There are two excited wavefunctions, singlet and triplet state. In the singlet state, the total spin is zero, while in the triplet state, the total spin is 1.

In any case, the Slater determinant is

For spin-singlet state, we set,

For triplet state,

Since the Hamitonian is spin independent, we only take care of the spatial part.

For the triplet state, if we put in the wavefucntion in the Hamiltonian. Using the basis set approximation, The Fock matrix is

The factor $latex \delta(\sigma_h, \sigma_k) $ is due to the spin component.

From the beginning, the Fock matrix is

We can see the exchange term is difference. We can interchange

and pull the out.

The calculation is straight forward. First, we calculate the 1s state due to the mean field created by the 2s orbital, then calculate the 2s state due to the mean field by the 1s orbital. And then repeat until converged.

Using the Hydrogen 1s, 2s, 3s, and 4s orbtial, we calculated

1s KE+PE = -2 Hartree = -54.42 eV

2s KE+PE = -0.4349 Hartree = -11.83 eV

The direct term = 0.2800 Hartree = 7.62 eV

The exchange term = 0.016 Hartree = 0.44 eV

The total energy = -2.171 Hartree = -59.07 eV.

Compare to the calculated ground state energy ( -77.11 eV), the excitation energy is 18.04 eV. And the ionization energy is 4.65 eV.

The experimental value for the triplet state excitation energy is 19.77 eV, and the ionization energy is 4.70 eV.

Consider the ground state energy is different from the experimental value by 2 eV. the calculation is fairly good.

For the singlet state, because the spin state of the two electron are opposite, the Fock matrix becomes Hartree matrix.

The calculated result is,

1s KE+PE = -2.000 Hartree = -54.42 eV

2s KE+PE = -0.400 Hartree = -10.89 eV

The direct term = 0.2531 Hartree = 6.89 eV

The exchange term = 0.040 Hartree = 1.09 eV

The total energy = -2.147 Hartree = -58.43 eV.

Compare to the calculated ground state energy ( -77.11 eV), the excitation energy is 18.68 eV. And the ionization energy is 4.00 eV.

The experimental value for the triplet state excitation energy is 20.55 eV, and the ionization energy is 3.92 eV.

It is interesting to compare with other formalism. Many other textbook use singlet wavefunction,

The triplet state is the same,

Since the Hamiltonian is spin independent, we can see, when we calculate the total energy of the singlet state and triplet state are

Thus, the energy difference between singlet and triplet state is 2 exchange term.

In our calculation, the wave function of the singlet state is difference. We directly calculate from the Slater determinant. And the Fock matrix is lack of the exchange term due to the opposite spin. However, the triplet state is the same, the exchange term is 0.44 eV. If we do not do Hartree-Fock on the singlet state, but calculate the singlet state using the exchange term, the singlet state energy would be -58.19 eV. Comparing with the Hartree-Fock result (-58.43 eV), I would say they agree. The small difference is due to the difference of the mean fields.