Lecture 26. Degenerate Fermi Gas

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Lecture 26. Degenerate Fermi Gas

• Ground State of a Fermi Gas, Fermi Energy
• Fermi Velocity, Fermi Momentum, Fermi Pressure
• Fermi Pressure and Stellar Evolution
• Non-Zero T, Fermi-Dirac Distribution
• Specific Heat of Degenerate Fermi Gas

When we consider ensembles of identical
particles, we need to distinguish two regimes
the low-density limit (the inter-particle
distance is much greater than the de Broglie
wavelength of these particles) and the
high-density limit (the opposite is true). In the
low-density limit, bosonic and fermionic
ensembles behave in a similar way (Boltzmann
statistics). In the high-density limit, fermions
and bosons are quite different!
high-density limit
nC depends on T !
The high-density limit is also referred to as the
degeneracy limit. The degeneracy requirement can
be formulated in terms of the so-called
degeneracy temperature of the gas, TC which
corresponds to the condition n nQ
For fermions TC EF/kB where EF is the Fermi
energy (see below), for bosons TC is the
temperature of BE condensation.
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Ground State of a Fermi Gas, Fermi Energy
In the next two lectures, well consider ideal
gases of fermions and bosons. In an ideal gas,
particles do not interact with one another
(unless they collide, and even these processes
are not likely because we consider point-like
particles). Thus, when we consider an ideal gas
of charged particles, well assume that their
charges are screened at short distances e.g.,
the negative charge of electrons in metals is
screened by the positive charge of ions (btw,
this is not the case for the protons in nuclei).
The ground state (T0) of a system of fermions is
governed by the Pauli Exclusion Principle all
the low-energy states are filled to a certain
maximum energy EF which depends on the density of
the fermionic gas (the so-called Fermi energy).
Occupancy ( the distribution function) the mean
number of fermions in a particular quantum state
While the occupancy f(?) is often less than
unity, it is not a probability
nN/V the average density of particles
Fermi energy the max energy of an individual
fermion in the ground state the increase of the
ground-sate energy when a single fermion is added
to the system.
Example N fermions in a 1D square well
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Fermi Energy of a 3D gas of Mobile Electrons
- at T 0, all the states up to ? EF are
filled, at ? gt EF empty
The electron density (the number of electron per
unit volume)
The DoS for a 3D free electron gas (m is the
electron mass)
?
?
?
? EF

the Fermi energy of an ideal 3D Fermi gas (the
non-relativistic case)
?
If EF gtgt kBT , the Fermi gas is strongly
degenerate.
? EF
The total (kinetic) energy of all electrons in
the conduction band (per unit volume)
- a very appreciable zero-point energy!
Average energy per particle in the degenerate
Fermi gas
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Degenerate Fermi Gas in Metals
empty states
?
When the metal atoms are brought together, their
outer electrons break away and can move freely
through the solid. In good metals with the
concentration 1 electron/ion, the density of
electrons in the conduction band n 1 electron
per (0.2 nm)3 1029 electrons/m3 .
EF
conduction band
0
valence band
electron states in an isolated atom
electron states in metal
the Fermi temperature for electrons in metals
vacuum
metal
- at room temperature, this Fermi gas is strongly
degenerate (EF gtgt kBT).
bottom of the conduction band
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Problem
Consider a non-relativistic Fermi gas in two
dimensions. The density of electronic states in
two dimensions per unit area (a) Find the
Fermi energy (in terms of the electron density
n). (b) Find the average energy per electron at T
0 (in terms of EF). (c) Calculate EF for the
electron density n11016 m-2 (a typical density
of electrons in a field-effect transistor),
assume that the effective mass is 0.2x(free
electron mass) (this is the effective mass of
charge carriers in Si MOSFETs). Is the electron
gas in a FET degenerate at room temperature?
(a) The Fermi energy
(b) The average energy per electron
?
2D
EF
g(?)
(c)
At 300K, the fas is not degenerate.
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The Fermi Gas of Nucleons in a Nucleus
Lets consider the system of nucleons in a large
nucleus (both protons and neutrons are fermions).
In heavy elements, the number of nucleons in the
nucleus is large and statistical treatment is a
reasonable approximation. We need to estimate the
density of protons/neutrons in the nucleus. The
radius of the nucleus that contains A nucleons
Thus, the density of nucleons is
For simplicity, we assume that the of protons
the of neutrons, hence their density is
The Fermi energy
EF gtgtgt kBT at room temperature, the system is
strongly degenerate (all nucleons are in their
ground state)
The average kinetic energy in a degenerate Fermi
gas 0.6 of the Fermi energy
- the nucleons are non-relativistic
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Fermi Velocity and Fermi Momentum
(non-relativistic gas)
In metals, the electrons at the Fermi level are
moving at a very high speed (at T0!)
- vF is the Fermi velocity
When an electron is confined in a very small
space, it "flies about its tiny cell at high
speed, kicking with great force against adjacent
electrons in their cells. This degenerate motion
... cannot be stopped by cooling the matter.
Nothing can stop it it is forced on the electron
by the laws of quantum mechanics, even when the
matter is at absolute zero temperature" (Thorne
1994).
This velocity is of the same order of magnitude
as the orbital velocity of the outer electrons in
an atom, and 10 times the mean thermal velocity
that a non-degenerate electron gas would have at
room temperature. Still, since vFltltc, we can
treat the mobile electrons as non-relativistic
particles.
The corresponding momentum
the Fermi momentum
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Pressure of a Degenerate (non-relativistic) Fermi
Gas
The pressure of a degenerate Fermi gas (Fermi or
degeneracy pressure) at T0 can be very high (in
stark contrast to the classical ideal gas
described by the equation of state PVRT)
density n
1m2
vF ? 1s
assuming all electrons have the same v and 1/6 of
them move towards the wall
exact result obtained by integration over all
angles and velocities
The degeneracy pressure depends on n (as n5/3)
and does not depend on T at T ltlt EF.
!
Lets estimate this pressure for a typical metal
In metals, this enormous pressure is counteracted
by the Coulomb attraction of the electrons to the
positive ions.
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Fermi Pressure and Stellar Evolution
How stars can support themselves against gravity

• gas and radiation pressure supports stars in
  which thermonuclear reactions occur
• pressure of a degenerate electron gas at high
  densities supports the objects with no fusion
  dead stars (white dwarfs) and the cores of giant
  planets (Jupiter, Saturn)
• pressure of a degenerate neutron gas at high
  densities support neutron stars

3Msun
1.4Msun
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Fermi Pressure of a Relativistic Fermi Gas
- scales with the density as n5/3 provided that
the electrons remain non-relativistic (speeds v
ltlt c). In a white dwarf or a neutron star, this
approximation breaks down.
For the relativistic degenerate matter, the
equation of state is softer
The number of electrons per unit volume with kltkF
At T0, all the states up to kF are occupied
In the ultra-relativistic case
The relativistic and non-relativistic expressions
for electron degeneracy pressure are equal at ne
1036 m-3, about that of the core of a 0.3 M?
white dwarf.
As long as the star is not too massive, the Fermi
pressure prevents it from collapsing under
gravity and becoming a black hole.
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White Dwarfs, Chandrasekhar Limit
By combining the ideas of relativity and quantum
mechanics, Chandrasekhar made important
contributions to our understanding of the star
evolution.
The Fermi pressure of a degenerate electron gas
prevents the gravitational collapse of the star
if the star is not too massive (the white dwarf).
Nobel 1983
An estimate the upper limit of the white dwarfs
mass (the Chandrasekhar mass)
Total number of electrons in the star ? the
number of protons
proton mass
Total potential (gravitational) energy of the
star
Total (kinetic) energy of the Fermi gas of
electrons (relativistic case)
-Ug
Et
energy
If MltMCh, the Fermi pressure is sufficient to
prevent the collapse.
M
MCh
(more accurately 1.4 M?)
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Neutron Stars M(1.4-3)M?
For a dead star more massive than 1.4 MSun, the
electron degeneracy pressure cannot prevent the
gravitational collapse. During the collapse an
extra energy liberated is sufficient to drive a
nuclear reaction
If the mass is not too high (lt3 MSun), the
further contraction is stopped by the degeneracy
pressure of neutrons (otherwise black hole).
By knowing the parameters of neutron stars, we
can explain why these stars are made of neutrons.
A typical neutron star has a mass like the Sun
(MSun 2 1033 g), but a much smaller radius R
10 km.
The neutron density n 1044 m-3
comparable to the nuclear density
The Fermi energy of the neutrons EF ? 30MeV.
Let consider a hypothetical star made of protons
and electrons, each with the concentration n
1044 m-3 (the of protons the of electrons,
the star is electrically neutral). The Fermi
energy for protons would be nearly the same as
that for the neutrons. On the other hand, the
electron EF would be greater by the ratio Mn/me
1838 in the non-relativistic case, but we need
to consider ultra-relativistic electrons. For
ultra-relativistic electrons with this density EF
280 MeV! Obviously, the energy difference
EF(electrons)-EF(neutrons) is sufficient to drive
the nuclear reaction
Note that in a white dwarf n 1036 m-3
EF(electrons) ? 0.5MeV, this reaction is still
energetically unfavorable
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Thermonuclear Runaway in Supernova
Thus, on one hand, the Fermi pressure helps to
stabilize dead stars against gravitational
collapse. On the other hand, reaching the state
of degeneracy in a still-alive star may have
dramatic consequences.
The structure of young stars (low densities,
non-degenerate core) includes a built-in
thermostat (negative feedback) that automatically
adjusts the temperature to just the value needed
to make the reaction go at the correct rate.
This negative feedback fails when the core
reaches the state of degeneracy the pressure of
degenerate electron gas does not depend on
temperature, therefore a slight rise in T is not
compensated by expansion.
burning H shell
Thermonuclear runaway
degenerate He core

• rise temperature
• accelerate nuclear reactions
• increase energy production

He core grows by H-shell burning until He-burning
sets in
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The outer layers start moving inward, crushing
the interior into a dense core. The neutron core
is formed at such a terrific speed, its surface
rebounds, meeting the matter from outer layers
slowly falling inwards.
neutron star
The shock wave travels outwards. When it meets
the outer layers, compresses them and heats them
up, the thermo-nuclear reactions are triggered.
The Crab Nebula, which was formed by a supernova
explosion recorded in 1054.
An explosion then results and a there is a
bright supernova flash.
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T ? 0 The Fermi-Dirac Distribution Function
What happens as we raise T, but keep kBTltltEF?
By heating a Fermi gas, we excite some electrons.
Only electrons with energies close to EF can be
excited empty states are available only above
the Fermi energy. Thus, some states above EF
(within a narrow energy range kBT around EF)
will be populated and some states below EF -
depleted.
- the Fermi-Dirac distribution function
(occupancy) the mean number of fermions in a
particular quantum state
Again, though the occupancy f(?) is less than
unity for fermions (but not necessarily for
bosons!), it is not a probability
1
kBT
At T 0, all the states with ? lt EF have the
occupancy 1, all the states with ? gt EF have
the occupancy 0 (i.e., they are unoccupied).
With increasing T, the step-like function is
smeared over the energy range kBT.
T 0
0
? EF
(with respect to EF)
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The Heat Capacity of a Cold Fermi Gas
One of the greatest successes of the free
electron model and FD statistics is the
explanation of the T dependence of the heat
capacity of a metal.
To calculate the heat capacity, we need to know
how the internal energy of the Fermi gas, Et(T),
depends on temperature. By heating a Fermi gas,
we populate some states above the Fermi energy EF
and deplete some states below EF. This
modification is significant within a narrow
energy range kBT around EF (we assume that the
system is cold - strong degeneracy).
kBT
The fraction of electrons that we transfer to
higher energies kBT/EF, the energy increase for
these electrons kBT. Thus, the increase of the
internal energy with temperature is proportional
to n?(kBT/EF) ?(kBT) n (kBT)2 /EF.
compare
for an ideal gas
The Fermi gas heat capacity is much smaller (by
kBT/EFltlt1) than that of a classical ideal gas
with the same energy and pressure. The small heat
capacity is a direct consequence of the Pauli
principle most of the electrons cannot change
their energy, only a small fraction kBT/EF of
the electrons are excited out of the ground
state.
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Comparison between Boltzmann and FD Distributions
Fermi-Dirac
Boltzmann
CV /NkB
3
1.5
2
1
zero-point energy, Pauli principle
T/TF
T/TF
1
0
1
2
0
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Problem
When the copper atoms form a crystal lattice with
the density of atoms of 8.51028 m-3, each atom
donates 1 electron in the conduction band. (a)
Assuming that the effective mass of the
conduction electrons is the same as the free
electron mass, calculate the Fermi energy.
Express your answer in eV.

• The electrons participate in the current flow if
  their energies correspond to the occupancy n(?)
  that is not too close to 1 (no empty states
  available for the accelerated electrons) and not
  too small (no electrons to accelerate). At
  T300K, calculate the energy interval that is
  occupied by the electrons that participate in the
  current flow, assuming that for these electrons
  the occupancy varies between 0.1 and 0.9.

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Problem (contd)

• Using the assumptions of (b), calculate the ratio
  n1/n where n1 is the density of
  current-carrying electrons, n is the total
  density of electrons in the conduction band.
  Assume that within the range where the occupancy
  varies between 0.1 and 0.9, the occupancy varies
  linearly with energy (see the Figure), and the
  density of states is almost energy-independent.
  The density of states for the three-dimensional
  Fermi gas

(c)
- important quantity that controls many
properties of metals (conductivity, electronic
specific heat, etc.)
Thus, at T300K, the ratio of the
current-carrying electrons to all electrons in
the conduction band is 0.012 or 1.2 .