I'm little confused about the maximal appropriate value for the SM Higgs quartic coupling. I know that the Higgs mass, $m_h= 125 \,\text{GeV}$ and that $ \lambda = m_h^2 / 2 v^2 \simeq 0.1 $ for $v = 246 \,\text{GeV}$.

But according to the perturbation theory, $\lambda $ enters into higher order corrections as $\lambda/(4 \pi)$. So that the very generic limit from perturbativity constraint is $ \lambda/(4 \pi) < 1 $ or $ \lambda < 4 \pi $.

Now after loop correction, say $\lambda = 10$, it won't any more account for the SM Higgs mass!


As you say, loop corrections change the value of the quartic coupling $\lambda$ is modified by loop corrections. If the renormalization group changed its value so $\lambda \gg 1$, the perturbative interpretation of the theory would break apart. Note that the condition $\lambda < 1$ (or $\lambda/(4\pi) < 1$) is only a hint of the perturbativity, the real condition is that the two-loop correction (order $\lambda^2$) must be neglible with respect to the one-loop correction.

Note also that the real danger isn't non-perturbativity, but the existence of a Landau pole, i.e., that the coupling constant goes to infinity at some finite energy. Of course, non-perturbativity is usually a warning of a Landau pole. But you could encounter "new physics" at some energy that modifies your beta function and makes the theory perturbative again. In that case, non-perturbativity only conveys our failure on doing the maths (well, the necessity of using other approaches such as lattice field theory).

But... in the SM that doesn't happen [the question asks specifically for SM, even though OP is concerned about other models in comments below]. The one-loop Renormalization Group equation is $$16\pi^2 \frac{d \lambda}{d(\ln \mu)} = 12 \lambda^2 + 12 y_t^2\lambda -12 y_t^4 $$
where $y_t$ is the Higgs-top Yukawa coupling, and the rest of Yukawa couplings and gauge group couplings have been neglected. The renormalization group equation is a first-order differential equation. So, after plugging the value of $\lambda$ at one energy scale, you can predict its value at any other energy. Usually we take that energy as the electroweak scale $\mu_{EW} \sim 100 \mathrm{GeV} \sim v$, where the spontaneous symmetry breaking occurs, and use the value you quote $\lambda(\mu_{EW}) = m_h^2/2v^2$.

As the energy scale $\mu$ grows, $\lambda(\mu)$ gets lower, and around $\mu\sim 10^{10}\mathrm{GeV}$, $\lambda(\mu)<0$. enter image description here

What does this mean? At large energies ($\mu \gg m_h$), $\lambda$ determines completely the Higgs potential, and if $\lambda<0$, then the potential is metastable, it can decay to a lower energy state of the Higgs field. enter image description here

Both figures are taken from G. Degrassi et al.: Higgs mass and vacuum stability in the Standard Model at NNLO arXiv:1205.6497 [hep-ph]

  • $\begingroup$ Could add why Higgs mass is calculated at renormalization scale of about 100 GeV $\endgroup$ – innisfree May 14 '16 at 15:15
  • $\begingroup$ Also the running of the quartic has a minima, then increases again to positive non perturbative vslues, doesn't it? $\endgroup$ – innisfree May 14 '16 at 15:17
  • $\begingroup$ @innisfree As per the first figure (calculated with two loops), $\lambda$ slightly increases at very high (Planck) energies, but not enough to make the potential stable again. $\endgroup$ – Bosoneando May 14 '16 at 15:22
  • $\begingroup$ Right, I saw the figure, but if you plot to beyond MP, it'll go positive, non perturbative, and increase monotonically $\endgroup$ – innisfree May 14 '16 at 15:24
  • $\begingroup$ @innisfree Well, I'm not sure if it makes much sense discussing about what happens at energies several orders of magnitude higher than the Planck scale $\endgroup$ – Bosoneando May 14 '16 at 15:39

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