# Why study SUSY?

1. Apr 26, 2015

### itssilva

Supersimmetry, by itself, is a neat, elegant concept; I've read somewhere else (think Griffiths' Introduction to Elementary Particles, if memory serves me right) that it allows the various running couplings of the Standard Model to converge to a single value at high enough energies; however, besides that, what are the theoretical/experimental motivations to study this complication of standard QT? I'm under the impression that this theory haven't been garnering much love from the non-string physics community, of late.

2. Apr 26, 2015

### Orodruin

Staff Emeritus
Supersymmetry is popular in large parts of the particle physics community, not only with string theorists. The main reason for this is that it would provide possible solutions to some problems with the Standard Model. For example, it would solve the hierarchy problem (assuming the SUSY scale is low enough) and it provides viable dark matter candidates.

3. Apr 27, 2015

### Ilja

One point of supersymmetry seems that the additional particles it introduces somehow "solve" the "problem" of a unification of couplings. If this is really a problem is a first question - it is, of course, a problem for GUTs, which unify all forces into some simple gauge group, which requires that at some fundamental scale of unbroken symmetry the interaction coupling has to be the same for all parts.

One can argue that it could be also a problem for an effective theory, which assumes a common critical length scale where the field theory breaks down into some sub-theory, say some "atomic ether" or so. In this case, one would reasonably expect that all the effects should have a similar order at this scale.

My question is a different one. Suppose one simply has some usual field theory, containing all what the SM contains but in different numbers - say, more fermions, more scalar fields, above with masses. Computing the corresponding functions for the couplings should be, I would guess, something already done in a quite general form, so that it could be sufficient to put in the numbers of the additional particles, their spin, their masses, I would guess also the representations of the SM gauge groups acting on them, to find out if in this theory the couplings unify or not.

Any suggestions where such things can be found?

4. Apr 27, 2015

### Orodruin

Staff Emeritus
You cannot simply add an arbitrary number of fermions to the SM without caring for anomaly cancellations. In the SM, the gauge couplings are such that the anomalies cancel generation by generation.

What you can do is to add extra Higgses. This does change the running behaviour of the gauge couplings and using an appropriate number of extra Higgs doublets (if I remember correctly, the required number was seven additional doublets, but do not quote me on that) you can get the couplings to unify.

5. Apr 27, 2015

### Ilja

Thanks, I know.

But I don't care much about this anomaly stuff, for a reason which is possibly completely wrong:

My understanding is that the problem they cause is non-renormalizability. But in this case, this would be a non-problem for an effective field theory with some explicit cutoff. There, the non-renormalizable terms would be those which descrease faster than the renormlizable ones in the large distance limit, so they would disappear automatically if the cutoff is sufficiently small, even if they would have comparable order at the critical distance. Roughly speaking, one could leave problems with non-renormalizability to the long distance limit. Or is this completely off, and there are other problems with them?

What I would like to add (for completely different reasons) is a single scalar field for each electroweak doublet. The triplets for quarks would be colored, those for leptons would not interact with gauge fields at all.

Now, naive counting gives equal numbers for bosons and fermions: 8+3+1 gauge bosons, 3 x (1+3) new scalars against the 3 x (1+3) electroweak doublets. So, it seems to make sense to look if some of the advantages of supersymmetry - whatever they are - would be present - by accident - in this theory too.

6. Apr 27, 2015

### Orodruin

Staff Emeritus
This is inconsistent. You cannot have an electroweak doublet that does not interact with gauge fields.

7. Apr 28, 2015

### Ilja

But I can have scalar fields which do not interact with gauge fields, and they can be associated with electroweak doublets.

This association would be a correspondence for numbers (there will be one scalar field for every electroweak doublet). Moreover, each scalar field borrows its gauge charges from one preferred component of "its" electroweak doublet. In the case of leptons, this preferred component will be the right-handed neutrino, for quarks the right-handed anti-down-type quark (EM charge 1/3).

8. Apr 28, 2015

### Orodruin

Staff Emeritus
Why dont you write down your intended Lagrangian? It will be more effective (haha) than trying to describe your idea in words.

9. Apr 28, 2015

### Ilja

I don't think this would simplify something. The Lagrangian of scalar fields interacting with gauge fields would be nothing, new, thus, only a misdirection of interest (one would think there is something interesting in this formula) and what matters would be anyway the description of the number of fields and their charges.

10. Apr 28, 2015

### Orodruin

Staff Emeritus
Exactly, and I am not getting a clear picture of exactly what you want to do, which fields you want to couple to the scalars and how. This would be much simpler if you just wrote down the (interaction) Lagrangian you had in mind.

11. Apr 28, 2015

### Ilja

If $$\mathcal{L}= \frac12 D_\mu \phi^*_{ga} D^\mu \phi_{ga} - \frac12 m^2_{ga} \phi^*\phi$$ with $$D_\mu = \partial_\mu + i A^b_\mu T^b$$ makes you happy. Here g is the generation index, a is from 0 (for leptons) to 3 (for the three quark colors), b goes over the SM gauge fields, and the $T^b$ describe the representation I have to describe in words anyway: It acts trivially on the $\phi_{g0}$ and by the standard three-dimensional representation of U(3) (obtained by factorizing out weak interactions $SU(2)_L$ and $\mathbb{Z}_3$ from the SM gauge group) on the three-dimensional space of the $\phi_{gi}$ with $i=1,2,3$, g fixed.

Last edited: Apr 28, 2015
12. Apr 28, 2015

### ChrisVer

That's true from my case since both Dark Matter scenarios (supersymmetric WIMPs) as well as the Hierarchy problem solutions of SUSY become less favorable as you go to larger energies. As already mentioned the hierarchy problem could be solved by SUSY if the scale of it was around the TeV scale. Obviously we haven't yet found anything in the LHC and the parameter space for the theory to exists gets tighter. On the other hand they always try to save SUSY models like MSSM, either by extending it or I don't know exactly , say that it might still be invisible at the LHC.

I still like SUSY but the normal things the theory could solve, have been cornered a lot.

The unification of the coupling constants is (for me) an aesthetic need of some people and I don't really like it (as a reason)
Check eg. this figure :
http://scienceblogs.com/startswithabang/files/2013/05/running_coupling.gif
The left is the SM and the right is the MSSM.
I don't personally find any reason to have the lines changing at only 1 given energy (at around 10^3 GeV ~ 1 TeV scale).

If you don't care about these problems and don't care about experiments (like string theorists) then you can allow for SUSY to exist at any energy.

13. Apr 28, 2015

### Orodruin

Staff Emeritus
So what are your Yukawa couplings?

If your fields transform non-trivially under any of the gauge groups, they are going to have gauge couplings so what is the point of having the covariant derivative there at all?

14. Apr 28, 2015

### Ilja

The coupling constants are the same as those of the quarks.

15. Apr 28, 2015

### Orodruin

Staff Emeritus
Which coupling constants? The Yukawa couplings are coupling constants between fermions and scalars. If you have gauge coupling constants your scalar fields interact with the gauge fields. Or are you saying that you want to have a scalar field that does not have any quantum number? What is going to stop this field from coupling to every fermion in the SM?

16. Apr 28, 2015

### Ilja

Yes, these scalar fields are not intended as Higgs fields which have to give any fermions any mass.

17. Apr 28, 2015

### Orodruin

Staff Emeritus
It is not a matter of having to give fermions mass, it is a matter of which terms are allowed in your Lagrangian. If they are allowed, they will be there or you need to explain why they are not. If you have a singlet scalar $\phi$, what stops me from writing down the interaction term $\phi \bar q_R q_R$?

18. Apr 28, 2015

### Ilja

So ok, you are free to add whatever you want or think it is somehow unpreventable. I have seen no justification or necessity to write down such terms, so I do not write them down by Ockham's razor, that's all.

edit: Thinking about this a little bit more, such a Yukawa term, which would connect the scalar field with the corresponding electroweak doublet, would be quite natural.

Last edited: Apr 28, 2015
19. Apr 28, 2015

### ChrisVer

Whatever you are allowed to... If you are allowed to write something, then you have to write it, otherwise explain how this thing went on missing.
If you could discard terms like this in the Lagrangian, then there would be no strong-CP-problem and there would be no "bad" proton decays in several other theories.

20. Apr 28, 2015

### Orodruin

Staff Emeritus
Ockham's razor would actually go in the other direction. If you do not prohibit the coupling using some symmetry, it is generally going to be generated by running at one scale even if it is zero at another scale.