Degenerate Fermi Gas

Fermi gas at low T

• Most applications are to electons, assume degeneracy g= 2 s +1 = 2

• Increase with decreasing T• Small enough T, wave functions overlap,

quantum statistics becomes important

mTpD2

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

1)( e

gn

T=0, beta = infinity

if

ifwhen

en

,0

,1)(

1

1)(

Distribution in Fermi gas at T=0

energy

Occupation number <n>

1

μ

What is μ?

mV

NF

V

NpF

dppV

N

dppV

e

gdnN

pF

2)3(

)3(

)2(

42

)2(

4

1

23/23/22

3/13/12

0

23

3

2

)(

Chemical potential ofFermi gas determinedby density only

Total energy

3/5

3/523/22

3/223/22

0

42

2

~

5

)3(

3

2

5

3

10

)3(3

2

2

nP

V

N

mEPV

NEE

NV

N

mdpp

m

VE

dm

pnE

F

pF

EoS of the type P ~ n^\gamma are called polytropic EoSgamma =5/3 for Fermi gas

Applicability

• T <<

• Recall, we talked about de Broglie wave length comparable to inter-particle distance. This is exactly the condition. Maxwell-Boltzmann is applicable for opposite inequality.

3/22

~

V

N

m

The Fermi Gas of Nucleons in a NucleusLet’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:

3/115 m103.1 AR

Thus, the density of nucleons is:

3-m m

44

315

101103.1

34

A

An

For simplicity, we assume that the # of protons = the # of neutrons, hence their density is

-3m 44105.0 np nn

The Fermi energy

MeV 27J 104.3J 105.03

106.18

106.6 12

3/2

4427

234

FE

The average kinetic energy in a degenerate Fermi gas = 0.6 of the Fermi energy

MeV 61E - the nucleons are non-relativistic

EF >>> kBT – the system is strongly degenerate. The nucleons are very “cold” – they are all in their ground state!

Empty states are available only above (or within ~ kBT ) of the Fermi energy, thus a very small fraction of electrons can be excited

The electrons with energies < EF -

(few) kBT cannot interact with anything unless this excitation is capable of

What happens as we raise T, but keep kBT<<EF so that EF?

T =0

~ kBT

(E-EF)

occ

up

ancy

raising them all the way to the Fermi energy.

Finite T<< εF

Only electrons near FERMI SURFACE participate in motion (thermal or, e.g., due to electric field