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This is a question about the self-consistency of the Standard Model - which I believe is the same as asking whether it is UV complete - in other words, can it be used to predict experimental results at arbitrarily high (and low) energy scales? Note I am asking for a rigorously defensible statement about the Standard Model, not a technical explanation of why said statement is true.

(An aside - I understand that the term "Standard Model" can include the original version (massless neutrinos), or various extensions that allow massive neutrinos.)

I understand that the Standard Model is certainly not correct at the Planck mass, and is not able to explain cosmological observations, so it is observationally falsified, but my question relates to internal self-consistency.

This note by Rubakov gives an attempted answer,

Standard Model is a well-defined theory, in the sense that everything is calculable, at least in principle, within this theory in terms of finite number of parameters (some quantities are hard and even impossible to calculate in practice because of strong coupling in the low-energy QCD). With $m_H \lesssim200 GeV$ this theory can be extended up to Planck energies.

In fact the Higgs mass $m_H$ is indeed less than 200 GeV.

The Wikipedia article equivocates:

The Standard Model is renormalizable and mathematically self-consistent(1)

(1) In fact, there are mathematical issues regarding quantum field theories still under debate (see e.g. Landau pole), but the predictions extracted from the Standard Model by current methods are all self-consistent. For a further discussion see e.g. R. Mann, chapter 25.

What is the best current answer?

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2 Answers 2

up vote 17 down vote accepted

The Standard Model is consistent in perturbative expansions.

It is inconsistent non-perturbatively but all these inconsistencies only show up "qualitatively" at energies well above the Planck energy – where we know the non-gravitational Standard Model to be inapplicable, anyway.

The inconsistencies of the Standard Model involve the Landau poles – the $U(1)_Y$ hypercharge coupling $g$ diverges at a certain energy scale, due to the renormalization group running – and a similar problem with the quartic Higgs self-coupling (it would be a problem at low, below-Planckian energies if the Higgs mass were higher than those 200 GeV).

A perhaps more serious problem related to the latter is the instability of the Higgs vacuum. The minimum at $v=246\text{ GeV}$ in the Standard Model isn't really a global minimum for certain values of the Higgs mass $m_H$. The observed $m_H$ is lower than 130 GeV or so and for these values, the potential isn't stable. It has lower minima at vevs $h\gg 246\text{ GeV}$. To say the least, the Standard Model potential is metastable (lower global minima exist but one can get there only through a tunneling which occurs rarely) with a dangerously short lifetime. Whether the metastability (intermediate situation) is an inconsistency – when all cosmological considerations are taken into account – is debatable.

But if one is satisfied with predictions at energies lower than the Planck scale and up to the accuracy of relative corrections of order $E/m_{Pl}$, then the Standard Model may pretty much be put on lattice and the continuum limit will agree with the perturbative expansions and produce a consistent theory for all these phenomena (limited by energies and the error margin). In particular, all the UV divergences may be consistently subtracted and all the IR divergences only encode real physical phenomena and the situation when one has asked a wrong or sloppy question.

This (limited) consistency doesn't mean that one should believe that the Standard Model is actually the exactly right theory of Nature up to the Planck scale. There are many reasons to think it is not the case.

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Clarification of your terms, 246 GeV is the vacuum expectation value of the Higgs field in the Standard Model. I agree that whether the metastability is an inconsistency may not be part of the question. It may be true (eg) that the Higgs field could not have survived inflation in its current state, but we already know that inflation requires new physics. –  akrasia Aug 25 '14 at 12:52
Key message: there are two internal inconsistencies ("Landau poles" and the "quartic Higgs self-coupling"). Still not sure whether the consistency of perturbative expansions is relevant. Can all calculations be done this way, including those in QCD? –  akrasia Aug 25 '14 at 12:53
Dear akrasia, the quartic Higgs coupling doesn't diverge assuming the (now) known value of the Higgs mass, below 200 GeV. So there are two inconsistencies, but they're the Landau pole and the Higgs instability. The latter is something else than a divergent quartic coupling - to some extent, it's the opposite problem because it occurs because the measured mass of the Higgs boson is too low for the Standard Model. ... I personally consider metastability of this kind to be an inconsistency. –  Luboš Motl Aug 25 '14 at 14:42
And yes, at low enough energies below the instability scale, and perhaps even above it, the Standard Model is consistent even nonperturbatively. I am sure that I have already answered that question. You may put the Standard Model on a lattice, for example (ignoring technical issues with fermion doublings etc. which are basically solvable). And yes, the perturbative consistency with a perturbative analysis of the RG flows etc. is enough to prove the nonperturbative consistency, too. The perturbative expansions know about "almost everything". –  Luboš Motl Aug 25 '14 at 14:44
Could you say a couple of sentences about neutrino masses in this context, please? Thanks in advance! –  CuriousOne Aug 26 '14 at 2:48

Indeed, the Standard Model is consistent in perturbative expansions, which acutally means that we do not really know if the Standard Model is consistent or not. So it is possible that the original Standard Model with 15 Weyl fermions per family is not consistent. In other words, there may not exist any well defined quantum model, whose low energy effective theory reproduce the original standard model. (Here a "well defined quantum model" has the following defining property: the dimension of the Hilbert space is finite for a space of finite volume, and the Hamiltonian operator acting on the Hilbert space contain only local interactions.)

The statements "The Standard Model is renormalizable and mathematically self-consistent" and "Standard Model is a well-defined theory, in the sense that everything is calculable" are not correct. The non-perturbative SU(2) instanton effects are not calculable at the moment. Perturbative expansions without instanton effects do not even conserve probability, and do not converge. Those fatal problems of the Standard Model are well known and have a name chiral fermion/gauge problem.

However, I have a recent work suggesting that a modified Standard Model with 16 Weyl fermions per family is consistent (ie UV complete). See arXiv:1305.1045 A lattice non-perturbative definition of an SO(10) chiral gauge theory and its induced standard model

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Thank you Xiao-Gang - but I suppose a GUT is not really the standard model. As for the instanton effects, I am not sure if you are agreeing with Lubos or not? –  akrasia Aug 25 '14 at 12:43
The modified Standard Model with 16 Weyl fermions per family is a low energy effective theory of SO(10) GUT. The SO(10) GUT is consistent (ie UV complete) -> The modified Standard Model with 16 Weyl fermions per family is consistent (ie UV complete) –  Xiao-Gang Wen Aug 25 '14 at 12:50
Could you please clarify how the consistency of a complete theory implyes the consistency of its low energy effective theory? –  firtree Aug 25 '14 at 13:25
I deleted an inappropriate comment and its responses. To the affected commenters: feel free to start the discussion over, keeping in mind our rules on civility, but you might want to do so in Physics Chat or another chat room in case it becomes extended. –  David Z Aug 25 '14 at 20:43
By "the Standard Model is consistent in perturbative expansions" I mean that perturbative expansions up to a small fixed order is well defined. But it is known that the perturbative expansion does not converge (the radius of convergince is zero). At a finite order, although the perturbative expansions is well defined, the results do not conserve conserve probability. –  Xiao-Gang Wen Sep 19 '14 at 5:34

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