PHYS 7440
Assignment 1
Due Jan. 22, 2003
Reading:
1. For those interested in the general history of solid state theory, see: Out of the Crystal
Maze: Chapters from the History of Solid-State Physics, by Lillian Hoddeson,
Ernest Braun, Jürgen Teichmann and Spencer Weart, Oxford University Press, Inc.
(1992).
2. Read Chapter 6 in Marder.
3. Related material on the Sommerfeld free-electron model is in most of the reserve
books e.g., Ashcroft and Mermin, Chap. 1, 2.
Comments:
Many of the most successful models of phenomena in solids are based upon 'educated
guesses' for the form of the low energy many-body wavefunctions as particular
expansions in terms of single particle wavefunctions. Upon selection of such an
approximate many-body wavefunction, a variational calculation (more on this later) is
applied to determine a 'self-consistent field' Hamiltonian which yields approximate eigen
energies useful in determination of thermal and transport properties. Examples include
the Hartree-Fock Equation and associated determinental wavefunctions, the BCS 'gapequation' and trial wavefunction for superconductivity and the Laughlin wavefunction for
the Fractional Quantum Hall Effect. The first problem involves looking at simple product
wavefunctions.
1. Marder 6.1 (This notation means Problem 1 in Marder's Chapter 6 problems).
2. Marder 6.2
Use whatever program you like (Mathematica is nice) to plot the energy versus electron
number.
3. Marder 6.4
Plot the density of states functions.
4. Marder 6.5
In this problem, determine i) the wavefunctions, ii) eigen energies, iii)the conditions on
wavevector, k, iv)the nature of k-space, v)the appropriate k-space density and vi) the
average energy per particle at zero temperature. What replaces the Fermi sphere in this
case?
Assume that the box is deformed so that the z-dimension is very short, but y=x and
retaining xyz=V. Sketch how the points in k-space are distributed and show how these
states are populated in the ground state.
Problem Set 1
Solutions
Phys. 4340
Spring 1995
This problem set emphasizes some of the simple pictures which we will use for
understanding the structures of real solids. The first two problems consider the simple
picture of a solid being composed of spherical atoms of a give radius. The third problem
deals with a specific pair potential and shows how you can take a simple form and adjust
the parameters to reproduce some of the measured properties of the solid.
Rosenberg 1.1.
The density of b.c.c. iron is 7900 kg-m-3 , and its atomic weight is 56. Calculate the
side of the cubic unit cell and the interatomic spacing.
The name of the game here is to figure out what the b.c.c. structure is and then
determine how large the unit cell must be to get the correct number of iron atoms per
unit volume to match the measured density. I’ll treat this problem in three steps:
1. Determine the weight of an iron atom.
2. Determine the number of atoms in the b.c.c. cubic unit cell.
3. Determine the size of the unit cell necessary to get the correct density.
Weight of an iron atom.
The atomic weight is 56. The atomic weight essentially tells us the number of
protons plus neutrons and these weight about the same amount, namely
1.67x10 -27 kg. Therefore, the atomic weight of an iron atom is:
Weight  56  1.67  10 27 kg
 9.4  10 26 kg
Number of atoms per bcc unit cell.
bcc unit cell from
Rosenberg.
Rosenberg shows the cubic bcc unit cell in
Fig. 1.2 (a) on page 3 of the text. I’ve
reproduced in here. As you can see, the
standard cubic bcc unit cell has atoms located
at the corners of the cube (there are eight
corner locations) and one atom at the center of
the cube where the body diagonals cross.
Remember that the crystal is constructed by
packing a huge number of these units together.
Therefore, you can see that, although the
central atom is entirely within a single cell, the
corner atoms are partially inside several cells which meet at the corner. In fact, each
corner atom is shared by eight cells. Thus, there is 1/8 of each corner atom inside
each cell. Since there are eight corners, the total contribution is one atom for the
corners and one atom at the center or two atoms per bcc cubic unit cell.
Density of bcc iron.
The cubic cell has a side length of L, so that the volume is L3. Therefore, since there
are two atoms per cell, the mass density is:

29.4  1026 kg
L3
Since we are told that   7900 kg - m-3 , then we can solve for L:
29.4  1026 kg
L  




13
 0.288 nm
OK, so the result is that the cell is a bit under 3 Angstroms on a side.
If you look at the structure, you can see that for any given atom, the nearest neighbor
atom is reached by traveling half-way along the body diagonal of the bcc cell. The
atoms located along the edges are somewhat further away. Therefore, the nearest
neighbor interatomic spacing is just half the length of the body diagonal. We can
calculate the body diagonal length via Pythagorus:
1
1
Body diag. length =
2
2
1

2
Face diag. length   Cubic side length 
2

2
2Cubic side length   Cubic side length 
2
2
3
Cubic side length 
2
 0.249 nm

Now here’s the real point as I see it: The atomic number of iron can be determined
through mass spectrometry in the gas phase i.e., it is a property of the single atom
which can be determined independently of any solid state property. Similarly, the
density of the iron is a measurable property. Therefore, we can predict what the atomic
spacing must be like depending upon which crystal structure the solid takes on. In the
case of iron, it actually does turn out to prefer the bcc structure. Furthermore, x-ray
diffraction data confirms that the bcc unit cell in iron has a cubic cell width of
0.287 nm. Thus, the atomic properties allow simple predictions about the solid state.
Rosenberg 1.2.
The density of fcc.gold is 19,300 kg-m-3 , and its atomic weight is 197. Calculate the
separation between the close-packed planes. On the assumption that the atoms may be
thought of as spheres which just touch one another, estimate the atomic radius.
This problem is similar to the last one, but now requires that you spend time thinking
about the fcc structure. As an aside, the bcc, fcc, and the hexagonal close-packed
structures are the crystal structures taken on by the vast majority of elemental solids.
Further, when you generalize the structures so that different atoms can reside around
the lattice points, you find that these structures are chosen by the majority of known
materials.
As Rosenberg explains on pages 3-6, the fcc and the hcp structures can both be
constructed by producing layers of close-packed spheres. You have all probably seen
this type of packing in fruit displays or in piles of cannonballs. Let’s proceed as we did
in 1.1, and use the standard cubic fcc unit cell, determine the number of atoms in the
cell, calculate the cubic side to get the correct mass density, and then figure out the
orientation of the close-packed planes (along the cubic body diagonal) to get the
spacing. Again, Rosenberg has provided the pictures, in this case, Fig. 1.5 on page 5.
For the fcc structure, there are eight atoms in the cube corners as for the bcc case, so
we have one atom total from the corner positions. Next, we have six atoms at the face
centers. In this case, each face atom is shared between only two unit cells, so each
contributes half an atom to the cubic fcc cell. Thus, there are four atoms in the fcc
cubic unit cell, one from the corners and three from the faces.
The gold atomic mass is 3.3x10-25 kg, so the cube has a size of:
43.3  10 25 kg
L  

-3
19,300 kg - m 
13
 0.409 nm
From here on, it’s ‘just’ solid geometry. Now, Fig. 1.5 (c) shows that the close-packed
planes are perpendicular to the body diagonal of the cubic fcc cell. Each close-packed
plane has the atoms arranged at the corners of equilateral triangles. Here’s a picture of
several of the equilaterals within the cubic fcc cell:
Further, the next plane up the stack is
an identical set of equilateral
triangles, but shifted over so that the
atoms lie above the centers of the
triangles in the plane below.
Therefore, the basic unit is a threesided pyramid constructed of
equilateral triangles. I’ve pulled out
this basic pyramid in the picture, so
that you can see it.
In any case, if you look at the way the
equilaterals are packed into the cube,
you can convince yourself that the
side of one of these triangles is just
the same length as half of the face
diagonal. Therefore, any edge of the
pyramid is of length 2L 2  L 2 .
What we want is the height of the top
atom of the pyrimid above the floor. Again, you can get the length by construction of a
right triangle. In this case, things are straight-forward because all the triangles we care
about have interior angles of 300. We find that the height is found by calculating the
base length of a right triangle with a side of length L 2 6 and a hypotenuse of
L 3 2 2 . Then,
Spacing between planes = L 3
 0.236 nm
If the atoms are spheres, then close packing means that they just touch at the middle of
the side of each equilateral triangle. Thus,
Atomic radius = L 2 2
 0.145 nm
Rosenberg 1.7.
9
The potential energy of a pair of atoms in a crystal is of the form A r  B r when
their separation is r. The equilibrium separation is 0.28 nm and the dissociation
energy is 8x10-19J. Calculate A and B. Find the effective modulus of elasiticity for the
pair of atoms and calculate the force which would be necessary in order to reduce
their spacing by 5%.
Here’re the main points which you should get from this problem: First, once you have
a few pieces of experimental information such as the atomic spacing and the
dissociation energy (the energy necessary to separate the atoms from each other) then
it is possible to take a hypothetical pair potential and adjust its parameters so that it
reproduces the experiment. Second, once you have fitted the pair potential, it is
possible to make predictions of other properties. To the extent that your predictions
work out well, you can assume that the potential is a good description of the bonding
strength of the atoms. We will see that the Lennard-Jones 6-12 potential turns out to
do an excellent job of describing the noble gas solids. The 1-9 potential which
Rosenberg has you think about does a pretty good job of describing solids like NaCl
where bonding is approximately due to Coulombic attraction between ions.
Fitting A and B to data.
First, let’s look at the potential and see what it is that we are trying to do. A sketch
of the function looks like this:
V
1/ r 9
ro
r
1/ r
-Vo
Sketch of the 1-9 pair potential.
This particular pair potential
is designed to describe the
potential energy between ions.
The attractive potential is of
the Coulombic form. The
strongly repulsive short range
piece is supposed to mimic the
repulsion between the electron
clouds due to the Pauli
exclusion principle. Between
the two terms, we see that the
function develops a potential
well.
The potential well causes the
two atoms which interact to reach a ground state in which they are separated by the
distance r0. Furthermore, it takes and energy V0 to pull the atoms apart. We fit the
function to the experimental data by adjusting A and B so that we get the correct
potential well depth and the correct position for the well minimum.
First, we generate an equation relating A and B to r0 by setting the derivative of the
potential to zero (true at the minimum point):
dV
dr
 9
r r 0
A
B
 2 0
10
r0
r0
Therefore, we have a relation between A and B, or we can think of it as an equation
for the equilibrium position.
 A
r0  9 
 B 
1/ 8
or
9A  Br08
Since we have two unknowns, we need a second relation to finally solve the
problem. Of course, the idea is that:
V r0   V0
This equation yields a second relation:
A B
  V0
r09 r0
or
B
A
 V0 r0
r08
We can combine these results to find that:
V0 r09
8
9
B  V0 r0
8
A
We can immediately plug in the experimental values for the equilibrium distance
and the dissociation energy to find:
A  1.06  10 105 J - m 9
B  2.52  10 28 J - m
A better way to look at it and to write it.
OK, so we found the values for A and B. We can press on from here to calculate
further properties, but these numbers don’t really help in understanding the physics.
There is a much better way to write the pair potential, which I believe helps to clean
things up. What you do is take the formal relations we found above between A and
B and the equilibrium position and dissociation energy. Plug those back into the
pair potential and do some minimal algebra and you get a new form for the
potential:
V r  
V0 r0 9
r0 

9
r 
8 
r 

It’s the same potential, still in terms of two parameters, but now we get some
physical insight into what the parameters are! Of course, the equilibrium position,
r0, now appears as a natural length scale for the potential, while the dissociation
energy, V0, appears as a natural energy scale. Much nicer!
Elastic forces.
Now we have a form form the potential, so we can begin predicting other
properties. As we get into the case of solids, we’ll find that elastic properties, sound
wave frequencies and velocities in particular, are among the most important of
these properties which are easiest to get out. Here, we only consider a pair of atoms,
so Rosenberg asks for the ‘elastic’ properties of the pair. The essential concept here
is that for small changes in separation (compared with the size of the equilibrium
spacing) the potential looks approximately parabolic i.e., the potential looks like a
harmonic oscillator potential. Defining the deviation from equilibrium spacing as:
r  r  r0 
what I’m saying is that it should be possible to write the potential as:
V r  r0   V0 
1
2
kr
2
where k is the modulus of elasticity.
The ‘better’ way to write the potential is also a better way to write it if you want to
derive this type of small oscillation approximation. To determine the approximate
potential, let’s do a Taylor expansion to second order around the equilibrium
position:
dV
V r  r0   V r0  
dr
1 d 2V
r 
2 dr2
r r0
r2  Or 3 
r r0
Evaluating the derivatives gives us:
dV
dr

r r 0
V0  r09
r 
9 10  9 02  0
8 
r0 
 r0

and
d 2V
dr2

r r 0
V0  r09
r 
90 11  18 03
8 
r0 
 r0

9
V0
r02
The first result is the standard observation that at the equilibrium spacing, there is
no net force on the system (so the first derivative of the potential is zero). The
second result says that, aside from the numerical factor of 9, the ‘natural’
parameters can only be combined in one way to yield a constant with the units of
the elastic modulus. Anyway, the potential is then:
1 V 
V r  r0   V0  9 02 r 2
2 r0 
so that the elastic modulus is:
k 9
V0
r02
 91.8 J/m 2
To get the force necessary to decrease their spacing by 5%, let’s continue to use our
small deviation approximation for the potential. Then, the force is determined by
the negative first differential of the potential:
F  V
 k r
To reduce the spacing by 5% means that r  0.05  0.28 nm  0.014 nm i.e., that
the required force is 1.29x10-9 N.
As an aside, Rosenberg’s elastisity modulus is defined just like our k, except she
has pulled in the factor of the equilibrium spacing so that she has a new modulus, ,
such that:
F    % length change 
or
  kr0
 2.57  10 8 N
Plotting the potential.
This part is simple once you have written the potential in the way I have above. As I
say, the natural scales of length and energy are the equilibrium spacing and
dissociation energy. Defining a dimensionless distance and potential, respectively
as, x  r r0 and v  V V0 , then the potential function to plot is:
v x  
1 1
1
 9 
9
8 x
x 
Here’s the Mathematica code necessary to plot it:
In[1]:=
v[x_]:=(1/8)((1/x)^9 -9/x)
In[4]:=
Plot[v[x],{x,0.1,6},
PlotRange -> {-2,6},
Frame -> True,
FrameLabel -> {"r/r0","V/V0"}]
Out[4]=
-Graphics-