H., - ES,.H12 - ES12=0SeculardeterminantH21 - ES21H22 - ES22(Hu - ESu)(H22 - ES22)-(H21- ES2i)(H12 - ESi2) = 0 (3The algebraic equation (3) has 2 roots, E, and E,Substituting E, into the secular equations, a set of ci, C, aswell as the corresponding Φ,= ciVi + c2V2 can be obtainedSubstituting E, into the seqular equations, a set of ci, C2/ aswell as the corresponding Φ2 can be obtained.Thus, the variational process gives two different energy E, andE,, and two different sets of ic, C2} Φ, and Φ?12
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 • The algebraic equation (3) has 2 roots, E1 and E2 . Secular determinant ( )( ) ( )( ) 0 (3) H1 1 ES1 1 H2 2 ES2 2 H2 1 ES2 1 H1 2 ES1 2 • Substituting E1 into the secular equations, a set of {c1 , c2 } as well as the corresponding 1= c11 + c22 can be obtained. • Substituting E2 into the seqular equations, a set of {c1 , c2 } as well as the corresponding 2 can be obtained. Thus, the variational process gives two different energy E1 and E2 , and two different sets of {c1 , c2 } 1 and 2
In general, for a linear variation function@ = Cy, + C,y, +... + CnyTwe have the secular equations (in matrix form)0H-ES.Hi2 -ESi2Hn -ESinCi0H21 -ES21H22 -ES22H.-C20HHn-ES,lH.. -ES.,SCandH., - EStH12 - ESi2Hin -ESinH21 - ES21H.r-ES2nH22 - ES22=0Secular determinantH.? -ES.H-ESH.-ES.102nnThis algebraic equation has n roots, which can be shown to be real12Arranging these roots in the order: E,≤ E,≤... ≤E
0 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 1 1 1 1 2 1 2 1 1 n n n n n n n n n n n n H E S H E S . H E S . . . . . . . . H E S H E S . H E S H E S H E S . H E S This algebraic equation has n roots, which can be shown to be real. Arranging these roots in the order: E1 E2. En . n n In general, for a linear variation function c1 1 c2 2 . c 0 0 0 2 1 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 1 1 1 1 2 1 2 1 1 n n n n n n n n n n n n n c c c H E S H E S H E S H E S H E S H E S H E S H E S H E S . . . . . . . . . . . we have the secular equations (in matrix form) and Secular determinant
Remarks on the linearvariational process: From the variation theorem, we know that the lowest value of root(W) is the upper bound for the system's real ground-state energy(E,), i.e., E, ≤Wi. Moreover, it is provable that the linear variation method providesupper bounds to the energies of the lowest n states of the systemE,≤W2. E,≤W3. ., En≤Wn: We use these roots (W as approximations to the energies of thelowest n states (E;}. If approximations to the energies of more states are wanted, we addmore functions fi (k > n) into to the trial function Φ. (= Zcf)? Addition of more functions fk can be shown to increase theaccuracy of the calculated energies (W.12
• From the variation theorem, we know that the lowest value of root (W1 ) is the upper bound for the system’s real ground-state energy (E1 ), i.e., E1 W1 • We use these roots {Wi } as approximations to the energies of the lowest n states {Ei }. • If approximations to the energies of more states are wanted, we add more functions fk (k > n) into to the trial function . ( = ci fi ) • Addition of more functions fk can be shown to increase the accuracy of the calculated energies {Wi }. E2 W2, E3 W3, . , En Wn, • Moreover, it is provable that the linear variation method provides upper bounds to the energies of the lowest n states of the system. Remarks on the linear variational process
H3. The approximate solution of0HR2rbBARV元=业1sAO ofA atom!1s=e-b/V元1s AO of B atom!=bLetp=CaVa+CpYbTrial functionfortheMOofH,(Linear combination of atomic orbitals into molecular orbitali.e.,LCAO-MO, widelyused!Nowbeginthevariationprocess![o"HddtaEaE=0E&12acOCbJoddta
3. The approximate solution of H2 A B e - r r b a R a r A a s e 1 / a a b b c c Trial function for the MO of H2 + b r B b s e 1 / Let d H d E * * ˆ & 0 a b c E c E Now begin the variation process! r r R H a b e 1 1 1 2 1 ˆ 2 1s AO of A atom! 1s AO of B atom! (Linear combination of atomic orbitals into molecular orbital i.e., LCAO-MO, widely used!)
Φ =c.y。 +cyp = The secular equations Secular determinant。-ESHHab - ESab(Haa-ESaa)c。+(Hab -ESab)c, =0aaaa= 0H.ba-ES,Hb - ES bb(Hba -ESba)ca +(Hbb -ESbb)c, = 0heho. y. has the same form as yb, ..Haa = Hbb, Hab = Hba=(Haa-ESaa)? =(Hab - ESab)2 Hag - ESa = ±(Hab - ESab)i) If Haa - ESaa = -(Hab - ES.ahNote:Saa= Sbb = 1Ha. +Habα+β& defineaa→E, S..+Sab1+S00Haa=Hbb= α (< 0)i) If H..-ES..= Hab - ESaHab =Hba=β (< 0)H.a- Habα-βSab = Sba = Sao→E2Saa-Sab1-S12
a b Ha a Hb b Ha b Hb a has the same form as , , 0 0 he secular quations : b a b a a b b b b b a a a a a a b a b b a a b b H E S c H E S c H E S c H E S c c c T e ( ) ( ) ( ) ( ) ( ) Ha a ESa a Ha b ESa b i) If & define Haa = Hbb = (< 0) Hab = Hba = (< 0) Sab = Sba = S 0 b a b a b b b b a a a a a b a b H ES H ES H ES H ES Secular determinant Ha a ESa a Ha b ESa b ii)If 2 2 ( ) ( ) Haa ESaa Hab ESab ( ) Ha a ESa a Ha b ESa b a a a b a a a b S S H H E 1 S 1 Note: Saa = Sbb = 1 S S S H H E a a a b a a a b 1 2