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Quantum statistics
Contents
1 Boltzmann distribution
1.1 Detailed balance for the Boltzmann distribution . . . . . . . .
1
4
2 Fermi-Dirac distribution
5
3 Bose-Einstein distribution
9
4 Density of states
9
4.1 DOS in a 3D uniform solid . . . . . . . . . . . . . . . . . . . . 11
4.2 DOS for a 2D solid . . . . . . . . . . . . . . . . . . . . . . . . 13
4.3 DOS for a 1D solid . . . . . . . . . . . . . . . . . . . . . . . . 14
1
Boltzmann distribution
Boltzmann distribution represents the simplest model of the energy distribution for distinguishable, identical, and non-interacting particles. The particles
are called distinguishable because they are distinct and their wavefunctions
do not overlap. They are called identical since two particles with the same
energy are taken to be similar. They are non-interacting, because the presence of a particle in a energy state does not affect the probability of another
particle occupying the same energy state. Consider a finite number of these
particles in an isolated system with a certain total energy. The Boltzmann
distribution predicts the energy distribution of these particles in the system.
In terms of probability, the Boltzmann distribution gives the probability of a
particle having a certain energy at a given temperature.
To derive the expression for the Boltzmann distribution consider an isolated
system with five particles. The system has quantized energy levels with the
lowest energy level equal to zero. For simplicity, the other energy levels are
equally distributed with energy difference equal to ∆E. Thus, the energy
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levels for this system are 0, ∆E, 2∆E, 3∆E, 4∆E .... and so on. A general
system can have continuous energy states. This is similar to a quantized
system, with ∆E → 0. The isolated system is at some finite temperature
T.
Consider a scenario when the total energy of the five particles is zero. This
is only possible, when all five particles are in the lowest energy state, and all
the other energy states are empty. Such a situation can be considered to be
the ground state of the isolated system.
Now, consider a system with total energy equal to ∆E. This is possible, if
there is one particle in the ∆E level, and the remaining four particles in the
zero energy level. Out of the five particles, any one can be in the energy
level ∆E. Thus, there are a total of five possibilities for this to happen,
since any of the five particles can be chosen and they are all identical. The
remaining four particles go to energy level zero. If n(E) is defined as the
number of particles with energy E, then for this scenario n(∆E) = 1 and
n(0) = 4. All the other energy levels are unoccupied. If p(E) is the probability of occupation of energy level E, then p(∆E) = 15 and p(0) = 45 ,
since one out of five particles occupy the energy level, ∆E, and four out of
five particles occupy zero energy level. Thus, higher the energy level lower is
the occupation probability.
Consider, a system with total energy 2∆E. There are two ways of achieving
this
Scenario 1: 1 particle with energy 2∆E and 4 particles with zero energy.
Scenario 2: 2 particles with energy ∆E and 3 particles with zero energy.
For scenario 1, there are 5 possibilities and for scenario 2 there are 10 possibilities (5 × 4 divided by 2, since we do not distinguish the order of choosing
two particles). There are a total of 15 possibilities, with 5 particles in each
possibility. The product works out to be 75. We can calculate n(E) for this
system, for each of the three energy states. The other states, above 2∆E,
have zero occupancy.
State zero: n(0) = 5 × 4 + 10 × 3 = 50
State ∆E: n(∆E) = 10 × 2 = 20
State 2∆E: n(2∆E) = 5 × 1 = 5
For each state, we have multiplied the number of possibilities with the number of particles. This can be converted into a probability, p(E). Both n(E)
and p(E) are listed in table 1. The number of particles and the corresponding
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Table 1: n(E) and p(E) for 5 particles in a system with total energy 2∆E.
Energy state n(E) p(E)
0
50
50/75
∆E
20
20/75
2∆E
5
5/75
Table 2: Different scenarios for arranging 5 particles with total energy 4∆E
and the number of ways of doing this.
Energy
Scenario 1
4∆E
1
3∆E
2∆E
1∆E
0
4
Possibilities
5
Scenario 2 Scenario 3
Scenario 4 Scenario 5
1
2
1
3
20
3
10
1
2
2
30
occupation probability decreases with increasing energy.
Consider a more general case for this system, where the total energy is 4∆E.
There are more possibilities now for distributing the five particles and different values for n(E) and p(E). Working with the same logic as the previous
example, the different scenarios are given in table 2. The total number of
ways of arranging this is 70 and there are 5 particles, so the product is 350.
Once again, it is possible to calculate n(E) and p(E) for the different energy
levels, table 3. The n(E) calculated in table 3 is obtained by multiplying the
number in the scenario column in table 2 with the number of possibilities
and adding all the scenarios for a given energy. The occupation probability decreases with increasing energy. For states above 4∆E, the occupation
probability is zero. It is possible to plot the probability as a function of energy, as shown in figure 1. It is possible to fit the data with an exponential
Table 3: No of particles and occupation probabilities for the various energy
levels
Energy
4∆E
3∆E
2∆E
1∆E
0
n(E)
5
20
50
100
175
3
p(E)
5/350
20/350
50/350
100/350
175/350
4
1
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Figure 1: The Boltzmann distribution as a function of energy. The dots
represent the data for p(E) from table 3 and the line is an exponential fit to
the data.
function of the form
p(E) = A exp(−βE)
(1)
where A and β are constants. Equation 1 shows that as the energy level
increases then the occupation probability drops exponentially. So for a system of identical non-interacting particles, most of the particle occupy the
lowest energy states. For atomic systems, like atoms in a solid that is a finite
temperature T the constant β is equal to kB T , which is the average thermal
energy of an atom that is treated as a simple harmonic oscillator. Also, the
system is defined such that the probability of occupation of the ground state
(E = 0) is 1, so that A = 1. Thus, the Boltzmann distribution can be written
as
(E − E0 )
]
(2)
p(E) = exp[−
kB T
where E0 is the ground state energy. For electrons in a metal, when the
Boltzmann approximation is used, E0 is taken to be equal to the Fermi
energy, EF .
1.1
Detailed balance for the Boltzmann distribution
Consider a system where the particles satisfy the Boltzmann distribution.
Consider two energy states, E1 and E2 , with n1 and n2 particles. Let R12 be
the rate of transition per particle from state 1 → 2. Similarly, R21 is the rate
of transition from 2 → 1. To get the total rate this needs to be multiplied
with the number of particles. For a system in dynamic equilibrium the rate
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of particles going from 1 → 2 equal to 2 → 1. This can be written as
n1 R12 = n2 R21
n2
R12
=
R21
n1
(3)
This dynamic equilibrium between individual energy states is called detailed
balance. Since the number of particles is directly proportional to the probability of occupation, p(E), using equation 2, the detailed balance, equation
3, can be written as
p2
(E2 − E1 )
∆E
R12
=
= exp[−
] = exp[−
]
R21
p1
kB T
kB T
(4)
Thus, the rates for the transitions are dependent on the energy difference
between the two levels. If E2 > E1 then ∆E in equation 4 is greater than
zero. This means, that R12 < R21 . Thus, it is easier for a particle to drop
down to a lower energy level than for a particle to occupy a higher energy
level. An external stimulus is needed for this transition to happen. Higher
the value of ∆E, lower is the rate for this transition.
2
Fermi-Dirac distribution
The Boltzmann distribution is used to describe the occupation statistics for
a system consisting of non-interacting particles. Electrons in a solid, on the
other hand, can interact with each other due to electrostatic repulsion. This
has an impact on the occupation of energy levels by electrons. Electrons are
also considered indistinguishable since when two electrons come close to each
other their wavefunctions can overlap. The consequence of electron interaction is the Pauli exclusion principle, which states that no two electrons
can have the same set of quantum numbers. A more general way of stating
the exclusion principle is that two electrons cannot occupy the same space.
We can use the Boltzmann statistics along with the exclusion principle to
derive a new statistics for electron distribution as a function of temperature
and energy, called the Fermi-Dirac distribution,
Consider a metal with N atoms. Corresponding, to the N atoms there are
N energy levels. These energy levels are formed by overlap of the electron
wavefunctions. Each of these energy levels can take two electrons, of opposite
spin, without violating the exclusion principle, so that there are a total of
2N energy states. If each atom contributes one electron then there are N
electrons and these electrons occupy energy states from 0 to N . Thus, half
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of the energy levels are full and half are empty. If the lowest energy level
is taken as zero (reference) the energy level separating the filled and empty
states is called the Fermi level and the corresponding energy is called the
Fermi energy. This is the scenario at T = 0 K. We are interested in what
happens to the electrons as temperature is increased. Also, we can take N
to be sufficiently large, so that the energy levels are very close to each other
i.e. continuous energy states. To give an idea of the magnitude of N , 1 gram
of Li contains around 1023 atoms!
Consider the detailed balance expression from equation 3. In this n1 and n2
are the average number of particle (electrons) in the two energy states given
by E1 and E2 . Because of the exclusion principle, the instantaneous values
of n1 and n2 can be either 0 or 1. Also, because of the exclusion principle
the presence of an electron in a energy states prevents another electron from
occupying this energy state. This is given by an inhibition factor, (1 − n), so
that when n = 0 there is no inhibition, while when n = 1 there is complete
inhibition. Hence, for electrons, the detailed balance equation 3 is modified
as
(5)
n1 (1 − n2 ) R12 = n2 (1 − n1 ) R21
The transition rate R1 2 is multiplied by the inhibition factor, (1 − n2 ), since
the rate is affected by the presence of electrons in energy state 2. Rearranging, and using equation 4 gives
exp(− kEB1T )
R21
n1 (1 − n2 )
=
=
n2 (1 − n1 )
R12
exp(− kEB2T )
(6)
Writing the terms corresponding to energy state 1 on one side and energy
state 2 on the other side gives
n1
E1
n2
E2
α
exp(
) =
exp(
) = exp(
)
1 − n1
kB T
1 − n2
kB T
kB T
(7)
Here, α is a constant. Considering only the energy state 1 in equation 7, it
can be rearranged to write n1 as
1
n1 =
1 + exp(
E1 − α
)
kB T
(8)
n1 is equivalent to the probability of finding the electron in energy state E1
and is written using the symbol f (E1 ). The constant α has to be evaluated.
For electrons in a metal, the Fermi level, EF , is taken to have a probability
of one-half (1/2) at all temperatures. This is possible only when α = EF .
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Figure 2: (a) The Fermi-Diract distribution as a function of energy. The
temperature is taken at 300 K and EF = 5 eV . (b) A close up of the region
around EF . Within 0.2 eV or 200 meV the Fermi function becomes close to
0 (for E > EF ) or 1 (for E < EF ). Plots were generated in MATLAB.
Thus, the Fermi-Dirac distribution for finding the probability of occupation
of energy level, E, is written as
f (E) =
1
(E − EF )
1 + exp
kB T
(9)
The distribution satisfies the following conditions for electron distribution in
a metal
Condition 1: At T = 0, f (E) = 1, when E < EF . All energy levels below
Fermi level are fully occupied.
Condition 2: At T = 0, f (E) = 0, when E > EF . All energy levels above
Fermi level are empty.
Condition 3: At any T , f (E) = 1/2, when E = EF . The Fermi level
occupation probability is always one-half.
The Fermi-Dirac distribution is plotted in figure 2. Since, the function has a
exponential term the occupation probability decreases rapidly when energy is
higher than the Fermi energy. In figure 2(b), it can be seen that above 0.2 eV
above EF the probability is close to zero. The behavior of the Fermi function
also depends on the temperature. f (E) at three different temperatures are
plotted in figure 3. At room temperature the probability decreases rapidly
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Fermi function
1
T = 30 K
T = 300 K
T = 3000 K
f(E)
0.8
0.6
0.4
0.2
0
4.7
4.8
4.9
5
Energy (eV)
5.1
5.2
5.3
Figure 3: (a) The Fermi-Diract distribution as a function of energy for different temperatures. EF = 5 eV and the temperatures are 30 K, 300 K, and
3000 K. The first represents cryogenic temperatures while the last is higher
than the melting points of most metals. Plot was generated in MATLAB.
Table 4: p(E) vs. f (E) for different values of (E − EF )/kB T .
Relative energy
(E − EF )/kB T
0.1
0.2
0.5
1
2
5
10
p(E)
f (E)
0.91
0.82
0.61
0.37
0.14
6.7 × 10−3
4.5 × 10−5
0.48
0.45
0.38
0.27
0.12
6.7 × 10−3
4.5 × 10−5
above EF . At all temperatures f (EF ) = 0.5. When, E − EF kB T , the
Fermi function can be approximated by the Boltzmann distribution. Table
4 lists the two functions at different values of (E − EF )/kB T . For small
values of (E − EF ) the Boltzmann distribution overestimates the occupation
probability compared to the Fermi Dirac function. But as (E − EF ) becomes
≈ 2kB T , both functions start to converge. To put this in energy units, at
room temperature, kB T is approximately, 0.025 eV or 25 meV . Thus, at
energies of 0.05 eV above EF the Fermi function can be replaced by the
Boltzmann approximation.
Electrons belong to a general class of particles called fermions. These are
particles, with a spin of one-half, and include electrons, positrons, protons,
neutrons, and muons. They obey the exclusion principle and are described
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by the Fermi-Dirac distribution.
3
Bose-Einstein distribution
1
Similar to the fermions, there are a set of particles called bosons, which
have integral spin i.e. spin of 0 or 1. Examples of bosons include the α particle, ground state of the He atom, π meson, photons, phonons, and deuterons.
Their behavior is opposite to that of fermions, in that the presence of a particle in an energy state increases the transition rate for another particle to
move to that energy state. Thus, instead of an inhibition factor, there is an
enhancement factor for these particles. Thus, the detailed balance for bosons
can be written as
n1 (1 + n2 ) R12 = n2 (1 + n1 ) R21
(10)
Rearranging equation 10 and using equation 4 gives
n2
α
E1
E2
n1
) =
) = exp(
)
exp(
exp(
1 + n1
kB T
1 + n2
kB T
kB T
(11)
where α is a constant. n1 represents the probability of occupation of a given
energy level by bosons, as a function of temperature and this is given by
n(E) =
1
(E − α)
exp
− 1
kB T
(12)
This distribution is called the Bose-Einstein distribution. This is usually
not applicable to electrons, since they obey the exclusion principle. But,
when materials exhibit superconductivity, electrons in them become paired.
The process is mediated through the lattice vibrations (phonons). These are
called Cooper pairs. While an electron has a spin of one-half, the Cooper
pairs have integral spin, ‘0’ or ‘1’. Hence, these are bosons and obey the BoseEinstein statistics. This, pairing is responsible for the negligible resistance
in superconductors.
4
Density of states
In a solid with N atoms there are a total of 2N energy states. These are
discrete states but for large values of N the spacing between them are so
1
Section 3 can be skipped without loss of continuity
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Figure 4: Calculated DOS of Cu around the Fermi level. The filled
states lie below EF while the empty states lie above EF . The broad
peak below EF corresponds to the filled 3d states.
Adapted from
http://www.personal.psu.edu/ams751/VASP-Cu/
small that they can be considered to be continuous. Density of states (DOS)
is defined as the number of available energy states per unit energy
per unit volume. The units are J −1 m−3 or eV −1 cm−3 and it provides information on how the energy states are distributed in a given solid. It is
typically denoted as g(E). The experimental density of states of a material
can be measured by photo electron spectroscopy or Scanning Tunneling Microscopy (STM) or Electron Energy Loss spectroscopy (EELS). These techniques probe the density of empty or filled states around the Fermi energy,
called Local DOS (LDOS). Density functional calculations can also be used
to calculate the density of states. These model the electron density distribution in a solid and also model the atomic potential within the solid. Figure
4 shows the calculated DOS for Cu around the Fermi energy.
The density of states can be used to calculate the total number of electrons.
If g(E) is the DOS, then the total number of electrons, S(E) is given by
Z
S(E) =
g(E)dE
(13)
E
where the limits of the integration is usually from 0 to the Fermi energy (at
0 K). This equation assumes that the probability of occupation of the energy
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state is 1. This is not always true, since at any temperature, the occupation
probability is defined by a Fermi function, f (E). An energy expression for
the density of states, g(E), will be useful for evaluation of the integral in
equation 13.
4.1
DOS in a 3D uniform solid
To simplify the calculation for g(E) consider a 3D box with uniform potential.
This is a simplification of a real solid but is a good approximation of the
valence band of metals where the electrons are loosely bound to the atom
and are delocalized. We will also use this approximation for electrons and
holes near the edge of the band. For simplicity the uniform potential can
be taken to be zero. For this solid, the electron is defined by 3 quantum
numbers (n1 ,n2 , n3 ) and its energy is given by
h2
(n21 + n22 + n23 )
E =
2
8me L
(14)
Equation 14 is for a cubic solid of length L with the 3 quantum numbers
for the 3 axes. me is the free electron mass. We can replace the 3 quantum
numbers by a single value n so that equation 14 is modified into
E =
h2 n2
8me L2
(15)
For small values of n these energy levels are quantized, but for large values
of n the spacing between them are close so that the energy levels can be
considered to be continuous. So n represents the radius of a sphere, where
the total number of states within the sphere is given by its volume. This
is shown schematically in figure 5. Since the quantum numbers can only be
positive (the quantum numbers represent the electron wavefunction and it
can be shown that to avoid exponential increasing functions the quantum
numbers have to be non-zero positive integers) we can only take the first
quadrant of the sphere.
Hence the total number of ‘orbitals’ (energy states), Sorb (n), is given by the
volume of the sphere in the first quadrant
Sorb (n) =
1
1 4 3
( πn ) = πn3
8 3
6
(16)
Since each orbital can take two electrons of opposite spin, the total number
of energy states (including spin), S(n), is given by
S(n) = 2Sorb (n) =
11
1 3
πn
3
(17)
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Figure 5: Energy states in a solid with uniform potential. The portion of the
sphere corresponds to the constant energy surface. Only the first quadrant is
to be considered since the quantum numbers are non-zero positive integers.
Adapted from Principles of Electronic Materials - S.O. Kasap.
We can relate n (quantum number of the electron in the solid) to the energy
E using equation 14 to write the total number of states in terms of energy,
S(E). This is given by
S(E) =
3
3 L
1
π (8me E) 2 3
3
h
(18)
Dividing by the volume of the cube will give total number of states per unit
volume, Sv (E). The density of states is the differential of the total number
of states, so that g(E) is given by
√ me 3 √
g(E) = 8π 2 ( 2 ) 2 E
h
(19)
Equation 19 gives the DOS in a solid with a uniform potential. At the reference point where energy is set to 0, DOS is zero. As the energy increases
g(E) also increases. The functional form is shown schematically in figure 6.
g(E) represents the density of available states. It does not provide information whether those states are occupied or not. The occupation is given by
the Fermi function and is usually a function of temperature.
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3D DOS
46
x 10
10
9
g(E) (J−1m−3)
8
7
6
5
4
3
2
1
0
0
1
2
3
Energy (eV)
4
5
Figure 6: Plot of DOS vs. E for a 3D solid using free electron mass. This is
a plot of equation 19 using MATLAB.
4.2
DOS for a 2D solid
The density of states function will change upon changing the dimensionality
of the solid. Consider the case of a 2D solid with a uniform potential. There
are 2 quantum numbers, n1 and n2 , which are related to the energy, similar
to equation 14.
h2 n2
h2
2
2
(n
+
n
)
=
(20)
E =
1
2
8me L2
8me L2
For a 2D case, n, represents the radius of a circle, shown in figure 7, and
only the first quadrant can be considered since the quantum numbers should
be positive. It is possible to calculate the density of states per unit area,
Sarea (n), including spin
πn2
Sarea (n) =
(21)
2
Using equation 20 it is possible to calculate the DOS in terms of energy,
Sarea (E)
4πme
Sarea (E) =
E
(22)
h2
Differentiating equation 22 gives the density of states in two dimensions
g(E) =
4πme
h2
(23)
The density of states function is independent of energy, unlike 3D where g(E)
increases with energy. It is represented as a step function at different energy
values.
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Figure 7: Energy states in a 2D solid with uniform potential. Instead of a
sphere states of constant energy form a circle and only the first quadrant is
relevant. Adapted from Principles of Electronic Materials - S.O. Kasap.
4.3
DOS for a 1D solid
The calculation for a 1D solid is similar to the earlier calculations except
that there is only one quantum number and spatially it is represented on a
line (instead of circle in 2D and sphere in 3D). The total number of states,
per unit length, S(n), is just 2n and this is related to energy by equation 24.
r
8me
E
(24)
n =
h2
The density of states is given by
r
g(E) =
8me 1
√
h2
E
(25)
In a 1D solid the density of states decreases with energy. For a zero dimensional solid, energy states are only discrete. Solids with two, one, and
zero dimensionality can be obtained by reducing the length in one or more
dimensions. A thin film is an example of a two dimensional solid (or a single
or bi layer of graphene ), while a quantum wire is a one dimensional solid. A
quantum dot is a zero dimensional solid. The density of states and hence the
electronic properties of these materials are different from a bulk solid. The
DOS for solids of different dimensionalities are shown in figure 8.
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Figure 8: DOS for bulk, 2D, 1D, and 0D solids. The y-axis is a
qualitative representation for the different dimensions. Adapted from
http://britneyspears.ac/physics/dos/dos.htm
References
The following source materials were used for preparing this handout
1. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles
by R. Eisberg and R. Resnick, Wiley Student edition, 2011
2. Principles of electronic materials and devices by S.O. Kasap, McGrawHill India, 2007
The PDF was prepared in LATEX, using Texmaker and MiKTeX compiler.
The plots were generated in MATLAB.
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