87.2.电子的自旋算符和自旋函数(不要求) 15/71 为求得G和0,在G表象中的矩阵表示,设 1lt12 11D12 21a22 b21b22 利用反对易关系: 10 11a12 11a12 10 TzOxtoroz= 21a22 a21 a 0-1 11a12 11-a12 2 a21-a22 21-a22 0-a 0 a 所以:a1=a22=0,即:Gx 21 0 0 a122 10 a210 12 0 0|a2l2 0 所以:|a12=|a212=1 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §7.2. >fg^ŽÎÚg^¼ê(؇¦) 15/71 ¦σbxÚσby3σbzL¥Ý L«§µ σbx = a11 a12 a21 a22! , σby = b11 b12 b21 b22! |^‡é´'Xµ σbzσbx + σbxσbz = 1 0 0 −1 ! a11 a12 a21 a22! + a11 a12 a21 a22! 1 0 0 −1 ! = a11 a12 −a21 −a22! + a11 −a12 a21 −a22! = 2 a11 0 0 −a22! = 0 ¤±µa11 = a22 = 0§=µσbx = 0 a12 a21 0 ! σb 2 x = 0 a12 a21 0 ! 0 a ∗ 21 a ∗ 12 0 ! = |a12| 2 0 0 |a21| 2 ! = 1 0 0 1! ¤±µ|a12| 2 = |a21| 2 = 1