厦门大学：《结构化学 Structural Chemistry》课程教学资源（课件讲稿）Chapter 4 The structure of diatomic molecules

=cΨ。+CΨn seqular equation Hoa-ESou Hab-ESab =0 Hba-ESBa Hob-ESpb .has the same form as '.Haa=Hbb. Hab=Hba (Hoa-ESaa)2=(Hab-ESab)2 Haa-ES a=(H ab-ESab) if Hoa-ESaa=-(Hab-ESab) E=H。+H地=a+B 1+Sab 1+S Haa-ESaa=Hab-ESab E2= Hoo-Hab. a-B 1-Sab 1-S

S S H H E if H ES H ES S S H H E if H ES H ES H ES H ES H ES H ES H H H H H ES H ES H ES H ES seqular equation c c ab aa ab aa aa ab ab ab aa ab aa aa ab ab aa aa ab ab aa aa ab ab a b aa bb ab ba ba ba bb bb aa aa ab ab a a b b                                     1 1 1 1 ( ) ( ) ( ) ( ) has the same form as 0 2 1 2 2 ,          

substituting E to seqular equation (Haa-ESda)Ca+(Hab-ESab)Cb=O () (Hba-ESpa)ca+(Hob-ESpb)Cp=O (2) Saa=Spb=1 顶H+Hc,+(H-H。+isc=0 1+Sab 1+Sab (Haa(1+Sab)-(Hoa +Hab))Ca+(Hab(1+Sab)-(Haa+Hab)Sab)Cp=O (Hoa(1+Sab)-(Haa+Hab))ca+(Hab(1+Sab)-(Haa+Hab)Sab)Cp=O (HSb-Hb)c。+(Hb-HmSb)c。=0 Ca-Ch=0 Ca=Cp p1=Ca"a+CbΨ%=Ca("a+Ψ6)

1 ( ) ( ) 0 (1) ( ) ( ) 0 (2) 1 ( )( )0 1 1 ( (1 ) ( )) ( (1 ) ( ) ) 0 aa aa a ab ab b ba ba a bb bb b aa bb aa ab aa ab aa a ab ab b ab ab aa ab aa ab a ab ab aa ab ab b substituting E to seqular equation H ES c H ES c H ES c H ES c S S HH HH H c H Sc S S H S H H c H S H HSc                       1 ( (1 ) ( )) ( (1 ) ( ) ) 0 ( )( )0 0 ( ) aa ab aa ab a ab ab aa ab ab b aa ab ab a ab aa ab b ab a b aa bb a a b H S H H c H S H HSc HS H c H HS c cc cc     ccc              

nomalization condition ∫4'4dr=1 「(c(y。+wb)'c(yn+)dr=l ∫[cyg2+2c,w,。+c,y2]dr=1 2cg21+Sb)=1 1 Ca= 2(1+Sab) 1 4= (y。+46) 2(1+S) substituting E2 to seqular equation Ca+Cp=0 Ca=-Cb 42=CaΨa+Cb"6=Ca(Ψa-Ψ6) 1 2= (w。-6) V2(1-Sab)

( ) 2(1 ) 1 ( ) 0 ( ) 2(1 ) 1 2(1 ) 1 2 (1 ) 1 [ 2 ] 1 ( ( )) ( ) 1 1 2 2 2 1 2 2 2 2 2 2 * 1 * 1 a b ab a a b b a a b a b a b a b ab ab a a ab a a a a b a b a a b a a b S c c c c c c c substituting E to seqular equation S S c c S c c c d c c d nomalization condition d                                                     

十。 E a+B 三 2G* 1+S E2= a-B H1s H1s 1-S 1o 1 (Ψ。+Ψb) V2(1+Sb) 1 中。= (9。-Ψ6) V2(1-Sb)

S E    1 1   S E    1 2   ( ) 2 ( 1 ) 1 ( ) 2 ( 1 ) 1 a b ab b a b ab a S S             + + -

Overlap Sw=∫ aWndr integral Coulombic Hn=∫4iydr integral i=-1v2-1_11 21 ”可gt-+ rR -Jvjv-wdrvvd-Jvtva =Rat∫y-E J≈5.5%EH Ha=EH+J≈E。=

Overlap integral                                                             aa H a H a H b H a a b a a a a a a a a b aa a a b aa a a ab a b H E J E J E d E J R r E d r d R d r d r r R H r r R H H H d S d 5.5% 1 1 1 1 ) 1 21 ( ) 1 1 1 21 ( 1 1 1 2 1 ˆ ˆ * * 2 * * * 2 2 * * Coulombic integral