Hartree Products

The simplest possible way of making t / as a combination of these molecular orbitals is by forming their Hartree product... 

However, such a function is not antisymmetric, since interchanging two of the r, s —equivalent to swapping the orbitals of two electrons—does not result in a sign change. Hence, this Hartree product is an inadequate wavefunction. 

In addition to the Schrodinger equation we have the antisymmetry requirement (Eq. II.2) connected with the Pauli principle and, by means of the antisymmetrization operator (Eq. 11.16), the Hartree product (Eq. 11.37) is then transformed into a Slater determinant ... 

This expression is just the one which obtains for the Hartree product wave-function. The difference between this Hartree wavefunction and the Fock wavefunction of Eq. (1) is the absence of the antisymmetrizer j4 in that equation. This means that in the Hartree wavefunction each electron can be identified with a specific molecular orbital, whereas in the Fock wavefunction all electrons make use of all orbitals. The Hartree wavefunction is of course not a proper quantum mechanical wavefunction, since it is not antisymmetric in the electrons. Moreover, for the Fock wavefunction, it is in general not possible to reduce the interorbital exchange energy to zero. But the localized molecular orbitals, as defined here, represent that set of molecular orbitals for which the energy expression comes closest to the Hartree form, i.e. they come closest to being identifiable with electrons which are not exchanged among different orbitals. 

The energy of this wave function is the sum of the spin orbital energies, E = Eji + " + EjH. We have already seen a brief glimpse of this approximation to the /-electron wave function, the Hartree product, in Section 1.3. 

The most simple ansatz is to approximate this IV-electron wave function by a product TIIP of N one-electron functions (r) (Hartree product of orbitals),... 

Every fermionic wave function has to be antisymmetric when exchanging the coordinates of any two particles (Pauli principle) which is not fulfilled for a simple product ansatz as in Eq. (3). Therefore, we explicitly have to antisymmetrize the Hartree product THP and obtain the Slater determinant ... 

In the most simple case of one pair of coordinates being interchanged the permutation operator Py acts as follows on a Hartree product ... 

This kind of wavefunction is called a Hartree Product, and it is not physically realistic. In the first place, it is an independent-electron model, and we know electrons repel each other. Secondly, it does not satisfy the antisymmetry principle due to Pauli which states that the sign of the wavefunction must be inverted under the operation of switching the coordinates of any two electrons, or... 

Part of the proof of equation 13 acknowledges this is not so for a Hartree Product. To remedy this, first consider a two-electron system, such as helium. Two equivalent Hartree Product wavefunctions for this system are... 

The term in square brackets is a Hartree product The numbers in round brackets refer to particular electrons, or more specifically, to the x, y, z, and spin coordinates of those... 

Answer. Orbitals are one-electron wave functions, ). The fact that electrons are fermions requires that each electron be described by a different orbital. The simplest form of a many-electron wave function, T(l, 2,..., Ne), is a simple product of orbitals (a Hartree product), 1(1) 2(2) 3(3) NfNe). However, the fact that electrons are fermions also imposes the requirement that the many-electron wave function be antisymmetric toward the exchange of any two electrons. All of the physical requirements, including the indistinguishability of electrons, are met by a determinantal wave function, that is, an antisymmetrized sum of Hartree products, ( 1,2,3,..., Ne) = 1(1) 2(2) 3(3) ( ). If (1,2,3,...,Ne) is taken as an approximation of (1,2,..., Ne), i.e., the Hartree-Fock approximation, and the orbitals varied so as to minimize the energy expectation value,... 

A wave function of die form of Eq. (4.35) is called a Hartree-product wave function. The eigenvalue of T is readily found from proving the validity of Eq. (4.35), viz.. 

As noted above, however, the Hamiltonian defined by Eqs. (4.32) and (4.33) does not include interelectronic repulsion, computation of which is vexing because it depends not on one electron, but instead on all possible (simultaneous) pairwise interactions. We may ask, however, how useful is the Hartree-product wave function in computing energies from the correct Hamiltonian That is, we wish to find orbitals that minimize (4 hp H I hp). By applying variational calculus, one can show that each such orbital i/f, is an eigenfunction of its own operator hi defined by... 

At this point it is appropriate to think about our Hartree-product wave function in more detail. Let us say we have a system of eight electrons. How shall we go about placing them into MOs In the Hiickel example above, we placed them in the lowest energy MOs first, because we wanted ground electronic states, but we also limited ourselves to two electrons per orbital. Why The answer to that question requires us to introduce something we have ignored up to this point, namely spin. 

Knowing these aspects of quantum mechanics, if we were to construct a ground-state Hartree-product wave function for a system having two electrons of the same spin, say a, we would write... 

Slater determinants have a number of interesting properties. First, note that every electron appears in every spin orbital somewhere in the expansion. This is a manifestation of the indistinguishability of quantum particles (which is violated in the Hartree-product wave functions). A more subtle feature is so-called quantum mechanical exchange. Consider the energy of interelectronic repulsion for the wave function of Eq. (4.43). We evaluate this as... 

---