# Modern Quantum Chemistry

Chapter04 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15

Exercise. This is the solution to exercise 4.1 in the book.

Solution. Remember that one electron matrix elements are zero if the two determinants deffer by two or more spin orbitals (see section 2.3.3 in the book).

The triple excitation term writes:

$\sum _{\begin{array}{c}c

which we can expand as sum of the following ${S}_{1}$ to ${S}_{49}$ terms. In all these terms $a$ and $r$ are not iteration variables.

${S}_{1}=\sum _{\begin{array}{c}c

${S}_{2}=\sum _{\begin{array}{c}c

${S}_{3}=\sum _{\begin{array}{c}c

${S}_{4}=\sum _{\begin{array}{c}c

${S}_{5}=\sum _{\begin{array}{c}c

${S}_{6}=\sum _{\begin{array}{c}a

${S}_{7}=\sum _{\begin{array}{c}a

Before we move on to further terms, note that in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all ${S}_{1}=\dots ={S}_{7}=0$.

${S}_{8}=\sum _{\begin{array}{c}c

${S}_{9}=\sum _{\begin{array}{c}c

${S}_{10}=\sum _{\begin{array}{c}c

${S}_{11}=\sum _{\begin{array}{c}c

${S}_{12}=\sum _{\begin{array}{c}c

${S}_{13}=\sum _{\begin{array}{c}a

${S}_{14}=\sum _{\begin{array}{c}a

All the above terms are zero except ${S}_{9}$. For the next 7 terms

${S}_{15}=\sum _{\begin{array}{c}c

${S}_{16}=\sum _{\begin{array}{c}c

${S}_{17}=\sum _{\begin{array}{c}c

${S}_{18}=\sum _{\begin{array}{c}c

${S}_{19}=\sum _{\begin{array}{c}c

${S}_{20}=\sum _{\begin{array}{c}a

${S}_{21}=\sum _{\begin{array}{c}a

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all ${S}_{15}=\dots ={S}_{21}=0$. For the next 7 terms

${S}_{22}=\sum _{\begin{array}{c}c

${S}_{23}=\sum _{\begin{array}{c}c

${S}_{24}=\sum _{\begin{array}{c}c

${S}_{25}=\sum _{\begin{array}{c}c

${S}_{26}=\sum _{\begin{array}{c}c

${S}_{27}=\sum _{\begin{array}{c}a

${S}_{28}=\sum _{\begin{array}{c}a

All the above terms are zero except ${S}_{25}$. For the next 7 terms

${S}_{29}=\sum _{\begin{array}{c}c

${S}_{30}=\sum _{\begin{array}{c}c

${S}_{31}=\sum _{\begin{array}{c}c

${S}_{32}=\sum _{\begin{array}{c}c

${S}_{33}=\sum _{\begin{array}{c}c

${S}_{34}=\sum _{\begin{array}{c}a

${S}_{35}=\sum _{\begin{array}{c}a

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all ${S}_{29}=\dots ={S}_{35}=0$. For the next 7 terms

${S}_{36}=\sum _{\begin{array}{c}c

${S}_{37}=\sum _{\begin{array}{c}c

${S}_{38}=\sum _{\begin{array}{c}c

${S}_{39}=\sum _{\begin{array}{c}c

${S}_{40}=\sum _{\begin{array}{c}c

${S}_{41}=\sum _{\begin{array}{c}a

${S}_{42}=\sum _{\begin{array}{c}a

All the above terms are zero except ${S}_{41}$. For the next 7 terms

${S}_{43}=\sum _{\begin{array}{c}c

${S}_{44}=\sum _{\begin{array}{c}c

${S}_{45}=\sum _{\begin{array}{c}c

${S}_{46}=\sum _{\begin{array}{c}c

${S}_{47}=\sum _{\begin{array}{c}c

${S}_{48}=\sum _{\begin{array}{c}a

${S}_{49}=\sum _{\begin{array}{c}a

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all ${S}_{43}=\dots ={S}_{49}=0$.

Therefore

$\sum _{\begin{array}{c}c

or (with $a$ and $r$ not being iteration variables)

$\begin{array}{rcll}\sum _{\begin{array}{c}c

After renaming the indices to consistent lettering and placing determinants in maximum coincidence we get:

$\sum _{\begin{array}{c}c

$\sum _{\begin{array}{c}c

$\sum _{\begin{array}{c}d

Therefore, two of the sum terms cancel and:

$\sum _{\begin{array}{c}c