# Can a symmetry breaking VEV lie far above the "symmetry breaking scale?"

The scale where some symmetry gets broken can be computed using the renormalization group equations for the gauge couplings. If there is only one Higgs VEV responsible for some breaking, can this VEV lie several magnitudes above the scale determined from the RGE running?

I know that VEVs that break a symmetry can lie far below the symmetry breaking scale. If there are multiple scalar fields that could break a given symmetry, we can always rotate such that only one linear combination gets a VEV. However, all other linear combination will get an induced VEV which is typically several orders magnitude below the "dominant" VEV.

I've never seen an argument how and why a VEV could lie far above the scale determined from the RGE running. However, in many papers, for example, the VEV that gives a mass to right-handed neutrino is computed to be far above the GUT scale or far above the intermediate scale, which is broken by exactly this VEV.

As an example take this paper. In Eq. 9 they fix the Pati-Salam scale to be $M_I = 10^{12}$ GeV. However in the paragraph below Eq. 37 the authors explain

The absolute neutrino mass scale can be inferred once the vev $v_R$ in Eq. (15) is determined by demanding that the small neutrino mass-squared differences [...] resulting from the fit procedure reproduces the experimental value [...]. It turns out that $v_R ≃ 1.3 \cdot 10^{14}$ GeV.

The VEV $v_R$ is exactly the VEV which is responsible for the breaking Pati-Salam $\rightarrow$ Standard model and thus, according to Eq. 9 should be around $10^{12}$.

The scale where some symmetry gets broken can be computed using the renormalization group equations for the gauge couplings.

It's the other way around. Once you know the scales of the model (masses of the fields) you can compute the RGE.

Since what ultimately matters is the mass of the representation, not its vev, if you accept some tuning (and have such a freedom), the vev could lie freely above or below that mass. Essentially, if the mass of a scalar is given $M^2 = \lambda v^2$, there's a priori no reason to expect $\lambda=1$; it's just a simplifying hypothesis that is commonly used in GUT (because of the enormous freedom).

• Thanks that makes sense. Normally a symmetry breaking scale is identified with the mass scale of the gauge bosons that get a mass through this symmetry breaking. Thus it makes sense, if for some reason the gauge bosons do not have a mass of order of the VEV as it is in the SM, that the VEV can be above this mass scale
– jak
May 6, 2016 at 11:36
• @JakobH for the gauge bosons you do not have much freedom, you actually know that the mass must be $M^2\sim g_M v^2$.. so it just depends on the value of the gauge coupling. Other effects in the scalar spectrum (a.k.a. threshold effects) can significantly change the value of $M_I$ and $M_{GUT}$.
– xi45
May 6, 2016 at 16:08
• @JakobH BTW, I have answered a couple of other questions by you on this platrform; no feedback. I'd be curious to see if that was helpful or if I need to expand and explain better.
– xi45
May 6, 2016 at 16:09