In Georgi's book (page 143), he calculates the tensor components of $3\otimes 8$ under the $\mathrm{SU(3)}$ explicitly using tensor components. Namely;
$u^{i}$ (a $3$) times $v^{j}_k$ (an $8$, meaning that it is also traceless, $v^j_j =0$), equal
$$
\dfrac{1}{2}\left(u^{i}v^{j}_{k}+u^{j}v^{i}_{k} – \dfrac{1}{4}\delta^{i}_{k}u^{l}v^{j}_{l} – \dfrac{1}{4}\delta^{j}_{k}u^{l}v^{i}_{l} \right) + \dfrac{1}{4}\epsilon^{ijl}\left(\epsilon_{lmn}u^m v^{n}_{k} + \epsilon_{kmn}u^m v^{n}_{l}\right) +\dfrac{1}{8}\left(3\delta^{i}_{k}u^l v^{j}_{l} – \delta^{j}_{k}u^{l}v^{i}_{l} \right).
$$
I understand that initially (the first term in the above equation) he symmetrizes with respect to the two upper indices while at the same time he subtracts the traces. In the second term above, I can see that he lowers two indices with the invariant $\epsilon_{lmn}$ but I don't understand why he then symmetrizes! ($\epsilon_{lmn}u^m v^{n}_k \propto A_{lk}$ and he symmetrizes $A_{lk}$). While at the last term he adds the traces he subtraced with a spurious 3 and signs I don't understand how he got.
Can someone elaborate on how exactly one can generally decompose tensor products into irreducible parts? Is there a connection with the Young Tableaux? With the Young tableaux one can easily conclude that $3\otimes 8 = 15 \oplus \bar{6} \oplus 3$, however how does one translate the Young tableaux into tensor components to get the above expression?
Thanks in advance.
Best Answer
The way I understand it, there is actually no shortcut method to write the explicit tensor components from the Young tableaux. We just need to symmetrize and antisymmetrize in all upper and lower indices, and whenever we come across mixed indices, remove traces as well.
Traceless symmetric tensors, antisymmetric tensors and traces transform irreducibly. To elaborate; we have $u^{i}v^{j}_k$, firstly we build the symmetric(in the upper indices) traceless part $$ S^{ij}_{k} = \dfrac{1}{2}\left(u^{i}v^{j}_k + u^{j}v^{i}_k - \alpha\delta_{k}^{i}u^lv^j_l- \beta\delta_{k}^{j}u^lv^i_l\right). $$ We calculate the $\alpha$ and $\beta$ coefficients from the requirement that $S^{ij}_k = S^{ji}_k$ and $S^{ij}_{i} = 0$, while keeping in mind that $v^i_i = 0$ as well. This gives $\alpha = \beta = \dfrac{1}{4}$.
Next we consider the antisymmetric part (thanks "Frederic Brünner" for pointing it out) $$ A^{ij}_k = \dfrac{1}{2}\left( u^{i}v^{j}_k - u^{j}v^{i}_k \right) = \dfrac{1}{2}\epsilon^{ijl}\epsilon_{lmn}u^{m}v^{n}_k. $$
Finally we should add the traces we subracted, getting $$ u^{i}v^{j}_k = \underbrace{S^{ij}_{k}}_{15} + \underbrace{A^{ij}_k}_{\bar{6}} +\underbrace{\dfrac{1}{8}\delta_{k}^{i}u^lv^j_l + \dfrac{1}{8}\delta_{k}^{i}u^lv^j_l}_{3} $$
However, I realised that the above procedure is almost complete, and actually leads to reducible tensors. Naively, I thought initially that antisymmetric tensors don't need to have the traces removed, but in fact $A^{ij}_k$ needs to have its two traces removed (otherwise one could contract with $\delta^{i}_j$ and get a reduced tensor). So to correct the above, we write $$ A^{ij}_k = \dfrac{1}{2}\left( u^{i}v^{j}_k - \lambda\delta^{i}_{k}u^lv_l^j - u^{j}v^{i}_k + \mu\delta^{j}_ku^lv^i_l \right). $$ Again by requiring antisymmetry and tracelessness; $A^{ij}_k = -A^{ji}_k$, $A^{ij}_i = 0$, we get $\lambda = \dfrac{1}{2}$ and $\mu = -\dfrac{1}{2}$.
In total then, to the right hand side, we have $-\dfrac{1}{8} - \dfrac{1}{4} = -\dfrac{3}{8}$ times the trace $\delta^{i}_{k}u^lv_l^j$, so to make the two sides of the original equations equal, we add $\dfrac{3}{8}$ of that trace (also for the other trace we need to subtract $\dfrac{1}{8}$).
The correct answer is, therefore, the one given by Georgi, if of course one immediately ``sees'' that symmetrizing $\epsilon_{lmn}u^mv^n_k$ with respect to $lk$ and then contracting with $\epsilon^{ijl}$ is equivalent to removing the traces of the $A^{ij}_k$!