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SU(3),
The Special Unitary
Group of Order 3

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).

    | 010 |
½ | 100 |
    | 000 |

    | 0−i0 |
½ | i  00 |
    | 0  00 |

    | 1  00 |
½ | 0−10 |
    | 0  00 |

    | 001 |
½ | 000 |
    | 100 |

    | 00−i |
½ | 00  0 |
    | i0  0 |

    | 000 |
½ | 001|
    | 010 |

    | 00  0 |
½ | 00−i |
    | 0i  0 |

              | 10  0 |
1/(2√3)  | 01  0 |
              | 00−2 |

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

λ1λ2 is equal to

    | i 00 |
¼ | 0-i0 |
    | 0 00 |

which reduces to
     | 1 00 |
¼i | 0-10 |
     | 0 00 |

which is ½iλ3

Likewise λ2λ1 is equal to

    | -i00 |
¼  | 0i0 |
    |  000 |

which reduces to
      | 1 00 |
-¼i | 0-10 |
      | 0 00 |

which is −½iλ3

Thus the commutation of λ1 and λ2 is equal to

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

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

j, λk] = iεjkmλm

where εjkm 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 εjkm is equal to zero.

The Formulation of Raising and Lowering Operators

(To be continued.)


(To be continued.)


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