Can a superpartner be less massive than its SM counterpart?

Question

Theoretically, can a superpartner be less massive than its standard model counterpart? I realize there are experimental constraints.

Answer 1

If you think about it, in a theory with supersymmetry (think about chiral supermultiplets of $\mathcal{N}=1 \; d=4$ for clarity) it's merely a convention to call some particles superpartner of the other particles.

Therefore yes, theoretically there aren't problems, even though we know that superpartner of the usual Standard Model particles must be heavier from experimental constraints.

EDIT: You may be interested in pag. 90 of A Supersymmetry Primer, Martin (2011).

As you can see in (8.1.1), the potential of the MSSM (Minimal Supersymmetric Standard Model, the most simple model!) is ridiculously complicated, but the point is that you now have two Higgs doublets:

$$H_u=(H^+_u,H^0_u) \;\;\; H_d=(H^0_d,H^-_d)$$

in which $H_u$ is the old Higgs doublet. You have the corresponding fermions too, usually denoted as $\tilde{H}_u$ and $ \tilde{H}_d$. Mass terms arise from typical Yukawa terms calculated in the minimum of the potential, for example $H\psi \psi$ for leptons and sleptons. There are many complications in this process: you have to identify the correct mass eigenstates, the supersymmetry can't be broken explicitly, therefore you have to mediate the breaking from an "hidden sector" to the "visible" one (many possible models: Gauge-mediated, Extra Dimensions-mediated , anomaly-mediated, and so on).

Anyway, both the masses of particles and sparticles involve the VEV of $H_u$ and $H_d$ in some complex way, for instance see (8.1.12) for $m^2_Z$ and (8.2.2) for neutralinos. These parameters (and other) are not fixed, you can know the value only from experiments. So you are free, at least in principle, to arrange the mass spectrum as you wish.

Answer 2

Yes, superpartners can have masses smaller than the mass they get from the Higgs mechanism.

For this the contribution to the squared mass parameter in the Lagrangian must be negative, which can be arranged but is very severely constrained experimentally.

In your comment I see a misconception though:

In supersymmetry, both the "regular" particle and its superpartner get the very same mass from the Higgs mechanism, SUSY breaking is just an additional source of mass for the partners. As Rexcirus mentions, thereotically all components of the superfield are treated with equal rights. It is only when we have supersymmetry breaking, that we can call the fields affected by the breaking the "superpartner" of the unaffected field.