- #1
lennyleonard
- 23
- 0
Hi everyone!
I would like to ask you a very basic question on the decomposition [tex]3\otimes\bar 3=1\oplus 8[/tex] of su(3) representation.
Suppose I have a tensor that transforms under the 8 representation (the adjoint rep), of the form [tex]O^{y}_{x}[/tex]
where upper index belongs to the $\bar 3$ rep and the lower ones to the 3 rep (x,y=1,2,3).
Now, I know that under $T_a$ (an element of the Lie algebra) this tensor transforms as
[tex](T_a O)^y_x\equiv (T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]
And that's by definition.
But it is also true that, for any s-dimensional representation of a Lie algrebra I should have
[tex][T_a,\mathcal O^{s}_i]=(T_a^{s})_{ij}\mathcal O_j[/tex]
where -s≤ i ≤s, as an operatorial equation.
This means that if we take the an operator that transforms under the adjoint rep (s=8) we should get
[tex][T_a,\mathcal O^{adj}_i]=(T_a^{\text{adj}})_{ij}\mathcal O_j[/tex]
where
[tex](T_a^{\text{adj}})_{ij}\equiv -if^{aij}[/tex]
are the matrix element of the adjoint representation of the su(3) Lie algebra, and the f^{abc}'s are the structure constant.My problem is that I can't explicitly constuct the correnspondence between the tensorial approach and the operatorial one, that is
[tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y\leftrightarrow (T_a^{\text{adj}})_{ij}\mathcal O_j=-if^{aij}\mathcal O_j [/tex]
In other words, I cannot find an identification
[tex] O^x_y\leftrightarrow \mathcal O_i[/tex]
where again (x,y=1,2,3), (i=1,8) that could make the algebra strcture constants appear into
[tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]
to check that the tensor actually transforms in the adjoint representation.I know that this may seem a rather cumbersome question, but it is in fact a ground question in group theory.
Please tell me if I have not been clear enough!
Thanks for your time!
I would like to ask you a very basic question on the decomposition [tex]3\otimes\bar 3=1\oplus 8[/tex] of su(3) representation.
Suppose I have a tensor that transforms under the 8 representation (the adjoint rep), of the form [tex]O^{y}_{x}[/tex]
where upper index belongs to the $\bar 3$ rep and the lower ones to the 3 rep (x,y=1,2,3).
Now, I know that under $T_a$ (an element of the Lie algebra) this tensor transforms as
[tex](T_a O)^y_x\equiv (T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]
And that's by definition.
But it is also true that, for any s-dimensional representation of a Lie algrebra I should have
[tex][T_a,\mathcal O^{s}_i]=(T_a^{s})_{ij}\mathcal O_j[/tex]
where -s≤ i ≤s, as an operatorial equation.
This means that if we take the an operator that transforms under the adjoint rep (s=8) we should get
[tex][T_a,\mathcal O^{adj}_i]=(T_a^{\text{adj}})_{ij}\mathcal O_j[/tex]
where
[tex](T_a^{\text{adj}})_{ij}\equiv -if^{aij}[/tex]
are the matrix element of the adjoint representation of the su(3) Lie algebra, and the f^{abc}'s are the structure constant.My problem is that I can't explicitly constuct the correnspondence between the tensorial approach and the operatorial one, that is
[tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y\leftrightarrow (T_a^{\text{adj}})_{ij}\mathcal O_j=-if^{aij}\mathcal O_j [/tex]
In other words, I cannot find an identification
[tex] O^x_y\leftrightarrow \mathcal O_i[/tex]
where again (x,y=1,2,3), (i=1,8) that could make the algebra strcture constants appear into
[tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]
to check that the tensor actually transforms in the adjoint representation.I know that this may seem a rather cumbersome question, but it is in fact a ground question in group theory.
Please tell me if I have not been clear enough!
Thanks for your time!