Hydrogenic ground state energy, variational

Figure 1. Variational ground state energy of a Z = 90 hydrogen-like atom obtained from the Dirac-Pauli equation as a function of a (abscissa) and (5 (ordinate) while a = b = s (left figure) and as a function of a (abscissa) and b while a = (5 = Z (right figure). The arrows are proportional to the gradient of . The saddle points correspond to the exact eigenvalues of the Dirac Hamiltonian.

As an example we apply the variational principle to the evaluation of the ground state energy of a hydrogen-like atom using a minimum basis set of two-component radial functions ... 

The exact wavefunction corresponds to a = b = s and a = (3 = Z. The variational ground state energy of Z = 90 hydrogen-like ion in the Dirac-Pauli... 

Ground state energy of the two-dimensional hydrogen atom confined with conical curves through the variational method... 

For consistency in this subsection, it is also necessary to question the validity of the trial variational function and the results of the calculation for the ground-state energy of the hydrogen atom confined by a hyperbolic boundary in [3]. In fact, the corresponding function in their notation is... 

Molecular orbital theory as applied to the hydrogen molecule ion is described fully in Computer Lab Vll. In the solution block below the variational expression for the energy is given along with the first derivative of the ene with respect to the variation parameters alpha and R. These three equations are solved to give the ground-state energy and the optimum values of alpha and R. 

Apply the variation function = e " to the hydrogen atom choose the parameter c to minimize the variational integral, and calculate the percent error in the ground-state energy. 

In 1971 a paper was published that applied the normalized variation function N exp(—brV o cr/ao) to the hydrogen atom and stated that minimization of the variational integral with respect to the parameters b and c yielded an energy 0.7% above the true ground-state energy for infinite nuclear mass. Without doing any calculations, state why this result must be in error. 

As can be seen from Table 10.1, there was a competition between theoreticians and the experimental laboratory of Herzberg. When, in 1964 Kotos and Wolniewicz obtained 36117.3 cm (see Table 10.1) for the dissociation energy of the hydrogen molecule, quantum chemists held their breath. The experimental result of Herzberg and Monfils, obtained four years earlier (see Table 10.1), was smaller, and this seemed to contradict the variational principle (Chapter 5 i.e., as if the theoretical result were below the ground-state energy), the foundation of quantum mechanics. There were only three possibilities either the theoretical or experimental results are... 

Using the following trial wavefunction, determine the ground-state energy of a hydrogen atom using variational theory. 

If we apply the variational procedure to hydrogen atoms at ground state, we find that the lowest orbitals in order of increasing energy are... 

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