Computational Chemistry

Basis functionf 2 gives

Xm

s¼ 1

cs 1 F 2 s¼e 1

Xm

s¼ 1

cs 1 S 2 s

Xm

s¼ 1

cs 2 F 2 s¼e 2

Xm

s¼ 1

cs 2 S 2 s

...

Xm

s¼ 1

csmF 2 s¼em

Xm

s¼ 1

csmS 2 s

ð 5 : 54 ‐ 2 Þ

Finally, basis functionfmgives

Xm

s¼ 1

cs 1 Fms¼e 1

Xm

s¼ 1

cs 1 Sms

Xm

s¼ 1

cs 2 Fms¼e 2

Xm

s¼ 1

cs 2 Sms

...

Xm

s¼ 1

csmFms¼em

Xm

s¼ 1

csmSms

ð 5 : 54 ‐mÞ

In themsets of equations5.54-1–5.54-meach set itself containsmequations (the
subscript ofe, for example, runs from 1 tom), for a total ofm'mequations. These
equations are the Roothaan–Hall version of the Hartree–Fock equations; they were
obtained by substituting for the MO’scin the HF equations a linear combination of
basis functions (f’s weighted byc’s). The Roothaan–Hall equations are usually
written more compactly, as

Xm

s¼ 1

Frscsi¼

Xm

s¼ 1

Srscsiei r¼ 1 ; 2 ; 3 ;...;m;

ðfor eachi¼ 1 ; 2 ; 3 ;...;mÞ

ð 5 : 56 Þ

We havem'mequations because each of themspatial MO’scwe used (recall
that there is one HF equation for eachc, Eqs.5.47) is expanded withmbasis
functions. The Roothaan–Hall equations connect the basis functionsf(contained in
the integralsFandS, Eqs.5.55, above), the coefficientsc, and the MO energy levels
e. Given a basis set {fs,s¼1, 2, 3,...,m} they can be used to calculate thec’s,
and thus the MOsc(Eq.5.52) and the MO energy levelse. The overall electron
distribution in the molecule can be calculated from the total wavefunctionC, which

200 5 Ab initio Calculations