Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 124

Assume we found a series of solutions for the Hartree-Fock equations, i.e. a set of N eigenfunctions φn(r) with the associated eigenvalues . We still have to compute the minimal variational energy of the N-particle system. This energy is given by the sum (26) of the three terms of kinetic, potential and interaction energies obtained by replacing in (15), (16) and (18) the θi(r) by the eigenfunctions φn(r):

(40)

(the subscripts HF indicate we are dealing with the average energies after the Hartree-Fock optimization, which minimizes the variational energy). Intuitively, one could expect this total energy to be simply the sum of the energies , but, as we are going to show, this is not the case. Multiplying the left-hand side of equation (38) by and after integration over d³r, we get:

(41)

We then take a summation over the subscript n, and use (15), (16) and (18), the θ being replaced by the φ:

(42)

This expression does not yield the stationary value of the total energy, but rather a sum where the particle interaction energy is counted twice. From a physical point of view, it is clear that if each particle energy is computed taking into account its interaction with all the others, and if we then add all these energies, we get an expression that includes twice the interaction energy associated with each pair of particles.

The sum of the does contain, however, useful information that enables us to avoid computing the interaction energy contribution to the variational energy. Eliminating between (40) and (42), we get:

(43)

where the interaction energy is no longer present. One can then compute 〈Ĥ₀〉_HF and using the solutions of the Hartree-Fock equations (38), without worrying about the interaction energy. Using (15) and (17) in this relation, we can write the total energy as:

(44)

The total energy is thus half the sum of the , of the average kinetic energy, and finally of the one-body average potential energy.