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

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.

• Why don't you try physicsoverflow.org which is more theoretically oriented? Jan 22, 2018 at 7:53
• Towards Addendum: If your $H$ gets a VEV, you can have something like $S^2 h$ after symmetry breaking where $h$ is for instance a real neutral scalar of $H$. Do note that this would not break your $\mathbb{Z}_2$ symmetry because this vertex preserves it: $(-1) \cdot (-1) = 1$ Mar 29, 2018 at 8:10
• Dear @image Thanks. So it is the $S^h$ term that is responsible for $SS\to h\to XX$. But wouldn't these dark matter annihilations continuously decrease the dark matter density? But I know that it's not possible because we have a constant relic density. What am I missing here?
– SRS
Mar 29, 2018 at 8:44
• Yes, they can decrease it. This is what governs how much relic density will be left after freeze-out. Before freeze-out, the temperature is high enough that XX can produce some (heavy) SS pairs. There will be a thermal equilibrium and the densities of S,h,X will be stable. During freeze-out the annhilation rates "SS $\rightarrow$ something" will govern how much of S will be left. After freeze-out, which is the period in time, where the expansion rate of the universe is so big that effectivly no S will meet another S, the relic density will be (roughly) stable, because S cannot decay by itself. Mar 29, 2018 at 10:07

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.
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