Are the wavefunctions calculated using scf and some dft functional already normalized?

- For Hartree-Fock, the question does make sense. The method computes occupied orbitals. The wavefunction constructed from those occupied orbitals is normalized (integral of the wavefunction times its complex conjugate over all space is 1), and the Hartree-Fock energy is the expectation value of the molecular Hamiltonian with that wavefunction.
- For DFT, the question does not make sense. The method computes occupied orbitals. The electronic density constructed from those occupied orbitals is normalized (integral of electronic density over all space is the number of electrons), and the DFT energy is the energy of that DFT functional with that density. No wavefunction intermediate is created to be normalized or not. While it is possible to use the same device (the Slater determinant) used in Hartree-Fock to convert the orbitals to a wavefunction, that wavefunction is unrelated to the computed energy in general. They are the same in the very special case that your DFT functional is Hartree-Fock.

Thank you very much, your answer is helping me a lot!

I guess I used some inaccurate words. What i meant was not the wavefunction of the entire system, but of the molecular orbitals.

Following your argumentation, these orbital wavefunctions are also created to fulfill Int <Phi_i | Phi_i> d^3r = 2, because each occupied orbital needs to have an occupancy of 2 electrons right?

Is there a similar argumentation for the unoccupied orbitals?

No, that integral is 1.

Remember that orbitals have not only spatial but spin coordinates. If Phi_i can be integrated over space to get a number, it only has spatial coordinates. What you can do instead is take the spinorbitals Phi_i * alpha and Phi_i * beta. If you integrate over spin and spatial coordinates, you get <Phi_i * alpha | Phi_i * alpha> + <Phi_i * beta | Phi_i * beta> = 1 + 1 = 2.

The same applied for virtual orbitals.

Okay, I think I understand. Does the same hold true if you calculate in restricted Kohn-Sham?

Yes. The same holds true for *all* flavors of Hartree-Fock and DFT in Psi4. I would be astounded if there were a mainstream electronic structure code for which this was not true.

Thanks a lot for the answers and that you took the time to answer my questions!