Next: Energies in the LCAO
Linear combination of atomic orbitals (LCAO) is a simple method of
quantum chemistry that yields a qualitative picture of the
molecular orbitals (MOs) in a molecule. Let us consider H again.
The approximation embodied in the LCAO approach is based on the notion
that when the two protons are very far apart, the electron in its
ground state will be a orbital of one of the protons. Of course,
we do not know which one, so we end up with a Schrödinger cat-like state in
which it has some probability to be on one or the other.
As with the HF method, we propose a guess of the true wave function
for the electron
is a hydrogen orbital centered
on proton A and
is a hydrogen orbital
centered on proton B. Recall
The positions and are given simply by the vectors
The explicit forms of
Now, unlike the HF approach, in which we try to optimize the shape of the
orbitals themselves, in the LCAO approach, the shape of the
orbital is already given. What we try to optimize here
are the coefficients and that determine the amplitude
for the electron to be found on proton A or proton B.
The guess wave function
is not normalized as
we have written down. Thus, our guess of the ground-state energy is
where is the electron's volume element, and
is the electronic Hamiltonian (minus the nuclear-nuclear
(we will account for the nuclear-nuclear repulsion later when
we consider the energies). Consider the denominator first:
Now, the wave function of hydrogen is normalized so
In the cross term, however, the integral
is not 1 because the orbitals are centered on different protons
(it is only one if the two protons sit right on top of each other,
which is not possible).
It is also not 0 unless the protons are very far apart. We can
calculate the integral analytically, however, it is not that
important to do so since there is no dependence on or .
Let us just denote this integral as . We know that
and this is good enough for now. Thus, the denominator is simply
As for the numerator
where we have defined a bunch of integrals I'm too lazy to do as
Again, these are integrals we can do, but it is not that important, so we will just
keep the shorthand notation. Note that since the two nuclei are the same
(they are both protons), we expect
. Since these are equal,
we will just call them both . Hence, the guess energy becomes
which is just a ratio of two simple polynomials. Since we know that , where
is the true ground-state energy, we can minimize with respect
to the two coefficients and . Thus, we need to take two derivatives
and set them both to 0:
Defining the denominator simply as , where
the two derivatives are
Thus, we have two algebraic equations in two unknowns and . In fact, these will not
determine and uniquely because they are redundant. However, we also have
the normalization of as a third condition, so we have enough information to
determine the coefficients absolutely.
The equations can be solved as follows: First we write them as
We now divide one equation by the other, which yields
Since we cannot solve for and independently, we
choose, instead, to solve for the ratio .
Dividing both sides of the above by equation by yields
To solve for , we cross multiply:
Multiplying this out, we obtain
which implies .
Thus, we have two solutions: or . So, to simplify the
notation, Let (drop the A subscript). We, therefore, have two possible
guess wave functions:
The constant can be determined for each guess wave function by imposing the normalization
condition. For , we have
Similarly, one can show that for ,
Let us now see how the two approximate solutions lead to bonding and antibonding orbitals.
. We will add these two wave functions pictorally, keeping
in mind that is a spherically symmetric orbital with an exponential decay. Thus,
The figure below shows the orbitals that result from the addition or subtraction
for or .
In the bottom left panel, we show the two orbitals centered
on protons A and B in . In the center on the left, the
result of adding them together is shown. It should be clear from the
figure that is a bonding orbital that is an approximation
to the true orbital that is the true ground state.
This approximation is often denoted . On the
other hand, the orbital is the difference
which is depicted in the top left panel. This orbital has
a node between the two nuclei and the amplitude between the
two nuclei is generally low. This is clearly an anti-bonding
orbital corresponding to . This approximate
orbital is often denoted
. Since this orbital
has a node, we expect that its energy will be higher than
the approximate ground-state energy .
LCAO approximation to the ground () and
first excited () orbitals in H.
Concerning the energies, the figure below shows the two
lowest energies curves as functions of , comparing the
results from the exact solution for H and the LCAO
approximation. Note, in particular, that, in accordance
with the variational principle, the exact energy curve
for the ground state is always lower than the LCAO approximation.
Next: Energies in the LCAO
Mark E. Tuckerman