I have a rather basic question about the (minimal) number of molecular orbitals that are needed for some computation work on the BeH2 molecule. To my understanding, STO-3G forms a minimal basis set, which I want to use for reasons of computational constraints.

For the BeH2 molecule in the ground state, we have 6 electrons, 4 in 1s orbitals and 2 in 2s orbitals. Thus I form the two different molecular orbitals in the standard way as a linear combination of 3 gaussian functions $$\chi = \sum_{i=1}^{N=3} c_i \phi_i$$

Now I have the two molecular orbitals $\chi^{1s}, \chi^{2s}$. Now I believe I am correct in saying that the product of 4 $\chi^{1s}$ and 2 $\chi^{2s}$ functions builds my basis set, for example the state could be $$|\chi_1^{1s} \chi_2^{1s} \chi_3^{1s} \chi_4^{1s} \chi_5^{2s} \chi_6^{2s} \rangle $$

Now if I want to use the occupation number representation, I get $$|b_1 b_2 \cdots b_6 \rangle$$ where $b_i$ is 1 if orbital i is occupied by an electron and 0 otherwise.

Here is what I have trouble understanding: It seems to me that the system will at all points in time be described by the ket $|111111 \rangle$ since any other one would mean that an electron disappeared somehow into nothingness.

Can anybody spot my issue? I am neither a chemist nor a physicist, unfortunately, so I might be wrong at about any point.

Thanks for any help


1 Answer 1


The STO-3G basis set not only has two contracted s functions for Be, but also one contracted p function, which can be used to construct the 2p orbital of Be. Therefore you have 6 empty spin-orbitals, which are approximately the alpha and beta Be 2p orbitals (with some admixture of the s orbitals of H and Be). It is the presence of these 6 empty orbitals that make the SCF/FCI/... solution (and in particular, the time evolution) of BeH2 not completely trivial when using the STO-3G basis set, since the electrons can excite into these orbitals.


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