equation, we might (eschewing a rigorous deductive approach) suspect that the
matrix form is the fairly obvious possibility
FC¼SCe *ð 5 : 57 ÞHereF,CandSwould have to bem'mmatrices, since there arem^2 F’s,c’s and
S’s, andewould be anm'mdiagonal matrix with the nonzero elementse 1 ,e 2 ,...,
em, sinceemust contain onlymelements, but has to bem'mto make the right
hand side matrix product the same size as that on the left.
This is easily checked: the left hand side of Eq.5.57is
FC¼
F 11 F 12 F 13 ((( F 1 m
F 21 F 22 F 23 ((( F 2 m
... ... ... ...Fm 1 Fm 2 Fm 3 ((( Fmm0
BB
B
B
@
1
CC
C
C
A
c 11 c 12 c 13 ((( c 1 m
c 21 c 22 c 23 ((( c 2 m
... ... ... ...cm 1 cm 2 cm 3 ((( cmm0
BB
B
B
@
1
CC
C
C
A
¼
F 11 c 11 þF 12 c 21 þF 13 c 31 ((( F 11 c 12 þF 12 c 22 þF 13 c 32 ((((((
F 21 c 11 þF 22 c 21 þF 23 c 31 ((( F 21 c 12 þF 22 c 22 þF 23 c 33 ((((((
...0
B
@
1
C
A
ð 5 : 58 ÞThe right hand side of Eq.5.57isSCe¼S 11 S 12 ((( S 1 m
S 21 S 22 ((( S 2 m
... ... ((( ...Sm 1 Sm 2 ((( Smm0
B
B
B
B@
1
C
C
C
CA
c 11 c 12 ((( c 1 m
c 21 c 22 ((( c 2 m
... ... ((( ...cm 1 cm 2 ((( cmm0
B
B
B
B@
1
C
C
C
CA
e 11 0 ((( 0
0 e 22 ((( 0
... ... ((( ...00 ((( emm0
B
B
B
B@
1
C
C
C
CA
¼
S 11 c 11 þS 12 c 21 þS 13 c 31 ((( S 11 c 12 þS 12 c 22 þS 13 c 32 ((((((
S 21 c 11 þS 22 c 21 þS 23 c 31 ((( S 21 c 12 þS 22 c 22 þS 23 c 33 ((((((
...0
B
B
@
1
C
C
Ae¼
e 1 ðS 11 c 11 þS 12 c 21 þS 13 c 31 (((Þ e 2 ðS 11 c 12 þS 12 c 22 þS 13 c 32 (((Þ(((
e 1 ðS 21 c 11 þS 22 c 21 þS 23 c 31 (((Þ e 2 ðS 21 c 12 þS 22 c 22 þS 23 c 33 (((Þ(((
...0
B
B
@
1
C
C
A
ð 5 : 59 ÞNow compareFC(5.58) andSCe(5.59). Comparing elementa 11 ofFC(multi-
plied out to give a single matrix as shown in Eq.5.58) with elementa 11 of
SCe(multiplied out to give a single matrix as shown in Eq.5.59) we see thatif
FC¼SCe, i.e.ifEq.5.57is true, then
202 5 Ab initio Calculations