The trial wave function o: dt ov,=0 φ=∑=1aVk dτ E=Jdψ/dφ=∑2aPE/2 Ja, E alal E
The trial wave function f: dt f * y1 = 0 f = k=1 ak yk dt f * y1 = |a1 | 2 = 0 Ef = dt f *Hf / dt f *f = k=2|ak | 2Ek / k=2|ak | 2 > k=2|ak | 2E2 / k=2|ak | 2 = E2
Application to H2 p=c y +c Y 2 1122 W=∫φHφdτ/∫φdτ (c12H1+2c1c2H1 12 +C2H2) /(c2+2c1c2S+c2) W(c12+2c1C2S+c2)=c12H1+2c1C2H12+c2H2
e + + y 1 y 2 f = c 1 y 1 + c 2 y 2 W = f *H f dt / f *f dt = (c1 2 H11 + 2c1 c2 H12 + c2 2 H22 ) / (c1 2 + 2c1 c2 S + c2 2 ) W (c1 2 + 2c1 c2 S + c2 2 ) = c1 2 H11 + 2c1 c2 H12 + c2 2 H22 Application to H2 +
Partial derivative with respect to C,OW/ac, =0) W(c1+Sc2)=c1H1+c2H12 Partial derivative with respect to c,(OW/ac,=0) W(Sc1+c2)=c1H12+c2H2 (H1-W)c1+(H12-SW)c2=0 (H12-SW)c1+(H2-W)c2=0
Partial derivative with respect to c1 (W/c1 = 0) : W (c1 + S c2 ) = c1H11 + c2H12 Partial derivative with respect to c2 (W/c2 = 0) : W (S c1 + c2 ) = c1H12 + c2H22 (H11 - W) c1 + (H12 - S W) c2 = 0 (H12 - S W) c1 + (H22 - W) c2 = 0
To have nontrivial solution H-W Hm-SW H-SW 22 For H2, Hu=H22, H12 0 Ground State: Eo=W,=(H1+H12/(1+S) 1=(v+V2)/V2(1+s)2 bonding orbital Excited State: Ee=W2=(Hu-H12)/(1-S 92=(V-V)/2(1-S)2 Anti-bonding orbital
To have nontrivial solution: H11 - W H12 - S W H12 - S W H22 - W For H2 + , H11 = H22; H12 < 0. Ground State: Eg = W1 = (H11+H12) / (1+S) f1 = (y1 +y2 ) / 2(1+S)1/2 Excited State: Ee = W2 = (H11-H12) / (1-S) f2 = (y1 -y2 ) / 2(1-S)1/2 = 0 bonding orbital Anti-bonding orbital
Results: D=1.76eVr=1. 32 A Exact: D =2.79eVR=1064 l ev=23.0605 kcal/mol
Results: De = 1.76 eV, Re = 1.32 A Exact: De = 2.79 eV, Re = 1.06 A 1 eV = 23.0605 kcal / mol