Computational Chemistry

simple H€uckel method (Section 4.3.3) and the extended H€uckel (Section 4.4.1) method. Here, however, we have seen (in outline) ...

The transformed Fock matrixF^0 satisfies F^0 ¼C^0 eC^0 #^1 ð 5 : 69 Þ (cf.Eq. 4.104). The overlap matrixSis readily calculated, ...

ofSection 5.2.2. The terms “Hartree–Fock calculations/method” and “SCF calcula- tions/method” are in practice synonymous.The key ...

5.2.3.6.3 Using the Roothaan–Hall Equations to do ab initio Calculations – the Equations in terms of thec’s andf’s of the LCAO E ...

Note that this is really a sixfold integral, since there are three variables (x 1 ,y 1 ,z 1 ) for electron 1, and three (x 2 ,y ...

which can be written more compactly as frð 1 ÞjK^jð 1 Þfsð 1 Þ ¼ Xm t¼ 1 Xm u¼ 1 c$tjcujðrujtsÞð 5 : 76 Þ where of course (cf. ( ...

P¼ P 11 P 12 P 13 .. . P 1 m P 21 P 22 P 23 ((( P 2 m ... ... ... ... Pm 1 Pm 2 Pm 3 ((( Pmm 0 B B B B @ 1 C C C C A ð 5 : 80 Þ ...

words, we are counting each repulsion twice. The simple sum thus represents properly the total kinetic and electron–nuclear attr ...

we get from Eq.5.86 EHF¼ Xn i¼ 1 eiþ Xm r¼ 1 Xm s¼ 1 Xn i¼ 1 c$ricsiHcorers ð 5 : 89 Þ Using Eq.5.81, Eq.5.89can be written in t ...

mode [ 28 ] (all this is done by standard programs, which print out the ZPE after the frequencies). Adding the HF electronic ene ...

5.2.3.6.5 Using the Roothaan–Hall Equations to do ab initio Calculations – an Example The application of the Hartree–Fock method ...

density around the helium nucleus falls off more quickly with distance than does that around the hydrogen nucleus (Fig.5.8). We ...

that will be used to transform the Fock matrixFtoF^0 and to convert the transformed coefficient matrixC^0 toC(Eqs.5.67–5.70). Th ...

VrsðH; 1 Þ¼ Z fr ZH rH 1  fsdv ð 5 : 102 Þ and VrsðHe; 1 Þ¼ Z fr ZHe rHe1  fsdv ð 5 : 103 Þ In Eq.5.102the variable is the d ...

in Eq.5.103the variable is the distance of the electron from the helium nucleus;ZH andZHeare 1 and 2, respectively. From Eq.5.10 ...

G 22 ¼ X^2 t¼ 1 X^2 u¼ 1 Ptu ð 22 jtuÞ# 1 2 ð 2 ujt 2 Þ  i.e: G 22 ¼ X^2 t¼ 1 Pt 1 ð 22 jt 1 Þ# 1 2 ð 21 jt 2 Þ  þPt 2 ð 22 ...

and the overlap matrix is S¼ 1 :0000 0: 5017 0 :5017 1: 0000  ð 5 : 112 Þ Step 3– Calculating the orthogonalizing matrix Calcu ...

Step 4– Calculating the Fock matrix (a) The one-electron matrices From Eq.5.100 F¼TþVðHÞþVðHeÞþG¼HcoreþG ð 5 : 116 Þ The one-ele ...

The total one-electron energy matrix,Hcore, is Hcore¼TþVðHÞþVðHeÞ¼ # 1 : 6606 # 1 : 3160 # 1 : 3160 # 2 : 3030  ð 5 : 120 Þ Th ...

Gmay now be calculated. From Eqs. (5.106)–(5.108), using the above values of Pand the integrals of Eq. (5.110), and recalling th ...