About the consequences of $\mathbb{Z}_2$ invariance in dark matter models

Question

This well-cited paper talks about is a minimal renormalizable extension to the Standard Model (SM) to incorporate particle dark matter (DM) into it by adding a real scalar field $S$ which (unlike the Higgs doublet $H$) is a singlet under the full SM gauge group. But we have to pay a price, we have to introduce three more free parameters in the theory (in addition to those already present in the Standard model): (i) the mass of the new scalar $m_0$, (ii) dimensionless self coupling of the scalar $\lambda_S$, and (iii) a dimensionless coupling to the Higgs $\lambda$.

The power-counting renormalizable Lagrangian of the model, is therefore, $$\mathscr{L}=\mathscr{L}_{\rm SM}+\frac{1}{2}(\partial_\mu S)^2-\frac{1}{2}m_0^2S^2-\frac{1}{4}\lambda_SS^4-\lambda S^2(H^\dagger H)\tag{1}$$ where a $\mathbb{Z}_2$ invariance is proposed under which all the SM fields are even but S is odd i.e. $S\to -S$ under $\mathbb{Z}_2$. This forbids linear ($\lambda_1 S$), cubic ($\lambda_3S^3$), and $\lambda^\prime S(H^\dagger H)$ terms in the Lagrangian.

As I can understand, by forbidding the term $\sim S(H^\dagger H)$ "one can prevent the decay of the dark matter $S$ into a pair of SM Higgs bosons", and maintain the constancy of the observed cosmological relic abundance. But that will anyway be kinematically forbidden if the mass of $S$ turns out to be less than twice the mass of the Higgs.

Addendum If the term $S(H^\dagger H)$ is absent, how is the following annihilation $SS\to H\to XX$, where $X=g,b, W,Z^0$ (as depicted in Fig. 1) become possible? Am I missing something? Moreover, if it happens, will it not continually deplete the relic abundance?

Can someone enlighten me keeping in mind that I'm no expert in this field.

Answer 1

The most common way to have dark matter candidates is to formulate a model which has some discrete symmetry for which you can assign new conserved quantum numbers. The easiest symmetry is $\mathbb{Z}_2$: assign $+1$ to all Standard Model (SM) particles and $-1$ to all dark sector particles. Hence, there will be no net destruction of $-1$ charges, i.e. dark matter, if the Lagrangian is invariant under this symmetry $\Rightarrow$ dark matter is stable.

As you have noticed, if there is no such symmetry, kinematic constraints can prohibit a certain decay path, too. However, kinematic constraints hold only for external legs. Nothing prevents your "non-symmetrically-(meta)-stabilzed" particle to decay into something much lighter than the SM Higgs boson on loop-levels, e.g. into a pair of electrons and positrons via internal SM Higgs boson propagators.

You can make your particle even lighter as neutrinos but this will be excluded by astrophysical observations. Something as light as a neutrino constitutes hot dark matter, which is heavily constrained by observations. Cold (heavy) dark matter has to be stabilized somehow. If it is stable, then you can always find a conserved quantum number and in turn a symmetry of your theory. Most probably a discrete one.

It doesn't have to be some $\mathbb{Z}_n$ symmetry. It can by anything really. Here is an example with dark matter candidates stabilized by a generalized CP symmetry: https://arxiv.org/abs/1512.09276