Physics 7440
Spring 2003
Solutions 8
1. Marder 16.1. Bloch oscillations.
Bloch oscillations, the motion of an electron in real space due to evolution through
the Brillouin zone due to an applied electric field, is one of the simpler predictions of
the semiclassical model of electron dynamics. Many groups have worked to observe
Bloch oscillations. Many have failed. This problem is designed to work out why it's
so hard to see the effect.
a) Marder's discussion of Bloch oscillations concentrates on behavior for a simple
1D band structure. In the case of a cosine band, he integrates the equations of
motion and shows that the electrons will move periodically in k-space and in real
space, with a period given by:
 Bloch 
aeE
Therefore, if you know the characteristic lattice spacing, the electronic charge,
and the electric field, you can work out the period or frequency of oscillation. One
of the reasons that Bloch oscillation would be so useful, if you could observe it, is
that it could be a source of oscillating electric current and associated
electromagnetic radiation. It could be great for making lasers and microwave
sources with frequencies that you can change at will by just changing the applied
dc electric field!
If we are going to observe the Bloch oscillation, it had better happen quickly
enough so that a cycle can finish before the electron scatters. Given some typical
time, , between scattering events, we would like to have:
   Bloch
All you need to do is turn up the electric field until you're in the right parameter
region. Right? So let's see how big a field you need for copper with the suggested
scattering time of 2x10-13 sec and a typical lattice spacing (from the front cover
of Marder) of 0.36 nm.:
E
ae 
 0.36  10
1034 J .sec
9
m 1.6  10
19
C  2  10
13
sec 
 8.7  106 V / m
b) Is this a large field or not? Well, if you want to apply this field to a copper bar that
is one meter long, just go out and find a 10 MeV voltage source... or go buy 10
million D cell batteries and hook them up. No problem. Put this way, it looks like
a large voltage. However, you could also use some of the photolithography tools
we have in the physics building and make a capacitor with parallel plates that are
separated by 1 micron. Then all you need to do is apply 10 volts across the
capacitor to get this field. That doesn't sound too bad.
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Solutions8.1
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Really, you need some criterion to decide whether this electric field is big or
small. In particular, 'big' or 'small' in this case mostly implies something about
whether the field is so large that the semiclassical model is likely to fail. For
example, if the electric field is so large that the band structure calculation is likely
to be incorrect, our semiclassical model cannot be trusted.
Marder asks you to consider the Zener tunneling process. Zener tunneling will
break the semiclassical model because the electron can jump to a new band. Then,
evolution is going to change and ruin the Bloch oscillation. Marder's rough
calculation of the tunneling amplitude is based on the WKB approximation and
yields an amplitude of:
 g
T  exp  
 eE
2m g 

2

You are told to assume a gap of 2eV. Further, you need some criterion for
deciding whether the tunneling probability is large or not. Just because the
function is exponential, let's look for the 'e-fold' point, where the amplitude is 1/e.
Then, we have:
g
2m g
2
eE
or
E
g
e
1
2m g
2
 1.5  1010 V / m
This electric field is similar to the field an electron feels at the Bohr radius around
a proton, so it's of the size of atomic fields. Therefore, we do not expect Zener
tunneling to be a big deal at the necessary 107 V/m of part a).
c) Of course, some of the electrons DO scatter or Zener tunnel, or otherwise fail to
Bloch oscillate. For those electrons, we expect behavior of the typical Ohm's Law
type. Then, the electric field causes a dc current (not an oscillating Bloch current)
and associated energy dissipated per unit volume per second of:
J E
and
Power
J2
J E
  E2
vol

Assume the entire sample did this. Using a typical resistivity of a microohm-cm
for copper, we find:
 E 2  108 107   1022
2
Watt
m3
This result looks like lots of power... is it? Well, let's us a specific heat from
Dulong-Petite to see what the associated temperature rise is like. Each second, we
Physics 7440
Solutions8.2
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put in 1022 joules per cubic meter. That change in energy causes a temperature
rise of:
T 
1
1
 

cV
3nk B
For copper, where the atomic density is8.49x1028 atoms/m3 we find that the
temperature rise in 1 second is:
T 
1
1
 
1022 J / m3  2.8  1015 K
28
3
23
3nkB
3 8.5  10 m 1.38  10 J / K 
Assuming that one 1 electron in a million suffers a collision, the Ohm's Law
current might be a million times smaller than this estimate. Then the power might
be 10-12 of our estimate, and the temperature rise might only be 1000K. Still, you
can tell that a little scattering can quickly destroy your sample at these electric
field strengths.
The point is that although the field looks like it's small enough that Zener
tunneling (and other processes that break the semiclassical model) looks to be
under control, other processes like heating can ruin your experiment.
d) In this section, you let the lattice constant become 10 nm and assume that the
scattering rate is much slower. The required electric field to get a Bloch
oscillation inside the scattering time is:
E
ae 
10  10
1034 J .sec
9
m 1.6  1019 C  3  1010 sec 
 208V / m
Now, this field is much smaller than we found for the case of copper. Associated
with the lower field strength and slower scattering is a better chance to avoid
heating. In fact, many groups have tried using GaAs heterostructures to observe
Bloch oscillations.
e) So, why don't we have Bloch oscillator lasers in every supermarket? One reason
is that the Zener tunneling problem comes back in the semiconductor superlattice.
The essential idea is to build a periodic structure with a lattice constant that is
very long. That causes a decrease in the electric field needed for a given period.
But, WHY does it lead to such a decrease? The reason is that the long lattice
constant means a much smaller Brillouin zone. The zone or range of k-vectors
that you need to evolve through decreases as 1/lattice constant. Big unit cells
mean small Brillouin zones.
Unfortunately, small Brillouin zones also imply small band gaps and increased
Zener tunneling. It's easy to see why by drawing a typical zone for a material with
some simple band structure. Here's a case with two bands:
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Solutions8.3
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E(k)
Eg
-/a
+/a
k
In this case, you only have two bands and the Zener tunneling rate is set by the
known size of the zone and by the single energy gap.
Now think about what happens if you design a periodicity that doubles the size of
the real space unit cell. Then, Brillouin zone decreases to half the linear
dimensions. Further, we can guess at the new band structure by folding the old
structure into the new zone. We get something like:
E(k)
Eg
-/a
-/2a
+/2
a
+/a
k
The original set of bands now yields twice as many bands. Also, the original band
gap is still roughly the same, but there are now two additional gaps that we need
to worry about. In the case of GaAs, you take a 4 angstrom unit cell and introduce
a periodicity every 100 angstroms or so. That means folding the zone back 25
Physics 7440
Solutions8.4
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times! Each folding doubles the number of bands and band gaps. Therefore, we
end up with 50 or so minibands and miniband gaps.
The details of the actual miniband gaps are not being handled, but we can say
that, given some initial energy scale for the band,  BAND , that after all the folding,
the minibands and the miniband gaps roughly add up to the original band scale. If
half the energy is bands and the other half is gaps, we're going to have some
approximate gap as a function of folding that looks like:
 GAP 

 BAND
1
2  number of folds 
 BAND
1
2  aSUPERLATTICE aSEMICONDUCTOR 
Therefore, the gap you need to worry about scales as the inverse of the unit cell
size in the superlattice.
If you go back to our Zener tunneling result, we found that tunneling becomes
important when the WKG exponent becomes too large. In the case of copper, we
found that the Zener effect limits the allowed electric field, but the limitation is
not severe:
g
2m g
2
eE
or
E
g
1
2m g
2
e
 1.5  1010 V / m
If we do the same calculation for 100 angstrom periodicity in GaAs, using a 4
angstrom basic unit cell for the GaAs semiconductor and the same rough scale of
2 V for the gap, we find that the electric field is limited now by:
g
2m g
2
eE
or
1
4
E
 g  4  2m g 100


e  100 
2
 1.2  108 V / m
The point here is that the Zener effect is moving down too. In fact, you can
combine the electric field you need:
E
ae
and the approximate gap scaling:
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Solutions8.5
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g 
Spring 2003
1  BAND aSEMICONDUCTOR
2
a
to bracket the electric field:
 BAND aGaAs
ea
2m BAND aGaAs
E
2
a
ae
One side drops as the inverse of lattice length, but the other side drops as the
length to a higher power. Eventually, for some long lattice parameter, the Zener
effect will catch up.
Finally, you can write the Zener relation at that point as:
g
2m g
eE
2


 BAND aGaAs 2m BAND aGaAs
2
e  ae  a
a
 BAND aGaAs
2m BAND aGaAs
2
a
1
This relationship shows that for a given material with some energy scale and unit
cell size, the Zener tunneling condition will depend upon the scattering rate and
the superlattice periodicity as  a
2. Marder 13.1.
For the diatomic chain, the form for the harmonic potential is:
2
1
1
U harm  K  u1 na  u2 na  G  u1 na  u2 {n  1}a
2 cells
2 cells
bg bg
bg b
g
2
In this form, the notation implies that we are conceptually dividing the chain into
N unit cells, each of which has two atoms. The atoms within the cell are
connected by a spring of strength, K; connections to neighbor cells are via springs
of strength, G.
Also, one atom moves with small deviation, u1, while the second atom in the cell
moves with small motion, u2. Assume a solution that looks like this:
u1  na   A exp ikna  i t 
u2  na   B exp ikna  i t 
Tossing this into the Hamiltonian and again working out the equations under the
harmonic plane-wave ansatz leads to two coupled equations for u1 and u2:
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Solutions8.6
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 M  2  ( K  G )  A   K  Ge  ika  B  0
 K  Geika  A   M  2  ( K  G )  B  0
These have the solutions:
1/ 2
K  G 1  2 2
2 

K  G  2 KG cos ka 

M
M
Thus, our dispersion relation is split into two branches, one acoustic and one optical
branch.
Physics 7440
Solutions8.7
Spring 2003