Degenerate Fermi Gas
Fermi gas at low T
• Most applications are to electons, assume
degeneracy g= 2 s +1 = 2


D  
p
2m T
• Increase with decreasing T
• Small enough T, wave functions overlap,
quantum statistics becomes important
Fermi gas at T=0
• No more than 1 (2 for g=2) electrons in
each state
• First “e” goes to lowest state, the second
must occupy higher energy state
• And so on until all the “e”s are put it
• There “motion” (non-zero energy) of
electrons even at T=0
Fermi gas at T=0
 n 
g
e
 (   )
1
T=0, beta = infinity
 1, if  
 n   (  )
when(  )  
e
1
 0, if  
1
Distribution in Fermi gas at T=0
Occupation number <n>
1
energy
μ
What is μ?
V 4p dp
N    n  d    (    )

3
e
 1 (2 )
2
g
4V
N 2
(2 ) 3
pF
2
p
 dp
0
1/ 3
N
pF  (3 )   
V 
2 1/ 3
F    (3 )
2 2/3
N
 
V 
2/3
2

2m
Chemical potential of
Fermi gas determined
by density only
Total energy
p2
E   n 
d 
2m
E
V
2m 2
pF
4
p
 dp 
0
3(3 )
10
2 2/3
2

m
N
 
V 
2/3
N
3
E  NE F
5
2
(3 2 ) 2 / 3  2  N 
PV  E 
 
3
5
m V 
P ~ n5 / 3
5/3
EoS of the type P ~ n^\gamma are called polytropic EoS
gamma =5/3 for Fermi gas
Applicability
• T <<  ~
 N
 
m V 
2
2/3
• Recall, we talked about de Broglie wave
length comparable to inter-particle distance.
This is exactly the condition. MaxwellBoltzmann is applicable for opposite
inequality.
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
Finite T<< εF
What happens as we raise T, but keep kBT<<EF so that   EF?
Empty states are available only above
(or within ~ kBT ) of the Fermi energy,
thus a very small fraction of electrons
can be excited
occupancy
~ kBT
T=0
(E-EF)
The electrons with energies  < EF (few) kBT cannot interact with anything
unless this excitation is capable of
raising them all the way to the Fermi energy.
Only electrons near FERMI SURFACE
participate in motion (thermal or, e.g.,
due to electric field