Quadratic Casimir Operator of SU(3)

[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
[T_i , T_j]= \sum f_{i,j,k} T_{k} ...eq2

[T² , T_i]=[\sum C_{i}_{j}T_{i}T_{j} , T_s]=\sum C_{i}_{j}T_{i}[T_{j}, T_{s}] + \sum C_{i}_{j}[T_{i}, T_{s}]T_{j}=0 ...eq3

[T_j,T_s]=\sum f_{j,s,m} T_{m} ...eq4
[T_i,T_s]=\sum f_{i,s,m} T_{m} ...eq5

[T² , T_i] = \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 T_i matrices but i need to compute C_i,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 C_i,j constants?

Thanks for you patience and advices.

ToreHan.

 

[nrqed]

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
[T_i , T_j]= \sum f_{i,j,k} T_{k} ...eq2

[T² , T_i]=[\sum C_{i}_{j}T_{i}T_{j} , T_s]=\sum C_{i}_{j}T_{i}[T_{j}, T_{s}] + \sum C_{i}_{j}[T_{i}, T_{s}]T_{j}=0 ...eq3

you mean T_s on the left side, not T_i. Otherwise it looks good.

[T_j,T_s]=\sum f_{j,s,m} T_{m} ...eq4
[T_i,T_s]=\sum f_{i,s,m} T_{m} ...eq5

[T² , T_i] = \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

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

** 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 T_i matrices but i need to compute C_i,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 C_i,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 Tr(T_a T_b) = C \delta_{ab} [/quote] where is an irrelevant constant. Then you shoul dbe able to show that the coefficients 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. <br /> <br /> At first sight, I don&#039;t see more restrictions on the coefficients but I haven&#039;t looked at the problem in detail.

 

[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?

 

[nrqed]

[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?

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}.

 

[torehan]

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

 

[nrqed]

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.

 

[torehan]

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 i \neq j \Rightarrow C_ij = -C_ji

That's enough to define the C coefficients.
Using [ T² , T_s ] = 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 C_ij = 0 for i \neq j and also I define that C₁₁=C₂₂=C₃₃=...=C₈₈.

 

[samalkhaiat]

As you said,

for i \neq j \Rightarrow C_ij = -C_ji

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

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

is zero, and you end up with

T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}

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

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

regards

sam

 

[torehan]

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

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

is zero, and you end up with

T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}

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

Indeed, for su(3) C_{ij} = c\delta_{ij}, 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 T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k} ?

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

for i /neq j C_ij = 0 that we have C₁₁=C₂₂=C₃₃=...=C₈₈.

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

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

It's QUADRATIC and CASIMIR OP.

By the way thanks for your interest.

ToreHan

 

[xepma]

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

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

is zero, and you end up with

T^{2} = \frac{1}{2} C_{ij} f_{ijk} T_{k}

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

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

regards

sam

You were a bit too hasty... He didn't say that C_{ij} 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:
\sum_{i,j}\kappa_{ij} T^iT^j
where \kappa_{ij} 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 \kappa_{ij} \propto \delta_{ij}. Your sum of squared generators is quite a general expression.

 

[samalkhaiat]

Sorry Sam,

I think there is a missunderstanding in definition of T².

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

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

for i /neq j C_ij = 0 that we have C₁₁=C₂₂=C₃₃=...=C₈₈.

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

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

It's QUADRATIC and CASIMIR OP.

By the way thanks for your interest.

ToreHan

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

C_{ij}=0, \ \ \mbox{for} \ i \neq j


sam

 

[samalkhaiat]

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

<br /> 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}<br />

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

 

[torehan]

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