For the helium atom, what happens if we try to use a product form
as a ``guess'' wave function? Ignoring spin for the moment,
we take as a guess of
the solution for the ground-state wave function
If we do this, we find that and satisfy a set of 2 coupled Schrödinger-like equations and that the potential energy in each equation corresponds to what the electron would experience if we averaged over all possible positions of the other electron (think back to the classical shell model!).
If we solve these equations, we find that both and bear a strikingly resemblance to the hydrogenic function , the 1 wave function. The difference, however, is that it decays a little faster, as if its value were between 1 and 2. In fact, the value one obtains from an actual calculation is . We will denote this as . We can imagine one of the electrons as ``occupying'' the state and the other as ``occupying'' the state , but of course, we cannot say which electron is in which state, which is why we need the above form of the guess wave function. We also obtain two energies and that are close in energy but that the energies are lower than the energy of the hydrogen atom.
Now, we see the problem with ignoring spin. If and are both , then the guess wave function becomes
As another shorthand notation, we can use symbols ``1'' and ``2''
to represent any of the variables of electrons 1 and 2. Thus, the
guess wave functions can be represented simply as
The lowest HF energy obtained for helium, Ry. Note how close this is to the formula for with ! The next energy is only slightly large than this value, so we see that , but strictly . However, the energies are so close that they can be considered as degenerate. In fact, there is a class of Hartree-Fock calculations (called ``closed-shell'' calculations), in which this degeneracy is assumed at the outset, and the pair of electrons is treated as a single unit or single quantum state.
Now, if we followed the same thing for lithium, we would need three functions to construct the guess wave function, which we could take as
For Beryllium, we would obtain four orbitals, two of which resemble 1s orbitals and two of which resemble 2s orbitals. Again, these decay as if they have an effective value between 3 and 4. We can, therefore, represent these orbitals as
For Boron, we would obtain 5 orbitals, two of which would resemble , two would resemble , and the fifth would resemble a orbital. Again, all of these would decay as if they have an effective between 4 and 5. We can thus represent the 5 orbitals as
For carbon, we obtain 6 orbitals, and this time, the 6th orbital does not resemble but rather . Hence, we would represent the 6 orbitals as
For Nitrogen, the 7th orbital would resemble a orbital, and we would just add . Now, for oxygen, the 8th orbital again resembles a orbital, and we would add this to the list but with a down spin wave function . For oxygen, we find that the 5-8 HF energies are approximately equal, , although strictly .
These examples illustrate a more general procedure known as Hartree-Fock theory or the Hartree-Fock approximation (named after Douglas Hartree and Vladimir Fock) developed in 1930. Hartree-Fock (HF) theory makes several important assumptions:
So how do we calculate probabilities in HF theory? What we are interested in is the probability of finding one of the electrons in a small volume about the point independent of the positions of the remaining electrons. This probability is given in terms of the electron density introduced earlier. That is
To give an idea of how well HF theory can predict the ground state energies of several atoms, consider the table below (all energies are in Ry):
The HF theory motivates the so-called ``aufbau'' principle for expressing/building up electron configurations in atoms. ``Aufbau'' is German for ``building up'' (Aufbauen = ``to build up''). We see that the HF orbitals are ordered according to their energies and that as the energy increases, their shapes follow the ``ladder'' of wave functions of the hydrogen atom. Thus, as we consider atoms with more and more electrons, we simply have to ``occupy'' more of the HF orbitals in order of increasing energy. Since groups of these orbitals are close in energy, there are two rules we need to introduce for this procedure of occupying orbitals:
For Helium (), HF theory is no longer exact, however as we discussed, HF theory gives two orbitals and , both of which resemble (but decays a little faster). The energies and are also very similar. Thus, we regard these orbitals as having roughly the same energy, and we place one electron into and one into . We regard this as placing two electrons into the single orbital and assigning the electrons the same set of H-like quantum numbers . To be consistent with Pauli's exclusion principle, they need to have one quantum number that is different, so we assign one electron and the other (see figure). We write the electron configuration as
For Lithium (), HF theory yields three orbitals , , and . The first two closely resemble , while the third resembles . HF theory also yields three energies , and with and . Each electron occupies on HF orbital, but we regard this as two electrons in a orbital and the third in a orbital (see figure). We write this as
Coming to Beryllium (), we have 4 HF orbitals, two of which resemble and two of which resemble , so we imagine that an electron in each HF orbital is equivalent to 2 in the and 2 in the . In each pair, the electrons have opposite spins to be consistent with the Pauli exclusion principle. We write this as
For Boron (), there are 5 HF orbitals. and resemble , and resemble , and resembles , i.e. a generic -orbital (we do not specify because the direction of the orbital does not matter). The energies are ordered such that , . Interestingly, is larger than all of these and not close to . Unlike in hydrogen, where orbitals with the same but different values of have the same energy, here orbitals with the same but different values of have different energies. Thus, is at a higher energy. Hence, when we fill, we start with , then and finally . We can choose any of the orbitals, so we just start with . We thus write the electronic configuration as
For carbon (), the situation is much the same. However, here we have and both with -like character and similar in energy. Here, however, we need Hund's rule. We first put a spin-up electron in the orbital and then another spin-up electron in the orbital. The electronic configuration is
For nitrogen (), we would place a spin-up electron in the orbital and write . When we get to oxygen (), we have now placed one electron in each of the three -like orbitals. We now have to return to and place another electron in it, which will now be spin-down to be consistent with the Pauli exclusion principle. Its electronic configuration is . For fluorine (), it is . Finally, when we get to Neon (), we have , which we can write simply as . With Neon, as with Helium, we have completed a shell, and the next HF orbital will be like and have a higher energy, thereby starting the next shell. Thus, we see that for Neon, 8 electrons are needed to complete a shell, the 2 2s electrons and the 6 2p electrons. The fact that 8 electrons close a shell for Neon lend this atom special stability and make it highly unreactive.