《量子化学》课程教学资源（PPT课件讲稿）第三章 矩阵与算符

矩阵和矩阵相乘 ll b2…bk l112 Ik C=AB=/q、、W 22 C 21022 2k nk n×m mxk nxk

矩阵和矩阵相乘             =                         = = n n n k k k m m m k k k n n n m m m c c c c c c c c c b b b b b b b b b a a a a a a a a C AB             1 2 2 1 2 2 2 1 1 1 2 1 1 2 2 1 2 2 2 1 1 1 2 1 1 2 2 1 2 2 2 1 1 1 2 1 nm mk nk

1,2,…,k)(3,17 CC a21a2…a2nb1b2…b

= = m p ij aip ppj c 1 (i = 1, 2, …, n, j= 1,2, …, k) (3.17)             =                         n n n k k k m m m k k k n n n m m m c c c c c c c c c b b b b b b b b b a a a a a a a a             1 2 2 1 2 2 2 1 1 1 2 1 1 2 2 1 2 2 2 1 1 1 2 1 1 2 2 1 2 2 2 1 1 1 2 1

21 例14 B=0 010 0 10 AB= 01 BA=01 010 0 010 10 101 般而言AB≠BA,即矩阵乘法不满足交换律 但满足结合律ABC=A(BC)=(ABC

例1       = 0 1 0 1 0 1 A           = 1 0 0 1 2 1 B       =                 = 0 1 3 1 1 0 0 1 2 1 0 1 1 0 0 1 AB           =                = 1 0 1 0 1 0 2 1 2 0 1 1 0 0 1 1 0 0 1 2 1 BA 一般而言 AB  BA, 即矩阵乘法不满足交换律， 但满足结合律 ABC = A(BC) =(AB)C

转置矩阵、共轭矩阵、转置共轭矩阵 A=四acmA=laJm如nA*=an]nxm(318) 例24 A 2i2 2 2i2 如果F=ABC…X 则FH=(ABC…XH=X, CHBHA(3,19)

转置矩阵、共轭矩阵、转置共轭矩阵 A = [aij] nm AT = [aji] mn A* = [aij*] nm 例2       = 2 1 2 i i A       = 2 2 1 i i A T       − − = 2 1 2 i i A*       − − = 2 2 1 i i A H (3.18) 如果 F = ABCX 则 FH = (ABCX)H = XHCHBHA (3.19)