# In SUSY why does electroweak symmetry breaking only happen in the SM sector?

This is a difficult question to phrase succinctly, so I hope the title makes sense. What I want to understand is what seems like a lack of symmetry (besides SUSY-breaking) between the SM sector and their superparters.

In SUSY we add a second higgs doublet, so we end up with 8 degrees of freedom. Three are eaten by the SM gauge bosons, leaving 5 higgs bosons. My questions is: why are the 3 degrees of freedom eaten only by SM particles rather than 3 SM and 3 SUSY particles: why the asymmetry? Would the situation be different without SUSY-breaking?

If it helps to visualize the problem, the asymmetry is most striking if you imagine that we lived in a world where the mass scales where reversed: the SM is off at some high SUSY-breaking scale, and our world consists of superparters. Would we not have gauge theory and electroweak symmetry breaking, or would the gaugino sector require electroweak symmetry breaking?

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Gauginos are spin-1/2 fermions, and they don't carry forces like the W and Z bosons do. They aren't connection coefficients, they don't superpose to macroscopic fields.

There is never a complete symmetry between bosons and fermions, even in a supersymmetric theory. The fermions are fermions and the bosons are bosons, they have completely different physical properties. The supersymmetry transformation is not like a spatial rotation--- it isn't as physical. If you rotate a sock, all the particles in the sock rotate. If you super-rotate a sock, it becomes a superposition of rotating one-particle at a time of the sock. Most of the sock stays the same, but one constituent is turned to its superpartner, and there is a quantum superposition over which constituents are flipped. The result is still mostly the original sock.

This is analogous to the notion of an infinitesimal generator, since an infinitesimal transformation acts on products one factor at a time. The SUSY transformations can be thought of as permanently infinitesimal, because their parameter squares to zero.

Supersymmetry tells you for each particle that the scattering amplitude of a boson is simply related to the scattering amplitude of the fermion. This relation is particle by particle. So supersymmetry just isn't a symmetry of objects, at least not in a useful classical sense. So in your example of the reversed heirarchy, the Higgs mechanism would still give W's and Z's a mass.

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Gauginos are fermions, but there are indeed force-like bosons (the stop, sbottom, etc) in the SUSY sector, right? In the reversed hierarchy alternative history, would there be no gauge theory at all (until we discovered SUSY)? In the absence of the SM (sitting at the SUSY-breaking scale) there would be no need to hypothesize any higgs or electroweak symmetry breaking? –  user1247 Feb 25 '12 at 16:27
@user1247: The stop and sbottom are scalars, so they aren't gauge fields, they wouldn't make a gauge theory, but a scalar/spinor Yukawa theory. They can have VEVs, but that's a stability issue. –  Ron Maimon Feb 25 '12 at 23:22
OK, I'm assuming then that there would be no gauge theory or higgs hypothesized if we lived in the reversed hierarchy world (at least until SUSY was discovered and taken seriously). Lucky for us, since without gauge theory the SM would be a lot less compelling! –  user1247 Feb 26 '12 at 1:33
Another question: in your example, if you were to super-rotate a sock, would the sock be physically identical to the SM sock (assuming that supersymmetry were unbroken)? If not, how is this a true symmetry (again, emphasizing that I'm assuming supersymmetry is unbroken so the superpartners have the same mass as their SM partners)? –  user1247 Mar 20 '12 at 9:34
@user1247: It would not be physically identical--- it would have opposite statistics, for one. A supersymmetry is a different kind of symmetry, and whether you want to call it a true symmetry is a question of nomenclature. If you super-rotate the super-rotated sock, it will just be a translated version of the original sock, except the translation parameter is a commuting product of anticommuting variables, so it still squares to zero, so it is still infinitesimal. The superalgebras are extensions of symmetry in the infinitesimal Lie sense, less so in the macroscopic transformation sense. –  Ron Maimon Mar 21 '12 at 0:36

If you want, from a symmetry breaking point of view, the only thing that happens is that 3 goldstone bosons, coming from the broken generators are eaten, only three, and by the higgs mechanism are "eaten" by the gauge bosons, in the theory. All this in very general grounds, only taking it into account as a QFT, without worrying about the particle content. Also the symmetry breaking effect and the fact that the bosons are massive is going to affect the superpartners, through the radiative corrections that these "feel" from the gauge bosons. So in a way, there is a symmetry breaking everywhere in the susy sector.

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An alternative way to think about it is that EWSB happens at the scale of $\sim M_W$. Above that scale the electroweak forces are unified. SUSY however must be broken at a much higher scale (call it $M_S$) and we know this is the case because we haven't observed any super-partners in experiments so far. Since $M_S>M_W$ the electroweak forces are unified in the supersymmetric theory. We only need to break the electroweak symmetry much later after the SUSY breaking and that's why EWSB only happens in the SM model sector.