October 28th, 2010
Class 15
Electrons in a three-dimensional infinite box
For the case of electrons in a box, the finite size box may be not the best way to go considering
the macroscopic dimensions of a metal that can be regarded, without much approximation, as
infinite in size. For a case like that, periodic boundary conditions may be more appropriate.
Thus for the three dimensional cases, periodic boundary conditions will be used.
By following the same mathematical procedure than the one we followed to determine the
number of vibrational modes in a crystal, we found that the solutions for an electron in a three
dimensional box with periodic boundary conditions are:


2 2
    n
2m
 2   2  2  2



2m  x 2 y 2 z 2

   n


Since  k r   exp ik  r 
d
d 2
2
2
 ik x expik x  x and
 k x expik x  x  k x 
dx
dx2
d 2
d
2
2
 k y expik y   y   k y 
 ik y exp ik y  y  and
dy2
dy


d
d 2
2
2
 ik z expik z  z  and
 k z expik z  z   k z 
2
dz
dz
2
2
2
       2 

   n becomes
Thus, 


2m  x 2 y 2 z 2 
2

 k x2  k y2  k z2    n
2m


or
2



k x2  k y2  k z2
2m
2 2
n 
k
2m
where k 2  k x2  k y2  kz2
n 
 


Our relationship between wavevector and wavelength is such that,
2n y
2nx
2nz
kx 
, ky 
, kz 
.
Lz
Ly
Lx
The quantization comes from the boundary conditions  n 0,0,0   n Lx , Ly , Lz 
Using the quantum mechanical definition of p


p k  i k  k  k
1
So the linear momentum of an electron with a wavevector k is ħk and the speed of such an
electron is v= ħk/m
The energy at the Fermi level (now in 3D called Fermi surface), is given by:
F 
 
2
2
kF
2m
where kF is the wavevector at the Fermi surface.
In the same way we did with the vibrational modes, we can count the number of electronic states
and since the solutions are the same, we can see that in the unit of volume there are
3
4 3
 L 

 electronic states in the k space. Thus in a sphere of volume k F (volume of the Fermi
3
 2 
3
V 3
 L  4
3
sphere) there are 2  
  3 k f  2 k F states (counting both spin states). If in the same
3
 2 
volume there are N atoms and each contributes with 1 electron, then that number above must be
equal to N from where
 3 2 N 

k F  
 V 
1
 2  3 2 N 


F 
2m  V 
3
2
3
2
k
   3 N 

and  F  F   
m
 m  V 
1
3
N/V is the electron density (equal to the atomic density assuming each atom contributes with 1
electron).
From the above equation it is useful to define the Fermi temperature TF as the temperature for
which kBTF=F. This temperature has nothing to do with the thermodynamical temperature,
TF~104 is the temperature Tg an electron gas need to be for its thermal energy be kBTg=F.
Density of States
2
 2  3 2 N  3

 , is valid for any value of energy not only for the Fermi energy,
The equation  
2m  V 
in that case Nε represents the number of electrons up to that energy. So inverting the equation we
obtain
V
N  2
3
 2m 
 2 
  
3
2
where N is the number of one electron states (with or without electrons) considering each energy
level as 2 states (one for spin up and the other for spin down) with energy between 0 and  for
any  larger or smaller than the Fermi energy.
so the density of states is
2
dN
V  2m 
D  



d 2 2   2 
3
2

1
2
3
3  V  2m  2  3N




2  3 2   2   2


Heat Capacity
As said before, only a few electrons near the Fermi level contribute to the heat capacity,
electrons with energy much smaller that the Fermi level cannot take advantage of the thermal
energy because energy levels above them are occupied and the Pauli principle does not allow
them to also occupy those energy levels.
Classically, the energy of an electron gas is 3/2kBT per electron, thus each electron contributes to
the heat capacity by a quantity 3/2kB (½ per degree of freedom). In the Drudes model (no Pauli),
all the N electrons contribute to the heat capacity and thus the electronic contribution to the heat
capacity was predicted to be 3/2NkB, or the specific heat capacity 3/2 N/V kB where N/V is the
electron concentration.
Considering the Pauli principle, electrons with energy within kBT of the Fermi energy can absorb
energy, thus only a fraction T/TF can actually contribute to the heat capacity and then
qualitatively the total thermal kinetic energy of electrons is
Uel≈(NT/TF)kBT
leading to a heat capacity of
Cel≈NkB(T/TF)
which is not the exact value but retains the T dependence found in the experiment. At room
temperature, T/TF~0.01 for TF~3x104K which is in the order of a typical value for the Fermi
temperature
For a more quantitative calculation consider the change in energy of the metal when the
temperature increases to a certain temperature T is given by

F
0
0
U    f  , T D d   D d
Multiplying the identity

F
0
0
N   f  , T D d   D d
by F

F
  F f  , T D d    F D d
0
what leads to
0
3
F

0
0
  F D d    F f  , T D d  0
Adding this zero to the equation above and reorganizing we get

F
0
0
U      F  f  , T D d 

 
F
  D d 
F

F
0
F
0
    F  f  , T D d      F  f  , T D d    F   D d
or
F

U 
     f  , T D d   
F
F
F
  1  f  , T D d
0
The first term is the product of the energy gained by one electron when it goes from the Fermi
level to the energy level , times the number of states at the energy  times the probability that
the state has been occupied integrated in all the level above the Fermi level. This is equal to the
total amount of energy invested to populate energy levels about the Fermi level. The second
term is the product of the energy gained by one electron when it goes from the energy level 
below the Fermi level to the Fermi level times the number of states at the energy  times the
probability that the state has indeed been vacated integrated in all the level below the Fermi level,
that is the total amount of energy invested to de-populate energy levels below the Fermi level.
So now the heat capacity can be obtained simply by taken the derivative of the last equation with
respect to T. Noticing that only f() depends on T
dU
df  , T 
Cel 
     F 
D d 
dT  F
dT

df  , T 
df  , T 
D d      F 
D d
  F   
F

dT
0
0
dT
Since we are mostly interested in temperatures around room temperature, then T/TF= kBT/F~0.01
where the density of states can be approximated by the density of states at the Fermi level and
the chemical potential can be considered independent of temperature and ~F
Then

Cel  D F     F 
0
f  , T  
df  , T 
d
dT
1
    F  
exp 
 1
 k BT 
    F  
exp 

 k BT 
df  , T     F 

k BT 2       F   2
dT
 exp 
  1

k
T
B

 

4
    F  
exp 

 k BT 

    F  
ex
2
2


Cel  k B D F  
d


k
TD

x
dx
B
F

2
2

x
k
T
e

1



B




0
 F
 exp     F    1
k BT


 k BT  

x=(-F)/ kBT
2



In order to integrate, we can replace the lower limit with -, which is approximately true for
kB<<F.
Thus

Cel  kB2TD F   x 2

e
ex
x
d  kB2TD F 
2

1
2
1
  2 D F k B2T
3 3
Using that
D  F  
3N
3N

2 F 2k B TF
1
3N
1
T
C el   2
k B2 T   2 Nk B
3 2 k B TF
2
TF
Experimentally Determined Heat Capacity
For very low temperatures, the total heat capacity can be considered the sum of the electron and
phonon contributions, thus
C=T+AT3
Where the electronic contribution is proportional to T and the phonon contribution is
proportional to T3. Dividing by T, we obtain a linear relationship between C/T and T2 with
intercept  and slope A. (see figure 9 chapter 6, in page 145)
The observed value of the intercept does not agree very well with the predicted value and
the reason is that the electron seems not to have the classical mass when in the solid but an
effective mass. Usually an effective mass is defined as
mth  measured 

m  calculated 
The reason for which the apparent mass is not the free electron mass involves three separate
effects
1) The interaction of electrons with positive ions. The resulting mass is called the band
effective mass
5
2) The interaction between electrons and phonons what distorts the lattice that in turn
interact with other electrons increasing their effective mass
3) Electron-electron interaction what also increases the effective mass
Electrical Conductivity
The momentum of a charge is given by mv=ħk, thus its equation of motion in an electric field
can be written as:
dv
dk
F m

 qE (where q=-e for electrons)
dt
dt
qEt
the same for ALL electrons, thus the entire Fermi sphere is moving!

What is consistent with a free charge in an electric field.
What leads to k 
In a real system, electrons are not really free, they interact with the ions that are hopping in place
(phonon) and they interact with imperfections in the crystal. A more appropriate model consists
in assuming that these electrons are free to accelerate just during a time t= after which they just
stop and start over. The overall effect is that these electrons find a steady state at a higher speed
than before the electric field is applied (sort of like a terminal velocity). So the shift in the
position of the Fermi sphere is qE/ħ while the speed increases an amount v=ħk/m= qE/m.
At equilibrium, the average speed of the electron gas is zero, thus v is the drift velocity.
Consider n electrons per unit of volume moving at a speed v, the number of electrons that will
cross a surface of unit area perpendicular to the direction of motion in a time to is equal to the
number of electrons in a cylinder of unit cross section area and length l=vto, thus, in the unit of
time, the total charge crossing a unit are is:
j=nqv=nq2E/m
j is the current density and the above is known as the Ohm’s law where
= nq2/m
is the conductivity and the inverse
= m/nq2
is the resistivity.
Another important property of charge carriers is their mobility which is defined as μq=v/E where
E is the electric field. Carriers with high mobility will pick up speed at small values of electric
field thus leading to larger currents. With this definition
j=nqv=nq μqE and
= nq μq or μq=q/m
The two main sources for resistivity are the electron-phonon interaction (interaction with the ions
that are hopping in their place) and the interaction with impurities and crystal defects. The
former is more relevant at high temperature while the later is relevant at all temperature. If the
6
electric field is then removed, collisions will make the electrons go back to their unbiased speed.
The rate of these collisions is independent from each other, thus
f=fL+fi
where f is the total number of collisions per unit of time fL is the number of collisions with the
lattice (charge-phonon collisions) per unit of time, and fi is the number of collisions with
impurities per unit of times
thus the characteristic time can be obtained as:
1 1 1


 L i
The equation above then indicates that the total number of collisions per unit of time is the sum
of the total number of collisions with the lattice (phonons) per unit of time plus the total number
of collisions with impurities per unit of time) or multiplying by m/ne2 =L+i
Hall Effect
If a magnetic field is applied perpendicular to a current, the electrons will suffer a force
perpendicular to both, the drift velocity and B, of a magnitude
F=qvxB
This force produces an accumulation of charge and an electric field that generates an
electrical force that opposes to the magnetic force. In equilibrium
-eEH=evB or EH=-vB
By defining the Hall coefficient RH as
EH=-RH JB
vB= RH JB or RH=v/J
(notice that RH can be measured by measuring the electric field perpendicular to the current and
it is a property of the material).
Using J=nqv RH=1/nq (where q is negative for electrons)
For Li, monovalent, the measured Hall coefficient at room temperature is -1.7x10-10 m3/C and
the calculated value -1.4x10-10 m3/C. For Al (trivalent) both values are ~ -0.3x10-10 m3/C. But
for Zinc, the calculated value is -0.5x10-10 m3/C
while the measured one is +0.3x10-10 m3/C. The
reason is that the carriers are positive. Although
electrons are always the ones moving, some
material, especially divalent, behave as if the
carriers were positive. We’ll talk about that later.
7