Transcript fermi dirac statistics in solids

Applications of statistical physics to
selected solid-state physics phenomena
for metals
“similar” models for thermal and electrical
conductivity for metals, on basis of free electron
gas, treated with Maxwell-Boltzmann statistics –
i.e. as if it were an ideal gas
Lorenz numbers, a fortuitous result, don’t be
fooled, the physics (Maxwell-Boltzmann statistics)
behind it is not applicable as we estimated earlier
But Fermi-Dirac statistics gets us the right physics
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Metals have high conductivities for both electricity and heat. To explain both
the high conductivities and the trend in this table we need to have a model for
both thermal and electrical conductivity, that model should be able to explain
empirical observations, i.e. Ohm’s law, thermal conductivity, WiedemannFranz law,
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300K
2.3
2.55
2.2
2
B
2
Wiedemann and Franz
Law, 1853, ratio K/σT =
Lorenz number = constant
 2.4 10-8 W Ω K -2
independent of the metal
considered !! So both
phenomena should be
based on similar physical
idea !!!
Classical from Drude (early 1900s)
theory of free electron gas
K 3k
8
2

 1.12  10 WK
T 2e
Too small by factor 2,
seems not too bad ???
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To explain the high conductivities and the trend we need to have a model for
both thermal and electrical conductivity, that model should be able to explain
Ohm’s law, empirical for many metals and insulators, ohmic
solids Conductivity , resistivity is its reciprocal value
J = σ E current density is proportional to
applied electric field
R = U / I for a wire R = ρ l / A
J: current density A/m2
σ: electrical conductivity Ω-1 m-1, reciprocal value of electrical
resistivity
E: electric field V/m
Also definition of σ: a single constant that does depend on the
material and temperature but not on applied electric field
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represents connection between I and U
Gas of classical charged particles, electrons, moves through
immobile heavy ions arranged in a lattice, vrms from equipartition
theorem (which is of course derived from Boltzmann statistics)
_2
1
3
me v  k BT
2
2
_
2
3k BT
vrms  v 
me
Between collisions, there is a
mean free path length: L = vrms τ
and a mean free time τ (tau)
Figure 12.11 (a) Random successive displacements of an electron in a metal without
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an applied electric field.
If there is an electric field
E, there is also a drift
speed vd (108 times
smaller than vrms) but
proportional to E, equal
for all electrons
eE
vd 
me
Figure 12.11 (b) A combination of random displacements and displacements produced
by an external electric field. The net effect of the electric field is to add together multiple
displacements of length vd  opposite the field direction. For purposes of illustration, this
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figure greatly exaggerates the size of vd compared with vrms.
neAvd dt
J
 nevd
Adt
Substituting for vd
ne 
J
E
me
2
Figure 12.12 The connection between current
density, J, and drift velocity, vd. The charge that
passes through A in time dt is the charge
contained in the small parallelepiped, neAvd dt.
So the correct
form of Ohm’s law
is predicted by the
Drude model !!
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ne 

me
2
2
With vrms according to
Maxwell-Boltzmann
statistics
With mean free time τ =
L/vrms
2
ne L

3k BTme
ne L

me vrms
Proof of the pudding: L should be on the order of magnitude of the inter-atomic
distances, e.g. for Cu 0.26 nm

8.49  1022 cm 3 (6.02  1019 C )2  0.26nm
3  1.381  10
23
1
31
JK  300 K  9.109  10 kg
σCu, 300 K = 5.3 106 (Ωm)-1 compare with experimental value 59 106
(Ωm) -1, something must we wrong with the classical L and vrms 8
Result of Drude theory one order of magnitude too small, so L
must be much larger, this is because the electrons are not
classical particles, but wavicals, don’t scatter like particles, in
addition, the vrms from Boltzmann-Maxwell is one order of
magnitude smaller than the vfermi following from Fermi-Dirac
statistics
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
1
3k BTme



2
ne L
So ρ ~T0.5 theory
for all temperatures,
but ρ ~T for
reasonably high T ,
so Drude’s theory
must be wrong !
Figure 12.13 The resistivity of pure copper as a function of temperature.
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Phenomenological similarity conduction of electricity and
conduction of heat, so free electron gas should also be the key
to understanding thermal conductivity
V
J  
x
Q
T
 K
A t
x
1
K  CV vrms L
3
k B nvrms L
K
2
Ohm’s law with Voltage gradient,
thermal energy conducted through area
A in time interval Δt is proportional to
temperature gradient
Using Maxwell-Boltzmann statistics,
equipartion theorem, formulae of Cv for
ideal gas = 3/2 kB n
Classical expression for K
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_
2
3k BT
vrms  v 
me
Lets
continue
For 300 K and Cu
23
1
3  1.381  10 JK 300 K
vrms 
9.109  1031 kg
k B nvrms L
K
2
1.381  1023 JK 1  8.48  1022 cm 3  1.1681  105 ms1  0.26nm
K
2
1.381  1023 JK 1  8.48  1028 m 3  1.1681  105 ms1  0.26  109 m
K
2
Ws
K  17.78
Kms
Experimental value for Cu at (300 K) = 390 Wm-1K-1,
again one order of magnitude too small, actually
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roughly 20 times too small
ne 2 L

me vrms
K
With
MaxwellBoltzmann
This was also one order of magnitude too
small,
2
e mrs
2
k B nvrms Lmevmrs k B m v


2

2ne L
2e
_
2
3k BT
vrms  v 
me
Lorenz number classical K/σ
K 3k B2
8
2
 2  1.12  10 WK
T 2e
2
3
k
K  BT
 2e 2
Wrong only by a
factor of about 2,
Such an
agreement is called
fortuitous
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replace Lfor_a_particle with Lfor_a_wavial and vrms with vfermi,
 classical 
2
ne L for _ a _ particle
me vrms
L for _ a _ wavical 
_
2
3k BT
vrms  v 
me
 quantum 
ne2 L for _ a _ wavical
mev fermi
mev fermi quantum
ne2
2 EF
v fermi 
me
For Cu (at 300 K), EF = 7.05 eV , Fermi energies have only
small temperature dependency, frequently neglected
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2 EF
v fermi 
me
v fermi,copper,300K
2  7.05  1.602  109 J
6
1


1
.
57

10
ms
9.109  1031 kg
one order of magnitude larger than classical vrms
for ideal gas
L for _ a _ wavical 
mev fermi quantum
ne2
L for _ a _ wavical _ cooper 
9.109  1031 kg  1.57  106 ms1  5.9  107  1m 1
8.49  1028 m 3 (1.602  1019 C )2
L for _ a _ wavical _ cooper
two orders of magnitude larger
 39nm than classical result for particle
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 classical 
2
ne L for _ a _ particle
me vrms
So here something two orders of two magnitude too small (L)
gets divided by something one order of magnitude too small
(vrms),
i.e. the result for electrical conductivity must be one order of
magnitude too small, which is observed !!
But L for particle is quite reasonable, so replace Vrms with Vfermi
and the conductivity gets one order of magnitude larger, which
is close to the experimental observation, so that one keeps the
Drude theory of electrical conductivity as a classical
approximation for room temperature
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effect, neither the high vrms of 105 m/s of the electrons
derived from the equipartion theorem or the 10 times higher
Fermi speed do not contribute directly to conducting a current
since each electrons goes in any directions with an equal
likelihood and this speeds averages out to zero charge
transport in the absence of E
.in
Figure 12.11 (a) Random successive displacements of an electron in a metal without an
applied electric field. (b) A combination of random displacements and displacements produced
by an external electric field. The net effect of the electric field is to add together multiple
displacements of length vd  opposite the field direction. For purposes of illustration, this figure
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greatly exaggerates the size of vd compared with vrms.
k B nvrms Lclassical
K classical 
2
Vrms was too small by one order of magnitude, Lclassical was
too small by two orders of magnitude, the classical
calculations should give a result 3 orders of magnitude
smaller than the observation (which is of course well
described by a quantum statistical treatment)
so there must be something fundamentally wrong with our
ideas on how to calculate K, any idea ???
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Wait a minute, K has something to do with the heat capacity
that we derived from the equipartion theorem
1
K classical  CV _ for _ ideal _ gasvrms L for _ particle
3
We had the result earlier that the contribution of the electron
gas is only about one hundredth of what one would expect
from an ideal gas, Cv for ideal gas is actually two orders or
magnitude larger than for a real electron gas, so that are two
orders of magnitude in excess, with the product of vrms and
Lfor particle three orders of magnitude too small, we should
calculate classically thermal conductivities that are one order
of magnitude too small, which is observed !!!
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2
e mrs
2
k
nv
Lm
v
k
m
v
B
rms
e
mrs
B
K 

2

2ne L
2e
K 3k B2
8
2
 2  1.12  10 WK
T 2e
fortunately L cancelled, but vrms gets squared, we are indeed
very very very fortuitous to get the right order of magnitude
for the Lorenz number from a classical treatment
(one order of magnitude too small squared is about two orders of
magnitude too small, but this is “compensated” by assuming that the
heat capacity of the free electron gas can be treated classically which in
turn results in a value that is by itself two order of magnitude too largetwo “missing” orders of magnitude times two “excessive orders of
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magnitudes levels about out

2
2
B
k T
K fermi 
(
)nL for _ a _ wavical
3 me v fermi
 quantum 
2
ne L for _ a _ wavical
mev fermi
That gives for the Lorenz number in a quantum treatment
K  k
8
2

 2.45  10 WK
T
3e
2 2
B
2
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Back to the problem of the temperature dependency of
resistivity
Drude’s theory predicted a dependency on square root
of T, but at reasonably high temperatures, the
dependency seems to be linear
This is due to Debye’s
phonons (lattice vibrations),
which are bosons and need to
be treated by Bose-Einstein
statistics, electrons scatter on
phonons, so the more
phonons, the more scattering
Number of phonons proportional to Bose-Einstein distribution function
n phonons 
1
e
 / k BT
1
Which becomes
for reasonably
large T
k BT
n phonons 
24
At low temperatures, there are hardly any phonons,
scattering of electrons is due to impurity atoms and
lattice defects, if it were not for them, there would
not be any resistance to the flow of electricity at
zero temperature
Matthiessen’s rule, the resistivity of a metal can be
written as
σ = σlattice defects + σlattice vibrations
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