Scalar in adjoint representation

In summary, the conversation discusses the breaking of an SU(3) gauge group to SU(2) x U(1) via a Higgs mechanism, with the scalar Φ responsible for the breaking transforming under the adjoint representation of SU(3). The most general potential for this scalar is constructed and it is noted that the cubic term should be discarded as it would destabilize the vacuum. The conversation also explores the invariants for SU(3) and the formation of cubic and quartic invariants. Useful relations and matrices for the N^{ 2 } – 1 dimensional adjoint representation are listed, including the Jacobi identities and contractions.
  • #1
MManuel Abad
40
0
Hello, people.

I'm studying (as an exercise) the breaking of an SU(3) gauge group to SU(2) x U(1) via a Higgs mechanism. The scalar responsible for the breaking is [itex]\Phi[/itex], who transforms under the adjoint representation of SU(3) (an octet). First of all I want to construct the most general potential for this scalar. So far I've got:

[itex]V= a Tr(\Phi^{4}) + b (Tr(\Phi^2))^2 + c Tr(\Phi^3) + d Tr(\Phi^2) [/itex].

Is the cubic term supposed to be there or is there a reason for it to disappear? I've been looking in the literature for examples of this kind of terms in potentials but I couldn't find any. I know that it would destabilize the vacuum, but there are quartic terms and therefore there are bound states and thus the vacuum is stable. If I want to minimize this potential my answer would depend on the sign of "c", wouldn't it?

Of course, I'm taking "a" and "b" positive while "d" is negative.

Cheers and thanks a lot.
 
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  • #2
MManuel Abad said:
[itex]V= a Tr(\Phi^{4}) + b (Tr(\Phi^2))^2 + c Tr(\Phi^3) + d Tr(\Phi^2) [/itex].
There's only two invariants for SU(3), one of them quadratic, the other cubic. Consequently your first term should be discarded. Tr(Φ4) and Tr(Φ2)2 are basically the same thing.
 
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  • #3
Are they really the same thing?
Only in the case in which:

Suppose the matrix to have diagonal elements [itex]h_{ii}[/itex]
They are the same if:
[itex]\sum_{i} h_{ii}^{4} = (\sum_{i}h_{ii}^{2})^{2} [/itex]
Maybe this holds? why?

As for the cubic term, it's not so common to take in account cubic terms in the lagrangians because they are somehow destroying the symmetry under reflections:
[itex] Φ \rightarrow -Φ [/itex]
In general if you want to break this symmetry, nothing stops you from taking this kind of terms into account.
 
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  • #4
The field Φ is said to be an SU(3) octet. That's Φi, i=1,...,8. There's one way to form an SU(3) invariant which is quadratic, namely δijΦiΦj. That's the Casimir invariant, which could be written Tr(Φ2).

How do you form a cubic invariant? In SU(3) there's one way to do that, namely dijkΦiΦjΦk.

Ok, now try to form an SU(3) invariant which is quartic. ΦiΦjΦkΦl times what? The only available things you can multiply this by are δijδkl or δikδjl or δilδjk. They all lead to the same thing: just the Casimir invariant squared.
 
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  • #5
Bill_K said:
Ok, now try to form an SU(3) invariant which is quartic. ΦiΦjΦkΦl times what? The only available things you can multiply this by are δijδkl or δikδjl or δilδjk. They all lead to the same thing: just the Casimir invariant squared.

What about terms with coefficients [itex]d^{ijk} d^{klm}[/itex]? Wouldn't they be able to produce [itex]\Phi^4[/itex] terms as well? Or does this contraction also have an expression in terms of [itex]\delta^{ij}[/itex]'s? If that is the case, do you happen to know it?
 
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  • #6
Almost all useful relations of the algebraic structure of [itex]SU(N)[/itex] are listed below:
[tex]
[ T_{ a } , T_{ b } ] = i f_{ a b c } T_{ c } , \ \ \ \ \ (1)
[/tex]
[tex]
\{ T_{ a } , T_{ b } \} = \frac{ 1 }{ N } \delta_{ a b } I_{ N } + d_{ a b c } T_{ c } , \ \ \ (2)
[/tex]
where [itex]I_{ N }[/itex] is the N-dimensional unit matrix. The [itex]f_{ a b c }[/itex] are antisymmetric and the [itex]d_{ a b c }[/itex] symmetric under the interchange of any two indices.

Traces:

[tex]
\mbox{ Tr } ( T_{ a } ) = 0, \ \ \mbox{ Tr } ( T_{ a } T_{ b } T_{ c } ) = \frac{ 1 }{ 4 } ( d_{ a b c } + i f_{ a b c } ) , \ \ \ (3)
[/tex]
[tex]
\mbox{ Tr } ( T_{ a } T_{ b } ) = \frac{ 1 }{ 2 } \delta_{ a b }, \ \ \mbox{ Tr } ( T_{ a } T_{ b } T_{ a } T_{ c } ) = - \frac{ 1 }{ 4 N } \delta_{ b c }, \ \ (4)
[/tex]
[tex]
2 T_{ a } T_{ b } = \frac{ 1 }{ N } \delta_{ a b } I_{ N } + ( d_{ a b c } + i f_{ a b c } ) T_{ c }, \ \ \ (5)
[/tex]
[tex]
2 ( T_{ a } )_{ i j } ( T_{ a } )_{ k l } = \delta_{ i l } \delta_{ j k } - \frac{ 1 }{ N } \delta_{ i j } \delta_{ k l } . \ \ \ \ (6)
[/tex]

Useful matrices for the [itex]N^{ 2 } – 1[/itex] dimensional adjoint representation:

[tex]
( F_{ a } )_{ b c } = - i f_{ a b c }, \ \ ( D_{ a } )_{ b c } = d_{ a b c } . \ \ \ (7)
[/tex]

The Jacobi identities:

[tex]f_{ a b e } f_{ e c d } + f_{ c b e } f_{ a e d } + f_{ d b e } f_{ a c e } = 0 , \ \ \ \ (8a)[/tex]
[tex]f_{ a b e } d_{ e c d } + f_{ c b e } d_{ a e d } + f_{ d b e } d_{ a c e } = 0 , \ \ \ (8b)[/tex]
or equivalently
[tex][ F_{ a } , F_{ b } ] = i f_{ a b c } F_{ c } , \ \ \ \ (9a)[/tex]
[tex][ F_{ a } , D_{ b } ] = i f_{ a b c } D_{ c } . \ \ \ \ (9b)[/tex]

Contractions and more traces:

[tex]
f_{ a b e } f_{ c d e } = \frac{ 2 }{ N } ( \delta_{ a c } \delta_{ b d } - \delta_{ a d } \delta_{ b c } ) + ( d_{ a c e } d_{ b d e } - d_{ b c e } d_{ a d e } ), \ \ (10)
[/tex]
[tex]f_{ a b b } = 0, \ \ \Rightarrow \ \ \mbox{ Tr } ( F_{ a } ) = 0 , \ \ (11a)[/tex]
[tex]d_{ a b b } = 0, \ \ \Rightarrow \ \ \mbox{ Tr } ( D_{ a } ) = 0 , \ \ (11b)[/tex]
[tex]f_{ a c d } f_{ b c d } = N \delta_{ a b }, \Rightarrow \mbox{ Tr } ( F_{ a } F_{ b } ) = N \delta_{ a b } , \ \ (12a)[/tex]
[tex]F_{ a } F_{ a } = N I_{ N^{ 2 } - 1 } , \ \ \ (12b)[/tex]
[tex]
f_{ a c d } d_{ b c d } = 0 , \ \mbox{ Tr } ( F_{ a } D_{ b } ) = 0 , \ \ F_{ a } D_{ a } = 0 , \ \ \ (13)
[/tex]
[tex]d_{ a c d } d_{ b c d } = \frac{ N^{ 2 } - 4 }{ N } \delta_{ a b }, \ \ \ (14a)[/tex]
[tex]\mbox{ Tr } ( D_{ a } D_{ b } ) = \frac{ N^{ 2 } - 4 }{ N } \delta_{ a b } \ \ (14b)[/tex]
[tex]D_{ a } D_{ a } = \frac{ N^{ 2 } - 4 }{ N } I_{ N^{ 2 } - 1 } , \ \ \ \ (14c)[/tex]
[tex]\mbox{ Tr } ( F_{ a } F_{ b } F_{ c } ) = \frac{ i N }{ 2 } f_{ a b c } \ \ \ (15a)[/tex]
[tex]\mbox{ Tr } ( D_{ a } F_{ b } F_{ c } ) = \frac{ N }{ 2 } d_{ a b c } , \ \ \ (15b)[/tex]
[tex]\mbox{ Tr } ( D_{ a } D_{ b } F_{ c } ) = i \frac{ N^{ 2 } - 4 }{ 2 N } f_{ a b c } , \ \ (15c)[/tex]
[tex]\mbox{ Tr } ( D_{ a } D_{ b } D_{ c } ) = \frac{ N^{ 2 } - 12 }{ 2 N } d_{ a b c } , \ \ (15d)[/tex]

And finally my favourite

[tex]\mbox{ Tr } ( F_{ a } F_{ b } F_{ a } F_{ c } ) = \frac{ N^{ 2 } }{ 2 } \delta_{ b c } . \ \ \ (16)[/tex]

Sam
 
  • #7
SU(3) is a rank two Lie group, and so we know there can only be two independent Casimir invariants. So it's not profitable to spend time looking for a third one. Any other invariant we form is guaranteed to be algebraically dependent on those two.

MManuel Abad said:
What about terms with coefficients [itex]d^{ijk} d^{klm}[/itex]? Wouldn't they be able to produce [itex]\Phi^4[/itex] terms as well? Or does this contraction also have an expression in terms of [itex]\delta^{ij}[/itex]'s? If that is the case, do you happen to know it?
See Eq. (3.10) of this paper.
 
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  • #8
Wooow! Thank you so much all, this was so complete!
 

1. What is a scalar in adjoint representation?

A scalar in adjoint representation refers to a type of representation used in mathematics and physics to describe certain physical quantities. It is a mathematical object that transforms under a specific group in a particular way, and is used to represent symmetries in physical systems.

2. How is a scalar in adjoint representation different from other types of representations?

Unlike other types of representations, such as vector or matrix representations, a scalar in adjoint representation is a single number that does not change under transformations. This means it is invariant under the group of transformations, making it a useful tool in studying symmetries.

3. What are some applications of scalar in adjoint representation?

Scalar in adjoint representation has various applications in physics and mathematics. It is commonly used in quantum field theory, where it describes the interactions between different particles. It is also used in Lie theory, as well as the study of geometric structures and symmetries in differential geometry.

4. How is a scalar in adjoint representation related to other mathematical concepts?

A scalar in adjoint representation is closely related to other mathematical concepts such as tensors, which are used to describe the properties of physical systems. It is also related to the concept of eigenvalues and eigenvectors in linear algebra, as it represents the eigenvalue of a transformation.

5. What are some properties of scalar in adjoint representation?

Some key properties of scalar in adjoint representation include its invariance under transformations, its ability to represent symmetries, and its use in describing the interactions between particles. It is also a fundamental concept in many branches of mathematics and physics, making it an important tool for understanding complex systems.

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