 We've seen that Hartree's method provides a good approximation to the energies and configuration of electrons and atoms. However, when we try to extend these ideas to molecules, we run into some challenges. The central field approximation is key to the success of Hartree's method. Treating the net force on each electron due to the nucleus and the other electrons as always directed towards the center of the atom greatly simplifies the problem. In a central field, the angular behavior of the wave function is the same as for the already solved case of the hydrogen atom. All that remains is to solve an equation in the single radial variable R. And, as in the case of hydrogen, each orbital is characterized by a set of quantum numbers. This allows us to enforce the exclusion principle by simply assigning different quantum numbers to each electron. However, the central field approximation is not generally applicable to molecules. The net attractive force on an electron, due to the pull of two or more nuclei, will not be spherically symmetric about some central point. And the repulsive force, due to the other electrons, even when treated in an average sense, is also unlikely to have spherical symmetry. Because of this, we will not be able to reduce Schringer's equation to a single radial variable R. This leaves us with a computationally much more challenging problem. Moreover, in a non-central field, the angular momentum of individual electrons will not be conserved. Without angular momentum quantum numbers, we cannot enforce the exclusion principle by simply assigning different quantum numbers to different electrons. This is a fundamental issue that we need to resolve before we can move forward. So far, we stated the Pauli exclusion principle along the lines of no two electrons can be in the same quantum state. But one of the postulates of quantum mechanics is that the state of a quantum system is specified by its wave function. So rigorously, the exclusion principle should be a statement about the wave function. Let's label the spin of an electron by a discrete coordinate S that has two values, up or down. Then for a two-electron system, the wave function would be a function of eight variables, the x, y, and z position coordinates plus the spin coordinate for each of the two electrons. To simplify the notation, let's represent the four coordinates of an electron by a single boldface coordinate x. Then our wave function is psi of x1, x2. And this gives the probability amplitude for finding electron 1 with coordinates x1 and electron 2 with coordinates x2. Consider the configuration where electron 1, which we color orange, has coordinates a, and electron 2, which we color green, has coordinates b. The probability of this configuration is specified by the magnitude squared of psi of a, b. If instead it was electron 1 that had coordinates b and electron 2 that had coordinates a, then the probability factor would be magnitude squared of psi of b, a. Mathematically, these could be different states of the system with different probabilities. And if instead of electrons we were keeping track of macroscopic objects like billiard balls or planets, these would definitely be different distinguishable states. The state of Venus being closer to the Sun than Earth is very different from that of Earth being closer to the Sun than Venus. But electrons, as elementary particles, are indistinguishable. We can't physically label or paint them, and given the uncertainty principle, there is no way to talk about the distinct path of electron 1 versus the path of electron 2 inside a molecule. The most that a measurement of a two electron system could tell us, even in principle, is that there is an electron with coordinates a and an electron with coordinates b. So, these probabilities must be equal. The behavior of a physical system cannot depend on how we arbitrarily assign subscripts to indistinguishable particles in a mathematical model. A valid two electron wave function must have the property that its magnitude squared does not change if the coordinates of the electrons are exchanged, since this is equivalent to simply relabeling indistinguishable particles. Now, what does this imply about the wave function itself? If u squared equals v squared, what can we say about u and v? One possibility is that u equals v. Certainly, if psi of x1, x2 equals psi of x2, x1, the probabilities will be equal. A wave function with this property is said to be symmetric with respect to the exchange of coordinates. But another possibility is that u equals minus v. After all, 2 squared equals 4, but so does minus 2 squared. So, we could have psi of x1, x2 equals minus psi of x2, x1. A wave function with this property is said to be anti-symmetric with respect to the exchange of coordinates. In nature, we find two types of elementary particles. The bosons, such as the photon, have symmetric wave functions and do not obey the exclusion principle. The fermions, such as the electron, have anti-symmetric wave functions and do obey the exclusion principle. The exclusion principle follows from this anti-symmetry. For example, suppose both sets of coordinates, x1 and x2, are equal to x. The two electrons have the same spin and are at the same location. Wave function anti-symmetry then requires that psi of xx equals its negative. But there's only one number that equals its negative, and that's zero. There is zero probability that two electrons with the same spin are at the same place, which is a particular case of two electrons in the same quantum state. In general, an an electron wave function must be anti-symmetric with respect to the exchange of coordinates of any pair of electrons. This is a rigorous statement of the exclusion principle, and it forms an additional postulate of quantum mechanics. But how do we enforce this condition? An elegant solution is given by the Slater determinant, due to John Slater. In Hartree's method, we represent the wave function of a two electron system by a product of two one electron wave functions, a so-called Hartree product. Slater pointed out that this form of wave function is not intrinsically anti-symmetric, hence it does not rigorously satisfy the exclusion principle. But we can make it anti-symmetric by subtracting the product with the coordinates exchanged. No matter what the one electron wave functions are, this two electron wave function is explicitly anti-symmetric. It is equivalent to the so-called determinant of a matrix in which the columns correspond to different one electron wave functions and the rows correspond to different electron coordinates. A normalizing factor, in this case one over the square root of two, is usually included. Let's see how this works by considering the example of two electrons with the same spin occupying hydrogen 1s and 2s orbitals. We take the 1s and 2s wave functions and form the Slater determinant two electron wave function. Consider the situation where one electron is at radius a. Let's arbitrarily take this to be electron two and replace r2 by a. This gives us a wave function for the other electron, a function of the single variable r1. When r1 equals a, this wave function is zero. Here we'll plot this wave function for different values of a, which we represent by a red dot. When the second electron is far from the nucleus, the first electron essentially occupies the 1s orbital. But as the second electron moves towards the nucleus, the wave function for the first electron changes dramatically, so as to always be zero at the location of the second electron. If we multiply the wave function by the radius and square the result, we get the probability density for electron one to be at a particular radius, which has more physical significance. We can see how electron two squeezes electron one's probability distribution, so as to always keep the two electrons separate. If we have three electrons, there are three different pairs of electrons we can swap, so we have to be careful about our bookkeeping. Let's represent the Hartree product shown as one, two, three. We can swap electrons two and three to get the permutation one, three, two. For anti-symmetry, a single pair swap term should have a negative sign. We could also swap electrons one and two to get the permutation two, one, three. Or we could swap electrons one and three to get the permutation three, two, one. We can get two more permutations by swapping a second pair of electrons. If in the two, one, three permutation, we swap electrons one and three, we get the permutation two, three, one. For two swaps, we have a minus minus sign, which is a plus sign. Finally, in the permutation three, two, one, we can swap electrons two and one to get the permutation three, one, two. We get the Slater determinant by adding the two positive permutations, two, three, one, and three, one, two, followed by subtracting the three negative permutations, one, three, two, two, one, three, and three, two, one. We end up with a total of six terms, three positive and three negative. Formally, we write this as the determinant of a three by three matrix in which the columns correspond to different one electron wave functions or orbitals, and the rows correspond to different electron coordinates. The normalizing factor in this case is one over the square root of six. This is starting to look unwieldy. For n electrons, there are n factorial terms in the Slater determinant because there are n factorial permutations of the numbers one through n. n factorial is n times n minus one times n minus two down to one. The water molecule has only 10 electrons, yet 10 factorial is more than 3 million. Fortunately, we can always take our single electron orbitals to be a so-called orthonormal set in which the projection of any orbital onto another is zero and the projection of an orbital onto itself is one. In this case, the vast majority of terms and calculations with Slater determinants are zero, and we end up with tractable expressions. Hartree had developed his method using intuitive concepts. Slater provided a theoretical foundation for the Hartree method. Indeed, if one starts with the assumption that the n electron wave function is a Hartree product of n one electron wave functions, then Hartree's method can be shown to be optimal. At the same time, Slater pointed out that the Hartree product wave function does not have the anti-symmetry required by the exclusion principle. What was needed was a way to extend the ideas behind Hartree's method to a wave function represented by a Slater determinant.