Physics 322a-2007 Solid State Physics Walter F. Smith Haverford College
Assignment 10 Due: Friday, Dec. 7 at 4 pm
Reading: Finish Ch. 13 if you’ve not already done so. Also read Ch. 14 and Ch. 15.
Assigned exercises (group problems unless otherwise
noted):
10A.
Individual Problem. Resistance of an
atomic wire In my research group, we
used to study one-atom-wide atomic chains which self assemble when a
sub-monolayer of tin is deposited onto a carefully prepared silicon
surface. Shown here are two images from
the paper that inspired our work (A. A. Baski, C. F. Quate, and J. Nogami,
Phys. Rev. B. 44, 11167 (1991)).
At left is an image taken 
by Scanning Tunneling Microscopy. The bright lines slanting from lower left to
upper right are the chains of tin atoms; each “pearl” in the chain is actually
a “dimer” (i.e. a pair of tin atoms).
The lines running from upper left to lower right are the rows of
underlying silicon atoms. At right is a
top-view schematic model for this structure; the tin atoms are shown in black,
the surface silicon atoms in grey, and the next-layer-down silicon atoms in
white. Tin has four valence electrons,
but as shown in the figure, each tin atom forms three covalent bonds (one with
its partner in the dimer and two with top-layer silicon atoms). This means there is one “free” electron per
tin atom. This is shown as a half-filled
orbital in the side view schematic below.
Assume the effect of the silicon is just to hold the tin atoms in place, and that it does not affect the electronic properties of the wire. Imagine we measure the resistance of one of the tin chains, and that this chain is very long. According to the nearly-free electron model, should this system show metallic behavior, or instead semiconducting/insulating behavior? Give a thorough justification for your answer, including a consideration of the band structure of this one-dimensional conductor.
10B. Explain why an hcp structure made of
monovalent atoms always forms a metal, no matter how strong the interaction
with the lattice (but assuming this interaction is still weak enough the
approximations we’ve been discussing apply; this isn’t much of a constraint in
practice). Hints: The hcp structure
is shown here. As you can see, it’s a hexagonal structure with a two-atom
basis. One of the atoms in the basis is at (0,0,0) and the other is at (2/3 ,
1/3 , 1/2). You should be able to argue
that there is no gap formed at the 1BZ boundaries perpendicular to the a3
direction, even when the interactions between the electrons and the lattice are
strong. (For ideas in formulating this
argument, you may wish to review your discussion of the above problem on why
there is no diffraction off the (100) planes of a bcc lattice.) Then figure out where the Fermi surface is
relative to this boundary, in the limit of very strong interaction between the
lattice and the electrons (which is the limit most likely to yield insulating
behavior). Finally, make your argument
about why there must be metallic behavior, at least in the a3
direction. (Since actual metals are made
of many tiny crystals with random orientations, this assures that a bulk sample
with dimensions greater than 0.1 mm will act as a metal in all directions.)
10C. Required energy gaps for a divalent fcc
material to be semiconducting. Recall that the reciprocal lattice for an fcc
lattice with conventional cell of side a is a bcc lattice (in reciprocal
space) with conventional cell of side length 4p / a. Shown here is the 1BZ for such a system. The origin is labelled G. The W point is the point on the 1BZ surface
which is farthest from the origin. The L
point is at the center of one of the pentagonal faces, and the X point is at
the center of one of the square faces, as shown.
a) Draw one conventional cell of the reciprocal lattice using the same orientation as the above figure.
b) Show that
the distance from the origin to the X point is 
.
c) Show that
the distance from the origin to the L point is 
.
The distance
from the origin to the W point is 
. (This is a bit
ickier to show, so I won’t ask you to do it.)
c) For a divalent, one-atom basis system, explain why it is harder to form a semiconductor from a system if the gap goes almost to zero at the W point (rather than having a larger gap there).
d) Assuming the gap at the W point is almost zero, what is the minimum gap at the L point to form a semiconductor for such a system?
e) Assuming the gap at the W point is almost zero, what is the minimum gap at the X point to form a semiconductor?
10D. Size of the gap for the
Nearly Free Electron Model. We’ll do this problem for a one-dimensional
system, and at the end discuss the generalization to 3D. We’ve discussed how, due to diffraction, the
electron states for Bloch wavevectors touching a BZ boundary are standing waves
given by
or
,
where (since k touches a
BZ boundary), 
, where n is an
integer. In this problem, we’ll
calculate the difference in the expectation value of the energy for these
states; this difference is equal to the energy gap.
a. The constant 
is determined by the condition that the electron in one of
these states must be somewhere within the sample. We consider a sample of length L, with x = 0 defined to be the point that is 
to the left of the leftmost lattice point, so that 
is the point that is to the right of the rightmost lattice point, where NL is the number of lattice
points. We will assume that the
wavefunctions go to zero beyond these limits.
Show that the correct normalization is achieved for
b. To start with, let’s assume a very simple form for the PE:
.
Show that, for n =
1 (i.e. at the 1BZ boundary), the
expectation value for the energy of 
lies below the
expectation value of the energy for the empty lattice (i.e. U = 0) by 
. (The potential is
attractive, so 
.) Hint: there are two contributions to the
energy expectation value: KE and PE.
Hopefully it’s obvious that the KE part would be the same for our model
potential as it is for the empty lattice, so you need only consider the
difference in the PE part. Use
Mathematica to evaluate the integral; this will save time in the next part. You may wish to use the “Assuming” command in
Mathematica to tell it that NL is an integer, so that it can simplify your result for you; here’s an
example of how to use this command:
c. Using the same
assumptions as in part b, show that
the expectation value for the energy of 
lies above the expectation
value of the energy for the empty lattice (i.e.
U = 0) by - . Thus, you’ve shown
(as I’ve often claimed in class) that the gap really is split evenly around the
free electron result, and that the total size of the gap (the difference in
energy expectation values for for and ) is simply U1.
d. Now let’s allow the potential energy to vary in a more realistic way. It must still have the periodicity of the lattice, and for simplicity we’ll assume that it’s symmetrical about the origin, so that we need only retain the cos terms in the Fourier expansion, i.e. we can write
We’ll continue to use n = 1 (corresponding to the 1BZ
boundary). You’ve already shown in parts
b and c that the term with j =
1 contributes U1 to the
difference in the expectation values of energy for and . Now show that all
the other terms contribute zero to the difference in energy expectation
values. Hint: For this, you will need to tell Mathematica to assume that more
than one thing is an integer. Here’s an
example of how to do this:
e. It turns out that has the same value for
higher values of n. Show that the energy gap for states with
Bloch wave vector touching the boundary of the nth BZ (i.e. for ) is simply Un
. Hint:
This should be pretty easy for you at this point, using Mathematica.
The three dimensional
generalization of this is that for 
(the 3D Fourier expansion of the PE function), the gap at any
point on the BZ boundary that is the bisector of G is simply 2UG , so long as you are not
close to an intersection of two BZ boundaries.
(Unfortunately, we’re often interested in exactly such intersections,
but the math gets quite icky there.) The
factor of 2 comes from the fact that we’re using the complex exponential
version of the Fourier expansion, instead of the cosine version.
Take home message: The gap size at a BZ
boundary equals the coefficient (in the Fourier expansion of U(r)) corresponding
to the G being bisected by that boundary, so long as you’re not close to the
intersection of two BZ boundaries.
(Presumably, the gap size varies smoothly from one value to the other as
you near the intersection and then move off along the other BZ boundary.)