Lecture 23. Degenerate Fermi Gas & Bose-Einstein
condensation (Ch. 7)
2
n n
nFD   
<n>
MB
BE
1
NVQ
FD
0
-6
-4
1
1
nBE   
  
  
  1
exp
exp 
 1
k
T
k
T
 B 
 B 
-2
0
2
()/kBT
4
6
V
  
nMB    exp  

k
T
B


Today’s plan: high density limit n ~ nQ
1. Fermion: Degenerate Fermi Gas (7.3)
2. Boson: Bose-Einstein condensation (7.6)
When will Maxwell-Boltzmann break down?
Answer: when nVQ
 V
1 doesn't hold.    k BT ln 
 NVQ
For a system with fixed density (n),  is a function of T.
 V  2 mk T  2 
B
MB  kBT ln  
 
2
 
 N  h
3
V  2 mk B T  2

0  
 1
2
N h


  0

2
MB
3
0
-2
2
h2
-4
0
1
2
3
3
 k BT  k B T * 
n
 
T/T*
2 m
T*: the temperature (energy) scale where quantum (exchange)
effect becomes pronounced.
TF (fermion)  degenerate fermi gas (T TF )
T* ~ 
 TC (Boson)  Bose-Einstein condensation (T  TC )
Degenerate fermi gas (T=0 limit)
T  0, n   =      F  , a step function. nFD   
 F     0  : fermi energy
1
  
exp 
1

 k BT 
At T=0, the total # of fermions in a degenerate fermi gas (spin S) is:
3

F
4
3D
3D
N   g    nFD   d    g   d    2S  1V 3  2m F  2
0
0
3h
2
2
h2 
3
N 3
h2
  FD  0    F 
kBTF   F k BT * 
n3



2m   2 S  1  4 V 
2 m
2


2
3
for electrons in metals: S  1 ,    h  3N  and T

300
K
F
F




2
8me  V 


The total (internal) energy of a degenerate fermi gas is:

U 0     g
0
3D
3
   nFD   d   N  F
5
U 3
 u 0    F
N 5
- a very appreciable zero-point energy!
Degenerate fermi gas (T=0 limit) (cont.)
The large (internal) energy U at 0 K is a consequence of Pauli
exclusion principle. Recall in ideal gas, UIG=3/2 PV, we expect
similar pressure generated by DFG. Indeed:
3   F 
3  2 F
 U 
P  


N


N 



5  V  N , S
5  3V
 V  N , S
 2U

 3V
U
3
PV
2
This pressure is called degeneracy pressure. It is the physical
mechanism that prevents white dwarf stars (electron) or neutron
stars (neutron) from collapse by gravity. (Pr. 7.23, 7.24)
Bulk modules
(compressibility-1)
 P 
B


V
of electron in metal:



V

T
5

2U
5P
10 U
 P 
P
V 3  





3V
3V
9 V2
 V T
 B  10 U
9 V
How stars can support themselves against gravity:
gas
Star Evolution
and radiation pressure supports
thermonuclear reactions occur
stars
in
which
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
Nobel 1983
Chandrasekhar
Degenerate fermi gas (T≠0, but T<<TF)
To be quantitative, we need to apply Sommerfeld expansion.
But the qualitative behavior can be captured by a back-ofenvelope calculation.

T>0
When T<<TF, the # of “excited” fermions is:  = EF
kBT
N
N  g  F   ~
k BT  N
F
F
The extra thermal energy
u
acquired by each fermion:
 U
N u  N
k BT
F
 k BT  N
k BT
 kBT 
2
g     1/ 2
F
The characteristic behavior of
 kB 
 U 
CV  
N
T T

electrons in metals.
F
 T V
2

1  k BT 
~



Slightly more complicate cal.: 
2

F
 F 
2
Chemical potential of fermi gas (with fixed n)
Sommerfeld expansion (T<<TF)
 T 
  k BT 
1


F
12   F 
FD/F
2
2
Maxwell-Boltzmann (T>>TF):
 V  2 mk T  3 2 
B
  kBT ln  
 
2
 
 N  h
1
0
-1
-2
0
1
kBT / F
2
The Fermi Gas of Nucleons in a Nucleus
Let’s apply these results to 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:


R  1.31015 m  A1/ 3
n
Thus, the density of nucleons is:
A


3
4
 1.3 1015 m  A
3
 11044 m-3
For simplicity, we assume that the # of protons = the # of neutrons, hence their
density is
np  nn  0.5 1044 m-3
6.6 10 

34 2
The Fermi energy
EF
 27
8 1.6 10
3

 0.5 1044 


2/3
J  4.31012 J  27 MeV
EF >>> kBT – the system is strongly degenerate. The nucleons are very “cold” – they
are all in their ground state!
The average kinetic energy in a degenerate Fermi gas = 0.6 of the Fermi energy
E  16 MeV
- the nucleons are non-relativistic
Bose-Einstein Condensation (Ch. 7 )
BEC and
related phenomena
BEC of photons
(lasers)
Townes
Basov Prokhorov
Nobel 1964
BEC in a stronglyinteracting system
(superfluid 4He)
BEC in a weaklyinteracting system
(atomic gases)
Nobel 2001
Einstein
described
the
phenomenon
of
Landau
condensation in an ideal gas of particles with
Kapitsa
Nobel 1962
Nobel 1978 nonzero mass in 1925. In the 1930’s Fritz London
realized that superfluity 4He can be understood on
terms of BEC. However, the analysis of superfluity
4He is complicated by the fact that the 4He atoms
in liquid strongly interact with each other.
70 years after the Einstein prediction, the BEC in weakly interacting Bose systems
has been experimentally demonstrated - by laser cooling of a system of weaklyinteracting alkali atoms in a magnetic trap.
Two types of Bosons
Two types of bosons:
(a) Composite particles which contain an even number
of fermions. These number of these particles is
conserved if the energy does not exceed the
dissociation energy (~ MeV in the case of the
nucleus).
(b) Particles associated with a field, of which the most
important example is the photon. These particles are
not conserved: if the total energy of the field
changes, particles appear and disappear. The
chemical potential of such particles is zero in
equilibrium, regardless of density.
Ideal Gas of Conserved Bosons
2
2
3
2
MB
h
k BT * 
n
2 m
An educated guess: something (BEC)
will happen below T* for bosons with
fixed density n.
  min    0
Recall:
0
-2
-4
0
1
2
3
T/T*
Q: Why bosons are special, i.e. forming BEC at low temperature?
A: The quantum nature, i.e. any # of bosons can occupy one
quantum state (energy level) .
A layman’s definition: Bose-Einstein Condensation (BEC) is a
special macrostate with macroscopic # of bosons occupying one
quantum state (often, the ground state) of a bosonic system.
As the system temperature is cooled below certain temperature TC,
BEC spontaneously forms. It is a phase transition purely driven by
quantum (exchange) effect.
BEC of Conserved Bosons
Let’s consider a simple but very special case: T = 0 K.
What is the macrostate of a bosonic system?
4
All the bosons (a macroscopic #!) occupy the
lowest energy level, i.e. the ground state, so
that the system has lowest energy.
3
2
1
the multiplicity of the system:   1. i.e. S  kB ln   0.
 S 
chemical potential   T 
  0 at T =0.
 N V ,U
On the other hand, 0 around T*
 (n,T)
according to Maxwell-Boltzmann
distribution. Indeed, =0 right at T=TC
and stay at 0 as T further decreases.
TC
T
Bose-Einstein Condensation (TC)
nBE   
Recall Bose-Einstein distribution
The total # of bosons:
N   ni
3/ 2
 3D

2 s  1  2m 
1/ 2
g


V







4 2  2 


i
Using density of state approximation

1
exp        1
g 3D  
2 S  1  2m k BT 
N 
d  2
V

2
h
exp





1







0


3/ 2 
x1/ 2
0 exp  x     1 dx
Let’s perform the integration at TC, i.e. =0.
N
2 S  1  2 mk BTC 
n 2


V
h2
 

3/ 2 
x1/ 2
0 e x  1 dx
Critical temperature of BEC
2
h2
Recall k BT * 
n3
2 m

x1/ 2
0 exp  x   1 dx  1.3   2.315


h
n
TC 
2 m kB  2.6122 s  1
2
Confirm: T*TC
2/3
Bose-Einstein Condensation (T<TC)
T>TC, the system can adjust n  2 2 S  1  2m k BT 


2
(<0) to satisfy the constraint:
  h

3/ 2 
x1/ 2
0 exp  x     1 dx
What happen at T<TC?  is already 0 at TC. The right hand side
decreases as T1.5. Something is wrong!
g()

g    

n()
=
g()n()
Resolving the paradox: The problem is caused by the behavior of
the 3D density of states and our use of the continuum
approximation. Because g()=0 at =0, our calculations of n
ignored all the particles in the ground (=0) state. At low
energies, we have to take into account the discreteness of the
quantum states.


Just excited states!
Bose-Einstein Condensation (T<TC) (cont.)
The eq. n(T) with  = 0 still works at T<TC for calculating
the number of particles not in the ground state (ignore spin):
g()n()
n 0
 2 mk BT 
 2.612 

2
h


3/ 2
T 
 n 
 TC 
3/ 2
T  TC 
The density of particles in
the ground state:
T < TC
n0  n  n 0

a tremendous number of
particles all sitting in the
very lowest available
energy state
  T 3/ 2 
 n  1     T  TC 
  TC  
n>0
n
n0
TC
T
Bose-Einstein Condensation (Summary)
We can discuss the ideal Bose gas in the same terms of a phase
transition. That is, at a critical value of temperature, TC, (n,T)
reaches the limit of  = 0 and stops increasing. Beyond this
3/ 2
point, the relation
2  S  1  2 m  
 d
n


2
  h  0 exp        1
is no longer able to keep track of all the particles – we miss the
particles in the ground state. Below TC, bosons begin to
condense into the ground state. The abrupt accumulation of
bosons in the ground state is called Bose-Einstein condensation.
 (n,T)
TC  2
TC
T
2

mk B


n
 2.62 s  1


2/3
0.527  h 2  2 / 3

 n 
TC S  0 
k B  2m 
Realization of BEC in a Dilute 87Rb Vapor
In principle, the lighter the bosons, the greater TC. For example, the BE condensation of
excitons (light-induced electron-hole pairs) in semiconductors has been observed before
the BE condensation in dilute gases (electron is a fermion, but an electron-hole pair has
an integer spin).
First observation of the BEC with weakly-interacting gases was observed with
relatively heavy atoms of 87Rb. 10,000 rubidium-87 atoms were confined within a
“box” with dimensions ~ 10 m (the density ~ 1019 m-3). The spacing between the
energy levels:
2
34 2
 ~  1 



3h
3
6.6 10

8m L2 8 87 1.7 10 27  105

2
 1.11032 J  8 1010 K
The transition was observed at ~ 0.1 K. This is in line with the estimate:
0.53  h 2   N 

  
TC 
k B  2 m   V 
2/3

 
2

2/3
0.53  6.6 10 34  1 1019

 8 10 8 K
 23
 27
1.38 10  2  87 1.7 10
kBTC 100  ! - again, it is worth emphasizing that the BEC occurs at kBT >>  :
2
h
 ~ 1 ~
m L2
h2
k BTC ~
m
N
 
V 
2/3
N 2/3
~  L 2 ~  N 2 / 3  
L
2
- the greater the total number of particles in the system, the greater this difference.
Realization of BEC in Dilute Vapor (cont.)
The atoms are not very close to each other in the classic sense - in fact, the average
density of this condensate is very low—one billionth the density of normal solids or
liquids. But at this temperature, the quantum volume becomes comparable to the
average volume per atom:


3
h3
6.6 1034
VQ 

2m kBT 3 / 2 1.381023  5 108  2  871.7 1027

At T=0.9TC, the number of atoms in the ground state:
degeneracy of the 1st excited state in a cube
For comparison, in the first excited state:
n>0
n
n0
TC
T
N1 

3/ 2
 6 1019 m3
 T
N 0  N 1  
  TC



3/ 2

  1,500

3
3

 300
exp / k BT   1 1  0.01 1
The ratio N0/N1, which is ~ 5 for N = 104, rapidly
increases with N at a fixed T/TC (it becomes ~
25 for N = 106).
Problem
(a)
(b)
Calculate the critical temperature for BE condensation of diatomic hydrogen H2
if the density of liquid hydrogen is 60 kg/m3. Would you expect superfluidity in
liquid hydrogen as well? Hydrogen liquefies around 20K and solidifies at 14K.
Above TC, the pressure in a degenerate Bose gas is proportional to T. Do you
expect the temperature dependence of pressure to be stronger or weaker at T
<TC ? Explain and draw a qualitative graph of the temperature dependence of
pressure over the temperature range 0 < T < 2 TC.
(a) Liquid hydrogen:
0.527  h   N 
TC 

  
k B  2 m   V 
2
2/3
6.626 1034 
2
0.527
60



23
27 
27 
1.38  10 2  2  1.67  10
 2  1.67 10 
2/3
 5.48 K
Since hydrogen solidifies at 14K, we do not expect to observe superfluidity in liquid hydrogen.
P(T)
TC
T
(b) The atoms in the ground state do not contribute to
pressure. At T < TC, two factors contribute to the fast
increase of P with temperature: (i) an increase of the
number of atoms in the excited states, and (b) an
increase of the average speed of atoms with temperature.
As the result, the rate of the pressure increase with
temperature is greater at T < TC than that at T > TC (in
fact , P~T5/2 at T < TC) .
How to cool the gas of Rb atoms down to ~0.1 K? The first stage – laser cooling,
the second stage – evaporative cooling.
Laser Cooling
E2
-works for a dilute gas of neutral atoms
(cannot be applied to cool solids)
E1
For photon absorption or
emission, the photon energy
h must be equal to E2-E1
If the laser frequency is tuned slightly
below E2-E1, an atom scatters (absorbs
and re-emits) photons only is it moves
towards the laser (Doppler effect). Atom
at rest or moving in the opposite
direction doesn’t scatter.
If a resonant photon is absorbed, the atom acquires momentum:

c
   2 eV 
p2
10
K


10
eV
2
9
2M 2M c
2  2310 eV
2
The corresponding energy:
Mv 
2
Na, A=23
An apparent limit on T achieved by laser cooling is reached when an atom’s recoil
energy from absorbing or emitting a single photon is comparable to its total K. The
single-photon recoil temperature limit (for Na):
T1  1010 eV 104 K / eV  1K
(for Na)
Laser Cooling and Trapping
By laser cooling, T ~ 10 K can be reached. At this
temperature, the atom’s speed is a few cm/s. These slowmoving atoms are relatively easy to confine in a non-uniform
magnetic trap. The magnetic field has a minimum value in
the center of the “magnetic bowl”. An atom with spin parallel
to the magnetic field (i.e., atomic magnetic moment antiparallel to the magnetic field), is attracted to the minimum; for
spin anti-parallel to the field, the atom is repelled from the
minimum.
Laser cooling has been used in the
experiments on BEC observation for precooling of the gas of alkali atoms. However, to
observe this phenomenon, even lower T are
necessary. Further reduction of T by ~3 orders
of magnitude (below 0.1 K) is required for the
exp. vapor densities ~ 1017 m-3. This is
achieved by the evaporation cooling after the
lasers are turned off.
Magneto-Optical Trap
Evaporative Cooling
Radio-frequency forced
evaporative cooling.
The resonance excitation
flips the spins and those
atoms
are
ejected
(evaporated). Reducing fr
frequency
evaporates
lower energy atoms.
Metastability is the Key
vapor-solid/liquid
phase boundary
ln T
vapor
liquid
He
BEC
~ 1010
lnn 
ln TC 
2
lnn   const
3
The experiments with Rb vapor were
aimed at realization of BEC in a
weakly-interacting system.
Though the interactions are weak in
the vapor of Rb atoms, they are
sufficiently strong for the phase
transition vapor-solid at ultra-low
temperatures. In conditions of thermal
equilibrium, one cannot get below the
blue line without phase separation.
How to cheat the Nature? The key is metastability. If the process of cooling is slow and
“gentle” enough, one can realize a “super-saturated” vapor below the coexistence line
without a condensed phase ever forming. For this, not only the interaction with walls
must be excluded, but also the three-particle collisions that assist forming molecules
and, eventually, condensed-matter phase – hence, very low densities.
Observation of BEC
To observe the distribution of velocities of atoms in the
system, the magnetic trap is turned off. The atoms find
themselves in free space, and, because they have some
residual velocity, they just fly apart.
After they have flown apart for some time, the cloud is much bigger, and it is easier to
take a snapshot of the atomic cloud (to make a snapshot, a laser beam is scattered by
the cloud).
The picture shows the velocity distribution of
atoms in the cloud at the time of its release,
instead of the spatial distribution.
For T > TC, atoms are distributed among many
energy levels of the system, and have a
Gaussian distribution of velocities. With cooling
of the cloud, a spike appears right in its middle.
It corresponds to atoms which are hardly moving
at all: for T < TC, the concentration of atoms in
the lowest state gives rise to a pronounced peak
in
the
distribution
at
low
velocities
(condensation in the momentum space).
This two-component cloud resembles the
situation in superfluid helium, where two
components coexist: normal and superfluid .
Each frame corresponds to the distance the
atoms have moved in about 1/20 s after
turning off the trap.
Thermodynamic Functions of a Degenerate Bose Gas
By integrating the heat capacity at constant volume, we can get the entropy:
U 5 U
CV 

T 2 T
The Helmholtz free energy ( = 0):

T
 
CV T * dT * 5 U
S

*
T
3T
0
2
F  U  TS   U
3
The pressure exerted by a degenerate Bose gas:
 F 
5/ 2
P  
 T
 V T
does not depend on volume!
This is due to the fact that, when compressing a degenerate Bose gas, we just force
more particles to occupy the ground state. The particles in the ground state do not
contribute to pressure – except of the zero-motion oscillations, they are at rest.