Modern Quantum Chemistry

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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:

c<d<e t<u<v Ψar||Ψ cdetuv

which we can expand as sum of the following S1 to S49 terms. In all these terms a and r are not iteration variables.

S1 = c<d<e<a t<u<v<r ccdetuvΨ ar||Ψ cdetuv

S2 = c<d<a t<u<v<r ccdatuvΨ ar||Ψ cdatuv

S3 = c<d<a<e t<u<v<r ccdetuvΨ ar||Ψ cdetuv

S4 = c<a<e t<u<v<r ccaetuvΨ ar||Ψ caetuv

S5 = c<a<d<e t<u<v<r ccdetuvΨ ar||Ψ cdetuv

S6 = a<d<e t<u<v<r cadetuvΨ ar||Ψ adetuv

S7 = a<c<d<e t<u<v<r ccdetuvΨ ar||Ψ cdetuv

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 S1 = = S7 = 0.

S8 = c<d<e<a t<u<r ccdeturΨ ar||Ψ cdetur

S9 = c<d<a t<u<r ccdaturΨ ar||Ψ cdatur

S10 = c<d<a<e t<u<r ccdeturΨ ar||Ψ cdetur

S11 = c<a<e t<u<r ccaeturΨ ar||Ψ caetur

S12 = c<a<d<e t<u<r ccdeturΨ ar||Ψ cdetur

S13 = a<d<e t<u<r cadeturΨ ar||Ψ adetur

S14 = a<c<d<e t<u<r ccdeturΨ ar||Ψ cdetur

All the above terms are zero except S9. For the next 7 terms

S15 = c<d<e<a t<u<r<v ccdetuvΨ ar||Ψ cdetuv

S16 = c<d<a t<u<r<v ccdatuvΨ ar||Ψ cdatuv

S17 = c<d<a<e t<u<r<v ccdetuvΨ ar||Ψ cdetuv

S18 = c<a<e t<u<r<v ccaetuvΨ ar||Ψ caetuv

S19 = c<a<d<e t<u<r<v ccdetuvΨ ar||Ψ cdetuv

S20 = a<d<e t<u<r<v cadetuvΨ ar||Ψ adetuv

S21 = a<c<d<e t<u<r<v ccdetuvΨ ar||Ψ cdetuv

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all S15 = = S21 = 0. For the next 7 terms

S22 = c<d<e<a t<r<v ccdetrvΨ ar||Ψ cdetrv

S23 = c<d<a t<r<v ccdatrvΨ ar||Ψ cdatrv

S24 = c<d<a<e t<r<v ccdetrvΨ ar||Ψ cdetrv

S25 = c<a<e t<r<v ccaetrvΨ ar||Ψ caetrv

S26 = c<a<d<e t<r<v ccdetrvΨ ar||Ψ cdetrv

S27 = a<d<e t<r<v cadetrvΨ ar||Ψ adetrv

S28 = a<c<d<e t<r<v ccdetrvΨ ar||Ψ cdetrv

All the above terms are zero except S25. For the next 7 terms

S29 = c<d<e<a t<r<u<v ccdetuvΨ ar||Ψ cdetuv

S30 = c<d<a t<r<u<v ccdatuvΨ ar||Ψ cdatuv

S31 = c<d<a<e t<r<u<v ccdetuvΨ ar||Ψ cdetuv

S32 = c<a<e t<r<u<v ccaetuvΨ ar||Ψ caetuv

S33 = c<a<d<e t<r<u<v ccdetuvΨ ar||Ψ cdetuv

S34 = a<d<e t<r<u<v cadetuvΨ ar||Ψ adetuv

S35 = a<c<d<e t<r<u<v ccdetuvΨ ar||Ψ cdetuv

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all S29 = = S35 = 0. For the next 7 terms

S36 = c<d<e<a r<u<v ccderuvΨ ar||Ψ cderuv

S37 = c<d<a r<u<v ccdaruvΨ ar||Ψ cdaruv

S38 = c<d<a<e r<u<v ccderuvΨ ar||Ψ cderuv

S39 = c<a<e r<u<v ccaeruvΨ ar||Ψ caeruv

S40 = c<a<d<e r<u<v ccderuvΨ ar||Ψ cderuv

S41 = a<d<e r<u<v caderuvΨ ar||Ψ aderuv

S42 = a<c<d<e r<u<v ccderuvΨ ar||Ψ cderuv

All the above terms are zero except S41. For the next 7 terms

S43 = c<d<e<a r<t<u<v ccdetuvΨ ar||Ψ cdetuv

S44 = c<d<a r<t<u<v ccdatuvΨ ar||Ψ cdatuv

S45 = c<d<a<e r<t<u<v ccdetuvΨ ar||Ψ cdetuv

S46 = c<a<e r<t<u<v ccaetuvΨ ar||Ψ caetuv

S47 = c<a<d<e r<t<u<v ccdetuvΨ ar||Ψ cdetuv

S48 = a<d<e r<t<u<v cadetuvΨ ar||Ψ adetuv

S49 = a<c<d<e r<t<u<v ccdetuvΨ ar||Ψ cdetuv

Again, in all these sums the triple excitations differ from the single excitation by more than two spin orbitals. Hence all S43 = = S49 = 0.

Therefore

c<d<e t<u<v Ψar||Ψ cdetuv = S 9+S25+S41

or (with a and r not being iteration variables)

c<d<e t<u<v Ψar||Ψ cdetuv = c<d<a t<u<r ccdaturΨ ar||Ψ cdatur + + c<a<e t<r<v ccaetrvΨ ar||Ψ caetrv + + a<d<e r<u<v caderuvΨ ar||Ψ aderuv

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

c<d t<u ccdaturΨ ar||Ψ cdatur = c<d t<u cacdrtuΨ ar||Ψ acdrtu

c<e t<v ccaetrvΨ ar||Ψ caetrv = c<d t<u cacdrtuΨ ar||Ψ acdrtu

d<e u<v caderuvΨ ar||Ψ aderuv = c<d t<u cacdrtuΨ ar||Ψ acdrtu

Therefore, two of the sum terms cancel and:

c<d<e t<u<v Ψar||Ψ cdetuv = c<d t<u cacdrtuΨ ar||Ψ acdrtu