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

+ 4 like - 0 dislike
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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?

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
asked Feb 25, 2012 in Theoretical Physics by user1247 (525 points) [ no revision ]

3 Answers

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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.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Ron Maimon
answered Feb 25, 2012 by Ron Maimon (7,295 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
@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.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Ron Maimon
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!

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
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)?

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
@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.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Ron Maimon
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The quick answer to this question is: Because of the experimental data.

Any theory beyond the SM must reproduce the results of the SM (at least to the order they have been verified by experiment). In particular, it must explain electroweak symmetry breaking (EWSB) because this is what we observe in nature. The method you describe is how EWSB is achieved when there are two higgs doublets. SUSY has two higgs doublets and therefore falls into this class of theories.

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.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Heterotic
answered Oct 23, 2013 by Heterotic (515 points) [ no revision ]
The heart of what I was getting at with my question (which was partly satisfied by Ron's answer) is that if supersymmetry is a good symmetry then I would expect to be able to super-rotate all of the SM particles and end up with the exact same physical theory. Even putting aside the high SUSY breaking scale, this does not appear to be the case. The physics would be different. But this leads me to conclude that supersymmetry is not a "symmetry" in the same sense as, say, translations or rotations. I'm still a bit puzzled by this.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
I see... There is indeed a distinction between spacetime symmetries like translations and rotations and internal symmetries like gauge symmetries (and SUSY). For a translation we think the system is the same because all the internal "charges" of the particles have not changed. This is of course not the case (by definition) for the internal symmetries. I do disagree however with saying "The physics would be different", although I understand we you might say that. I prefer the viewpoint that the physics is the same (because the lagrangian is the same), but the system is different.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Heterotic
What I said above about spacetime symmetries applies just as well to internal symmetries (like gauge symmetries) that I actually consider symmetries. For example if you rotate every particle globally by a complex phase the physics is unchanged. This does not appear to be the case of you globally rotate every fermion into a boson and vice versa. The physics is different, but the couplings are unchanged (according to Ron's answer above), so assuming he is right, I don't really like putting SUSY into the same "symmetry" category as most others we like to talk about.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user user1247
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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.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Ayfel
answered Jul 4, 2013 by Ayfel (25 points) [ no revision ]

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