The abstract group SU(3) is represented by a set of eight 3×3 matrices of complex elements which have determinant of unity. These elements of the group can be generated by eight special matrices. These matrices must be Hermitian; i.e., the transpose of their complex conjugates is the same as the matrix. These matrices do not have determinants of unity; instead all have traces (sums of elements on the principal diagonal) of zero. For a proof these proposition see Group Generators.

In the following i represents the imaginary unit √-1. Here are the generators for the matrix representation of SU(3).

| 0	1	0 |
½ | 1	0	0 |
| 0	0	0 |

| 0	−i	0 |
½ | i	0	0 |
| 0	0	0 |

| 1	0	0 |
½ | 0	−1	0 |
| 0	0	0 |

| 0	0	1 |
½ | 0	0	0 |
| 1	0	0 |

| 0	0	−i |
½ | 0	0	0 |
| i	0	0 |

| 0	0	0 |
½ | 0	0	1|
| 0	1	0 |

| 0	0	0 |
½ | 0	0	−i |
| 0	i	0 |

| 1	0	0 |
1/(2√3)  | 0	1	0 |
| 0	0	−2 |

Let these matrices be labeled λ₁, λ₁, …, λ₈. Now consider the computation of the commutations of them.

λ₁λ₂ is equal to

| i	0	0 |
¼ | 0	-i	0 |
| 0	0	0 |

which reduces to

| 1	0	0 |
¼i | 0	-1	0 |
| 0	0	0 |

which is ½iλ₃

Likewise λ₂λ₁ is equal to

| -i	0	0 |
¼  | 0	i	0 |
|  0	0	0 |

which reduces to

| 1	0	0 |
-¼i | 0	-1	0 |
| 0	0	0 |

which is −½iλ₃

Thus the commutation of λ₁ and λ₂ is equal to

[λ₁, λ₂] = λ₁λ₂ −λ₂λ₁
= ½iλ₃ − (−½λ₃) = iλ₃

Similarly the commutations for the other matrices can be found and the results can be represented as

[λ_j, λ_k] = iε_jkmλ_m

where ε_jkm is +1 if jkm is an even permutation of values of j,k,m written in the order of their sizes and −1 if jkm is an odd permutation. If there are repeated indices then ε_jkm is equal to zero.

The Formulation of Raising and Lowering Operators

(To be continued.)

(To be continued.)