Advanced Quantum Chemistry spring 2012, 3rd exercise

The first problem is to be returned in the exercise session, or to the grey metal cabinet in Physicumbefore the exercise session. To be discussed in the exercise session on Fri Mar 30 2012 in Physicumlinux class room.

Abstract

We proceed with the introduction to quantum chemistry by examining simple many-electron sys-tems, paying attention to basis set and correlation effects.

To be done before the exercise session

1. Compute and plot the energy curve of the linear BeH2 molecule at Hartree-Fock level of theoryfrom RBH = 0.5 Å to R = 10 Å, using the following basis sets:

(a) cc-pVDZ

(b) cc-pCVDZ

(c) cc-pVTZ

(d) cc-pCVTZ

Repeat the calculation using full configuration interaction (FCI). Comment on the differences.

To be solved during the exercise session

2. We consider the total energies of the singlet and triplet states obtained by promoting one electronfrom the occupied canonical orbital φi to the virtual canonical orbital φa of a closed-shell Hartree-Fock state. Using the singlet and triplet excitation operators

S0,0pq =

1√2

(a†pαaqα + a†pβaqβ

)T 1,0pq =

1√2

(a†pαaqα − a

†pβaqβ

)the total energies of the singlet and triplet states can be written as

1Eai =⟨

HF∣∣∣S0,0ia HS

0,0ai

∣∣∣HF⟩

3Eai =⟨

HF∣∣∣T 1,0ia HT

1,0ai

∣∣∣HF⟩.

(a) Show that the singlet total energy can be expressed as

1Eai =⟨

HF∣∣∣H∣∣∣HF

⟩+

1

2

⟨HF∣∣∣Eia [H, Eai]∣∣∣HF

⟩(b) Use

[H, Eai

]|HF〉 =2 IFia |HF〉+

∑b

IFbaEbi −∑j

IFijEaj +∑bj

LbjiaEbj

+

∑bjc

gbjcaEbjEci −∑bjk

gbjikEbjEak

|HF〉 ,

where IF is the so-called inactive Fock matrix

IFmn =hmn +∑i

(2gmnii − gmiin) ,

to show that the singlet energy may be written in the form

1Eai =⟨

HF∣∣∣H∣∣∣HF

⟩+ εa − εi + 2gaiia − gaaii

(c) Show that the triplet and singlet energies are related as

3Eai =1Eai − 2

⟨HF∣∣∣EβiaHEαai∣∣∣HF

⟩where

Eαai =a†ααaiα

Eβai =a†αβaiβ

(d) Show that the energy of the triplet state becomes

3Eai =⟨

HF∣∣∣H∣∣∣HF

⟩+ εa − εi − gaaii.

2

To be solved and returned by the end of the course

3. The electronic gradient is given by

E(1)mn =2 (Fmn − Fnm) ,

where m and n are orbital indices. To go from the AO basis to the MO basis use the equation

FMOmn =

∑bfs µ,ν

CµmFAOµν Cνn.

Solve the problem below.

3