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The Helium Atom:

The helium atom is an example of a polyelectronic system. It is comprised of a nucleus with a atomic number of Z=2 and two electrons. This system cannot be treated in the same manner that the hydrogenic model was due to the interactions among the two electrons. One begins by writing the hamiltonian for the atom.

eq.1

The total hamiltonian is simply comprised of the one electron hamiltonains devised earlier and also an operator to account for electron-electron repulsion, called the coulomb operator.

An eigenfunction which will commute with the helium hamiltonian constructed is a product of two one-electron atomic orbitals, one for each of the electrons in ²He.

eq.2

Two possible wavefunctions are possible, depending on which electron occupies which one-electron orbital. Caution is warrented here, these wavefunctions are obviously not the exact solutions due to electron-electron interactions. However in the limiting case of non-interaction, these are correct and they provide a good starting point anyway. As will be shown, for states other than the ground state, the two one-electron orbitals will differ causing the atom to be subject to additional electronic phenomena. The one-electron hamiltonians are of the form:

eq.3

These consist of the kinetic energy operator and the nuclear electrostatic potnetial operator. Since there exist two possible wavefunctions, all possible combinations of integrals need to be accounted for when operating with the hamiltonian.

eq.4

An easy way of keeping track of the integrals is to treat the hamiltonian as a matrix.

eq.5

The elements of this matrix corrsespond to operations of various combinations of the possible wavefunctions. We assume the two helium wavefunctions are orthogonal since they are centered on the same nucleus, meaning the overlap matrix is of the form:

eq.6

Continuing, one turns to the Schrodinger Equation, which is of the form:

eq.7

This can be rearranged into the familiar eigenvalue equation:

eq.8

Unique solutions exist for when the determinant is zero. Knowing the overlap matrix, all one needs to determine is the form of the hamiltonian matrix. Each of the matrix elements needs to be solved in turn. For element H₁₁, one has:

eq.9

Each of the one-electron atomic orbitals will commute with its respective one-electron hamiltonian and the coulomb operator. As a result of the integral, one obtains the following.

eq.10

One sees the result is just the energy of each electron with reference to the bare nuclear field plus a third term which is the coulomb term. The coulomb term is always denoted as J in literature, which will also be done here.

eq.11

The solution to the coulomb integral is difficult to derive and can only be solved for through computational methods covered later in this site. For now it is merely important to recognize its physical interpretation, that being the interelectron repulsion. As one can prove independently, the H₂₂ term is equal to the H₁₁ term.

eq.12

Next, one needs to consider the other two terms of the hamiltonian matrix, H₂₁ and H₁₂. Setting up the integral for H₂₁ results in:

eq.13

Solving results in a solution which looks similar to the ones derived for the first two terms solved for, but with very important differences.

eq.14

Since the orthogonality restriction has been placed on this system, one immediately recognizes the first two terms are zero. The remaining term is called the exchange integral. It represents the ability of two electrons with parallel spins to exchange places. In the ground state of helium, the two electrons occupy the same space orbital and therefore have opposing spins. This means the two cannot exchange places. However, in the case of excited states of helium, the electrons will occupy different space orbitals and therefore have parallel spins. This means the two can in fact exchange places with each other. This term is always denoted K in literature.

eq.15

Again, one can prove independently that the matrix elements H₂₁ and H₁₂ are equal.

eq.16

Now one has everything to solve for the energy of the atom. Filling in the eigenvalue equation (eq.8) with the terms one has derived, the result is:

eq.17

Taking the determinant leaves one with a simple polynomial equation.

eq.18

Rearranging this leaves one with the final expression of the energy for the helium atom.

eq.19

There exist two solutions, however only the one containing the minus sign is physically important, since the universe has a tendency to keep energy at a minimum. The energy of the helium atom is therefore the sum of the two electron energies as influenced only by the nucleus, the energy of the interelectron repulsion, and the stabalization energy dervied from electron exchange.