How to calculate the structure constants $f^a_{bc}$ in SU(3) for the gluon field $G^a_{\mu\nu}$?

Question

The gluon field strength tensor $G^a_{\mu\nu}$ as found in

$$L_{QCD} = \sum_i^3 \bar \Psi_i(i\gamma^\mu(D_\mu)_{ij} - m\delta_{ij})\Psi_j - \frac{1}{4}\color{red}{G^a_{\mu\nu}G^{\mu\nu}_a}$$

is computed as:

$$G^a_{bc} = \frac{dA^a_c}{db} - \frac{dA^a_b}{dc} \mp \sum_{i,j}^8 f^a_{ij}A^i_bA^j_c$$

where $A^{floor}_{room}$ is the 8 high, 4 wide gluon vector field.

$f^a_{ij}$ the structure constant of SU(3) is further computed as:

I am not clear as to how these answers can be calculated from the generator matrices,

1. What does 458 in $f^{458}$ refer to?
2. How does $f^{458}$ evaluate to $\frac{\sqrt3}{2}?$
3. How does $f^{abc}$ relate to the $f^a_{bc}$ which is the required ordering of indices combination in $L_{QCD}$?

Answer 1

By definition

$$[T_a, T_b] = f_{abc}iT_c$$

eg.

$$[T_1, T_4] = T_1T_4 - T_4T_1=f_{147}iT_7$$ Expanding the T matrices: $$=\begin{bmatrix} 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} 0&0&\frac{1}{2} \\ 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0\end{bmatrix}-\begin{bmatrix} 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0\end{bmatrix}\begin{bmatrix} 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=\begin{bmatrix} 0&0&0 \\ 0 & 0 & \frac{1}{4} \\ 0 & -\frac{1}{4} & 0\end{bmatrix}$$ $$=if_{147} \begin{bmatrix} 0&0&0 \\ 0 & 0 & -\frac{i}{2} \\ 0 & \frac{i}{2} & 0\end{bmatrix}=f_{147} \begin{bmatrix} 0&0&0 \\ 0 & 0 & \frac{1}{2} \\ 0 & -\frac{1}{2} & 0\end{bmatrix}$$ Therefore $$f_{147} = \frac{1}{2}$$