\documentclass[12pt]{article}
\usepackage{wrapfig}
\usepackage{amsmath}
\usepackage{graphicx}
\message{
 	Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of 
 Physics and Atronomy, Louisiana State University, LA 70803).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
 permission of the author listed above.	
}

\title{Chapter 1: Chemical Bonding}
\author{Linus Pauling (1901--1994)}
\input{../defs}
\def\baselinestretch{1.4}
\bdo
%\large
\maketitle
\tableofcontents

\pagebreak


\leavevmode

\begin{figure}[htb]
\centerline{\includegraphics[width=6.0in,keepaspectratio,clip=true]{periodic_fancy.pdf}}
\caption[]{\em{Periodic Chart}}
\label{fig:PeriodicChart}
\end{figure}

\pagebreak


     Solid state physics is the study of mainly periodic systems
(or things that are close to periodic) in the thermodynamic limit
$\approx 10^{21}$ atoms/cm$^3$.  At first this would appear to be a 
hopeless task, to solve such a large system.
\begin{figure}[htb]
\begin{center}
\centerline{\includegraphics[width=6.0in,keepaspectratio,clip=true]{fig1.pdf}}
\caption[]{\em{The simplest model of a solid is a periodic array of 
valance orbitals embedded in a matrix of atomic cores.}}
\label{fig:simple_solid}
\end{center}
\end{figure}

However, the self-similar, translationally invariant nature of the 
periodic solid and the fact that the core electrons are {\em{very}} tightly
bound at each site (so we may ignore their dynamics) makes approximate 
solutions possible.  Thus, the simplest model of a solid is a periodic array 
of valance orbitals embedded in a matrix of atomic cores.
Solving the problem in one of the irreducible elements of
the periodic solid (cf.\ one of the spheres in Fig.~\ref{fig:simple_solid}), 
is often equivalent to solving the whole system.  For this reason we must 
study the periodicity and the mechanism (chemical bonding) which binds 
the lattice into a periodic structure.  The latter is the emphasis of this 
chapter.


\section{The development of Bands and their filling}

\begin{table}[htb]
\begin{tabular}{|l|l|}\hline
nl  & elemental solid \\\hline
1s  & H,He \\
2s  & Li,Be \\
2p  & B$\to$Ne \\
3s  &  Na,Mg  \\
3p  & Al$\to$Ar   \\
4s  & K,Ca \\
3d  & transition metals Sc$\to$Zn \\
4p  & Ga$\to$Kr\\
5s  & Rb,Sr\\
4d  & transition metals Y$\to$Cd\\
5p  & In-Xe \\
6s  & Cs,Ba\\
4f  & Rare Earths (Lanthanides) Ce$\to$Lu \\
5d  & Transition metals La$\to$Hg  \\
6p  & Tl$\to$Rn \\ \hline
\end{tabular}
\caption[]{\em{Orbital filling scheme for the first few atomic orbitals}}
\label{tab:orb-fill}
\end{table}

     We will imagine that each atom (cf.\ one of the spheres in 
Fig.~\ref{fig:simple_solid}) is composed of Hydrogenic
orbitals which we describe by a screened Coulomb
potential
\beq
V(r)=\frac{-Z_{nl}e^2}{r}
% + \frac{l(l+1)}{2mr^2}
\eeq
where $Z_{nl}$ describes the effective charge seen by each electron (in
principle, it will then be a function of $n$ and $l$).
As electrons are added to the solid, they then fill up the
one-electron states 1s 2s 3s 3p 3d 4s 4p 4d 4f$\cdots$, where the 
correspondence between spdf and $l$ is $s\to l=0$, $p\to l=1$, etc.
The elemental solids are then made up by filling these orbitals
systematically (as shown in Table~\ref{tab:orb-fill}) starting with the
lowest energy states (where $E_{nl}=\frac{me^4 Z_{nl}^2}{2\hbar^2 n^2}$
\begin{figure}[htb]
\centerline{\includegraphics[height=2.8in,keepaspectratio,clip=true]{potential2.pdf}}
\caption[]{\em{Level crossings due to atomic screening. The potential felt 
by states with large $l$ are screened since they cannot access the nucleus. 
Thus, orbitals of different principle quantum numbers can be close in 
energy.  I.e., in elemental Ce, ($4f^15d^16s^2$) both the $5d$ and 
$4f$ orbitals may be considered to be in the valence shell, and form metallic 
bands.  However, the 5d orbitals are much larger and of higher symmetry than 
the 4f ones.  Thus, electrons tend to hybridize (move on or off) with the 5d 
orbitals more effectively.  The Coulomb repulsion between electrons on
the same  4f orbital will be strong, so these
electrons on these orbitals tend to form magnetic moments.}}
\label{potential}
\end{figure}

	Note that for large $n$, the orbitals do not fill up simply as
a function of $n$ as we would expect from a simple Hydrogenic model 
with $E_n=\frac{mZ^2e^4}{2\hbar^2n^2}$ (with all electrons seeing the same 
nuclear charge $Z$).  For example, the 5s 
orbitals fill {\em{before}} the 4d!  This is because the situation is 
complicated by atomic screening.  
I.e.\ s-electrons can sample the core and 
so are not very well screened whereas d and f states face the angular 
momentum barrier which keeps them away from the atomic core so that they
feel a potential that is screened by the electrons of smaller $n$ and $l$.  
To put is another way, the effective $Z_{5s}$ is larger than $Z_{4d}$.
A schematic atomic level structure, accounting for screening, is shown in
Fig.~\ref{potential}.

Now let's consider the process of constructing a periodic
solid.  The simplest model of a solid is a periodic array of valence 
orbitals embedded in a matrix of atomic cores (Fig.~\ref{fig:simple_solid}).
As a simple model of how the eigenstates of the individual atoms
are modified when brought together to form a solid, consider a 
pair of isolated orbitals.  If they are far apart, each orbital has
a Hamiltonian $H_0=\epsilon n$, where $n$ is the orbital occupancy
and we have ignored the effects of electronic correlations (which
would contribute terms proportional to $n_\uparrow n_\downarrow$).
\begin{figure}[htb]
\centerline{\includegraphics[height=0.9in,keepaspectratio,clip=true]{isolated_orbs.pdf}}
\caption[]{\em{Two isolated orbitals.  If they are far apart, each has
a Hamiltonian $H_0=\epsilon n$, where $n$ is the orbital occupancy.}}
\label{fig:isolated_orbs}
\end{figure}
If we bring them together so that they can exchange electrons, i.e.
hybridize, then the degeneracy of the two orbitals is lifted.
\begin{figure}[htb]
\centerline{\includegraphics[height=0.9in,keepaspectratio,clip=true]{hybridize_orbs.pdf}}
\caption[]{\em{Two orbitals close enough to gain energy by hybridization.  
The hybridization lifts the degeneracy of the orbitals, creating bonding
and antibonding states.}}
\label{fig:hybridize_orbs}
\end{figure}
Suppose the system can gain an amount of energy $t$ by moving
the electrons from site to site (Our conclusions will not depend upon 
the sign of $t$.  We will see that $t$ is proportional to the 
overlap of the atomic orbitals).  Then
\beq
H=\epsilon (n_1+n_2) -t(c_1^{\dag} c_2 + c_2^{\dag} c_1 )\,.
\eeq
where $c_1$ ($c_1^{\dag}$) destroys (creates) an electron on orbital 1.
If we rewrite this in matrix form
\beq
H=\left( c_1^\dag , c_2^\dag\right) 
\left[\begin{array}{rr}
\epsilon & -t \\
-t & \epsilon \\
\end{array}\right]
\left( \begin{array}{r}
c_1 \\
c_2 \\
\end{array} \right)
\eeq
then it is apparent that system has eigenenergies $\ep \pm t$.  Thus the two 
states split their degeneracy, the splitting is proportional to $|t|$, and 
they remain centered at $\epsilon$

	If we continue this process of bringing in more isolated orbitals 
into the region where they can hybridize with the others, then a band
of states is formed, again with width proportional to $t$, and centered
around $\epsilon$ (cf.\ Fig.~\ref{band_form}).
\begin{figure}[htb]
\centerline{\includegraphics[width=6.0in,keepaspectratio,clip=true]{fig2.pdf}}
\caption[]{\em{If we bring many orbitals into proximity so that they
may exchange electrons (hybridize), then a band is formed centered around
the location of the isolated orbital, and with width proportional to the
strength of the hybridization}}.
\label{band_form}
\end{figure}
This, of course, is an oversimplification.  Real solids are
composed of elements with multiple orbitals that produce multiple
bonds.  Now imagine what happens if we have several orbitals on
each site (ie
s,p, etc.), as we reduce the separation between the orbitals and
increase their overlap, these bonds increase in width and may
eventually overlap, forming bands.

     The valance orbitals, which generally have a
greater spatial extent, will overlap more so their bands will
broaden more.  Of course, eventually we will stop gaining energy
(\~t) from bringing the atoms closer together, due to overlap of
the cores.  Once we have reached the optimal point we fill the
states 2 particles per, until we run out of electrons. 
Electronic correlations complicate this simple picture of band formation
since they tend to try to keep the orbitals from being multiply occupied.  

\section{Different Types of Bonds}

     These complications aside, the overlap of the orbitals is bonding.
The type of bonding is determined to a large degree by the amount
of overlap.  Three different general categories of bonds form
in solids (cf.\ Table~\ref{tab:bond_types}).

\begin{table}[htb]
\begin{tabular}{|l|l|l|l|}\hline
Bond	& Overlap	     & Lattice	                  & constituents \\\hline
Ionic	&very small ($<a$)   & closest unfrustrated       &	dissimilar \\
    	&                    & packing                    &	\\
Covalent& small ($\sim a$)   & determined by the          & similar \\
        &                    & structure of the orbitals  &  \\
Metallic&very large ($\gg a$)& closest packed		  & unfilled valence \\
	&		     &			          & orbitals\\\hline
\end{tabular}
\caption[]{\em{The type of bond that forms between two orbitals is dictated 
largely by the amount that these orbitals overlap relative to their separation
$a$.}}
\label{tab:bond_types}
\end{table}

\subsection{Covalent Bonding}

     Covalent bonding is distinguished as being orientationally
sensitive.  It is also short ranged so that the interaction between
nearest neighbors is of prime importance and that between more distant
neighbors is often neglected.  It is therefore
possible to describe many of its properties using the chemistry
of the constituent molecules.

     Consider a simple diatomic molecule O$_2$ with a single electron and
\beq
H=-\frac{\lap}{2m} -\frac{Ze^2}{r_a}-\frac{Ze^2}{r_b}+\frac{Z^2e^2}{R} 
\eeq
We will search for a variational solution to the the problem of the
molecule ($H\Psi_{mol}=E\Psi_{mol}$), by constructing a variational 
wavefunction from the atomic orbitals $\psi_a$ and $\psi_b$.
Consider the variational molecular wavefunction
\beq
\Psi'=c_a\psi_a+c_b\psi_b
\eeq
\beq
E'=\frac{\int \Psi'^* H \Psi'}{\int \Psi'^* \Psi'} \ge E
\eeq
The best $\Psi'$ is that which minimizes $E'$.  We now define
the quantum integrals
\beq
S=\int \psi_a^*\psi_b \;\;\;H_{aa}=H_{bb}=\int \psi_a^* H\psi_a \;\;\;H_{ab}=\int \psi_a^* H \psi_b \,.
\eeq
Note that $1>S_r>0$, and that $H_{abr}<0$ since $\psi_a$ and $\psi_b$
are bound states [where $S_r={\rm{Re}} S$ and $H_{abr}={\rm{Re}} H_{ab}$]b.  
With these definitions,
\beq
E'=\frac{(c_a^2+c_b^2)H_{aa}+2c_ac_bH_{abr}}{c_a^2+c_b^2+2c_ac_bS_r}
\eeq
and we search for an extremum $\pde{E'}{c_a}=\pde{E'}{c_b}=0$.
From the first condition,$\pde{E'}{c_a}=0$ and after some simplification, and 
re-substitution of $E'$ into the above equation, we get the condition
\beq
c_a(H_{aa}-E')+c_b(H_{abr}-E'S_r)=0
\eeq
The second condition, $\pde{E'}{c_b}=0$, gives 
\beq
c_a(H_{abr}-E'S)+c_b(H_{aa}-E')=0\,.
\eeq
Together, these form a set of secular equations, with eigenvalues
\beq
E'=\frac{H_{aa}\pm H_{abr}}{1\pm S_r}\,.
\eeq

\begin{figure}[htb]
\centerline{\includegraphics[height=3.3in,keepaspectratio,clip=true]{dioxygen.pdf}}
\caption[]{\em{Two oxygen ions, each with charge $Ze$, bind and electron 
with charge $e$.  The electron, which is bound in the oxygen valence 
orbitals will form a covalent bond between the oxygens}}
\label{fig:dioxygen}
\end{figure}
Remember, $H_{abr}<0$, so the lowest energy state is the $+$ state.
If we substitute Eq.~10 into Eqs.~8 and 9, we find that the
$+$ state corresponds to the eigenvector $c_a=c_b=1/\sqrt{2}$;
i.e.\ it is the bonding state.
\beq
\Psi'_{bonding}=\frac{1}{\sqrt{2}}\left(\psi_a+\psi_b\right)
\;\;\;
E'_{bonding}=\frac{H_{aa}+ H_{abr}}{1+ S_r}\,.
\eeq
For the $-$, or antibonding state, $c_a=-c_b=1/\sqrt{2}$.  Thus, 
in the bonding state, the wavefunctions add between the atoms,
which corresponds to a build-up of charge between the oxygen 
molecules (cf.\ Fig.~\ref{fig:dioxygen}).  In the antibonding state, there is a 
deficiency of charge between the molecules.

Energetically the bonding state is lower and if there are two electrons, 
both will occupy the lower state (ie., the molecule gains energy by 
bonding in a singlet spin configuration!).  Energy is lost if there are 
more electrons which must fill the antibonding states.  Thus the covalent 
bond is only effective with partially occupied single-atomic orbitals.  
If the orbitals are full, then the energy loss of occupying the 
antibonding states would counteract the gain of the occupying the bonding 
state and no bond (conventional) would occur.  Of course, in reality it 
is much worse than this since electronic correlation energies would also 
increase.

     The pile-up of charge which is inherent to the covalent bond
is important for the lattice symmetry.  The reason is that the
covalent bond is sensitive to the orientation of the orbitals.
For example, as shown in Fig.~\ref{spbonding} an S and a P$_\sigma$
orbital can bond if both are in the same plane; whereas an S and a 
P$_\pi$ orbital cannot.  I.e., covalent bonds are directional!
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{spbonding.pdf}}
\caption[]{\em{A bond between an S and a P orbital 
can only happen if the P-orbital is oriented with either its plus
or minus lobe closer to the S-orbital.  I.e., covalent bonds are
directional!}}
\label{spbonding}
\end{figure}  
An excellent example of this is diamond (C) in which the
(tetragonal) lattice structure is dictated by bond symmetry. 
However at first sight one might assume that C with a 1s$^2$2s$^2$2p$^2$
configuration could form only 2-bonds with the two electrons in the
partially filled p-shell.  However, significant energy is gained from 
bonding, and 2s and 2p are close in energy (cf. Fig.~\ref{potential})
so that sufficient energy is gained from the bond to promote one of the 
2s electrons.  A linear combination of the 2s 2p$_{\rm{x}}$, 2p$_{\rm{y}}$
and 2p$_{\rm{z}}$ orbitals form a sp$^3$ hybridized state, and
C often forms structures (diamond) with tetragonal symmetry.  

Another example occurs most often in transition metals where the d-orbitals 
try to form covalent bonds (the larger s-orbitals usually form metallic
bonds as described later in this chapter).  For example, consider
a set of d-orbitals in a metal with a face-centered cubic (fcc)
structure, as shown in Fig.~\ref{dorbs}.  The $xy$, $xz$, and $yz$ orbitals
all face towards a neighboring site, and can thus form bonds
with these sites; however, the $x^2-y^2$ and $3z^2-r^2$ orbitals
do not point towards neighboring sites and therefore do not participate
in bonding.  If the metal had a simple cubic structure, the situation
would be reversed and the $x^2-y^2$ and $3z^2-r^2$ orbitals, but not
the $xy$, $xz$, and $yz$ orbitals, would participate much in the bonding.
Since energy is gained from bonding, this energetically favors an
fcc lattice in the transition metals (although this may not be the
dominant factor determining lattice structure).
\begin{figure}[htb]
\centerline{\includegraphics[height=3.0in,keepaspectratio,clip=true]{dorbs.pdf}}
\caption[]{\em{In the fcc structure, the $xy$, $xz$, and $yz$ orbitals
all face towards a neighboring site, and can thus form bonds
with these sites; however, the $x^2-y^2$ and $3z^2-r^2$ orbitals
do not point towards neighboring sites and therefore do not participate
in bonding}}
\label{dorbs}
\end{figure}  

     One can also form covalent bonds from dissimilar atoms, but
these will also have some ionic character, since the bonding electron 
will no longer be shared equally by the bonding atoms.

\subsection{Ionic Bonding}


     The ionic bond occurs by charge transfer between dissimilar atoms 
which initially have open electronic shells and closed shells afterwards.
Bonding then occurs by Coulombic attraction between the ions.  The energy 
of this attraction is called the cohesive energy.  This, when added to the
ionization energies yields the energy released when the solid is formed
from separated neutral atoms (cf.\ Fig.~\ref{nacl}).  The cohesive energy is 
determined roughly by the ionic radii of the elements.  For example, for
NaCl
\beq
E_{cohesive}=\frac{e^2}{a_o}\frac{a_o}{r_{Na}+r_{Cl}}=5.19 eV\,.
\eeq
Note that this does not agree with the experimental figure given
in the caption of Fig.~\ref{nacl}.  This is due to uncertainties
in the definitions of the ionic radii, and to oversimplification
of the model.  However, such calculations are often sufficient to
determine the energy of the ionic structure (see below).  Clearly, ionic
solids are insulators since such a large amount of energy $\sim 10$eV 
is required for an electron to move freely.
\begin{figure}[htb]
\centerline{\includegraphics[height=4.0in,keepaspectratio,clip=true]{nacl.pdf}}
\caption[]{\em{The energy per molecule of a crystal of sodium chloride
is (7.9-5.1+3.6)~eV=6.4eV lower than the energy of the separated neutral
atoms.  The cohesive energy with respect to separated ions is 7.9eV
per molecular unit.  All values on the figure are experimental.  This
figure is from Kittel.}}
\label{nacl}
\end{figure}  

     The crystal structure in ionic crystals is determined by balancing 
the needs of keeping the unlike charges close while keeping like charges 
apart.  For systems with like ionic radii (i.e.\ CsCl, 
$r_{Cs}\approx1.60\AA$, $r_{Cl}\approx1.81\AA$) 
this means the crystal structure will be the closest unfrustrated packing.  
Since the face-centered cubic (fcc) structure is frustrated (like charges 
would be nearest neighbors), this means a body-centered cubic (bcc) structure 
is favored for systems with like ionic radii (see Fig.~\ref{3dcubic}).  For 
systems with dissimilar 
radii like NaCl (cf.\ Fig.~\ref{nacl}), a simple cubic structure 
is favored.  This is because the larger Cl atoms requires more room.    
If the cores approach closer than their ionic radii, then
since they are filled cores, a covalent bond including both
bonding and anti-bonding states would form.  As discussed before,
Coulomb repulsion makes this energetically unfavorable.

\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{3dcubic.pdf}}
\caption[]{\em{Possible salt lattice structures.  In the simple cubic
and bcc lattices all the nearest neighbors are of a different species
than the element on the site.  These ionic lattices are unfrustrated.  
However, it not possible to make an unfrustrated fcc lattice using
like amounts of each element.}}
\label{3dcubic}
\end{figure}  

\subsubsection{Madelung Sums}

     This repulsive contribution to the total energy requires a
fully-quantum calculation.  However, the attractive Coulombic
contribution may be easily calculated, and the repulsive potential
modeled by a power-law.  Thus, the potential between any two
sites i and j, is approximated by 
\beq
\phi_{ij}=\pm \frac{e^2}{r_{ij}}+\frac{B}{r_{ij}^n}
\eeq
where the first term describes the Coulombic interaction and the
plus (minus) sign is for the potential between similar (dissimilar) 
elements.  The second term heuristically describes the repulsion due to 
the overlap of the electronic clouds, and contains two free parameters
$n$ and $B$ (Kittel, pp.~66--71, approximates this heuristic term with 
an exponential, $B\exp{(-r_{ij}/\rho)}$, also with two free parameters).  
These are usually 
determined from fits to experiment.  If $a$ is the separation of 
nearest neighbors, $r_{ij}=ap_{ij}$, and their are $N$ sites in the 
system, then the total potential energy may be written as 
\beq
\Phi=N\Phi_i=N\leb-\frac{e^2}{a}\sum_{i\neq j}\frac{\pm}{p_{ij}}
+\frac{B}{a^n}\sum_{i\neq j}\frac{1}{p_{ij}^n}\rib\,.
\eeq
The quantity $A=\sum_{i\neq j}\frac{\pm}{p_{ij}}$, is known as the
Madelung constant.  $A$ depends upon the type of lattice only
(not its size).  For example $A_{NaCl}=1.748$, and $A_{CsCl}=1.763$.
Due to the short range of the potential $1/p^n$, the second term
may be approximated by its nearest neighbor sum.

\subsection{Metallic Bonding}

	Metallic bonding is characterized by at least some long ranged and 
non-directional bonds (typically between s orbitals), closest packed lattice 
structures and partially filled valence bands.  From the first characteristic, 
we expect some of the valance orbitals to encompass many other lattice sites,
as discussed in Fig.~\ref{metallicbonds}.
\begin{figure}[htb]
\centerline{\includegraphics[width=6.0in,keepaspectratio,clip=true]{metallicbonds.pdf}}
\caption[]{\em{In metallic Ni (fcc, 3d$^8$4s$^2$), the 4s and 3d bands 
(orbitals) are 
almost degenerate (cf. Fig.~\ref{potential}) and thus, both participate
in the bonding.  However, the 4s orbitals are so large compared to the
3d orbitals that they encompass many other lattice sites, forming 
non-directional bonds.  In addition, they hybridize weakly with 
the d-orbitals (the different symmetries of the orbitals causes 
their overlap to almost cancel) which in turn hybridize weakly with 
each other.  Thus, whereas the s orbitals form a broad metallic
band, the d orbitals form a narrow one.}}
\label{metallicbonds}
\end{figure}  
Thus, metallic bonds lack the directional sensitivity of the covalent bonds
and form non-directional bonds and closest packed lattice structures
determined by an optimal filling of space.  In addition, since the
bands are composed of partially filled orbitals, it is possible
to supply a small external electric field and move the valence electrons
through the lattice.  Thus, metallic bonding leads to a relatively high
electronic conductivity.  In the transition metals (Ca, Sr, Ba) the d-band 
is narrow, but the s and p bonds are extensive and result in conduction.  
Partially filled bands can occur by bond overlap too;  ie., in Be and Mg 
since here the full S bonds overlap with the empty p-bands.

\subsection{Van der Waals Bonds}

      As a final subject involving bonds, consider solids formed
of Noble gases or composed of molecules with saturated orbitals. 
Here, of course, there is neither an ionic nor covalent bonding
possibility.  Furthermore, if the charge distributions on the atoms were 
rigid, then the interaction between atoms would be zero, because 
the electrostatic potential of a spherical distribution of electronic 
charge is canceled outside a neutral atom by the electrostatic potential of 
the charge on the nucleus.   Bonding can result from small quantum fluctuations
in the charge which induce electric dipole moments.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{vanderwaals.pdf}}
\caption[]{\em{Noble gasses and molecules with saturated orbitals
can form short ranged van der Waals bonds by inducing fluctuating electric 
dipole moments in each other.  This may be modeled by two harmonic 
oscillators binding a positive and negative charge each.}}
\label{vanderwaals}
\end{figure}  

As shown in Fig.~\ref{vanderwaals} we can model the constituents as
either induced dipoles, or more correctly, dipoles formed of harmonic 
oscillators.  Suppose a quantum fluctuation on 1 induces a dipole moment
$\p_1$.   Then dipole 1 exerts a field
\beq
\E_1= \frac{3\nn(\p_1\cdot\nn)-\p_1}{r^3}
\eeq
which is felt by 2, which in turn induces a dipole moment 
$\p_2\propto E_1\propto 1/r^3$.  This in turn, generates a dipole field $E_2$
felt by 1 $\propto \p_2/r^3\propto 1/r^6$.  Thus, the energy of the interaction
is very small and short ranged.
\beq
W=-\p_1\cdot \E_2 \propto 1/r^6
\eeq


\subsubsection{Van der Waals-London Interaction}

Of course, a more proper treatment of the van der Waals interaction
should account for quantum effects in induced dipoles modeled as
harmonic oscillators (here we follow Kittel).

     As a model we consider two identical linear harmonic
oscillators 1 and 2 separated by R.  Each oscillator bears
charges $\pm e$ with separations $x_1$ and $x_2$, as shown in  
Fig.~\ref{vanderwaals}.  The particles oscillate along the x axis
with frequency $\omega_0$ (the strongest optical absorption line of the 
atom), and momenta $\P_1$ and $\P_2$. If we ignore
the interaction between the charges (other than the self-interaction
between the dipole's charges which is accounted for in the harmonic 
oscillator potentials),  then the Hamiltonian of the system is
\beq
H_0=\frac{P_1^2+P_2^2}{2m} + \frac12 m\omega_0^2 (x_1^2+x_2^2)\,.
\eeq
If we approximate each pair of charges as point dipoles, then
they interact with a Hamiltonian
\beq
H_1\approx \frac{-3(\p_2\cdot\nn)(\p_1\cdot\nn) + \p_1\cdot\p_2}
{\left|\x_1+\R-\x_2\right|^3} = -\frac{-2p_1p_2}{R^3}=-\frac{2e^2 x_1x_2}{R^3}
\,.
\eeq

     The total Hamiltonian $H_0+H_1$ can be diagonalized a normal 
mode transformation that isolates the the symmetric mode (where both
oscillators move together) from the antisymmetric one where they move
in opposition
\beq
x_s=(x_1+x_2)/\sqrt{2}\;\;\;\;x_a=(x_1-x_2)/\sqrt{2}
\eeq
\beq
P_s=(P_1+P_2)/\sqrt{2}\;\;\;\;P_a=(P_1-P_2)/\sqrt{2}
\eeq
After these substitutions, the total Hamiltonian becomes
\beq
H= \frac{P_s^2+P_a^2}{2m} 
+\frac12\left(m\omega_0^2-\frac{2e^2}{R^3}\right) x_s^2
+\frac12\left(m\omega_0^2+\frac{2e^2}{R^3}\right) x_a^2
\eeq
The new eigenfrequencies of these two modes are then
\beq
\omega_s=\left( \omega_0^2-\frac{2e^2}{mR^3}\right)^{1/2}
\;\;\;\;\;
\omega_a=\left( \omega_0^2+\frac{2e^2}{mR^3}\right)^{1/2}
\eeq

     The zero point energy of the system is now 
\beq
E_0=\frac12\hbar(\omega_s+\omega_a)\approx 
\hbar\omega_0\left(1-\frac14\left(\frac{2e^2}{m\omega_0^2R^3}\right)^2
+\cdots\right)
\eeq
or, the zero point energy is lowered by the dipole interaction
by an amount
\beq
\Delta U \approx \frac{\hbar\omega_0}{4}\left(\frac{2e^2}{m\omega_0^2R^3}\right)^2
\eeq
which is typically a small fraction of an electron volt.


     This is called the Van der Waals interaction, known also as
the London interaction or the induced dipole-dipole interaction. 
It is the principal attractive interaction in crystals of inert
gases and also in crystals of many organic molecules.  The
interaction is a quantum effect, in the sense that  $\Delta U\to 0$ as 
$\hbar\to 0$.  
\edo
