Condensed Matter Physics: Lecture 7 (Normal)
In the last lecture we encountered the Kronig-Penney model, a simple onedimensional system in which the Schrödinger equation contains a periodic
square-well potential. While you are familiar with the solutions of the
individual square-well potential, the periodic case is significantly different.
Today’s computer exercises are designed to illustrate some of those
differences, in particular, the development of energy bands. The picture
below shows the Kronig-Penney model and the solutions of the Schrödinger
equation in the two regions of the system.
As we saw previously, by applying both the continuity condition and Bloch’s
theorem we arrive at a somewhat intimidating expression at the top of the
right-hand slide.
To make life easier, we take the delta-function limit in which the area of the
barrier was set to some constant value before taking the limit V0→∞ and t→0.
This procedure greatly simplifies the right-hand side, yet leaves most of the
important Physics. We thus have the comparatively simple equation
which describes the relationship between the Bloch wave-vector k and the
energy E (the latter is related to Q as per the left-hand slide). The crucial
point to realise is that the expression on the RHS can sometimes fall outside
of the range -1 to +1, and thus there are particular energies E for which the
equation cannot be solved. These “forbidden” energies are central to
understanding the physics of metals, insulators and semiconductors.
Finally, we comment on the quantity µ which appears in the equation above.
This number is directly proportional to the area of the barrier (see right-hand
slide), and thus is a measure of the strength of the barrier which separates
adjacent square-wells. You will consider the effect of varying µ all the way
from zero (corresponding to no barrier at all), up to ∞ (corresponding to
isolated, non-interacting atoms). For the intermediate cases a variety of
interesting behaviour is observed, all of which is very useful for understanding
real solid-state systems.
In today’s exercises you will use a pre-written MATLAB function which will
help you explore the Kronig-Penney model in the delta-function limit.
The Windows computers should automatically launch the MATLAB program.
If this doesn’t happen, launch MATLAB from the Start menu. Once MATLAB
is running, change the current directory to “E:\CMP” using the dialog box at
top of screen. In this directory you will find the MATLAB function delta.m,
which is shown below. The program is fairly easy to understand and includes
some comments. Spend a couple of minutes convincing yourself that the
code is equivalent to the expression on the previous page.
During the class you will need to edit delta.m, and so you need to make a
copy of the file in E:\CMP. In the E:\CMP directory, create a directory of your
own (perhaps using your surname or given name), and copy delta.m into that
directory. Now change the current MATLAB directory to this location.
function delta(mu)
% Disable error messages
warning off
% Choose sensible values for a and Emax
a=2;
% in Angstrons
Emax=100;
% in eV
% Define some fundamental constants
hbar=1.055e-34;
mass=9.11e-31;
% Convert eV and Angstroms into SI units
a=a * 1e-10;
Emax=Emax * 1.6e-19;
% Create a large vector of energies ranging from 0 to Emax
E=linspace(0,Emax,1e5);
% Compute the quantity Q and the right-hand-side
Q=sqrt(2*mass*E)/hbar;
rhs=cos(Q*a) + mu*sin(Q*a)./(Q*a);
% Plot the rhs vs Energy (using eV for the latter)
subplot(2,1,1)
plot(E/1.6e-19,rhs)
xlabel('Energy (eV)')
ylabel('Right-hand side')
title(['Allowed and Forbidden Solutions for mu=' num2str(mu)])
%axis([-Inf +Inf -1 1])
% only plot y-values between -1 and +1
% Compute and plot the dispersion relation
ka=acos(rhs);
subplot(2,1,2)
plot(ka,E/1.6e-19,-ka,E/1.6e-19)
xlabel('ka')
ylabel('Energy (eV)')
title('Dispersion relation')
axis([-3.5 3.5 0 +Inf])
Part 1: The Free-Electron Limit
To begin with, type ‘delta(0)’ at the MATLAB prompt and hit return. This will
call the function with the value mu=0, which corresponds to no barrier at all
between adjacent wells.
Now take a close look at the two plots:
•
The upper graph is the plot of the RHS in the “green equation”. In the
case of mu=0 the RHS is particularly simple, as the sin(Qa)/Qa term
vanishes. In this (unusual) instance the RHS always falls in the range 1 to +1, and thus we have solutions for all energies.
•
The lower graph shows the relationship between the quantity ‘ka’ and
the energy E. It is obtained by taking the inverse cosine, which you
can check for yourself by noting the ‘acos’ function in the source code.
Questions:
1. In the space below, quickly sketch the dispersion relation (the plot of E∝k2)
for the free-electron gas as covered in earlier lectures.
2. Explain the appearance of the dispersion relation in our calculation using
mu=0. What is the significance of the Bloch phase-factor exp[ika]?
Hint: recall that ka is only uniquely defined over an interval of length 2π.
Part 2: The Nearly-Free Electron Regime
Now consider some finite values of mu. This corresponds to have a small, yet
finite, interaction between the electrons and the positively-charged ion cores.
Begin by considering mu=0.1, ie. type ‘delta(0.1)’. Try a few other values,
such as 0.2, 0.5 and 1. Take a close look at the upper panel which plots the
the RHS. The term involving sin(Qa)/Qa no longer vanishes, and as a
consequence the RHS sometimes makes excursions outside of the range -1
to +1. As a result there some energies for which a Bloch-vector k cannot be
found. Within such energy ranges the crystal contains no solutions.
To clarify the regions over which no solutions exist, uncomment (ie. remove
the percent sign) from the ‘axis’ command immediately following the first ‘plot’
command. Rerun the function and note the forbidden energies.
Questions:
1. Using mu=1, sketch the dispersion relation below for the first three ‘energy
bands’. Note that MATLAB draws vertical lines which bridge the forbidden
region. To correct this would require some complex coding, so we won’t
worry about that today.
2. Using the zoom tool (the magnifying glass in the Figure window), take a
close look at what is happening at the points k=+/- π/a. For the free-electron
case (mu=0) a sharp discontinuity (and positive gradient) is present at such
points. Sketch and describe what you find in this instance (mu=1).
Part 3: The Tight-Binding Regime
As the value of mu is increased, the strength of the barrier which separates
adjacent wells becomes stronger. This means the behaviour of the system no
longer resembles the free-electron gas (which was a useful reference point
when describing the NFE regime). In the case of large mu our system is more
atomic-like, with the electrons being more “tightly bound” to their local well.
Explore the effect of increasing mu: e.g. try 5, 10, 20, 100.
Questions.
1. Comment on the ‘shape’ of the allowed solutions. Would you say that the
dispersion relation still resembles a quadratic?
Can you guess the mathematical form which describes the energy bands?
Hint: think about fundamental trigonometric functions
2. In what sense is the behaviour at the band-edges (k=+/- π/a), the same as
in the NFE regime.
3. Take a close look at the upper panel which plots the RHS expression. How
is this graph qualitatively different to the NFE case (with small mu) seen
earlier? Have a look with and without the ‘axis’ clipping of the y-axis to
appreciate what is going on.
4. Recalling that the vertical lines in the dispersion relation are a MATLAB
artefact, sketch the dispersion relation for mu=10 in the box below.
Part 4: The Free-Atom Limit
Now consider the case of a very large value of mu (e.g. 1000). The barrier
between adjacent wells is effectively infinite, and so the energy levels are
those of isolated atoms (ie just a single square well potential).
Note that only discrete energies are allowed, and that the Bloch wave-vector k
is no longer a good quantum number (i.e. it doesn’t tell us anything, as all the
states are degenerate).
Questions:
1. Again neglecting the vertical lines, sketch the “dispersion relation” for
mu=1000 in the box below.
2. Using the upper panel for accuracy, write down the exact energies for the
three solutions shown. The energies are related by particular ratios, and can
be expressed as a simple relationship involving the band-index n. Write down
the ratios and determine the relationship.
Part 5: Bandstructure as a function of mu
Finally, copy the file ‘loop.m’ from E:\CMP, and run the m-file by typing ‘loop’
at the MATLAB prompt. This m-file is very simple: it repeatedly calls the
function ‘delta’ using a list of values of mu which are evenly spaced (on a
logarithmic scale) between 10-1 and 103.
We thus see the complete evolution of the system from free-electron bands
when the barrier area is zero, all the way through to free-atom solutions when
the barrier is infinite.