
In the free electron gas model an
electron had the mass m_{e} (always
written straight and not in italics because it is not a variable but a
constant of nature (disregarding relativistic effects for the moment), and from
now on without the subscript _{e}) – and that was all to
it. 


If a force F acts on it, e.g. via
an electrical field E, in classical
mechanics Newtons laws applies and we can write 


F 
= – e · E
= m · 
d^{2}r
dt^{2} 




With r = position vector of
the electron. 


An equally valid description is possible using
the momentum p which gives us 




Quantum
mechanics might be different from classical mechanics, so let's see
what we get in this case. 


The essential relation to use is the identity of
the particle velocity with the the
group velocity
v_{group} of the wave package that describes the particle
in quantum mechanics. The equation that goes with it is 


v_{group} 
= 
1

· Ñ_{k} E(k) 




Let's see what we get for the free electron gas
model. We had the following
expression for the energy of a quasifree particle: 


E(k) = E_{kin} 
= 
^{2} · k^{2}
2m 




For v_{group} we then
obtain 


v_{group} 
= 
1
_{ } 
·
_{k} 
^{2} · k^{2}
2m^{ } 
= 
·
k
m 
= 
p
m 
= 
v_{classic} 




Since k was
equal to the momentum
p = mv_{classic} of the particle, we have
indeed v_{group} = v_{classic} =
v_{phase} as it should be. 


In
other words: As long as the E(k) curve
is a parabola, all the energy may be
interpreted as kinetic energy for a
particle with a (constant) mass m. 


Contrariwise, if the dispersion curve is not a parabola, not all the energy is kinetic energy
(or the mass is not constant?). 

How does this apply to an electron in
a periodic potential? 


We still have the wave vector
k, but
k is no longer identical to
the momentum of an electron (or hole), but is considered to be a
crystal
momentum. 


E(k) is no longer a
parabola, but a more complicated function. 

Since we usually do not know the
exact E(k) relation, we seem to be stuck. However,
there are some points that we still can make: 


Electrons at (or close to) the Brillouin zone edge in
each band are diffracted and form standing waves, i.e. they are described by
superpositions of waves with wave vector k and
–k. Their group
velocity is necessarily close to zero! 


This implies that Ñ_{k}E(k) at the
BZ edge must be close to zero too, which demands that the dispersion curve is
almost horizontal at this point. 


The most important point is: We are not
interested in electrons (or holes) far away from the band edges. Those
electrons are just "sitting there" (in kspace)
and not doing much of interest; only electrons and holes at the band edges
(characterized by a wave vector k_{ex})
participate in the generation and recombination processes that are the hallmark
of semiconductors. 

We are therefore only interested in the
properties of these electrons and holes – and consequently only that part of the
dispersion curve that defines the maxima or
minima of the valence band or conductance
band, respectively, is important. 


The thing to do then is to expand the
E(k) curve around the points
k_{ex} of the extrema into a Taylor series,
written, for simplicities sake, as a scalar equation and with the terms after
k^{2} neglected. 


E_{n}(k)
= E_{V,C} + k · 
¶E_{n}
¶k_{ } 
÷
÷ 
k_{ex} 
+ 
k^{2}
2 
· 
¶^{2}E_{n}
¶k^{2} _{ } 
÷
÷ 
k_{ex} 
+ .... 




Since we chose the extrema of the dispersion
curve, we necessarily have 


¶E_{n}
¶k_{ } 
÷
÷ 
k_{ex} 
= 
0 

E_{n}(k_{ex}) =
E_{V} or
E_{C} 




i.e. we are looking at the top of the valence and
the bottom of the conductance band. 

This leaves us with 


E_{n}(k) =
E_{V,C} + 
k^{2}
2 
· 
¶^{2}E_{n}
¶k^{2}_{ } 
÷
÷ 
k_{ex} 



If we now look at the conduction band
and consider only the deviation from its
bottom (the zero point of the energy scale usually is at the top of the valence
band), we have the same quadratic relation in k
as for the free electron gas, provided we change the definition of
the mass as follows: 


¶^{2}E_{C}
¶k^{2}_{ } 
÷
÷ 
k_{ex} 
=: 
^{2}
m^{*} 




which permits to rewrite the Taylor expansion
for the conduction band as follows 


E_{C}(k) –
E_{g} = 
^{2}
2m^{*} 
· k^{2} 




And this is the same form as the dispersion
relation for the free electron
gas! 

However, since ¶^{2}E_{V,C}/¶k^{2} may have arbitrary values, the effective mass m* of the quasifree
particles will, in general, be different from the regular electron rest mass m. 


We therefore used the symbol
m^{*} which we call the effective mass of the carrier and write it in
italics, because it is no longer a constant
but a variable. It is defined by 


m^{*} = 
^{2} · 
1^{ }
¶^{2}E_{n}/¶k^{2} _{kex} 




The decisive factor for the effective mass is
thus the curvature of the dispersion curve at the
extrema, as expressed in the second derivative. Large curvatures (=
large second derivative = small radius of
curvature) give small effective masses, small curvatures (= small second
derivative = large radius of curvature)
give large ones. 

Let's look at what we did in a simple
illustration and then discuss what it all means. 





Shown is a band diagram not unlike Si. The
true dispersion curve has been approximated in the extrema by the parabola
resulting from the Taylor expansion (dotted lines). The red ones have a larger
curvature (i.e. the radius of an inscribed circle is small); we thus expect the
effective mass of the electrons to be smaller than the effective mass of the
holes. 

The
effective mass has nothing to do with a real mass; it is a
mathematical contraption. However, if we know the dispersion curves (either
from involved calculations or from measurements), we can put a number to the
effective masses and find that they are not too different from the real
masses. 


This gives a bit of confidence to the following
interpretation (which can be fully justified theoretically): 


If we use the effective mass
m* of electrons and holes instead of their real mass m, we
may consider their behavior to be identical to that of electrons (or holes) in
the free electron gas model. 

This applies in particular to their response to forces. In this
case, the deviation from the real mass takes care of the influence of the
lattice on the movement of the particle. 

Taken to the extremes, this may even
imply zero or negative effective masses.
(Yes, even zero effective mass is possible, but this is a special case of its
own and will not be discussed here; for example, graphene has some
kpoints where m* = 0.) 


As one can see from the dispersion
curves, the effective mass of the electrons in the valence band,
m*_{V}, is always negative. (This means that a force in
+x direction will cause such an electron to move in
–x direction.) 


Since this is counterintuitive,
usually one doesn't consider the electrons in the valence band, but just the
unoccupied places – i.e., the holes; however, in their proper definition
(which we have disregarded so far), holes have a positive mass by setting 





Note (again) that "holes"
only exist in the valence band; this is due to their proper definition as
quasiparticles contributing to the electrical conductivity. Thus, all other
empty states – in the conduction band, or on additional levels (e.g. of
dopant atoms) – are not holes; this
is clear since they do not contribute to the electrical conductivity. 

We will not go into
more detail but give some (experimental) values for effective masses: 

Holes: m*/m 
IV; IVIV 
IIIV 
IIVI 
IVVI 
C 
0.25 
AlSb 
0.98 
CdS 
0.80 
PbS 
0.25 
Si 
0.16
(0.49) 
GaN 
0.60 
CdSe 
0.45 
PbTe 
0.20 
Ge 
0.04 (0.28) 
GaSb 
0.40 
ZnO 
? 


SiC
(a) 
1.00 
GaAs 
0.082 
ZnS 
? 




GaP 
0.60 






InSb 
0.40 






InAs 
0.40 






InP 
0.64 





Electrons: m*/m 
IV; IVIV 
IIIV 
IIVI 
IVVI 
C 
0.2 
AlSb 
0.12 
CdS 
0.21 
PbS 
0.25 
Si 
0.98
(0.19) 
GaN 
0.19 
CdSe 
0.13 
PbTe 
0.17 
Ge 
1.64
(0.082) 
GaSb 
0.042 
ZnO 
0.27 


SiC
(a) 
0.6 
GaAs 
0.067 
ZnS 
0.40 




GaP 
0.82 






InSb 
0.014 






InAs 
0.023 






InP 
0.077 





