A general higgs potential

  • Thread starter timb00
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  • #1
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Hi guy's,

this is my first thread at Physics-forums. I am currently working on GUT's and try to learn
the Higgs mechanism. Therefore my supervisor gave me a toy model. This model is a SU(4) gauge theory with a infinite series of 4+ \bar 4 as Matter fields (fermions) and a Higgs sector which consist of a singlet and a 15.

I tried to break the symmetry down to a SU(3)xU(1) by inserting VEV into the lagrangian and to minimize the higgs potential. Group theory tolds me that the representations decompose as

[tex] 15 = 8 + 3 +\bar 3 +1 [/tex]
[tex]4 = 3 +1 [/tex]
[tex]\bar 4 = \bar 3+ 1 [/tex]

I choose a vev which forces the 8 and the 1 of the 15 to be heavy. These are
[tex] <8>_{SU(3)} = l 1_{3x3} , <1>_{SU(3)} = -3 l => <15>_{SU(4)} = l ~diag(1,1,1,-3)[/tex]
as well as

[tex] <1>_{SU(4)}= w~ diag(1,1,1,1) [/tex]

The Result I got for the minima of the potential is that
l = 0 and w < 0.

I Think my Higgspotential is wrong. It is of the Form

[tex]Tr[(w_1 1_R /4 + w_2 15_R /3 + v~1_{4x4}/(\mu 4))^4 ][/tex]

Here 1_R and 15_R are the Representations of the SU(4) Higgs sector and the rest are constants.

I hope you could help me or enjoined the reading.

Thank you for reading

timb00
 

Answers and Replies

  • #2
Bill_K
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timb00, I'm not real sure either, so forgive me if these remarks are off the mark. SU(4) has 15 real parameters while SU(3) x U(1) has 9. So 9 of the symmetries will remain unbroken, i.e. you will have 9 gauge bosons that remain massless. The other 6 acquire mass, and will each need to steal their third component from Higgs. You will have 9 Higgses left, φa, a = 1..9 to participate in the potential, which you have taken to be the 8 and 1.

Now what will the potential look like? It must include a term for every possible way to couple two φ's or four φ's together to make a scalar. I think you have written the quadratic terms but not the quartic.
 
  • #3
15
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Hello everybody ,

and first of all thanks four reading to everybody and a thank you Bill_K for your reply.
I'm not real sure either, so forgive me if these remarks are off the mark.
I think all your remarks where right. Yesterday I found a dissertation where they gave an
example of a Higgs potential. In this paper they used for every term of the potential a independent constant . It look like
[tex] \frac{a_1}{2} tr(15^3)^2 + \frac{b_1}{2} tr(15^4) + c_1 ~tr(15^3) - m_{15} tr(15^2) + \frac{l}{4}(1^{*}1)^2 - m_1 ~1^{*}1+a_2 1^{*}1~ tr(15^2)[/tex]

I hope your agree with me that this might be the most general Higgs potential one could write
down, in this case.

thank you for reading.

Timb00
 
  • #4
Bill_K
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Not quite what I was expecting. Your Higgs field is down to 8 and 1, and I was thinking the potential would have terms like 82, 84, etc
 
  • #5
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Hi Bill_k,

Your Higgs field is down to 8 and 1, and I was thinking the potential would have terms like 82, 84, etc
It actually has, because if your write the 15 as
[tex] 15 = \begin{pmatrix}8 & 3 \\ \bar 3 & 1 \end{pmatrix} [/tex]
they will appear, e.f.
[tex]Tr(15^2)^2 = (Tr(8^2) +2~3\bar3 + 1^2)^2 =Tr(8^2)^2 + 4~Tr(8^2) 3\bar3 +
2 Tr(8^2) ~1^2 + 2~3\bar3~1^2 + 1^4 +4~(3\bar3)^2[/tex].
This is due to the underlying SU(4) gauge theory, which is spontaneously broken to SU(3)xU(1).

Thanks for reading,

Timboo
 
  • #6
Bill_K
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I still think there's more to it than that. Granted you want to begin with a Lagrangian that's manifestly SU(4) symmetric, and that's why you build your potential from 152, 154, etc, which is the most general SU(4) potential possible. But after the breaking you want to be left with nine components of the Higgs field, and they must see a SU(3) x U(1) potential built from 82, 84, etc.

The question is, how do you choose the potential so your model will break as SU(3) x U(1) and not something else?
 
  • #7
Bill_K
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Aha, insight. You want the Lagrangian to be SU(4) symmetric, but that does not mean the Higgs field φ has to be 15 + 1. You can take the Higgs field to be any representation of SU(4) you want. In particular if you take φ to be an SU(4) spinor, then everything works. Six of the SU(4) transformations are needed to rotate it into (0, 0, 0, φ0), and these get broken. The remaining transformations that leave φ alone do not get broken, and are exactly SU(3) x U(1).
 
  • #8
15
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Hi Bill_K,

and sorry for my late replay. I'm not quite sure why we are using the 15+1. I had a
very long discussion with my Professor and my second supervisor About that, and
I'm not familiar with the Higgs breaking scheme because the main part of my work
is on Orbifold breaking an Extra dimensions.

So sorry I'am that can not explain why we are using the 15 + 1 as Higgs-sector. But our
main interest is the construction of the smallest possible anomal gauge theory.

But I just realized that one can add an additional term to the potential with is given by
[tex] b_1 \phi_1 Tr(15^3). [/tex]
And I think that this is important, because otherwise, the Lagrangian would serve a
SU(4)xU(1) or maybe a SU(4)xSU(4) symmetry.

I will think about the question why we are using a 1+15 Higgs-sector and if i got an answer
i will post it.

thanks for reading and relaying,

Timb00
 

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