Quadratic Casimir Operator of SU(3)

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

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

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

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

[T2 , Ti] = [tex]\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} [/tex]=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.
 
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Answers and Replies

  • #2
nrqed
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Hi all,
I need to construct the Casimir op. of group SU(3).
I have these relations;

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

[T2 , Ti]=[[tex]\sum C_{i}_{j}T_{i}T_{j}[/tex] , Ts]=[tex]\sum C_{i}_{j}T_{i}[T_{j}, T_{s}] + \sum C_{i}_{j}[T_{i}, T_{s}]T_{j} [/tex]=0 ...eq3
you mean T_s on the left side, not T_i. Otherwise it looks good.
[Tj,Ts]=[tex]\sum f_{j,s,m} T_{m}[/tex] ...eq4
[Ti,Ts]=[tex]\sum f_{i,s,m} T_{m}[/tex] ...eq5

[T2 , Ti] = [tex]\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} [/tex]=0 ...eq6
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 [itex] Tr (T_a T_b) = \delta_{ab}/2 [/itex]

** 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.

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 [itex] 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} [/itex] must be antisymmetric when [itex] i \neq j [/itex] and there is no restriction when [itex] i = j [/itex]. So a possible choise is [itex] C_{ij} =0 [/itex] for [itex] i \neq j [/itex] and [itex] C_{ii} = 1 [/itex] 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
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[Ti , Tj]=[tex]\sum i f_{i,j,k} T_{k}[/tex] 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?
 
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  • #4
nrqed
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[Ti , Tj]=[tex]\sum i f_{i,j,k} T_{k}[/tex] 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?

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
[itex] Tr (T_a T_b) = C \delta_{ab} [/itex] where C is an irrelevant constant, then one can show that for [itex] i \neq j [/itex] we must have [itex] C_{ij} = - C_{ji} [/itex].
 
  • #5
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I'm done with it.
Thanks for the help.
 
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  • #6
nrqed
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I'm done with it.
Thanks for the help.

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
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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.

As you said,

for [itex] i \neq j [/itex] [tex]\Rightarrow[/tex] 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 [itex] i \neq j [/itex] and also I define that C11=C22=C33=.......=C88.
 
  • #8
samalkhaiat
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As you said,

for [itex] i \neq j [/itex] [tex]\Rightarrow[/tex] Cij = -Cji

You are making a very big mistake! If [itex]C_{ij}[/itex] is antisymmetric, then the second term in

[tex]T^{2} = \frac{1}{2} C_{ij} [T_{i},T_{j}] + \frac{1}{2} C_{ij}\{T_{i},T_{j} \}[/tex]

is zero, and you end up with

[tex]T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}[/tex]

Obviously, this is neither QUADRATIC nor Casmir operator, because it is LINEAR in the generators and [itex][T^{2},T_{n}] \neq 0[/itex]

Indeed, for su(3) [itex]C_{ij} = c\delta_{ij}[/itex], I will explain this to you later.

regards

sam
 
  • #9
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You are making a very big mistake! If [itex]C_{ij}[/itex] is antisymmetric, then the second term in

[tex]T^{2} = \frac{1}{2} C_{ij} [T_{i},T_{j}] + \frac{1}{2} C_{ij}\{T_{i},T_{j} \}[/tex]

is zero, and you end up with

[tex]T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}[/tex]

Obviously, this is neither QUADRATIC nor Casmir operator, because it is LINEAR in the generators and [itex][T^{2},T_{n}] \neq 0[/itex]

Indeed, for su(3) [itex]C_{ij} = c\delta_{ij}[/itex], I will explain this to you later.

regards

sam


Sorry Sam,

I think there is a missunderstanding in definition of T2.

how do you get this [tex]T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}[/tex] ?

[tex]T^{2} = \sum C_{ij} T_{i}T_{j}[/tex] that is the definition of Quad. Casimir Op.

for [itex] i /neq j [/itex] Cij = 0 that we have C11=C22=C33=.......=C88.

and Quadratic Casimir Operator becomes [tex]T^{2} = \sum C_{ii} T_{i}^{2}[/tex]

Expanded version of quadretic Casimir Op. is , [tex]T^{2} = C_{11} T_{1}^{2} + C_{22} T_{2}^{2} + C_{33} T_{3}^{2} + ....... + C_{88} T_{8}^{2}[/tex]

It's QUADRATIC and CASIMIR OP.

By the way thanks for your interest.

ToreHan
 
  • #10
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You are making a very big mistake! If [itex]C_{ij}[/itex] is antisymmetric, then the second term in

[tex]T^{2} = \frac{1}{2} C_{ij} [T_{i},T_{j}] + \frac{1}{2} C_{ij}\{T_{i},T_{j} \}[/tex]

is zero, and you end up with

[tex]T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}[/tex]

Obviously, this is neither QUADRATIC nor Casmir operator, because it is LINEAR in the generators and [itex][T^{2},T_{n}] \neq 0[/itex]

Indeed, for su(3) [itex]C_{ij} = c\delta_{ij}[/itex], I will explain this to you later.

regards

sam

You were a bit too hasty... He didn't say that [itex]C_{ij}[/itex] was completely antisymmetric, only that his off-diagonal terms are. This matches your definition of the structure constants.

On a sidenote, the more general identity for the quadratic Casimir operator is:
[itex]\sum_{i,j}\kappa_{ij} T^iT^j[/itex]
where [itex]\kappa_{ij}[/itex] is the Killing metric. This holds for finite dimensional, semisimple Lie algebra's. I think it's also possible (at least in SU(N), but maybe for more general cases as well) to choose a basis of generators such that [itex]\kappa_{ij} \propto \delta_{ij}[/itex]. Your sum of squared generators is quite a general expression.
 
  • #11
samalkhaiat
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Sorry Sam,

I think there is a missunderstanding in definition of T2.

how do you get this [tex]T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}[/tex] ?

[tex]T^{2} = \sum C_{ij} T_{i}T_{j}[/tex] that is the definition of Quad. Casimir Op.

for [itex] i /neq j [/itex] Cij = 0 that we have C11=C22=C33=.......=C88.

and Quadratic Casimir Operator becomes [tex]T^{2} = \sum C_{ii} T_{i}^{2}[/tex]

Expanded version of quadretic Casimir Op. is , [tex]T^{2} = C_{11} T_{1}^{2} + C_{22} T_{2}^{2} + C_{33} T_{3}^{2} + ....... + C_{88} T_{8}^{2}[/tex]

It's QUADRATIC and CASIMIR OP.

By the way thanks for your interest.

ToreHan

Sorry for the confusion. So you ment to say that [itex]C_{ij}[/itex] is DIAGONAL. You did not need to write [itex]C_{ij} = - C_{ji}[/itex]. It is more informing to write

[tex]C_{ij}=0, \ \ \mbox{for} \ i \neq j[/tex]


sam
 
  • #12
samalkhaiat
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A very important concept in the study of Lie algebras is the symmetric Cartan metric, defined by

[tex]
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}
[/tex]

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

[tex]g_{ij}f_{jnk} = - g_{kj}f_{jni} \ \ \ \ (R)[/tex]


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

[tex]C_{(2)} = g_{ij} X_{i}X_{j}[/tex]

Now

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

Changing the dummy indices in the second term leads to

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

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

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

Thus, according to Schurs lemma, [itex]C_{(2)}[/itex] 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.,

[tex]g_{ij} = c \delta_{ij}[/tex]

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

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

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

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

[tex]\mathcal{L} = - \frac{1}{2} g_{ij}F_{i\mu \nu}F_{j}^{\mu\nu}[/tex]

for some real symmetric and constant matrix [itex]g_{ij}[/itex]. In order for this Lagrangian to be locally gauge invariant, the matrix [itex]g_{ij}[/itex] 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
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Thanks for the detailed information. As a newbie of Group Theory, I need all info that i can get.
 

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