As I know, for Hydrogen atom a $1s$ Slater Type Orbital (STO) equation is (I can get it from here):

\begin{equation}\label{STO} \mathrm{STO} = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta r}. \end{equation}

With remark from Szabo A., Ostlund N.S. Modern Quantum Chemistry: Intro to Advanced Electronic Structure Theory, Dover, 1996:

For example, the standard exponent for the $1s$ basis function of hydrogen is , $\zeta = 1.24$. This is larger than the $\zeta = 1.00$ exponent of the hydrogen atom, since the hydrogen $1s$ orbital in average molecules is known to be "smaller" or "denser" than in the atom.

Gaussian Type Orbital (GTO) equation for $1s$-orbital (where did this coefficient $\left(\frac{2\alpha}{\pi}\right)^{3/4}$ come from?):

\begin{equation} \mathrm{GTO} = \left(\frac{2\alpha}{\pi}\right)^{3/4} e^{-\alpha r^2} \end{equation}

So, I have good analitical form of Hydrogen STO and GTO, and using formula \begin{equation}\label{} \sqrt{\frac{\zeta^3}{\pi}}e^{-\zeta r} \approx C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} + C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} + C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_2 r^2}. \end{equation} and copyng $\alpha_i$'s and $C_i$ from basissetexchange.org

# Basis Set Exchange
# Version v0.8.13
# https://www.basissetexchange.org
#   Basis set: STO-3G
# Description: STO-3G Minimal Basis (3 functions/AO)
#        Role: orbital
#     Version: 1  (Data from Gaussian09)

BASIS "ao basis" PRINT
#BASIS SET: (3s) -> [1s]
H    S
      0.3425250914E+01       0.1543289673E+00
      0.6239137298E+00       0.5353281423E+00
      0.1688554040E+00       0.4446345422E+00

I can plot Hydrogen orbital

enter image description here

Now I want to do the same for a Lithium atom using data

BASIS "ao basis" PRINT
#BASIS SET: (6s,3p) -> [2s,1p]
Li    S
      0.1611957475E+02       0.1543289673E+00
      0.2936200663E+01       0.5353281423E+00
      0.7946504870E+00       0.4446345422E+00
Li    SP
      0.6362897469E+00      -0.9996722919E-01       0.1559162750E+00
      0.1478600533E+00       0.3995128261E+00       0.6076837186E+00
      0.4808867840E-01       0.7001154689E+00       0.3919573931E+00

But, I do not understand where I can get a analitical formula for GTO's primitives. All of references, I saw: \begin{equation}\label{GTO} \mathrm{GTO}(x,y,z;\alpha,\ell_1,\ell_2,\ell_3) = N x^{\ell_1} y^{\ell_2} z^{\ell_3} e^{-\alpha r^2} \end{equation}
with unknown normalizing factor $N$. Sometimes I saw for $1p$-orbital \begin{equation} \mathrm{GTO}(p_x) = \left(\frac{128\alpha^5}{\pi^3}\right)^{1/4} x \exp(-\alpha r^2) \end{equation} but I don't understand where the factor $\left(\frac{128\alpha^5}{\pi^3}\right) $ comes from. So, my general question: where can I see the analytical view of the GTO primitive (like for STO's), and how can you find $\zeta$ for atom obital? (how to read such data?)

  • 1
    $\begingroup$ I believe this website (ccl.net/cca/documents/basis-sets/basis.html ) might be helpful, even if it does not answer all the questions. For the 1s STO-3G orbital, the $\mathrm{(\frac{2\alpha}{\pi})^{3/4}}$ is in fact, the normalization constant. So I believe that for the p orbitals (or other orbitals) you only need the C and $\alpha$, then you can normalize that. The 'factor' that you see is in fact the algebraic form of the normalization constant. $\endgroup$ – Shoubhik R Maiti Feb 8 at 19:00

Main differenece betweet STO's and GTO's is a $r$-expotetnt. Normalizing factor of GTO one can get from common procedure: \begin{equation} N^2 \int_0^\infty \left(r^{n-1}e^{-\zeta r^2}\right)^2 r^2 dr =1 \end{equation} Of course, computation is not so simple for $n > 1$, but we can use online WolframAlpha (clickable calculation) and we get the formulas from the question post.

For a Lithium atom we have contraction (6s,3p) -> [2s,1p], which means the first 3s of 6s GTO's cotract to one s "1s"-STO $$\mathrm{STO}(1s) = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta_1 r},$$ second 3s of 6s GTO's contract to another s "2s"-STO
$$\mathrm{STO}(2s) = \sqrt{\frac{\zeta^5}{3\pi}} r e^{-\zeta_2 r},$$ 3p GTO's contract to one p - "2p"-STO $$\mathrm{STO}(2p) = \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r}$$ (notation in orbitals in quotation marks "1s", "2s" means the same as for the hydrogen atom). So, as we can see, every atomic STO contract by only 3 GTO's (which justifies their name).

For a Litium atom we have for "1s" atomic orbital \begin{equation} \sqrt{\frac{\zeta_1^3}{\pi}} e^{-\zeta_1 r} \approx C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} + C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} + C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_3 r^2} \end{equation} where \begin{align} \alpha_1 = 0.1611957475E+02; C_1 = 0.1543289673E+00;\\ \alpha_2 = 0.2936200663E+01; C_2 = 0.5353281423E+00;\\ \alpha_3 = 0.7946504870E+00; C_3 = 0.4446345422E+00; \end{align} enter image description here

For "2s" atomic orbital \begin{equation} \sqrt{\frac{\zeta_2^5}{3\pi}} r e^{-\zeta_2 r} \approx C_4 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_4 r^2} + C_5 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_5 r^2} + C_6 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_6 r^2} \end{equation} where \begin{align} \alpha_4 = 0.6362897469E+00; C_4 = -0.9996722919E-01;\\ \alpha_5 = 0.1478600533E+00; C_5 = 0.3995128261E+00; \\ \alpha_6 = 0.4808867840E-01; C_6 = 0.7001154689E+00; \end{align} enter image description here For "2p_x" atomic orbital \begin{equation} \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r} \approx D_1 \left(\frac{128\alpha_4^5}{\pi^3}\right)^{1/4} xe^{-\alpha_4 r^2} + D_2 \left(\frac{128\alpha_5^5}{\pi^3}\right)^{1/4} xe^{-\alpha_5 r^2} + D_3 \left(\frac{128\alpha_6^5}{\pi^3}\right)^{1/4} xe^{-\alpha_6 r^2} \end{equation} where \begin{align} \alpha_4 = 0.6362897469E+00; D_1 = 0.1559162750E+00;;\\ \alpha_5 = 0.1478600533E+00; D_2 = 0.6076837186E+00; \\ \alpha_6 = 0.4808867840E-01; D_3 = 0 0.3919573931E+00; \end{align} enter image description here (For $p_y$ and $p_z$ --- do the same)

Also, for STO's best fit $\zeta_1 = 2.69$ and $\zeta_2 = 0.81$ are consistent with those given in the article Hehre, W. J., Ditchfield, R., Stewart, R. F., & Pople, J. A. (1970). Self‐Consistent Molecular Orbital Methods. IV. Use of Gaussian Expansions of Slater‐Type Orbitals. Extension to Second‐Row Molecules. The Journal of Chemical Physics, 52(5), 2769–2773. doi:10.1063/1.1673374


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.