《结构化学 Structural Chemistry》课程教学资源（课件讲稿）Chapter 4 Structures of Diatomic Molecules

Example: Devise a trial variation function for the ground state of theparticle in a one-dimensional box of length lV=0A simple function that has theproperties of the ground state istheparabolic functionΦ= x(l -x)for 0<x<lh?13hdHd6m2m1h?Φ ddt =["sx(l - x)dx = [5 /30 H2m dxSh2h2<E>=J0*Hpdt8m/2Ja*dt4元m12Parabolic抛物线

Example: Devise a trial variation function for the ground state of the particle in a one-dimensional box of length l. 0 l A simple function that has the V=0 properties of the ground state is the parabolic function:   x(l  x) for 0< x < l m l lx x dx dx d lx x m H d l 6 ( ) ( ) 2 ˆ 2 3 2 2 2 0 2 2 *              ( ) /30 2 5 0 2 * d x l x dx l l         2 8 2 2 2 4 2 5 * * ˆ ml h ml h d H d E              Parabolic—抛物线) 2 ˆ ( 2 2 2 dx d m H    

3. Linear Variation Functionsfi, f2, ...f, are linearlyindependent,butnotnnecessarily eigenfunctions of=cfi +cf +...+cnf, = Zcfanyoperatorsi=1Based on the variation theorem, thecoefficients are regulated by the minimizationroutine so as to obtain the wavefunction thatHdaDITZE<E>corresponds to the minimum energy. This isJa*pdttaken to be the wavefunction that closely=<E>approximates the ground stateTo minimize , make O/ Oc, = 0Real Ground-state Energy(Os / c, = 0) (a total of n equations of (c)={8j,g, =Zc f (j = 1,2,.,n)12i=1

To minimize  , make  / ci  0 Based on the variation theorem, the coefficients are regulated by the minimization routine so as to obtain the wavefunction that corresponds to the minimum energy. This is taken to be the wavefunction that closely approximates the ground state. * 0 * ˆ E d H d E              E  3. Linear Variation Functions       n i n n i i c f c f c f c f 1 1 1 2 2  . f1 , f2 , .fn are linearly independent, but not necessarily eigenfunctions of any operators. ( 1,2,.,n) 1     { c f j n i i j j j i  , } Real Ground￾state Energy { 0} (a total of n equations of {c })    i  i  / c

g*HpdtCr, C, and E to beE:ExampleΦ=Cii+C2solved by theIo*pdtvariation theorem!x = [g*pdt = [(cii +C22)"(ciyi +C22)dt= /(ciyi y1 +CiC2yiV2 +CC22 y1 +c2y2 y2)dtOverlap(cyiy+2cc2w, +c,y2y2)dtintegral!= c2Si + 2cic2S12 +c, S22Let S, =Jyiwdt = Sj=c2 +2cC2S12 +C2(: Sti = S22 = 1)Yi and y, arenormalized functionsy= [ Hdt = [(c +C22)"H(ci + C22)dtJ (cy,Hy++CC2y,'Hy2 +CC2V,Hyi +c2y2'Hy2)dt=cH +2cC,H12 +c2H22(H, = Hj,= [y,Hy,dt)12

y  Hd c c H( c c )d ˆ ( ) ˆ * *     1 1  2 2 1 1  2 2 Example c   c c  c c  c   d   (    ) * * * * 2 2 2 1 1 1 2 1 2 1 2 2 1 2 2 1          d H d c c E      * * ˆ 1 1 2 2 c1 , c2 and E to be solved by the variation theorem! x  d ( c c ) ( c c )d * *     1 1  2 2 1 1  2 2 c   c c  c   d   (   ) * * * 2 2 2 1 1 1 2 1 2 2 2 1 2 22 2 11 1 2 12 2 2  c1 S  2c c S  c S j i Let Si j  i j d  S     * Overlap integral! 2 1 2 12 2 2  c1  2c c S  c ( 1) S11  S22  1 and2 are normalized functions 2 2 2 2 1 1 1 2 1 2 2 2  c1 H  c c H  c H c  H c c H c c H c  H )d ˆ ˆ ˆ ˆ ( * * * * 2 2 2 1 1 1 2 1 2 1 2 2 1 2 2  1     ) ˆ ( *  H  H   H d i j j i i j

[o'Hpdtc’Hn + 2cc,Hi2 +c,H22 = y:EJo'ddtc2St + 2c,C2S12 + C2S22xaE1 Qyy axTo make E = Eo, we have Ox2 c,acx Oc,-(2c,Hi+2c,H/2)(2c,Su +2c2S12)= 0xx=(c,H1 +c,H12)- 二(cSt +c,Si2)= 0x→(cH +C2H12)-E(cSu +C2S12)= 0(1)(H-ESu)c, +(H2 -ES2)c, =0Similarly, by making aE/c, = 0, we have(2)(H21 - ES21)Ci +(H22 - ES22)c2 = 012

2 2 2 2 0 1 1 we have 0 1 1 1 2 1 2 2 1 1 1 2 1 2 1 2 1 1 0                ( ) ( ) , c S c S x y c H c H x c x x y c y c x E To make E E ( ) ( ) 0 (2) H2 1  ES2 1 c1  H2 2  ES2 2 c2  x y c S c c S c S c H c c H c H d H d E           2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 1 1 1 2 1 2 2 2 1 * * 2 2 ˆ       Similarly, by making E/c2 = 0, we have ( c1 H1 1  c2 H1 2 ) E( c1 S1 1  c2 S1 2 )  0  ( 1 1 1  2 1 2 )  ( c1 S1 1  c2 S1 2 )  0 x y c H c H ( ) ( ) 0 (1)  H1 1  ES1 1 c1  H1 2  ES1 2 c2 

Φ=CV+C2V2Trial functionNow we have two secular equations(1)(Hu -ES)c +(Hi2 - ESi2)c, = 0Secular equations(2(H21 - ES2)c +(H22 -ES22)c, = 0that can be express in the matrix form:H- ESi1H2 - ES,DH21 - ES21H22 - ES2As C,C, +O, the secular equations thus demand thecorresponding secular determinant to be zero, i.e.,Hi - ESi1H12 - ES12:0H22 - ES22H21 - ES210

As c1 ,c2  0, the secular equations thus demand the corresponding secular determinant to be zero, i.e., Secular equations  1 1 2 2  c  c ( ) ( ) 0 (2) ( ) ( ) 0 (1) 2 1 2 1 1 2 2 2 2 2 1 1 1 1 1 1 2 1 2 2         H ES c H ES c H ES c H ES c Trial function                             0 0 2 1 2 1 2 1 2 2 2 2 1 1 1 1 1 2 1 2 c c H ES H ES H ES H ES that can be express in the matrix form: 0 2 1 2 1 2 2 2 2 1 1 1 1 1 2 1 2      H ES H ES H ES H ES Now we have two secular equations