## Mass generation and supersymmetry

I have uploaded a paper on arxiv with a new theorem of mine. I have already exposed the idea in this blog but, so far, I have had no much time to make it mathematically sound.  The point is that the mechanism I have found that gives mass to Yang-Mills and scalar fields implies supersymmetry. That is, if I try to apply it to the simplest gauge theory, in a limit of a strong self-interaction of a massless Higgs field, all the fields entering into the theory acquire identical masses  and the couplings settle down to the proper values for a supersymmetric model. Being this result so striking, I was forced to produce a theorem at the classical level, as generally done with the standard Higgs mechanism, and let it widely known. My next step is to improve the presentation and extend this result after a fully quantum treatment. This is possible as I have already shown in the case of a Yang-Mills theory.

My view is that just a mechanism could be seen in Nature to produce masses and I expect that this is the same already seen for QCD. So, supersymmetry is mandatory. This will imply a further effort for people at work to uncover Higgs particle as they should also say to us what kind of self-interaction is in action here and if it is a supersymmetric particle, as it should.

The interesting point is that all the burden of the spectrum of the standard model will rely, not on the mechanism that generates masses but on the part of the model that breaks supersymmetry.

Interesting developments are expected in the future. Higgs is always Higgs but a rather symmetric one. So, stay tuned!

### 12 Responses to Mass generation and supersymmetry

1. Daniel de França MTd2 says:

Shouldn’t you at least find the need for a Higgsino? I mean, a Higgs is just supersymmetric if it has a susy companion…

• mfrasca says:

Dear Daniel,

My aim was just to show that interaction of other fields with a massless scalar field having an infinitely large self-interaction induces masses being all equal and couplings as expected for supersymmetry. Specifically, the simplest model of interaction of a massless spinor field and a massless scalar field with an infinitely large self-interaction is supersymmetric and with massive solutions. In this case the spinor field is the higgsino.

Marco

2. Ulla says:

Is it a magnetic monopole?

• mfrasca says:

As far as massive quantum electrodynamics is considered, magnetic monopole solutions should be expected. But this is just a model to show the mechanism of mass generation as generally done also for the Higgs model, nothing else.

Marco

3. vorpal says:

I didn’t get it. I see that you wrote down a non-linear eigenvalue equation, so clearly you will get eigenvalues.

I see a lot of words, but I don’t see how you get something that leads to m^2=E^2-p^2 . By the, way the terms from the Lagrangian you use come from the mass-energy equation. So are you saying that if you use the mass-energy equation with an eigenvalue problem, then you get mass?

I don’t get it.

4. mfrasca says:

Dear Vorpal,

I did not solve any eigenvalue equation. I solve pertubatively the classical equations of motion of the theory in the limit $\lambda\rightarrow\infty$. These solutions have the property to satisfy the same dispersion relation

$p^2=\mu^2\sqrt{\frac{\lambda}{2}}$

and so represents massive solutions all with the same masses as required by supersymmetry. Consistency reasons for these solutions to hold require also constraints on couplings again in agreement with supersymmetry requirements.

The starting point is the exact solution of the classical $\lambda\phi^4$ theory with equation of motion

$\partial^2\phi+\lambda\phi^3=0$

given by

$\phi(x)=\mu\left(\frac{2}{\lambda}\right)^\frac{1}{4}{\rm sn}(p\cdot x+\theta,i)$

with $\mu$ and $\theta$ two integration constants, provided the dispersion relation given above holds. So, this exact solution of the classical theory is massive notwithstanding the equation of motion is for a massless field. When you couple this field, in the limit $\lambda\rightarrow\infty$, to a spinor field and an abelian gauge field the leading order solution imposes supersymmetry.

Please note that the mass formula is just $m=\mu\lambda^\frac{1}{4}2^{-\frac{1}{4}}$. You need a full supersymmetric standard model with a proper mechanism to break sypersymmetry to uncover standard relations between masses in the standard model.

Marco

5. vorpal says:

when I said eigenvalue, I thinking along the lines of this paper:

http://arxiv.org/abs/1001.0377

I still don’t see how a model (a la Standard) that presumes mass can explain mass, but that’s my problem.

Kudos to you for being able to understand supersymmtry. It’s beyond my skill level.

• mfrasca says:

Dear Vorpal,

I do not presume mass anywhere in my equations. This is a consequence of a strong self-interaction term in the equations of motion otherwise massless. This same mechanism happens in a Yang-Mills theory with a strong coupling as it maps on a massless quartic scalar field theory that displays this effect. Mass appears in the solution of the equations of motion when the asymptotic limit of an infinitely large coupling is taken.

Marco

6. Marco, what about the matching between fermionic and bosonic degrees of freedom? Is it also implied?

• mfrasca says:

Dear Alejandro,

This paper contains a theorem. This states that for a U(1) theory with a Fermion coupled, as also the gauge boson is, to a massless scalar field with a finite self-interaction (not a self-interaction going to zero as usually assumed), all the particles of the theory acquire mass dynamically and all these masses must be equal. So, the theory fulfills the criteria to be supersymmetric completed and you get a way to provide masses to all the particles of the theory with a proviso of a full supersymmetric theory and a corresponding breaking pattern. The breaking pattern for supersymmetry is a must if you want a non-degenerate spectrum.

From standard quantum field theory I cannot go farther than this. But it is interesting to note that, if this mechanism applies, you will get a clear signature for the scalar field as a single excitation has superimposed a spectrum of excitations as if it would be structured (but it is not). So, you will observe several Higgs particles with masses separated by a set of golden numbers, the same seen in QCD.

Marco

• Thanks. It is only that I think that the matching is an important point too… without it, the statement of the result sounds mostly as an extension or a variant of S. Coleman and E. Weinberg http://prd.aps.org/abstract/PRD/v7/i6/p1888_1

For instance, a trivial point, but that I havent heard in the recent post-LHC discussions, is that a odd number of higgs bosons automatically rules out susy.

• mfrasca says:

No, this has nothing to do with Coleman and Weinberg argument. But I think you should read my paper. I do not work in supersymmetry, I just prove that, for this mass generating mechanism, is needed.

Current supersymmetry models use standard Higgs models and I cannot compare them with my approach. My personal view is that the Higgs model with its formulation of the Standard Model in the sixties is something to forget about. But the best approach from now, without further results from LHC, is just to sit and wait.

Marco