# Quadratic Casimir Operator of SU(3)

• torehan
In summary: I'm sorry for these mistakes. I didn't use the commutation relations for [Ti,Tj] and it was my big mistake.So if I'm right, C_{ij}=C_{ji} and eq6 is right for the coefficients.[quote="torehan, post: 2060904"][QUOTE]You are making a very big mistake! If C_{ij} is antisymmetric, then the second term inT^{2} = \frac{1}{2} C_{ij} [T_{i},T_{j}] + \frac{1}{2} C_{ij}\{T_{i},T_{j} \}is zero
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:
torehan said:
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
you mean T_s on the left side, not T_i. Otherwise it looks good.
[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
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 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 $Tr(T_a T_b) = C \delta_{ab} [/quote] where is an irrelevant constant. Then you shoul dbe able to show that the coefficients [itex] C_{ij}$ 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.

[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:
torehan said:
[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 $C_{ij} = - C_{ji}$.

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

Last edited:
torehan said:
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.

nrqed said:
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 $i \neq j$ $$\Rightarrow$$ 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.

torehan said:
As you said,

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

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

samalkhaiat said:
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 $i /neq j$ Cij = 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 , $$T^{2} = C_{11} T_{1}^{2} + C_{22} T_{2}^{2} + C_{33} T_{3}^{2} + ... + C_{88} T_{8}^{2}$$

By the way thanks for your interest.

ToreHan

samalkhaiat said:
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.

torehan said:
samalkhaiat said:
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 $i /neq j$ Cij = 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 , $$T^{2} = C_{11} T_{1}^{2} + C_{22} T_{2}^{2} + C_{33} T_{3}^{2} + ... + C_{88} T_{8}^{2}$$

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

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

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

## 1. What is the Quadratic Casimir Operator of SU(3)?

The Quadratic Casimir Operator of SU(3) is a mathematical concept used in the study of quantum mechanics and group theory. It is a Hermitian operator that measures the total angular momentum of a system. In the context of SU(3), it is used to calculate the total number of particles in a system and their corresponding quantum states.

## 2. How is the Quadratic Casimir Operator of SU(3) calculated?

The Quadratic Casimir Operator of SU(3) is calculated by taking the sum of the squared generators of the SU(3) group, which represent the symmetry transformations of the system. This sum is then multiplied by a constant factor, known as the Casimir eigenvalue, to obtain the final value of the operator.

## 3. What are the applications of the Quadratic Casimir Operator of SU(3)?

The Quadratic Casimir Operator of SU(3) has various applications in physics, particularly in the study of subatomic particles and their interactions. It is used in the calculation of energy levels and quantum states of particles, as well as in the analysis of symmetries in physical systems.

## 4. Can the Quadratic Casimir Operator of SU(3) be extended to other groups?

Yes, the concept of the Quadratic Casimir Operator can be extended to other Lie groups, which are mathematical structures that describe symmetries in physics. However, the specific form and calculation method may vary depending on the group and its generators.

## 5. How does the Quadratic Casimir Operator of SU(3) relate to the other Casimir Operators?

The Quadratic Casimir Operator of SU(3) is one of the three Casimir Operators of the SU(3) group, along with the Linear and Cubic Casimir Operators. These operators are all used to characterize the symmetries and properties of a system, but they differ in their mathematical form and the information they provide about the system.

• Special and General Relativity
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
5
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
930
• General Math
Replies
6
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
4
Views
1K
• Cosmology
Replies
1
Views
896
• High Energy, Nuclear, Particle Physics
Replies
23
Views
4K
• Introductory Physics Homework Help
Replies
16
Views
1K
• Classical Physics
Replies
3
Views
2K
• Quantum Physics
Replies
2
Views
1K