Sorry but I VTC on that basis. it can be said it is an EFT up to the electroweak scale It's basically all we have, what else can be said except personal opinions on where it's headed and how we use limited resources to best improve on it. $\endgroup$ – user108787 Nov 2 '16 at 23:47
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    $\begingroup$ How do you define a "not-effective field theory", given our current inability to construct rigorously defined non-effective theories? $\endgroup$ – ACuriousMind Nov 3 '16 at 0:09
  • $\begingroup$ I am with ACuriousMind. According to the traditional understanding EFT is a perturbative QFT that uses cut off at high energies and renormalization to produce finite answers. SM is obviously that. Since you expect arguments "from each side" you must mean something else by "EFT". What is it? $\endgroup$ – Conifold Nov 3 '16 at 0:35
  • $\begingroup$ I don't agree. Lattice QFTs are definitely effective. The gold standard, really. $\endgroup$ – user1504 Nov 3 '16 at 0:36
  • $\begingroup$ @user1504 I am not sure we use "effective" the same way, in this context it just means "not fundamental", "phenomenological": "An effective field theory includes the appropriate degrees of freedom to describe physical phenomena occurring at a chosen length scale or energy scale, while ignoring substructure and degrees of freedom at shorter distances (or, equivalently, at higher energies)" says Wikipedia on EFT. $\endgroup$ – Conifold Nov 3 '16 at 0:44
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Well, the Standard Model is definitely an effective field theory description of physical reality since it neglects both gravity and neutrino masses.

It's also an effective QFT in the sense that there's a sensible definition of it on the lattice. (Not trivial, because there are chiral fermions.)

But I think what you're really asking is: Does the specific QFT we currently call the Standard Model have a continuum limit? That's an open question in theoretical physics. I think most theorists lean towards "no", because hypercharge and quartic Higgs couplings have beta functions with the wrong sign in the perturbative regime. Absent a miracle, this means that those sectors have Landau poles, making it impossible to find a continuum limit with interactions in the IR. (A Landau pole means that if you adjust the lattice coupling as you refine the lattice in order to keep the long-distance physics fixed, you'll see the lattice coupling run off towards infinity at a finite lattice scale.)

But this isn't proof: the perturbative approximation becomes unreliable as one approaches the region where the Landau pole would happen. Numerical computation with lattice $\phi^4$ theory also suggests that there's a Landau pole. However, those calculations aren't under analytic control; we don't know that we're not missing something.

On the other hand, I don't know of any compelling evidence that the Standard Model might not have a Landau pole.

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Before this question gets closed, let me point out that there is something called "Standard Model Effective Field Theory" (SMEFT) which is distinct from the Standard Model (SM) as you find it in textbooks. Basically, the SM consists of all the renormalizable terms consistent with its defining symmetries and representations, whereas SMEFT also includes all the non-renormalizable terms, multiplied by inverse powers of a cutoff scale.

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