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How to get Gell-Mann matrices $f_{i}$ (more or less strictly)? What are the requirements for getting them, excluding $||f_{i}|| = 1$, commutational law $[f_{i}, f_{j}] = if_{ijk}f_{k}$ and hermitian nature? Is that enough?

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1 Answer 1

It may be got by a simple properties of SU(n) group matrix representation.

First, by representing the matrix $\hat {U}$ of group near identity matrix, $\hat {U} = \hat {E} + \hat {A}$, you may (keeping the linearity by $\hat {A}$) get properties of $\hat {A}$: $$ \hat {U}\hat {U}^{+} = \hat {E} \Rightarrow \hat {A}^{+} = -\hat {A}, \quad det \hat {U} = 1 \Rightarrow Tr(\hat {A}) = 0. $$ The first condition leads to the absence of the real part of diagonal components of $\hat {A}$ and to representation of non-diagonal components in form of $A_{ij} = a_{ij} + ib_{ij}, A_{ji} = -a_{ij} + ib_{ij}$. The second leads to condition of $\sum_{i} A_{ii} = 0$, but except this the parametrization of the diagonal components is arbitrary.

So for $SU(3)$ representation is not hard to see that corresponding matrice is $$ \hat {A} = i\begin{vmatrix} a_{3} + a_{8} & a_{1} - ia_{2} & a_{4} - ia_{5} \\ a_{1} + ia_{2} & a_{8} - a_{3} & a_{6} - ia_{7} \\ a_{4} + ia_{5} & a_{6} + ia_{7} & -2a_{8} \end{vmatrix}. $$ It remains only to expand matrix in sum of $\sum_{i}a_{i}\hat {R}_{i}$, where $\hat {R}_{i}$ are Gell-Mann matrices.

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It should be $det \hat U=1$. –  Adam Dec 3 '13 at 18:50
@Adam : yes, thank you. –  Andrew McAddams Dec 3 '13 at 18:59

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