Helium ground state energy calculations

Figure J Helium ground state energy computed by two different methods. The Rivelino and Vianna calculations [108] vs the results of reference [117] considered here as the exact ones.

To go further, we turn to so-called all-order methods. We start by considering relativistic calculations of type carried out by Lindroth [38] and Blundell et al. [39], which account for all possible single and double excitations of the ground-state helium HF wave function. These calculations have the potential of giving an exact helium ground-state energy. 

There are several commonly used approximation schemes that can be applied to the electronic states of multielectron atoms. The first approximation scheme was the variation method, in which a variation trial function is chosen to minimize the approximate ground-state energy calculated with it. A simple orbital variation trial function was found to correspond to a reduced nuclear charge in the helium atom. This result was interpreted to mean that each electron in a helium atom shields the other electron from the full charge of the nucleus. A better variation trial function includes electron correlation, a dependence of the wave function on the electron lectrcm distance. ... 

Before the individual parts of this function are discussed, the energy eigenvalue will be considered. The ground state energy g of the helium atom is just the energy value for double-ionization which can be determined accurately by several different kinds of experiments. Before the experimental value can be compared with the calculated one, some small corrections (for the reduced mass effect, mass polarization, relativistic effects, Lamb shift) are necessary which, for simplicity, are... 

The result of such a calculation for Pls(r) of helium and how it compares with the hydrogenic case are shown in Fig. 7.6, and the differences can be seen clearly. For the total energy of the helium ground state the HF approach yields Eg = — 2.862 au. This is rather close to the experimental value, Eg(exp.) =... 

Abstract. With an eye on the high accuracy ( 10 MHz) evaluation of the ionization energy from the helium atom ground state, a complete set of order ma6 operators is built. This set is gauge and regularization scheme independent and can be used for an immediate calculation with a wave function of the helium ground state. 

Calculation of the ground state energy of the helium atom is a critical case as well because it is the first example of correlation energy, the difference between the Hartree-Fock energy and the exact value. The energy required to remove one electron from a neutral He atom is the first ionization potential... 

The Hall-Roothaan equations for the case of the helium ground state appear, when equation 5.39 is written out in an appropriate linear combination of functions upon which the variation principle procedure can be applied to return the best energy in a calculation. For example, for the double-zeta basis of Slater functions used in the previous section, we have... 

Figure 5.4b The output of each iteration to self-consistency for the Clementi double-zeta Slater basis calculation of the ground state energy of the helium atom and the final converged result of the worksheet hfs in fig5-4.xls. Further iterations lead to no improvement of the results and the energy of the helium atom is found to be 2.86167 a.u.

Exercise 5.4. Formation of a 4-31) split-basis set for helium using the data of Table 5.1 and the calculation of the energy of the helium ground state. 

Figure 5.10 Determination of the best ground state energy for helium by varying the independent coefficient of the split-basis linear combination from the sto-4g) Gaussian set of Table 5.1. The energy of helium is found to be —2.85516 Hartree for the coefficients of equation 5.40 equal 0.51380 and 0.59189. The orbital energy, is, is calculated to be —0.91412 Hartree and compares well with Huzinaga s result. Table 5.2.

Table 1 Ground-state energies of the helium atom calculated with the g function given by Eq. (10) and the initial function xjro given by Eq. (11) ...

Notice that O5 and Og are two-electron functions, which cannot be factorized into one-electron functions. By calculating all matrix elements and solving the 6 x 6 eigenvalue problem, Hylleraas, in 1928, obtained, without comparison, the best description of the helium atom with the energy -2.903329 H, compared to the earlier best value of -2.86 H. With the help of modern computers, it was recently possible to determine the ground state energy with more than accurate 20 decimal places (-2.903724 H) using essentially the Hylleraas method. 

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