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Invariant tensors of SU(3)

  1. Nov 12, 2015 #1

    CAF123

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    'Using the following normalization in the su(3) algebra ##[\lambda_i, \lambda_j] = 2if_{ijk}\lambda_k##, we see that ##g_{ij} = 4f_{ikl}f_{jkl} = 12 \delta_{ij}## and, by expanding the anticommutator in invariant tensors, we have further that $$\left\{\lambda_i, \lambda_j\right\} = \frac{4}{3}\delta_{ij} + 2d_{ijk}\lambda_k.$$
    The first statement about ##g_{ij}## I understand but how did the one about the anticommutator come about?
    I can reexpress ##\left\{\lambda_i, \lambda_j\right\} = [\lambda_i, \lambda_j] + 2 \lambda_j \lambda_i = 2if_{ijk}\lambda_k + 2 \lambda_j \lambda_i##. Now, ##\lambda_j \lambda_i## is a second rank tensor so can be written as ##a \delta_{ij}##, for some a. I was thinking I could then consider a single case to determine a (i.e i=j=1) but this didn't work.

    Any tips would be great!

    Thanks!
     
  2. jcsd
  3. Nov 12, 2015 #2
    You are incorrect to say that ##\lambda_j \lambda_i## can be written as ##a \delta_{i j}##, which explains why your approach didn't work. Unfortunately, I don't know how to prove your identity. What is ##d_{i j k}##?
     
  4. Nov 12, 2015 #3

    CAF123

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    Hi Jackadsa,
    Yup, I saw that I was incorrect in that after I posted my thread. ##\delta_{ij}## and ##d_{ijk}## are supposed to be the two invariant tensors for SU(3). In my notes it also says that ##d_{ijk} = \frac{1}{4}\text{Tr} \lambda_i \left\{\lambda_j, \lambda_k\right\}##, so I guess I can use this fact. So then $$d_{ijk} = \frac{1}{4}\text{Tr} \lambda_i ([\lambda_j, \lambda_k] + 2 \lambda_k \lambda_{\ell}) = \frac{1}{4} \text{Tr} \lambda_i (2if_{jk \ell }\lambda_{\ell} + 2 \lambda_k \lambda_{j}) = \frac{i}{2} f_{jk\ell}\text{Tr} \lambda_i \lambda_{\ell} + \frac{1}{2}\text{Tr} (\lambda_i \lambda_k \lambda_j)$$ using the given normalisation of the algebra. Any ideas how to continue?

    Thanks!
     
  5. Nov 12, 2015 #4

    samalkhaiat

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    Any [itex]n \times n[/itex] hermitian matrix [itex]M[/itex] can be expanded in terms of the [itex]n \times n[/itex] hermitian traceless matrices [itex]\lambda^{a}[/itex] , [itex]a = 1,2, \cdots , n^{2}-1[/itex], and the [itex]n \times n[/itex] identity matrix [itex]I_{n}[/itex] as follow
    [tex]
    M = \frac{1}{n} \mbox{Tr}(M) \ I_{n} + \frac{1}{2} \sum_{c=1}^{n^{2}-1} \mbox{Tr}(M \lambda^{c}) \ \lambda^{c} . \ \ \ (1)
    [/tex]
    Now, take [itex]M = \{ \lambda^{a} \ , \lambda^{b} \}[/itex] and define the symmetric invariant tensor
    [tex]d^{abc} \equiv \frac{1}{4}\mbox{Tr}\left( \{ \lambda^{a} \ , \lambda^{b} \} \lambda^{c} \right) .[/tex] With the [itex]\lambda[/itex]’s normalized according to [itex]\mbox{Tr}(\lambda^{a}\lambda^{b}) = 2 \delta^{ab}[/itex], equation (1) becomes
    [tex]\{\lambda^{a} , \lambda^{b} \} = \frac{4}{n} \delta^{ab} \ I_{n} + 2 d^{abc} \lambda^{c} .[/tex]
    Adding this to the algebra [itex][\lambda^{a} , \lambda^{b}] = 2 i f^{abc} \lambda^{c}[/itex], and multiplying by another [itex]\lambda[/itex], you get
    [tex]\lambda^{a} \lambda^{b} \lambda^{e} = \frac{2}{n} \delta^{ab} \lambda^{e} + (i f^{abc} + d^{abc} ) \ \lambda^{c} \lambda^{e} .[/tex] Taking the trace, you get
    [tex]\frac{1}{2} \mbox{Tr} (\lambda^{a} \lambda^{b} \lambda^{c}) = i f^{abc} + d^{abc} .[/tex]
     
    Last edited: Nov 13, 2015
  6. Nov 13, 2015 #5

    CAF123

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    Hi samalkhaiat,
    I see, thanks. The only thing I didn't understand was this statement above^^. In general for any irreducible representation we have that ##\text{Tr}T_a T_b = C(R) \delta_{ab}## where ##C(R)## is the Casimir of the representation. In this case, the ##\lambda_a## constitute the fundamental representation for SU(3) so are indeed irreducible. I am just not sure how to get ##\text{Tr}\lambda_i \lambda_j = 2\delta_{ij}##, i.e showing the casimir of the defining rep of SU(3) is 2. Any ideas on this?

    I think I could also obtain the result by writing ##\left\{\lambda_i, \lambda_j\right\} = a\delta_{ij} + b d_{ijk}\lambda_k##, which computing some traces gives me a and b.
    Thanks.
     
  7. Nov 13, 2015 #6

    samalkhaiat

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    Don’t confuse normalization (i.e., orthogonal transformation plus scalling) of generators with the Dynkin’s index of irreducible representations.
    The quadratic Casmir of Lie algebra is given by
    [tex]C_{2}(r) = g^{ab} \ t^{(r)}_{a} \ t^{(r)}_{b} = d_{(r)} \ I_{r}[/tex] where [itex]d_{r}[/itex] is a representation-dependent number called the Dynkin’s index, and [itex]I_{r}[/itex] is the identity matrix in the irreducible representation, i.e., [itex]\mbox{Tr}(I_{r})[/itex] is the dimension of the representation space [itex]\mbox{dim}(r)[/itex]. Taking the trace, you get [tex]g^{ab} \ \mbox{Tr}(t^{(r)}_{a} \ t^{(r)}_{b}) = d_{(r)} \ \mbox{dim}(r) .[/tex] In the Adjoint representation, you can normalize the generators [itex]A_{a}[/itex] so that [tex]\mbox{Tr}(A_{a} \ A_{b}) = g_{ab}.[/tex] So, for any Lie algebra you have [tex]g^{ab} \ g_{ab} = d_{A} \ \mbox{dim}(A) , \ \ \Rightarrow \ \ d_{A} = 1.[/tex] This is because [itex]g^{ab}g_{ab} = \delta^{c}_{c} = \mbox{dim}(A)[/itex].

    For [itex]SU(n)[/itex] and [itex]SO(n)[/itex] you can always make the following normalization convention [itex]\mbox{Tr}(T_{i} \ T_{j}) = \lambda \ \delta_{ij}[/itex], because [itex]\mbox{Tr}(T_{i} \ T_{j})[/itex] is a real symmetric matrix and can be diagonalized by taking an appropriate real linear combination of the generators, with diagonal coefficients set to a constant [itex]\lambda[/itex]. With this basis of the algebra, the structure constants are given by [tex]C^{k}_{mn} = - \frac{i}{\lambda} \ \mbox{Tr}(T_{k}[T_{m},T_{n}]) ,[/tex] which implies that [itex]C^{k}_{mn}[/itex] is totally antisymmetric in all three indices.
    For example [itex]SU(2)[/itex], where [itex]g^{ab}= \frac{1}{2}\delta^{ab}[/itex], we choose, for the Fundamental representation, [tex]\mbox{Tr}(T_{a}^{(F)} \ T_{b}^{(F)}) = \mbox{Tr}( \frac{\sigma_{a}}{2} \ \frac{\sigma_{b}}{2} )= \frac{1}{2} \ \delta_{ab} ,[/tex] So, [tex]d_{(F)} \ \mbox{dim}(F) = \frac{1}{2} \ \delta^{ab} \ \frac{1}{2} \ \delta_{ab} = \frac{3}{4} .[/tex] From this we find the Dynkin’s index [itex]d_{(F)} = \frac{3}{8}[/itex], because [itex]\mbox{dim}(F) = 2[/itex].
     
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