I'm setting up a simple Hartree-Fock problem in pyscf with ECPs to test things out. However, the numbers I get do not make sense to me as the energies are very different. I have never used ECPs before so I don't know exactly what to expect. Here is my code and outputs. Without ECPs (using sto-3g) I get

mol = gto.M(atom="Li 0. 0. 0.; H  0.  0.  1.", basis={'Li':'sto3g', 'H':'sto3g'}, verbose=0)
mf = scf.RHF(mol)

Output: -7.76736

and with ECP for lithium I get

mol = gto.M(atom="Li 0. 0. 0.; H  0.  0.  1.", basis={'Li':'stuttgart', 'H':'sto3g'}, ecp = {'Li':'stuttgart'}, verbose=0)
mf = scf.RHF(mol)

Output: -0.63098

As you can notice the energies are very different in these two cases. Is this normal or is something wrong here?


1 Answer 1


This is perfectly normal.

In ab initio calculations, the energy zero point is usually chosen as the state where all explicitly described particles are infinitely far from each other and each possessing zero kinetic energy. Thus, in an all-electron calculation of LiH, the calculated energy is versus the state where the lithium nucleus, the hydrogen nucleus, and the 4 electrons are infinitely separated from each other.

However, in your ECP calculation, the core electrons of lithium are not described explicitly. So the energy zero point is the state where a $\ce{Li+}$ cation, a $\ce{H+}$, and two electrons are infinitely far away from each other. This is of course a considerably lower reference energy than the all-electron case, because the two core electrons of lithium are not ionized. Therefore, in the ECP calculation, the total energy of LiH is much higher in number.

  • $\begingroup$ Okey thanks. As these energies are not comparable, which metric would be a good comparison between these two approaches? In other words how do I know how much โ€worseโ€ it is to use the ECP? As usually I compare basis sets by looking at the ground state energy. $\endgroup$
    – NPHA
    Nov 14, 2022 at 18:23
  • 3
    $\begingroup$ @NPHA You should look at energy differences instead (e.g. bond energy), and/or valence properties (e.g. excitation energies), and compare them against the results of near-complete all-electron basis sets. Comparing total energies (and then invoking the variational principle) is not the only way to compare the qualities of basis sets; it can be sometimes misleading (e.g. a basis set that is very good for the core orbitals but bad for the valence orbitals may be erroneously ranked as good), and as you see now, it is not always applicable. $\endgroup$
    – wzkchem5
    Nov 14, 2022 at 19:49
  • $\begingroup$ Thanks! Great answer ๐Ÿ‘๐Ÿป $\endgroup$
    – NPHA
    Nov 14, 2022 at 20:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .