 Before we consider the details, recall from the hydrogen atom video that it's convenient to work in so-called spherical coordinates, where r is the distance from the nucleus to the electron, and the angles theta and phi are basically the longitude and latitude of the electron. Hydrogen orbitals are characterized by three quantum numbers. n specifies energy, l specifies angular momentum, and m specifies the z-component of angular momentum. Now we'll sketch out a derivation of the Hartree equation. We start with the one-electron Schrodinger equation. We assume the solution can be written as a radial part that we write as 1 over r times a function p of r times an angular part y of theta and phi. The quantum number n will be associated with the radial function. The quantum numbers l and m are associated with the angular function. This part is what specifies the angular variations associated with the spd and f orbitals. Substituting this factored form into the Schrodinger equation, we arrive at an equation for the p function. Roughly speaking, it looks like minus one-half the curvature, or second derivative, of the p function times the potential function in brackets, times the p function equals the energy times the p function. The potential energy term, depending on quantum number l, represents a centrifugal force that increases with increasing angular momentum. The remaining potential energy consists of a nuclear attraction term and an electron repulsion term. Writing out all terms, we have the Hartree equation. This is the Schrodinger equation for a one-electron orbital, assuming the angular part is identical to that of a hydrogen orbital, while the radial part includes the average potential energy of the electron repulsion computed using the central field approximation. We can rearrange this into the form shown here, where all the potential terms are combined into an effective potential. This equation can't be solved analytically, but it's simple to solve numerically. For a small step size delta, the curvature term can be approximated as shown. Putting this into the Hartree equation and rearranging, we get an expression that tells us how to calculate p at r plus delta, given p at r and r minus delta. p is zero at zero, and p at delta can be set to an arbitrary small number since at the end we can scale the entire result as needed. Then we can run out this formula to any r value we choose. There are just two things missing. First, we don't know what energy value to put into the equation. Second, we can't compute the effective potential without knowing the electron orbitals, but we can't solve for the orbitals without knowing the effective potential, a real chicken and egg problem. To solve the first problem, Hartree took a trial and error approach to calculating energies based on the following fact. If we set up the Hartree equation for the hydrogen atom and put in a correct energy, running our formula produces a correct wave function, but an energy that is a bit too high gives a wave function that eventually diverges to infinity. For an energy that is a bit too low, it diverges to infinity in the opposite direction. If we run our formula for two energy values and get results that diverge in opposite directions, then we know that the correct energy value is in between. This is sometimes called the shooting method for solving equations of this form. The shooting method works for all energy levels and wave functions. Combining with the hydrogen 1s energy, we get the 1s wave function. As we increase the energy, our result initially diverges, but eventually converges back to zero when we reach the energy of the 2s orbital. From there, increasing the energy causes divergence in the other direction, but then eventually converges back to zero when we reach the energy of the 3s orbital. In this manner, we can find the energy and wave function of any orbital. If that problem is solved, we turn to the chicken and egg issue of the effective potential. Hartree named his solution the method of the self-consistent field. The Hartree equation for the i-th electron depends on the repulsion it feels from each of the other electrons. For the j-th electron, this force depends on an effective charge Q sub j. This effective charge depends on the j-th wave function, and this contributes to the i-th potential. So the i-th wave function depends on the i-th potential. The i-th potential depends on the j-th wave function. The j-th wave function depends on the j-th potential. But the j-th potential depends on the i-th wave function. So we have a circular set of problems. Hartree found that if he started with some gas for the orbitals, and then ran through this circular process of enough times, the orbitals eventually converged. The final orbitals formed a self-consistent set. Every orbital ends up being consistent with the equations for the other orbitals. Let's see how this works by performing a self-consistent calculation of the helium atom orbitals. Our initial guess for the orbitals is nothing. The first electron sees only the nucleus, and we obtain a hydrogen-like 1s orbital. Now we freeze this orbital and use it to calculate the potential that will be seen by the second electron. Solving the Hartree equation for the second electron's orbital, we get something slightly different than for electron 1. Now we freeze electron 2's orbital and recalculate electron 1's. The repulsion due to electron 2 causes electron 1's orbital to spread out a bit. Now we continue freezing one orbital and solving for the other until we see no significant change in the orbital energies. The result is a self-consistent field in which the two electrons have identical wave functions. We find, as did Hartree, that the orbital energies are close to minus 25 electron volts. The ionization energy of helium is about 24.6 electron volts. The close agreement of these values suggests that Hartree's method might provide a good approximation to the actual wave functions of multi-electron atoms. For lithium, with three electrons, we have to consider the exclusion principle that we discussed in video 8. Only one electron can be in a given quantum state, including spin, so only two electrons can occupy a given orbital, one with spin up and the other with spin down. The third electron of lithium, therefore, has to go into an n equals 2 orbital, either an s orbital with l equals 0 or p orbital with l equals 1. The s orbital has lower energy, so it represents the ground state of lithium. The 2s orbital energy of minus 4.79 electron volts is in rough agreement with the observed lithium ionization energy of 5.39 electron volts. Once we've solved for all the electron orbitals, we can combine the squares of the radial parts of the wave functions to get the total electron density of the atom as a function of radial distance from the nucleus. These values for helium are shown in this table from his paper. Here we plot those values as red dots and our calculations as a solid line. It's interesting to compare this to the hydrogen 1s orbital density and the helium ion 1s density. We see that the two electrons of helium are considerably more tightly bound in the atom than the single electron of hydrogen is. So although helium has two electrons, it actually forms a smaller atom than hydrogen does with a single electron. Hartree also presented a self-consistent field calculation of the orbitals of the rubidium ion with 36 electrons. Here we plot his total electron density results as red dots and our calculation is a solid line. We find that the outer electrons are broadly spread out to radii a few times that of the hydrogen atom. Zooming in we see how the electrons arrange themselves into shells, corresponding primarily to the quantum number n. The innermost shell of rubidium containing the two 1s electrons is very tightly bound, constrained to radii of less than about one-twentieth the radius of the electron of the hydrogen atom. By firing free electrons at a collection of atoms and observing how the electrons are scattered, it's possible to experimentally determine an atom's actual radial electron density. Observations such as those seen here for argon generally show good agreement with the Hartree theory. This gives us confidence that even with its simplifying approximations, the method of the self-consistent field is capturing the main features of electron interaction. The energies of electrons in different orbitals can also be measured experimentally. For the inner core electrons with very large binding energies, this can be done by illuminating the atom with x-ray photons of the corresponding energy. Here we see experimental values for the rubidium ion compared to the results of Hartree's theory. The level of agreement between Hartree's theory and experiment led physicist John Slater to conclude that, with it, a completely theoretical calculation of the energy levels of all the atoms became possible, so that the theory of the periodic table was made quantitative.