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Originally, Hartree–Fock atomic calculations were done by using numerical methods to solve the Hartree–Fock equations, and the resulting orbitals were given as tables of the radial functions for various values of r. Since 1951, Roothaan proposed representing the Hartree–Fock orbitals as linear combinations of a complete set of known functions, called basis functions: \begin{equation*} \phi = \sum c_i \chi_i, \end{equation*} where the $\chi_i$ functions are some complete set of functions, and where the $c_i$'s are expansion coefficients that are found by the SCF iterative procedure.

But in textbooks always says that "A commonly used set of basis functions for atomic Hartree–Fock calculations is the set of Slater-type orbitals (STOs)": \begin{equation*} \chi_i = \frac{(2\zeta_i)^{n + 1/2}}{[(2n)!]^{1/2}} r^{n - 1}e^{-\zeta_i r} \cdot Y_{lm}(\theta, \phi). \end{equation*}

As a rule, no further details follow. And this creates some confusion. It seems to me that it would be more logical to use hydrogen-like orbitals as a basis, since they are the exact solution of the Schrödinger equation.

All I could find was a sparing mention of Born and Hylleraas:

In 1928, it was already recognised by Born and Hylleraas that the He atom could not be described by a CI expansion using the H-like bound-state eigenfunctions.

Source: https://www.esqc.org/lectures/WK3.pdf

In this connection, I have a question: what are the main advantages/benefits of Slater-type orbitals compared to hydrogen-like ones?

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    $\begingroup$ @Grag May be you mean, the advantage of STO because it nodless? Then, yes. What is the advantage of nodeless functions? $\endgroup$
    – Sergio
    Commented Apr 15, 2022 at 7:33
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    $\begingroup$ @Greg "nodeless" - radial part of wave-function without nodes. $\endgroup$
    – Sergio
    Commented Apr 15, 2022 at 7:37
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    $\begingroup$ I don't know the answer but I will observe that it seems to me that you can construct hydrogenic functions as a linear combination of STOs. You can thus express an STO is a linear combination of hydrogenic functions. They are thus equivalent, and the simpler form of STO is probably easier to handle. $\endgroup$
    – Ian Bush
    Commented Apr 15, 2022 at 8:23
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    $\begingroup$ @IanBush One reason mentioned in Slater article Phys. Rev. 36, 57: "The nodes in the wave function are found [...] to be unimportant, [...] they come much nearer the nucleus than for hydrogen wave functions, and are therefore less important. Consequently we neglect them entirely, taking as the radial part Of the wave function of one electron simply". $\endgroup$
    – Sergio
    Commented Apr 15, 2022 at 8:28
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    $\begingroup$ This may sound naive, but aren't Slater orbitals simply easier to calculate? $\endgroup$ Commented Apr 18, 2022 at 11:24

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There are two criteria for a good set of basis functions:

  1. The solution of the problem has to be in the basis, i.e. some combination of the basis functions has to approximate the real wave function reasonably well.
  2. Using the set of basis functions has to be sufficiently computationally efficient that you can find the solution (or an approximation of it).

If it is easy to map the basis functions on something we are familiar with (like our mental model of the atomic wave functions 1s, 2s, 2p etc.), interpreting the results might be a bit easier.

The choice of basis set concerns the radial functions only. For the angle-dependent part, it seems like all calculations use the spherical harmonics.

From first principles, the Laguerre polynomials (so the solution to one-electron atoms) and the Slater orbitals individually model the wavefunction well at large distances from the nucleus, while a Gaussian function does not.

The Laguerre polynomials are the only ones that model the radial nodes in a single function.

[OP] It seems to me that it would be more logical to use hydrogen-like orbitals as a basis, since they are the exact solution of the Schrödinger equation.

They are not the exact solution of the Schrödinger equation for multi-electron atoms. In particular, the hydrogen-like orbitals are a bad model for the valence electrons in period 2 and higher elements, which are shielded quite a bit from the charge of the nucleus (in the one-electron model).

[OP] In this connection, I have a question: what are the main advantages/benefits of Slater-type orbitals compared to hydrogen-like ones?

In a paper on computational difficulties using Slater-type orbitals, the authors made the following statement:

Due to that fact, the hydrogen-like orbitals do not form a complete set (for finite n), they need orbitals of the continuum to be complete.

So according to this statement, hydrogen-like orbitals seem to fail criterion #1 (complete basis set) in a practical way. With current algorithms and hardware, Slater-type basis sets seem to fail criterion #2. Thus, the commonly used basis sets are built from Gaussian functions, with integrals that are easier to solve analytically.

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