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Quadratic Casimir Operator of SU(3)

  1. Jan 31, 2009 #1
    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.

    Last edited: Jan 31, 2009
  2. jcsd
  3. Feb 1, 2009 #2


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    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
    by the way;
    * I normalized the system. i=1
    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]

    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.
  4. Feb 1, 2009 #3
    [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?
    Last edited: Feb 1, 2009
  5. Feb 1, 2009 #4


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    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].
  6. Feb 2, 2009 #5
    I'm done with it.
    Thanks for the help.
    Last edited: Feb 2, 2009
  7. Feb 2, 2009 #6


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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.
  8. Feb 3, 2009 #7
    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.
  9. Feb 3, 2009 #8


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  10. Feb 4, 2009 #9
  11. Feb 4, 2009 #10
  12. Feb 4, 2009 #11


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  13. Feb 4, 2009 #12


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    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***

    [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]


    [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.


  14. Feb 4, 2009 #13
    Thanks for the detailed information. As a newbie of Group Theory, I need all info that i can get.
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