# Can a superpartner be less massive than its SM counterpart?

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

• Though I am nowhere near an expert, this would require the SUSY breaking scale to be very small, right? – jwimberley Jan 13 '15 at 19:41
• Yeah it would. But I don't know whether a low SUSY breaking scale implies that the superpartners' masses are absolutely small, or just that their mass increase relative to their SM partner's is small. – user1247 Jan 13 '15 at 21:56
• Without SUSY breaking, the masses would be equal. Naively, any SUSY breaking seems like it would add mass, but I don't know that the mass matrix from SUSY breaking wouldn't include negative eigenvalues, so that for bosons, e.g., m^2 => m^2 - \lambda, and for fermions m => m - \lambda, reducing the masses. But maybe there is some requirement that the mass eigenvalues from SUSY breaking be positive? – jwimberley Jan 14 '15 at 14:36
• Exactly, this is what I would like to know! Thanks for helping to clarify this question! – user1247 Jan 14 '15 at 19:28

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.

• Experimentally it's a convention, but theoretically they are distinguished, right? On one side they get their mass from only the Higgs mechanism, and on the other side they get their mass from SUSY breaking. So I'm asking whether the side that get's its mass from only the Higgs can be more massive than the corresponding partner that additionally gets mass from some SUSY breaking mechanism. – user1247 Jan 6 '15 at 16:06
• I'm going to start a bounty because, without addressing this comment, you haven't actually answered my question. Thanks! – user1247 Jan 13 '15 at 18:56

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.

• I don't see what my misconception is. I pointed out that supersymmetry breaking breaks the symmetry, which is what you are repeating. So my question comes down to whether the mass parameter from SUSY breaking must be positive. You say it can be negative, which effectively answers my question. Do you have a citation that says as much, so I can be sure you are right about this? Thanks! – user1247 Jan 14 '15 at 19:31
• Or wait, maybe I do have a misconception: are you saying that some of our SM particles could in fact get mass from SUSY breaking, and what we call its (as yet undiscovered) "superpartner" only gets mass from the Higgs? – user1247 Jan 14 '15 at 19:40
• @user1247 "On one side they get their mass from only the Higgs mechanism, and on the other side they get their mass from SUSY breaking." This sentence sounded to me like the misconception that "superpartners" get their mass ONLY through susy breaking, which is not true. – Neuneck Jan 15 '15 at 12:52
• If you read my sentence immediately after that one you see I meant "additional" mass from SUSY breaking. In any case I still would like a citation (or equation, or some argument rather than just a statement) that shows the additional mass from SUSY breaking could be negative. – user1247 Jan 15 '15 at 15:27
• @user1247 True, that was too shallow reading of mine. But yes, there are models with negative susy breaking contributions to the masses, but those are uninteresting since they are basically ruled out. The only possibility for a "sparticle" mass smaller than the "particle" mass is the stop, where a stop with a mass slightly lower than the top mass can evade collider constraints because its decay possibilities are very constrained. – Neuneck Jan 19 '15 at 14:46