Van der Waals Forces Between Atoms
Michael Fowler 2/28/07
Introduction
The perfect gas equation of state PV = NkT is manifestly incapable
of describing actual gases at low temperatures, since they undergo a
discontinuous change of volume and become liquids. In the 1870’s, the Dutch physicist Van der
Waals came up with an improvement: a gas law that recognized the molecules
interacted with each other. He put in
two parameters to mimic this interaction.
The first, an attractive intermolecular force at long distances, helps
draw the gas together and therefore reduces the necessary outside pressure to
contain the gas in a given volume—the gas is a little thinner near the
walls. The attractive long range force
can be represented by a negative potential -aN/V
on going away from the walls—the molecules near the walls are attracted
inwards, those in the bulk are attracted equally in all directions, so
effectively the long range attraction is equivalent to a potential well
extending throughout the volume, ending close to the walls. Consequently, the
gas density N/V near the walls is
decreased by a factor 
. Therefore, the
pressure measured at the containing wall is from slightly diluted gas, so 
becomes 
, or 
. The second parameter
van der Waals added was to take account of the finite molecular volume. A real gas cannot be compressed
indefinitely—it becomes a liquid, for all practical purposes
incompressible. He represented this by
replacing the volume V with 
, Nb is referred to as the “excluded volume”, roughly speaking the
volume of the molecules, to give his famous equation
This rather crude approximation does in fact give sets of isotherms representing the basic physics of a phase transition quite well. (For further details, and an enlightening discussion, see for example Appendix D of Thermal Physics, by R. Baierlein.)
Ground State Hydrogen Atoms
Our interest here is in understanding the van der Waals long-range attractive force between electrically neutral atoms and molecules in quantum mechanical terms.
We begin with the simplest possible example, two hydrogen atoms, both in the ground state:
We label the atoms A
and B, the vectors from the protons
to the electron position are denoted by 
and 
respectively, and 
is the vector from
proton A to proton B.
Then the Hamiltonian 
, where
and the electrostatic interaction between the two atoms
.
The ground state of 
is just the product of
the ground states of the atoms A, B, that is,
.
Assuming now that the distance between the two atoms is much
greater than their size, we can expand the interaction V in the small parameters 
. As one might suspect
from the diagram above, the leading order terms in the electrostatic energy are
just those of a dipole-dipole interaction:
Taking now the z-axis
in the direction , this interaction energy is
Now the first-order correction to the ground state energy of
the two-atom system from this interaction is 
, where here 
and 
. Beginning with the first term 
in V
is clearly zero since the ground states are spherically symmetric. Similarly, the other terms in V are zero to first order.
Recall that the second-order energy correction is 
That is,
.
A typical term here is
,
so the single-atom matrix elements are exactly those we
discussed for the Stark effect (as we would expect—this is an electrostatic
interaction!). As before, only 
contribute. To make a
rough estimate of the size of 
, we can use the same trick used for the quadratic Stark
effect: replace the denominators by the constant 
(the other terms are a
lot smaller for the bound states, and continuum states have small overlap terms
in the numerator) . The sum over
intermediate states 
can then be taken to
be completely unrestricted, including even the ground state, giving
,
the identity operator. In this approximation, then, just as for the Stark effect,
where E1 = - 1 Ryd., so this is a lowering of energy.
In multiplying out 
, the cross terms will have expectation values of zero. The ground state wave function is
symmetrical, so all we need is 
, where 
is the Bohr radius.
This gives
using 
. Bear in mind that
this is an approximation, but a pretty good one—a more accurate calculation
replaces the 6 by 6.5.
Forces between a 1s Hydrogen Atom and a 2p Hydrogen Atom
With one atom in the 
and the other in 
, say, a typical leading order term would be
,
and this is certainly zero, as are all the other leading terms. Baym (Lectures on Quantum Mechanics) concluded from this that there is no leading order energy correction between two hydrogen atoms if one of them is in the ground state. This is incorrect: the first excited state of the two-atom system (without interaction) is degenerate, so, exactly as for the 2-D simple harmonic oscillator treated in the previous lecture, we must diagonalize the perturbation in the subspace of these degenerate first excited states. (For this section, we follow fairly closely the excellent treatment in Quantum Mechanics, by C. Cohen-Tannoudji et al.)
The space of the degenerate first excited states of the two noninteracting atoms is spanned by the product-space kets:
The task, then, is to diagonalize in this
eight-dimensional subspace.
We begin by representing V
as an 8‰8
matrix using these states as the basis.
First, note that all the diagonal elements of the matrix are zero. Second, writing 
, it is evident that V
is unchanged if the system is rotated around the z-axis (the line joining the two protons). This means that the commutator 
, where 
is the total angular
momentum component in the z-direction,
so V will only have nonzero matrix
elements between states having the same total . Third, from parity
(or Wigner-Eckart) all matrix elements in the subspace spanned by 
must be zero.
This reduces the nonzero part of the 8‰8
matrix to a direct product of three 2‰2
matrices, corresponding to the three values of = m. For example, the m = 0 subspace is spanned by 
. The diagonal
elements of the 2‰2 matrix are zero, the off-diagonal elements are equal to 
, where we have kept the unnecessary labels A, B
to make clear where this term comes from.
(The xA and yA terms will not contribute
for m = 0.)
This is now a straightforward integral over hydrogen wave functions. The three 2‰2 matrices have the form
(following the notation of Cohen-Tannoudji) where 
, and the energy eigenvalues are 
, with corresponding eigenkets 
.
So for two hydrogen atoms, one in the ground state and one
in the first excited state, the van der Waal interaction energy goes as 
, much more important than the 
energy for two hydrogen atoms in the ground state. Notice also that the can be positive or negative, depending on
whether the atoms are in an even or an odd state—so the atoms sometimes repel
each other.
Finally, if two atoms are initially in a state 
, note that this is not
an eigenstate of the Hamiltonian when the interaction is included. Writing the state as a sum of the even and
odd states, which have slightly different phase frequencies from the energy
difference, we find the excitation moves back and forth between the two atoms
with a period 
.