
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 lets 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) 




Lets see what we get for the free electron gas
model. We had the following
expression for the energy of a 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 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 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  recombination process that is 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 consider only the conduction band and use its bottom as the
zero point of the energy scale, we have the same
quadratic relation in k as for the free electron gas,
provided we change the definition of the mass as follows: 


1
2 
· 
¶^{2}E_{n}
¶k^{2}_{ } 
÷
÷ 
k_{ex} 
:= 
^{2}
m^{*} 




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


E_{C}(k) = 0
+ 
^{2}
2m^{*} 
· k^{2} 
= 
^{2}
2m^{*} 
· k_{ }^{2} 




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

However, since ¶^{2}E_{n}/¶k^{2} may have arbitrary values, the effective mass m* of the particle
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}^{ }
k_{ex} 
· 
1^{ }
¶^{2}E_{n}/¶k^{2} 




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. 