Molecules are of course atoms that are held together by shared valence electrons. That is, most of each atom is pretty much as it would be if the atom were isolated, but one or a few electrons are located in regions (for example, between two atomic cores) where they lead to an overall attractive effect. To see roughly how this works, you might consider the case of two protons and two electrons at the corners of a square. If you calculate (classically) the total electrostatic energy of this arrangement, you will find that it is negative - the system is bound.
Molecules have several kinds of motion, with quite different energy
scales. Assume that a couple of nuclei and a couple of electrons are
confined to a volume with characteristic size (a couple of Å,
typically). Then the uncertainty principle requires that at a minimum,
, so the typical electron energies will be of order
The nuclei may also participate in vibrations, in which they interact
directly with the electron cloud. The energies of such motions are
larger than , and may be estimated by treating the
interaction between electrons and nuclei as simple harmonic motion
with a spring constant , the same (of course) for both parties. In
this case the ratio of electronic () to nuclear vibrational
energies () will be of order
Consider the Schrödinger equation for a system of two identical
nuclei and two electrons, in the centre of mass system of the nuclei,
We will solve this system by a separation and some approximations.
Because the electrons are much lighter than the nuclei, the
nuclei will hardly move at all in the time that an electron takes to
``orbit'' once, so let's suppose that we can solve the simpler problem
of electron motion in the presence of motionless nuclei separated by a
fixed , for which the electronic wave equation is
To find equations satisfied by the 's, put into
the exact Schrödinger equation and projects using each of the
We now make the adiabatic or Born-Oppenheimer
approximation by assuming that we can neglect
. In this case, we get the nuclear
The equation for is in the form of a wave equation with the function playing the role of potential energy. Let us consider the specific case of the electrons in a state of zero orbital angular momentum. In this case, is a function only of the radial variable , and the wave equation for becomes one for a spherically symmetric potential. In this case, the Hamiltonian commutes with and . The simultaneous eigenfunctions of these operators are the spherical harmonics, and have eigenvalues and .
Since there is no privileged direction in space, the total energy of
the system cannot depend on , but it does depend on . (You may
wonder why the direction of is not a privileged direction
now. It is for the electron cloud, but in a spherically symmetric
electron potential, it is just a coordinate as far as the nuclei are
concerned. Thus can rotate in space, and the nuclei may have
non-zero angular momentum.) It will be found that there is also
another quantum number in the system, , which acts as a principal
quantum number and will be found to number vibrational states. Thus we
may write (dividing the radial function by to get a simple form
for the resulting equation)
Now in a bound molecular state, the energy varies from the
sum of the energies of the two atoms separately for large , through
a local minimum, to a large positive value as the two nuclei get close
to one another. We may approximate the minimum region of by a
second order expansion:
For the electron cloud, the (slowly rotating) internuclear axis picks out a direction in space which we take as the -axis. The
electronic structure of the molecule is invariant under rotation
around this axis, so commutes with . However, ,
and do not commute with . The electronic eigenstates
of may also be made to be eigenstates of , so that
Diatomic molecules are symmetric to reflections through planes
including , such as the plane. Thus if operator
If is 0, the state is not degenerate, and can only multiply it by a constant. Since , such states are either symmetric or anti-symmetric to reflection through a plane, and one distinguishes and states.
If the molecule has two nuclei with identical charge (a homonuclear molecule), the centre is also a symmetry point, and states have parity. States of even (g) and uneven (u) parity are denoted by , , etc. A homonuclear diatomic molecule has four non-degenerate sigma states, , , , and .
Finally, each electronic eigenstate has a total spin , with eigenvalues of of . The value of is given as a left superscript (the multiplicity) on the designation. Thus, since most molecular ground states (often labelled X) are states of zero total spin, the complete label for a ground state could be X or X .
We may start our study of electronic wave functions with the simplest
molecule of all, H, with two protons and one electron. To
simplify writing, we will switch to Hartree's dimensionless atomic units, in which the units of mass, charge, angular momentum,
and length are chosen to be , , , and
the result (you may check this) that the units for time, velocity, and
where is the ground state binding energy of H. In these units,
the electronic wave equation becomes
We will solve this equation using the simple approximation of a linear
combination of atomic orbitals (LCAO). If the nuclei are far apart,
the electron will be attached to one (say A), with wave function
The numerator is
This problem may be solved exactly (although numerically, not analytically) through the use of confocal elliptic coordinates, which are also used in the evaluation of the two-centre integrals.
With two electrons, we must consider the effects of the exclusion principle, which states that eigenstates must be antisymmetric to exchange of two identical fermions. The eigenstates of the two-electron cloud are products of spatial wave functions and spin functions. How does the spin affect the situation?
The total spin operator is
It is easily shown that has eigenvalues of and of and , while has and . We now have four normalized and orthogonal spin eigenstates that we will denote by for clarity. is antisymmetric to particle exchange, while the other three states are symmetric. We call the one state a spin singlet, while the other three states form a spin triplet.
In solving (approximately) the problem of the electron cloud structure for the two-electron system H, we have the obvious choice of using as a basis set the molecular orbitals found for H, or the atomic orbitals of the individual H atoms. Either provides a useful basis for solution. We will start with the molecular orbitals.
Recall that we have two molecular orbitals and
(which are even and odd under reflection through the
midpoint of ). The product
is symmetric under exchange of electrons 1 and 2, as are the
the other hand, the function
is antisymmetric under
exchange of the electrons. Thus we may form (only) four orthogonal
eigenfunctions that are antisymmetric under exchange:
To get a bonding state, we need the two electrons to prefer to be between the two protons. If they both occupy the bonding orbitals of the H system, and have opposite spins (so that the exclusion principle will not require them to occupy different spatial states), we guess that we should get maximum bonding. Thus we guess that may a reasonable approximation to the ground state of H.
We now need to solve the equation
If one uses the approximate LCAO molecular orbitals found above, the binding energy is calculated to be 2.68 eV (compared to an accurate value of 4.75 eV), with a computed equilibrium of 0.8 Å, compared to an accurate value of 0.74 Å.
The trial wavefunction that we have used may be written as the sum of
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