# Quadratic Casimir Operator of SU(3)

1. Jan 31, 2009

### torehan

Hi all,
I need to construct the Casimir op. of group SU(3).
I have these relations;

T2=$$\sum C_{i}_{j}T_{i}T_{j}$$ i,j=1,2....,8 ...eq1
[Ti , Tj]= $$\sum f_{i,j,k} T_{k}$$ ...eq2

[T2 , Ti]=[$$\sum C_{i}_{j}T_{i}T_{j}$$ , Ts]=$$\sum C_{i}_{j}T_{i}[T_{j}, T_{s}] + \sum C_{i}_{j}[T_{i}, T_{s}]T_{j}$$=0 ...eq3

[Tj,Ts]=$$\sum f_{j,s,m} T_{m}$$ ...eq4
[Ti,Ts]=$$\sum f_{i,s,m} T_{m}$$ ...eq5

[T2 , Ti] = $$\sum C_{i}_{j} \sum f_{j,s,m} T_{i} T_{m} + \sum C_{i}_{j} \sum f_{i,s,m} T_{m}T_{j}$$=0 ...eq6

by the way;
* I normalized the system. i=1
** i'm still improoving my mathematical skills because of that i might use more sum operator that i need. sorry for that.
*** All the indexes are in [1,8] range, i,j,k,s,m = 1,2,....,8
**** f coefficients are known!

I have the Ti matrices but i need to compute Ci,j constants to construct the quad. Casimir op.

1. Is the eq6 right?
2. If it's right , can i collect all of components in ONE sum operator? how the indexes change?
3. How could i compute the Ci,j constants?

Thanks for you patience and advices.

ToreHan.

Last edited: Jan 31, 2009
2. Feb 1, 2009

### nrqed

you mean T_s on the left side, not T_i. Otherwise it looks good.
Again, it's T_s pn the left side nsteas of T_i
[/quote]
by the way;
* I normalized the system. i=1
[/quote]
I am not sure what you mean by this. i is an index ranging over the number of generators.
Normalization has to do with defining the trace of a product of two generators to have some specific value, for example $Tr (T_a T_b) = \delta_{ab}/2$

To make progress, I would take the trace of you rfinal result and use the fact that the trace of two generators is chosen to be $Tr(T_a T_b) = C \delta_{ab} [/quote] where is an irrelevant constant. Then you shoul dbe able to show that the coefficients [itex] C_{ij}$ must be antisymmetric when $i \neq j$ and there is no restriction when $i = j$. So a possible choise is $C_{ij} =0$ for $i \neq j$ and $C_{ii} = 1$ which is the conventional choice.

At first sight, I don't see more restrictions on the coefficients but I haven't looked at the problem in detail.

3. Feb 1, 2009

### torehan

[Ti , Tj]=$$\sum i f_{i,j,k} T_{k}$$ i is a complex number.

I use the word "normalization" in simplification meaning. I took the complex i=1 to simplify the equation.

Last part of your post is not so clear. could you please repost it again?

Last edited: Feb 1, 2009
4. Feb 1, 2009

### nrqed

Sorry if my post got garbled.
I was saying that if we take the trace on both sides of your result and we use that
$Tr (T_a T_b) = C \delta_{ab}$ where C is an irrelevant constant, then one can show that for $i \neq j$ we must have $C_{ij} = - C_{ji}$.

5. Feb 2, 2009

### torehan

I'm done with it.
Thanks for the help.

Last edited: Feb 2, 2009
6. Feb 2, 2009

### nrqed

You're welcome. What did you find? What conditions did you find the C_ij must obey? I am wondering if I missed some trick.

7. Feb 3, 2009

### torehan

As you said,

for $i \neq j$ $$\Rightarrow$$ Cij = -Cji

That's enough to define the C coefficients.
Using [ T2 , Ts ] = 0 [s=1,2,....,8] commutation relations I get the coefficient equations like -C12 - C21= 0 and more of them. With theese equations I define all Cij = 0 for $i \neq j$ and also I define that C11=C22=C33=.......=C88.

8. Feb 3, 2009

9. Feb 4, 2009

### torehan

10. Feb 4, 2009

### xepma

11. Feb 4, 2009

### samalkhaiat

12. Feb 4, 2009

### samalkhaiat

A very important concept in the study of Lie algebras is the symmetric Cartan metric, defined by

$$g_{ij} = f_{ikl}f_{jlk} = \left( ad(X_{i})\right)_{kl} \left( ad(X_{j})\right)_{lk} = Tr \left( ad(X_{i}) ad(X_{j}) \right) = g_{ji}$$

By using the Jacobi identity for the structure constants, we can derive the following (very) important identity***

$$g_{ij}f_{jnk} = - g_{kj}f_{jni} \ \ \ \ (R)$$

We use the Cartan metric to construct the second-degree Casimir "operator"

$$C_{(2)} = g_{ij} X_{i}X_{j}$$

Now

$$[C_{(2)} , X_{n} ] = g_{ij}f_{jnk}X_{i}X_{k} + g_{ij}f_{ink} X_{k}X_{j}$$

Changing the dummy indices in the second term leads to

$$[C_{(2)},X_{n} ] = ( g_{ij}f_{jnk} + g_{jk}f_{jni} ) X_{i}X_{k}$$

It follows from Eq(R) that the LHS is identically zero, i.e.,

$$[C_{(2)}, X_{n}] = 0, \ \ \forall {n}$$

Thus, according to Schurs lemma, $C_{(2)}$ is a multiple of the identity in any irreducible representation.

For compact, semi-simple or simple Lie algebras like su(3), the Cartan metric can be diagonolized, i.e.,

$$g_{ij} = c \delta_{ij}$$

This means that the structure constant is totally antisymmetric. In this case, it is trivially obvious that

$$[C_{(2)}, X_{n}] = c ( f_{ink} + f_{kni}) X_{i}X_{k} = 0$$

***************

***:
In the non-abelian gauge theories, one takes the Lagrangian for gauge field as

$$\mathcal{L} = - \frac{1}{2} g_{ij}F_{i\mu \nu}F_{j}^{\mu\nu}$$

for some real symmetric and constant matrix $g_{ij}$. In order for this Lagrangian to be locally gauge invariant, the matrix $g_{ij}$ must satisfy the identity of Eq(R).
We can also show that this identity restricts the possible gauge group to be compact (semi-)simple Lie group. This is why I described that identity as "very" important.

regards

sam

13. Feb 4, 2009

### torehan

Thanks for the detailed information. As a newbie of Group Theory, I need all info that i can get.