# Scalar in adjoint representation

1. Jan 31, 2014

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 $\Phi$, 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:

$V= a Tr(\Phi^{4}) + b (Tr(\Phi^2))^2 + c Tr(\Phi^3) + d Tr(\Phi^2)$.

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.

2. Jan 31, 2014

### Bill_K

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.

Last edited: Jan 31, 2014
3. Jan 31, 2014

### ChrisVer

Are they really the same thing?
Only in the case in which:

Suppose the matrix to have diagonal elements $h_{ii}$
They are the same if:
$\sum_{i} h_{ii}^{4} = (\sum_{i}h_{ii}^{2})^{2}$
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:
$Φ \rightarrow -Φ$
In general if you want to break this symmetry, nothing stops you from taking this kind of terms into account.

Last edited: Jan 31, 2014
4. Jan 31, 2014

### Bill_K

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.

Last edited: Jan 31, 2014
5. Jan 31, 2014

What about terms with coefficients $d^{ijk} d^{klm}$? Wouldn't they be able to produce $\Phi^4$ terms as well? Or does this contraction also have an expression in terms of $\delta^{ij}$'s? If that is the case, do you happen to know it?

Last edited: Jan 31, 2014
6. Feb 1, 2014

### samalkhaiat

Almost all useful relations of the algebraic structure of $SU(N)$ are listed below:
$$[ T_{ a } , T_{ b } ] = i f_{ a b c } T_{ c } , \ \ \ \ \ (1)$$
$$\{ T_{ a } , T_{ b } \} = \frac{ 1 }{ N } \delta_{ a b } I_{ N } + d_{ a b c } T_{ c } , \ \ \ (2)$$
where $I_{ N }$ is the N-dimensional unit matrix. The $f_{ a b c }$ are antisymmetric and the $d_{ a b c }$ symmetric under the interchange of any two indices.

Traces:

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

Useful matrices for the $N^{ 2 } – 1$ dimensional adjoint representation:

$$( F_{ a } )_{ b c } = - i f_{ a b c }, \ \ ( D_{ a } )_{ b c } = d_{ a b c } . \ \ \ (7)$$

The Jacobi identities:

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

Contractions and more traces:

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

And finally my favourite

$$\mbox{ Tr } ( F_{ a } F_{ b } F_{ a } F_{ c } ) = \frac{ N^{ 2 } }{ 2 } \delta_{ b c } . \ \ \ (16)$$

Sam

7. Feb 1, 2014

### Bill_K

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.

See Eq. (3.10) of this paper.

Last edited: Feb 1, 2014
8. Feb 1, 2014