Ground state energy equations

Since the constant first-order contribution to the energies of all the states is not physically interesting, let us examine the second-order contribution to the ground-state energy. Equation (12-23) shows that this is related to the first-order correction to the wavefunction,, which, as we have already seen, causes the wavefunction to become skewed toward the low-potential end of the box. It is clear from Eq. (12-22) that, in calculating the coefficients for we have already done most of the work needed to... 

The HF-LCAO calculation follows the usual lines (Figure 11.10) and the frozen core approximation is invoked by default for the CISD calculation. CISD is iterative, and eventually we arrive at the improved ground-state energy and normalization coefficient (as given by equation 11.7) — Figure 11.11. 

This proof shows that any approximate wave function will have an energy above or equal to the exact ground-state energy. There is a related theorem, known as MacDonald s Theorem, which states that the nth root of a set of secular equations (e.g. a Cl matrix) is an upper limit to the n — l)th excited exact state, within the given symmetry subclass. In other words, the lowest root obtained by diagonalizing a Cl matrix is an upper limit to the lowest exact wave functions, the 2nd root is an upper limit to the exact energy of the first excited state, the 3rd root is an upper limit to the exact second excited state and so on. 

The ground-state energy E of a hydrogen atom (Z = 1) as given by equation (6.48) is... 

The variation method gives an approximation to the ground-state energy Eq (the lowest eigenvalue of the Hamiltonian operator H) for a system whose time-independent Schrodinger equation is... 

We next substitute equation (9.3) into the integral for in (9.2) and subtract the ground-state energy Eq, giving... 

If the quantity is identieal to the ground-state energy Eq, which is usually non-degenerate, then the trial funetion 0 is identical to the ground-state eigenfunetion 0o- This identity follows from equation (9.5), which becomes... 

As a simple application of the variation method to determine the ground-state energy, we consider a particle in a one-dimensional box. The Schrodinger equation for this system and its exact solution are presented in Section 2.5. The ground-state eigenfunction is shown in Figure 2.2 and is observed to have no nodes and to vanish at x = 0 and x = a. As a trial function 0 we select 0 = x(a — x), 0 X a... 

The exact ground-state energy Ei is shown in equation (2.39) to be 7i h /2ma. Thus, we have... 

The reason why we obtain the exact ground-state energy in this simple example is that the trial function 0 has the same mathematical form as the exact ground-state eigenfunction, given by equation (4.39). When the parameter c is evaluated to give a minimum value for S , the function 0 becomes identical to the exact eigenfunction. 

The first-order perturbation correction to the ground-state energy is obtained by evaluating equation (9.24) with (9.80) as the perturbation and (9.82) as the unperturbed eigenfunction... 

The first step in the solution of equation (10.28b) is to hold the two nuclei fixed in space, so that the operator drops out. Equation (10.28b) then takes the form of (10.6). Since the diatomic molecule has axial symmetry, the eigenfunctions and eigenvalues of He in equation (10.6) depend only on the fixed value R of the intemuclear distance, so that we may write them as tpKiy, K) and Sk(R). If equation (10.6) is solved repeatedly to obtain the ground-state energy eo(K) for many values of the parameter R, then a curve of the general form... 

The molecular constants o , B, Xe, D, and ae for any diatomic molecule may be determined with great accuracy from an analysis of the molecule s vibrational and rotational spectra." Thus, it is not necessary in practice to solve the electronic Schrodinger equation (10.28b) to obtain the ground-state energy o(R). 

Let us consider a case in which an ion (donor, D ) and a solvent (acceptor. A) form a CT complex. The ground state energy (see Fig. 5) can be obtained as a solution of the secular equation ... 

Let us summarize what we have shown so far once N and Vext (uniquely determined by ZA and Ra) are known, we can construct H. Through the prescription given in equation (1-13) we can then - at least in principle - obtain the ground state wave function, which in turn enables the determination of the ground state energy and of all other properties of the system. Pictorially, this can be expressed as... 

Stated in still other words this means that for any trial density p(r) - which satisfies the necessary boundary conditions such as p( ) - 0, J p( ) dr = N, and which is associated with some external potential Vext - the energy obtained from the functional given in equation (4-6) represents an upper bound to the true ground state energy E0. E0 results if and only if the exact ground state density is inserted into equation (4-8). The proof of the inequality (4-11) is simple since it makes use of the variational principle established for wave functions as detailed in Chapter 1. We recall that any trial density p(r) defines its own Hamiltonian H and hence its own wave function. This wave function can now be taken as the trial wave function for the Hamiltonian generated from the true external potential Vext. Thus, we arrive at... 

Thus, once we know the various contributions in equation (5-15) we have a grip on the potential Vs which we need to insert into the one-particle equations, which in turn determine the orbitals and hence the ground state density and the ground state energy by employing the energy expression (5-13). It should be noted that Veff already depends on the density (and thus on the orbitals) through the Coulomb term as shown in equation (5-13). Therefore, just like the Hartree-Fock equations (1-24), the Kohn-Sham one-electron equations (5-14) also have to be solved iteratively. 

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