exercise06

Hartree-Fock course for the Han-sur-Lesse Winterschool of 2021::Exercise 06

Problem description

With a working implementation for H2, it is logical that we are going to expand towards non-trivial molecules. Let us stick with two atoms to make matters not overly complicated and simulate the CO molecule. In this exercise, you are tasked to modify our earlier implementation to accomplish this goal. Furthermore, we are going to focus a bit more on the energies of the molecular orbitals and the linear combination of atomic orbitals that constitute them.

In the past exercises, we build the CGFs corresponding to the atomic orbitals by explicitly specifying the coefficients. Such practices will become quickly very impractical. For example, for the CO molecule, we have to define ten sets of two coefficients to set-up the basis set. As such, we are going to make use of a custom routine that does this process automatically for us. There is no magic behind this procedure: the values are stored in a JSON file and the method simply extracts these from the JSON file and builds the basis set for us. The code is shown below:

# build molecule
mol = Molecule('CO')
mol.add_atom('C', 0.0, 0.0, -2.116/2.0)
mol.add_atom('O', 0.0, 0.0,  2.116/2.0)

cgfs, nuclei = mol.build_basis('sto3g')
nelec = np.sum([n[1] for n in nuclei])
N = len(cgfs)

Questions

Take the result of the previous exercise and modify the script to simulate CO rather than H2. Take a critical look at the for loops and modify the number of CGFs they need to iterate over. The result of your calculation should produce an electronic energy of -111.223473 HT.

Once you have this working, extract the eigenvalues and -vectors from the simulation and take a critical look at these. Try to answer the following questions:

1. How many molecular orbitals are there in total? How does this correspond to the number of basis functions?
2. How many of the molecular orbitals are occupied?
3. Which of the orbitals are non-bonding?
4. Which of the orbitals are degenerate?
5. Which 1s orbital lies lower in energy, the one of C or the one of O?

Solution

The solution is given in ex06_solution.py

The answers to the questions are as follows:

1. Given a basis set of 10 AOs (basis functions), all integral matrices will be 10x10 matrices and thus we can expect to obtain 10 solutions. In general, the number of solutions always matches the number of basis functions.
2. The CO molecule has in total 14 electrons (8 for O, 6 for C) and thus the number of occupied molecular orbitals is 14/2 = 7.
3. Non-bonding orbitals can be identified by checking for solution vectors where there is only one coefficient with a value very close to 1.0 while all other values are very close to 0.0. For example, the first and second molecular orbitals correspond to this pattern and it can be said that these molecular orbitals are synonymous to their corresponding atomic orbitals.
4. Degenerate orbitals can be identified by them having the same energy. Molecular orbitals (5,6) and (8,9) are two pairs of double-degenerate orbitals. These are known as the 1pi bonding and 2pi* anti-bonding orbitals.
5. The first molecular orbital corresponds to the 1s atomic orbital of oxygen and thus lies lower in energy than the 1s atomic orbital of carbon. This result is to be expected as O has a higher nuclear charge than C and thus the electron experiences a stronger interaction for O than for C.