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Chapter 9: Problem 4

In a Hartree-Fock calculation on atomic hydrogen using four primitive s-type Gaussian functions (S. Huzinaga, J. Chem. Phys., \(1293,42(1965)\) ), optimized results were obtained with a linear combination of Gaussians with coefficients \(c_{j}\) and exponents \(\alpha\) of 0.50907,0.123317 0.47449,\(0.453757 ; 0.13424,2.01330 ;\) and 0.01906 \(13.3615 .\) Describe how these primitives would be utilized in a \((4 s) /[2 s]\) contraction scheme.

Short Answer

Expert verified
In a (4s)/(2s) contraction scheme, four given primitive Gaussians effectively contract into two new Gaussians. The first two primitives combine to form the first contracted Gaussian and the next two form the second contracted Gaussian. These two Gaussians are then utilized in the process.

Step by step solution

01

Explain the Concept

The (4s)/(2s) contraction scheme is a method to represent or approximate the wave function of an atom or molecule in quantum chemistry. This notation means that a combination of four primitives (s-type Gaussians) are used to approximate one contracted function. However, out of these, only two will remain as linearly independent functions in further calculations. In other words, four primitive functions are used, but they get 'contracted' into two hence the term contraction scheme.
02

Applying the Concept

Given coefficients \(c_{j}\) and exponents \(\alpha\) can be used in the contraction scheme as follows: First two Gaussians form the first contracted function and the next two form the second contracted function. Thus the first two primitives i.e. with coefficients and exponents 0.50907,0.123317 and 0.47449,0.453757 are combined to form the first contracted s-type Gaussian. Similarly, the next two primitives i.e. with coefficients and exponents 0.13424,2.01330 and 0.01906,13.3615 are combined to form the second contracted s-type Gaussian.
03

Illustrating the Result

So in a (4s)/(2s) contraction scheme, these four primitive s-type Gaussians effectively contract to form two new Gaussians. Hence, two Gaussians are utilized, which is a more manageable and simplified system for further computations and calculations in the Hartree-Fock procedure.

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Most popular questions from this chapter

Use the AM1 and PM3 semiempirical methods to compute the equilibrium bond lengths and enthalpies of formation of (a) ethanol, (b) 1,4 -dichlorobenzene.

Prove Brillouin's theorem; that is, show that hamiltonian matrix elements between the HF wavefunction \(\Phi_{0}\) and singly excited determinants are identically zero. Hint. Use the Condon-Slater rules.

(a) For a CASSCF calculation of the ground-state wavefunction of diatomic \(\mathrm{C}_{2},\) describe a reasonable choice for the distribution of \(\sigma\) and \(\pi\) molecular orbitals into active, inactive and virtual orbitals. (b) How many inactive and active clectrons arc there in the calculation? (c) In an RASSCF calculation, how might the set of active orbitals be further divided?

A single Slater determinant is not necessarily an cigenfunction of the total clectron spin operator. Therefore, even within the Hartree-Fock approximation, for the wavefunction \(\Phi_{0}\) to be an eigenfunction of \(S^{2},\) it might have to be expressed as a linear combination of Slater determinants. The linear combination is referred to as a spin-adapted configuration. As a simple example, consider a two-electron system with four possible Slater determinants: $$\begin{array}{l} \Phi_{1}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1) \psi_{2}\left(r_{2}\right) \alpha(2)\right| \\ \Phi_{2}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1) \psi_{2}\left(r_{2}\right) \beta(2)\right| \\ \Phi_{3}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1) \psi_{2}\left(r_{2}\right) \alpha(2)\right| \\ \Phi_{4}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1) \psi_{2}\left(r_{2}\right) \beta(2)\right| \end{array}$$ (a) Show that the Slater determinants \(\Phi_{1}\) and \(\phi_{4}\) are themselves cigcnfunctions of \(S^{2}\) with cigenvaluc \(2 \hbar^{2}\) (corresponding to \(S=1\) ). (b) From \(\Phi_{2}\) and \(\Phi_{3}\), determine two linear combinations, one of which corresponds to \(S=1, M_{s}=0\) and the other of which corresponds to \(S=0\) \(M_{s}=0\)

Consider two Slater determinants \(\Phi_{1}\) and \(\Phi_{2}\) that differ by only one spinorbital; that is, $$\begin{array}{l} \Phi_{1}=\left\|\cdots \varphi_{m} \varphi_{i} \cdots\right\| \\ \Phi_{2}=\left\|\cdots \varphi_{p} \varphi_{i} \cdots\right\| \end{array}$$ Derive the following Condon-Slater rule: $$\begin{aligned} \left\langle\Phi_{1}|H| \Phi_{2}\right\rangle=&\left\langle\varphi_{m}(1)\left|h_{1}\right| \varphi_{p}(1)\right\rangle+\sum_{i}\left\\{\left[\varphi_{m} \varphi_{i} | \varphi_{p} \varphi_{i}\right]\right.\\\ &\left.-\left[\varphi_{m} \varphi_{i} | \varphi, \varphi_{p}\right]\right\\} \end{aligned}$$ where we have used the notation $$\left[\varphi_{4} \varphi_{b} | \varphi_{c} \varphi_{d}\right]=j_{0} \int \varphi_{a}^{*}(1) \varphi_{b}^{*}(2)\left(\frac{1}{r_{12}}\right) \varphi_{c}(1) \varphi_{d}(2) \mathrm{d} r_{1} \mathrm{d} r_{2}$$
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