5
$\begingroup$

Currently, I'm trying to get into the field of quantum chemistry and was reading about minimal basis sets. As I understood this is the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons.

For example, H has one electron, so a minimal basis set would be $\mathrm{1s}$ (STO, STO-nG, or something else).

Li has 3 electrons, so my guess would be that we need 1s and 2s. But in the literature, also the $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$ orbitals are included in the minimal basis. Why is this the case? Together with the spin degrees of freedom, the three electrons already fit in the 1s and $\mathrm{2s}$ orbitals?

$\endgroup$
5
$\begingroup$

A minimal basis set is a set that employs just one function to describe each orbital. For $\ce{H}$ you take into account one orbital ($\mathrm{1s}$), so you use one function, and for $\ce{Li}$ you consider five orbitals ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$, $\mathrm{2p_z}$) and you use five functions, one for each.

What you described ("the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons") would use a function for each occupied orbital - leaving unoccupied orbitals, even in a partially occupied subshell, out of the picture. This is a very bad idea - shells are a set of degenerate solutions for the hydrogen atom, so reducing the number of functions to describe them will seriously distort the depiction of the atom.

The usual approach is, instead, to always provide functions at least for the entire valence shell of the atom (and its inner shells). So for all 2nd period elements, you would use all 2nd shell orbitals ($\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$), even in a minimal basis set, no matter how many electrons actually occupy that shell. What still makes that basis set a minimal one is that you are only using one function for each orbital.

Similarly, a minimal basis set calculation for $\ce{Na}$ would include one function for each orbital in the 1st, 2nd and 3rd shells, and so on.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hi, thanks for this answer! :-) So, a minimal basis set include one function for all possible atomic orbitals for a given principal quantum number, right? The definition would be clear for me, but I find it a bit surprising, that the minimal basis of Li is the same as for Ne. Intuitively, I would guess that Li is much better approximated by this minimal basis then Ne is. $\endgroup$ – QuantumMechanics Mar 14 '18 at 9:03
  • $\begingroup$ Take into account that the functions will still represent the same orbitals - so both your basis set and the number of orbitals in your output will be the same for $\ce{Li}$ and for $\ce{Ne}$ - what will change is the number of occupied vs. unoccupied orbitals. $\endgroup$ – user41033 Mar 14 '18 at 10:12
  • $\begingroup$ Thanks again! But having more unoccupied orbitals increases the "goodness" of the result, or? So, the ratio of unoccupied vs. occoupied orbitals is better for Li than for Ne, meaning that the minimal basis set is a better approximation for Li than for Ne? $\endgroup$ – QuantumMechanics Mar 14 '18 at 10:57
  • $\begingroup$ Having a larger basis set definitely improves the variational approximation of the results - as more functions can be combined to find a better solution within the functional space defined by the basis set - that's the rationale behind multi-zeta (i.e. non-minimal) basis sets, diffusion functions, etc. So a calculation for $\ce{Li}$ including third shell orbitals would definitely be more accurate than a calculation not involving them. However, it's not immediately obvious to me than the same reasoning can be applied to two calculations involving the same functions but different occupancies. $\endgroup$ – user41033 Mar 14 '18 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.