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!


closed as unclear what you're asking by AccidentalFourierTransform, Kostya, user36790, Gert, CuriousOne May 16 '16 at 0:10

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.