
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 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 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 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 
   

