Fermi-Dirac and Bose-Einstein - University of …cosmology.berkeley.edu/Classes/S2012/Physics_112/...Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein 1 B.Sadoulet Fermi-Dirac and Bose-Einstein

Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet1

Fermi-Dirac and Bose-Einstein Beg. chap. 6 and chap. 7 of K &KQuantum Gases

Fermions, BosonsPartition functions and distributionsDensity of states

Non relativisticRelativistic

Classical LimitFermi Dirac Distribution

Fermi Energy Electrons in solids

Nuclear matterWhite Dwarf

Bose Einstein DistributionBose-Einstein CondensationLiquid Helium, SuperconductivityModern experiments with Bose Einstein Condensates

B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein

In class

Mean number of particles in state of energy

Which is which? A: Fermi Dirac Bose Einstein

B: Fermi Dirac Bose Einstein

Fermi Dirac and Bose Einstein

2

ε

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1


In class




3

ε

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

<s(ε)>

µ ε

1


Quantum GasesBosons, Fermions

Consider quantum system and its quantum states Quantum Field Theory =>

Integer spin= Boson = number of particles in a given state is arbitraryHalf Integer spin=Fermion= at most one on each state:

Pauli exclusion principlePartition Functions and mean occupation numbers

Thermodynamical system=state of energy εUse Gibbs method and grand partition function Fermion

At most one

Boson Sum on all integers

Z = 1+ exp µ − ε

τ⎛⎝⎜

⎞⎠⎟

s ε( ) =τ ∂ logZ

∂µ=

exp µ − ετ

⎛⎝⎜

⎞⎠⎟

1+ exp µ − ετ

⎛⎝⎜

⎞⎠⎟

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

Z= exp s µ − ετ

⎛⎝⎜

⎞⎠⎟

s=0

∞

∑ = 1

1− exp µ − ετ

⎛⎝⎜

⎞⎠⎟

s ε( ) = τ ∂ logZ

∂µ=

exp µ − ετ

⎛⎝⎜

⎞⎠⎟

1− exp µ − ετ

⎛⎝⎜

⎞⎠⎟

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1


Ideal quantum gases“Ideal gas” approximation

States are not modified by presence of other particlesNo correlation between states

“Density of states”multiplicity x density in phase space change of variable to energy

Non relativistic

Ultra-Relativistic

Note: 2 notations either with or without volume included. These notes do not include the volume, and use the density of states per unit volume.

gi ⋅d3x d3p

h3p→ ε

ε = p2

2m⇒ p2 = 2mε

dp = mdεp

=m2dεε

D ε( )dε = 4πgip2dph3

=gi4π2

2m2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ 32 εdε

ε = pc

D ε( )dε = 4πgi

p2dph3

=gi

2π23c3ε2dε

gi ⋅

p2dp dΩh3Ω∫ = D(ε)dε where D(ε) is the density of states per unit volume


In class


How do we determine µ?A: Given by the reservoirB: We compute the mean number of particles

How to determine µ?

6

εs ε( ) =

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1


In class


How do we determine µ?B: We compute the mean number of particles

A1:

B1

C1


7

ε

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

s ε( )0

∞

∫ dε

s ε( ) D ε( )0

∞

∫ dε

V s ε( ) D ε( )0

∞

∫ dε


In class


How do we determine µ?B: We compute the mean number of particles

NOT

B1 If you use density of state with volume in it

C1 If you use density of state per unit volume

Implicit equation for µ


8

ε

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

s ε( )0

∞

∫ dε

N = s ε( ) D ε( )0

∞

∫ dε

N = V s ε( ) D ε( )0

∞

∫ dε


BehaviorSign is critical for (ε-µ)/τ small

Fermi-Dirac

Bose-EinsteinBose condensation

For (ε-µ)/τ large, classical limitOccupation number << 1

Same old results of Boltzmann !

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + 1

=12

for ε = µ

= 1 for ε << µ

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

→∞ for ε → µ

Prob ε( ) ≈ s ε( )N

≈ s ε( )N

=exp − ε

τ⎛⎝⎜

⎞⎠⎟

VnQ=exp − ε

τ⎛⎝⎜

⎞⎠⎟

Z1

< N >= V s ε( )∫ D ε( )dε = V exp µτ

⎛⎝⎜

⎞⎠⎟exp −

ετ

⎛⎝⎜

⎞⎠⎟∫d 3ph3

= V exp µτ

⎛⎝⎜

⎞⎠⎟nQ ⇒ µ = log

nnQ

⎛

⎝⎜⎞

⎠⎟

<s(ε)>

µ ε

1

s ε( ) ≈ exp µ − ε

τ⎛⎝⎜

⎞⎠⎟= λ exp −

ετ

⎛⎝⎜

⎞⎠⎟

independent of F.D. or B.E.

nQ =

mτ2π

⎛⎝⎜

⎞⎠⎟3/2


Thermo functions for ideal quantum

Number of Particles

µ(τ) set by requirement that N=total number of particles≠ case of Black Body Nγ α τ3

Energy

- Bose- Einstein + Fermi-Dirac

Entropy

+ Bose- Einstein - Fermi-Dirac

N = V s ε( ) D ε( )dε0

∞

∫ = VD ε( )dε

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

0

∞

∫

U = ε = VεD ε( )dε

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

0

∞

∫

σ ε( ) = ∂ τ logZ( )

∂τ =ε − µτ

s + logZ = − log prob s =< s >( )⎡⎣ ⎤⎦

�

= ε − µτ

s ± log 1± s( )= ± 1± s( ) log 1± s( )− s log s


Fermi Gas Ground State (Non relativistic)

Fermi EnergyCalculation :

Watch out:! here n=densityEx: Electrons in metal Kittel+Kroemer do this calculation with integer

n ≠density

Energy

Free Energy

Pressure Repulsive!

s ε( )

εεF

τ << µ

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + 1

→1 for ε < µ

→ 0 for ε > µ

N =V s ε( ) D ε( )dε0∞∫ = V D ε( )dε0

ε F∫εF = µ τ = 0( )

Spin 12⇒ gi = 2

density of states gi

d3xd 3ph3 and ε =

p2

2m⇒ N =

V2π2

2m2

⎛ ⎝

⎞ ⎠

32

εdε0

ε F

∫ ⇒εF =

2

2m3π2n( )

23 with n =

NV

�

ε F = 5 eV ⇒ vF = 108 cm/s = 3 ⋅10 −3 c

U = V εD ε( )dε0ε F∫ =

35NεF

F =V F ε( )D ε( )dε0ε F∫

�

F ε( ) =U at τ = 0! σ = 0 for s ε( ) = 0 or 1F =U =

35NεF

p = −∂F∂V τ

=25nεF ∝V

− 53

Spin 1/2


Fermi Gas Ground State (Relativistic)Note that even at zero temperature very large kinetic energies: in

some case ultra-relativistic

Fermi Energy

Energy

Pressure

τ << µ

εF = µ τ = 0( )

N =V D ε( )dε0

ε F∫ =V

3π23c3εF3

⇒εF = 3π2n( )

13 c

U = F = V εD ε( )dε0ε F∫ =

34NεF

p = −∂F∂V τ

=14nεF ∝V

−43

�

Note : same general relation P = 1/3u (ultra relativistic)


InterpretationPauli exclusion principle

2 fermions cannot be in the same quantum state! In particular not at the same position=> fermion does not have the

full volume available but only

Heisenberg uncertainty principle

=> large random momenta

Fermi energy: Non relativistic

Relativistic

=>pressure

VN

=1n

ΔpxΔx ≥

Δpx ≥ n1/ 3

εF ≈ E ≈

Δp2

2m≈2

2mn2 /3

εF ≈ E ≈ cΔp ≈ cn1/3


In class

When we increase the temperature, what happens?A: The distribution remains square and the chemical potential changesB: The chemical potential stays constant and the distribution is

rounded offC: The distribution is rounded off and the chemical potential changes

slightly

What is the effect of temperature

14


In class

When we increase the temperature, what happens?C: The distribution is rounded off and the chemical potential

changes slightly (goes down)

Why does the chemical potential decreases slightly in 3 dimensions?A: Do not knowB: The density of states increases with energyC: The density of states decreases with energy


15

<s(ε)>

µ ε

1


In class

When we increase the temperature, what happens?C: The distribution is rounded off and the chemical potential

changes slightly (goes down)

Why does the chemical potential decreases slightly in 3 dimensions?B: The density of states increases with energyMore states are gained in the high energy tail than in the region just below the

chemical potential. We have to reduced slightly the chemical potential to keep this constant.

This is at the origin of the Thermocouple Effect!


16

<s(ε)>

µ ε

1


In class

A hole is a missing electron in the “filled sea” of the Fermi Dirac distribution at zero temperature

Holes

17

<s(ε

)>

εF

ε1

Holes


In class


What apparent charge? A: negative B: positive C: no charge

Holes

18

<s(ε

)>

εF

ε1

Holes


In class


What apparent charge? B: positive If we impose an electric field

towards the right, electrons will move to left, the vacancies towards the right

Holes

19

<s(ε

)>

εF

ε1

E


In class


What energy? A: 1 has energy greater than 2 B: 2 has energy greater than 1

Holes

20

<s(ε

)>

εF

ε1

12

Holes


In class


What energy?

B: 2 has energy greater than 1 You need a greater energy to excite the electron!

In some sense, holes act as anti-electrons (Positrons). This is how Dirac interpreted negative energy states in the solution of his equation for spin 1/2 particles.

Holes

21

<s(ε

)>

εF

ε1

12

Hole like excitations: increasing energy

Holes

Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet

Symmetry of Fermi Dirac distribution

Basic symmetry (except for lower bound at δ =-µ ). The distribution can be described either as the presence of electrons or the absence of electrons (=holes).

Decomposition into hole-like and electron-like excitations

22

Hole and Electron Excitations

�

1− f ε ,τ( ) = 1

exp −ε − µ( )τ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + 1

describes Fermions of energy -ε and chem.pot. - µ

Writing ε = µ +δ

f ε ,τ( ) = 1

exp δτ⎛ ⎝

⎞ ⎠

+ 1 1− f ε ,τ( ) = 1

exp −δτ

⎛ ⎝

⎞ ⎠

+ 1

�

NV

= D ε( ) f ε ,τ( )0∞∫ dε = D ε( )0

ε F∫ dεD ε( ) f ε ,τ( )0

∞∫ dε = D ε( ) f ε ,τ( )0ε F∫ dε + D ε( ) f ε ,τ( )ε F

∞∫ dεD ε( ) f ε ,τ( )ε F

∞∫ dε

B ="electrons"

= D ε( ) 1− f ε ,τ( )( )0ε F∫ dε

A ="holes"

�

f ε ,τ( )

�

ε

f ε,τ( ) = s ε( ) τ


Electron and hole energyChange of energy with temperature

Note that as referenced to the Fermi energy, the energy of holes are opposite to that of the corresponding missing electrons and are positive

�

u τ( )− u 0( ) = εD ε( ) f ε ,τ( )0∞∫ dε − εD ε( )0

ε F∫ dε= εD ε( ) f ε ,τ( )0

εF∫ dε + εD ε( ) f ε ,τ( )ε F

∞∫ dε − εD ε( )0ε F∫ dε

+ε F D ε( ) f ε ,τ( )0εF∫ dε + D ε( ) f ε ,τ( )ε F

∞∫ dε − D ε( )0ε F∫ dε⎛

⎝ ⎜ ⎞

⎠ ⎟

= 0

= ε −ε F( )D ε( ) f ε ,τ( )ε F

∞∫ dε

"electrons"

+ ε F −ε( )D ε( ) 1− f ε ,τ( )( )0ε F∫ dε

"holes"

�

ε

k

Electron like excitation: increasing energy

Hole like excitation: increasing energy

�

ε F

�

ε F

�

f ε ,τ( )

�

ε


Energy Band structure Periodicity of the lattice (e.g., spacing a) • Periodic zones in momentum space

• Resonant tunneling: free propagation of specific modes(Bloch Waves)

These relationships do not necessarily overlap in energy

=> gap (cf. Kittel, introduction to Solid States Physics Chap. 7))

≈ 1 eV

Seen in projection on the energy axis: energy bands• Valence band• Conduction band

Metal: Fermi level = chemical potential in conduction band => conduction can be described by free Fermi gasInsulator: Fermi level in gap

Electrons in crystals: Quantum States

k ↔

k +

G with G = 2πa

�

Discrete E k( )

E

k

εc

εv

conduction band

valence band

Gap


Electrons in metalsOrder of magnitudeτ=o : very good approximation at room temperature and below

Common notation:

Heat Capacity

but conservation of number of electrons

Then use:• df /dt only large close to • µ does not vary fast with τ

• large negative

Change variable

εF ≈ 5 eV ⇒ TF ≈ 5 104 K >> Tlab

�

Cel = kBdUdτ

= kBddτ

VεD ε( ) f ε,τ( )0

∞∫ dε = kBV εD ε( ) ddτ

f ε ,τ( )0

∞∫ dε

ddτ

D ε( ) f ε,τ( )0

∞

∫ dε = 0⇒εF D ε( ) ddτ

f ε,τ( )0

∞

∫ dε = 0

�

Cel = kBV ε −ε F( )D ε( ) dfdτ0

∞∫ dεε = εF

−εFτ

µ ≈ εF

f ε,τ( ) = s ε( ) τ

df

dτ≈

ε − εF( )τ 2

expε − ε Fτ( )

expε − ε F

τ( ) + 1( )2

�

Cel ≈ kBVD εF⎛ ⎝ ⎜ ⎞

⎠ ⎟ ε −εF⎛ ⎝ ⎜ ⎞

⎠ ⎟ dfdτ0

∞∫ dε ≈ kBVD εF⎛ ⎝ ⎜ ⎞

⎠ ⎟ ε −ε

F( )2τ 2

expε −ε

F

τ

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

expε −ε

F

τ

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ +1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2dε

0

∞∫

ε → x = ε − εFτ

�

Cel ≈ kBVD ε F( )τ x 2ex

ex +1( )2−

εF

τ

∞∫ dx ≈ kBVD ε F( )τ x 2ex

ex +1( )2−∞∞∫ dx = π2

3kBVD ε F( )τ


In class

Why does the heat capacity increase linearly with temperature?

Why ?

26


In class


A: Because the behavior of the density of statesB: Because the number of electrons involved is proportional to TC: Because of the dependence of the chemical potential on T

Why ?

27

D ε( )dε ∝ εdε


In class


Why ?

28

B: Because the number of electrons involved is proportional to T

! " τ = 1

df

dτ≈

ε − εF

( )τ

2

expε − ε

F

τ( )

expε − ε

F

τ( ) + 1( )2

εF

s ε( ) =1

exp ε − εFτ

⎛⎝⎜

⎞⎠⎟ +1

≈ 3τ

5 10 15 20

-0.2

0.2

0.4

0.6

0.8

1.0

ε − εFτ

⎛⎝⎜

⎞⎠⎟2 exp ε − εF

τ⎛⎝⎜

⎞⎠⎟

exp ε − εFτ

⎛⎝⎜

⎞⎠⎟ +1

⎛⎝⎜

⎞⎠⎟2


Heat capacity of metals

FinallyHence taking into account phonons

Ctot = Cel +Cϕ = γT + AT3

Cel <<32NkBImportant puzzle historically

Very few electrons affected

�

ε F

�

ε�

f ε ,τ( )


Example Ge,Si,GaAsfrom Sze “Physics of Semiconductor devices” p13

Wiley-InterScience 1981

30


Insulators: Density of states cf. K&K chap 13

In particular intrinsic semiconductors (no role of impurities)

Statistical distributionStill good approximation to consider free electrons as quantum ideal gas=> occupation number

Density of states

We can then get the total number of electrons

f ε( ) = 1exp ε − µ( ) /τ( ) +1

neT = f ε( )D ε( )dε0

∞

∫ = f ε( )Dh ε( )dε0

εv

∫ + f ε( )De ε( )dεεc

∞

∫

= 1− 1exp µ − ε( ) /τ( ) +1

⎛

⎝ ⎜

⎞

⎠ ⎟ Dh ε( )dε

0

ε v

∫ +1

exp ε − µ( )/τ( ) +1De ε( )dε

εc

∞

∫

Dh(ε)dε =

24π 2

2mh*

2⎛

⎝ ⎜

⎞

⎠ ⎟

32(εv −ε )dε

De(ε)dε =

24π 2

2me*

2⎛

⎝ ⎜

⎞

⎠ ⎟

32(ε − εc)dεεc

εv

ε ε

Gap

D (ε)

conduction band

valence band

2 spin states

Parabolicat gap edge

Electron state density below the gap


Electrons and HolesThis can be rewritten as

Therefore we can describe the electron population by two non relativistic “gases”: holes and electrons (cf. what we did with metals).

The equality of the number of holes and electrons fixes the chemical potential: Charge neutrality!

Fermi level: in the middle of the gap if mh*=me*Measuring from the edge of the valence and conduction band respectively

�

neT = nvTtotal invalence bandat zero temperature=total number of elctrons

− 1exp µ −ε( ) /τ( ) + 1

Dh ε( )dε0

εv

∫holes

+ 1

exp ε − µ( ) /τ( ) + 1De ε( )dε

εc

∞

∫electrons

⇒ 1exp µ −ε( ) /τ( ) + 1

Dh ε( )dε0

ε v

∫holes

= 1

exp ε − µ( ) /τ( ) + 1De ε( )dε

εc

∞

∫electrons

ne =2

4π 2

2me*

2

⎛⎝⎜

⎞⎠⎟

3/21

exp ε '− µ − εc( )( ) / τ( ) +1ε 'dε '

0

∞

∫ ≈ nQe exp −εc − µτ

⎛⎝⎜

⎞⎠⎟

with nQe = 2 me*τ

2π2

⎛⎝⎜

⎞⎠⎟

32

for τ << εv = nh =2

4π 2

2mh*

2

⎛⎝⎜

⎞⎠⎟

3/21

exp ε '− εv − µ( )( ) / τ( ) +1ε 'dε '

0

∞

∫ ≈ nQh exp −µ − εvτ

⎛⎝⎜

⎞⎠⎟

with nQh = 2 mh*τ

2π2

⎛⎝⎜

⎞⎠⎟

32


from Sze “Physics of Semiconductor devices” p850Wiley-InterScience 1981

33

Example Ge,Si,GaAs


In class

How does the number of holes in the valence band depend on µ?

Dependence on µ?

34


In class


A: Increases with increasing µB: Decreases with increasing µ

Dependence on µ?

35


In class


B: Decreases with increasing µ

Dependence on µ?

36

εv εcµ

lognh µ( )

εv εcµ

logne(µ)

εv εcεµ

f

1

1/2

Blue=holes

1exp µ − ε( ) / τ( ) +1

holes

= 1− 1exp ε − µ( ) / τ( ) +1

electrons

nh =1

exp µ − ε( ) / τ( ) +1Dh ε( )dε0

εv

∫holes

ne =1

exp ε − µ( ) / τ( ) +1De ε( )dεεc

∞

∫electrons


In class How do we determine µ?

37


In class

A: Given by the µ of the reservoirB: Impose that the occupation numbers of electrons and holes are

equal at µC: Impose charge neutrality

How do we determine µ?

38


In class

C: Impose charge neutrality


39

No impurities: intrinsic semiconductors occupation number

Yes do intersect at 1/2but does not fix position! charge neutrality does!

εv εcεµ

f

εv εcµµ

logne(µ)lognh µ( )

1

1/2


Determining the chemical potentialNo impurities: intrinsic semiconductors occupation number

Yes do intersect at 1/2but does not fix position! charge neutrality does!

εv εcεµ

f

εv εcµµ


1

1/2


SemiconductorsLarge role of impurities: localized states (Not band !) in gap

If they are shallow (≈ 40meV (Si) ≈10meV (Ge)), such states can be excited at room temperature. This modifies totally the behavior!

Donors

Acceptors

note: 2 A0 state because a bond is missing and the missing electron can be spin up or down,

A- bond established (pair of electrons of antiparallel spins) : 1 state

⇒The number of free electrons(holes) is no more constant

Number of electrons can be increased by donors and decreased by acceptorsBut we need to keep charge neutrality = method to compute the Fermi level⇒For large enough impurities concentration, the Fermi level can move close

to the edge of the gap⇒(Thermally generated) conductivity either dominated by • electron like excitation: negative carriers (n type) • hole like excitation: positive carriers (p type)

do ↔ d+ + e− nd = nd + + nd o

a− ↔ ao + e− na = na− + nao

k

εc

εv

εDεA


Donors

negative carriers (n type)

Acceptors

positive carriers (p type)

Semiconductors cf. K&K fig 13.6

εv εcµ

µ


d 0 ↔ d+ + e−

nd+

= ndexp − ε+

τ⎛⎝⎜

⎞⎠⎟

exp − ε+

τ⎛⎝⎜

⎞⎠⎟ + 2exp − ε0 − µ

τ⎛⎝⎜

⎞⎠⎟

= nd1

1+ 2exp µ − εdτ

⎛⎝⎜

⎞⎠⎟

with εd ≡ ε0 − ε+ ≈ εc −0.04 eV Si0.01 eV Ge

lognd+

µ( )

εd

Electric neutralityne = nd+ + nh ≈ nd+

εv εcµ

µ

logne(µ)lognh µ( )logn

d+µ( )

εa

Electric neutralityne + na− = nh ≈ na−

a0 + e− ↔ d+

na−= na

exp − ε− − µτ

⎛⎝⎜

⎞⎠⎟

2exp − ε0τ

⎛⎝⎜

⎞⎠⎟ + exp − ε− − µ

τ⎛⎝⎜

⎞⎠⎟

= na1

1+ 2exp εa − µτ

⎛⎝⎜

⎞⎠⎟

with εa ≡ ε− − ε0 ≈ εv +0.04 eV Si0.01 eV Ge

+

-


Other examples of degenerate Fermi gas

3HeSpin 1/2 cf. Problem setVery different behavior from 4He: phase separation

Basis for dilution refrigerators (K&K Chapter 12: evaporation of 3He into superfluid 4He which acts as an excellent pump: Works down down to ≈10mK)

Nuclear Matter

⇒Fermi momentum1948 sub threshold production of @184” LBL

Np ≈ Nn ≈A2

R ≈ 1.3 10−13 A13 cm

np ≈ nn ≈ 5 1037 cm-3 ⇒εF = 4 10−12 J ≈ 30 MeV

TF ≈ 3 1011 K

α 380 MeV( )+12 C→ π− + .....


White Dwarfs and Neutron StarsWhite dwarf stars (and core of massive stars)

=> Degenerate Fermi gasFermi pressure balances gravity => Condition for equilibrium

Look at total energy of system: Assume constant density Gravitational potential energyKinetic energy (neglect angular momentum)

Non relativistic

minimum of total energy: stable!

Ultra Relativistic UFD has the same R dependence as gravitational energy

Degeneracy pressure cannot balance gravity if M too big! Chandrasekhar limit

Neutron starsSame story for neutrons (uncertainty of equation of state)Similar Chandrasekhar limit if larger => black hole

ρ ≈ 106 g/cm3 n ≈ 1030 cm-3 εF ≈ 0.5 10−13 J ≈ 3 105eV TF ≈ 3 109 K >> Tstar

�

ρ = M4 /3πR 3 ⇒ n∝ M

R 3

UG ≈ −GM2

RUFD ≈ nVεF ∝MεF

εF ∝ n2 / 3 ∝ M2 / 3

R2

εF ∝ n1/ 3 ∝ M1/3

R

UT =dM 4 /3 − bM 2( )

R

1.4 M

εF ≈ 300 MeV 3 M

R

U

UG

UFD Non Relativistic

UT NR

R

U

UG

UFD Relativistic

UT Rel

UT =aM 5 /3

R2−bM 2

R


White Dwarf Explosions: SN Ia

time

Lum

inos

ity

Distance

Ωm=1ΩΛ=0

Fain

ter

An accelerating universe?

Supernovae Type Ia at high redshift (2 groups) Ωm-ΩΛ

Distant supernovae appear dimmer than expected in a flat universe

Potential problemsAre supernova properties

really constant?Dust?

The Uninvited Guest: Dark Energy

Large negative energy

aacceleration

=Gr2

1c2 u

energy density + 3 p

pressure

⎛

⎝⎜

⎞

⎠⎟

GR gravitational mass

VGravity becomes repulsive!


In class



B: Fermi Dirac Bose Einstein


46

ε

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1


In class




47

ε

s ε( ) =1

exp ε − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ ±1

<s(ε)>

µ ε

1


In class


Can we put every particle on the ground state ?A: No, because of the Pauli exclusion principleB: Yes, but only at zero temperatureC: Yes, to an excellent approximation at finite temperature

Bose Einstein

48

ε

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1


In class


Can we put every particle on the ground state ?C: Yes, to an excellent approximation at finite temperature

A is wrong because Pauli exclusion principle applies only to FermionsB: Slightly wrong. Naively we would expect it would be only true if

but

Bose Einstein

49

ε

s ε( ) =1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

τ << Energy second level( ) − Energy ground state( )

f 0,τ( ) = s 0( ) =1

exp ε0 − µτ

⎛⎝⎜

⎞⎠⎟ −1

≈ −τµ

for µ<<τ

≈ N

⇒ we only need µ ≈ ε0 −τN

1

exp ε s − µτ

⎛⎝⎜

⎞⎠⎟ −1

≈1

exp ε s − ε0

τ⎛⎝⎜

⎞⎠⎟ −1

<< N or exp ε s − ε0

τ⎛⎝⎜

⎞⎠⎟>> 1+ 1

N⇔τ << ε s − ε0( )N


In class

A:

B:


50

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

0

∞

∫ D ε( )dε = n = NV

1

exp ε0 − µτ

⎛⎝⎜

⎞⎠⎟ −1

+1

exp ε s − µτ

⎛⎝⎜

⎞⎠⎟ −1

s>1∑ = N


In class

A:

not accurate enough!B is correct: too many particles can land in the ground state


51

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

0

∞

∫ D ε( )dε = n = NV

1

exp ε0 − µτ

⎛⎝⎜

⎞⎠⎟ −1

+1

exp ε s − µτ

⎛⎝⎜

⎞⎠⎟ −1

s>1∑ = N


Bose-Einstein CondensationCalculation of chemical potential

Let us take the origin of energy at the ground stateAs usual the chemical potential can be computed by imposing the

average number of particles in the systemSeparating between the ground state s=1 and the other states

At low enough temperature, we can solve the equation by having

and put nearly all the particles in the ground state, with the occupation of the higher states given by

�

1

exp −µτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

+ 1

exp ε s − µτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1s>1

∑ = N

�

exp −µτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ≈ 1

�

1

exp ε s − µτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

≈ 1

exp ε s

τ⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

<< N

orexp ε s

τ⎛ ⎝ ⎜

⎞ ⎠ ⎟ >> 1+ 1

N⇔τ <<ε sN

Bose EinsteincondensationFor large densitiesnot a very low temp.phenomenon


B-E Condensation: Quantitative ApproachChemical potential at low temperature

With the energy origin at ground orbital. The occupation number of the ground state is

(µ<0 because <εo=0)

for 4He

Comparison with 2nd stateCannot use continuous approximation

For 4He

=> the occupation number of 2nd state is much smaller

f 0,τ( ) = s 0( ) =1

exp − µτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

≈ −τµ

for µ << τ

but f 0,τ( ) = N0 ⇒ µ ≈ −τN0

N0 ≈ N =102 2 cm-3 ⇒ µ = 1.4 10−45 J at T =1K and V = 1cm3

ε nx ,ny ,nz( ) =

2

2m

π

L( )2⎡ ⎣ ⎢

⎤ ⎦ ⎥ nx

2 + ny2 + nz

2( )

ε 1,1,1( ) = 3

2

2m

π

L( )2⎡ ⎣ ⎢

⎤ ⎦ ⎥

ε 2,1,1( ) = 6

2

2m

π

L( )2⎡ ⎣ ⎢

⎤ ⎦ ⎥

⇒ NΔε = 2.48 10−15 J >> τ = 1.38 10−23at T=1K ⇒ f ε 2,1,1( ),τ( ) =1

expΔε − µ

τ( ) − 1 <<

1

exp −µ

τ( ) − 1 = N

0

Δε = ε 2,1,1( )− ε 1,1,1( ) = 3

2

2mπL( )2⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2.48 10−37 J for L = 1cm>> µ


In class

A:

B:


54

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

0

∞

∫ D ε( )dε = n = NV

1

exp ε0 − µτ

⎛⎝⎜

⎞⎠⎟ −1

+1

exp ε s − µτ

⎛⎝⎜

⎞⎠⎟ −1

s>1∑ = N


In class

A:

not accurate enough!B is correct: too many particles can land in the ground state


55

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

0

∞

∫ D ε( )dε = n = NV

1

exp ε0 − µτ

⎛⎝⎜

⎞⎠⎟ −1

+1

exp ε s − µτ

⎛⎝⎜

⎞⎠⎟ −1

s>1∑ = N


Bose-Einstein Condensation(2)Temperature dependence

Calculate separately condensed phase and normal phase

Replace sum by integral

Defining the Einstein condensation temperature

For large densities, τE is not very small

N = f 0,τ( )condensed + f ε,τ( )

ε 2 ,1,1( )

∞∑excited

N ≈1

exp − µτ

⎛⎝⎜

⎞⎠⎟ −1

+V4π2

2m2

⎛⎝⎜

⎞⎠⎟3/2 ε1/2dε

exp − µτ

⎛⎝⎜

⎞⎠⎟ exp

ετ

⎛⎝⎜

⎞⎠⎟ −1

0

∞

∫

exp −µτ

⎛⎝⎜

⎞⎠⎟≈ 1 , x1/2dx

ex −10

∞

∫ ≈ 1.306 π

⇒ Nexc = 2.612mτ2π2

⎛⎝⎜

⎞⎠⎟3/2

V = 2.612nQV

does not depend on N if µτ≈ 0

�

τ E = 2π 2

mN

2.612 V⎛ ⎝ ⎜

⎞ ⎠ ⎟

2/3

⇒ N exc = N ττ E

⎛

⎝ ⎜

⎞

⎠ ⎟

3/2


Liquid Helium 4Properties of 4He

Loose coupling =>liquid (4.2K at 1 Atm) ≈ ideal gas 4He has spin 0 => boson ≠3He spin 1/2

=> expect condensation at ≈3.1KExperimentally “lambda point” 2.17K (Landau temp.)

Phase transition => peculiar propertiesMacroscopic quantum state

cf energy of electromagnetic field

=> Quantization phenomenae.g. Vortex Equivalent of Josephson effect

=>SuperfluidityNo resistance to flow <= impossibility to transfer energy to a quantum liquid

Ψ = n1/2eiϕ t, x ( )

E ∝ nω ˆ E ei ωt−

k ⋅ x ( )

Δϕ = 2nπ where n is integer

C 4He

T2.17K


In class

Collision of an object of mass M with another of mass m at rest

When do we have a maximum energy transfer

A: When m is much smaller than M?B: When m is much bigger than M?C: When the two masses are equal?

Ordinary Kinematics

58


In class

Collision of an object of mass M with another of mass m at rest

When do we have a maximum energy transferC: When the two masses are equal

e.g., billiard ballif M<<m

Ordinary Kinematics

59

If the mass of an atom is m

1 / 2MV2 = 1 / 2MV '2+1 / 2mv2

M

V = M V '+m v

⇒ MV − mv( )2 = M

V '( )2

⇒ M 2V2 − mMv2 = M 2V '2

−2mM v ⋅ V + m m + M( )v 2 = 0

v =2M

m + MV cosθ θ = recoil angle in the lab

Energy transfer

12mv2 = 2mM 2

m + M( )2V 2 cos2θ

12mv2 → 0

E

p


In class

Consider an object of mass M with velocity V

Similar equations

Phonon Emission

60

E

p

1 / 2MV2

= 1 / 2MV'2 +ω k

M

V = M V '+

k

M

V − k ( )2 = M

V '( )2

M2V2 − 2ω kM = M2V'2

−2Mk ⋅V + 2k2 + 2ωkM = 2

ω k2

cs2 − 2M ω k

csV cosθ + 2ωkM = 0

⇒ ωk = 2Mcs V cosθ − cs( ) > 0 only possible if V is large enough (>velocity of sound)


Liquid Helium SuperfluidityConsider an object of mass M going through helium at

velocity VBecause coherent quantum state, cannot transfer energy to single atom:

only possibility is transfer to excitations ≈ phonons Single atom (no superfluidity) phonons

Conservation energy and momentum in collision

=>subtracting

only possible if V is large enough (>velocity of sound)

If the mass of an atom is m

1 / 2MV2 = 1 / 2MV '2+1 / 2mv2 1 / 2MV2

= 1 / 2MV'2 +ω k

M

V = M V '+m v M

V = M

V '+

k

M

V − m v ( )2 = M

V '( )2 M

V − k ( )2 = M

V '( )2

M2V2 − mMv 2 = M2V'2 M2V2 − 2ω kM = M2V'2

−2mM v ⋅ V + m m + M( )v 2 = 0

v =2M

m + MV cosθ θ = recoil angle in the lab

Energy transfer

12mv2 = 2mM 2

m + M( )2V 2 cos2θ → 0 m→∞

ωk

k

V cosθ

−2Mk ⋅V + 2k2 + 2ωkM = 2

ω k2

cs2 − 2M ω k

csV cosθ + 2ωkM = 0

⇒ ωk = 2Mcs V cosθ − cs( ) > 0


Much cleaner system: Alcali VaporsBE condensation for atoms demonstrated in 1995 => 2001 Nobel Prize in Physics

awarded jointly to Eric A. Cornell of NIST / JILA; Wolfgang Ketterle of MIT; and Carl E. Wieman of CU / JILA.

Time sequence of images showing one cycle of the ringing of a Bose-Einsteincondensate (BEC) in the JILA TOP (time-averaged orbiting potential) trap after being driven by strong oscillationsof trap potential.


BEC AtomsCooling in a trap

See:http://bec.nist.gov/

Images of the velocity distributions of the trapped atomsLeft: just before the appearance of the Bose-Einstein condensateCenter: just after the appearance of the condensateRight: after further evaporation.


Pairing of Fermions

SuperconductivityLow Tc: Pairing of electrons s=0 (Cooper pairs) <= phonon interaction

But condensation theory bad approximation (not free)τsuperconductivity << τcondensation

Similar effects• Zero resistance• Quantization of flux : Vortices

3HeSpin 1/22 phases of pairing s=1

similar to superconductivity but magnetic properties

τcondensation = 0.95mK and 2.5mK


Energy Density of Ultra Relativistic GasesGeneralizationImportant for behavior of early universe (energy density

=>expansion)Suppose that particles are non degenerate (µ<<τ)

Density of energy with

where g is the spin multiplicity

Bosons Fermions

Effective number of degrees of freedom for a relativistic

f ε( ) ≈ 1

exp ετ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ±1

u = εD ε( ) f ε( )dε0∞∫

D ε( )dε =g2

ε2dεπ 2c33

u = g

2τ4

π 2c33x3dxex ±10

∞∫

x3dxex −10

∞∫ =

π 4

15x3dxex +10

∞∫ =

78π 4

15

g2

π2

153c3kBT( )4 = g2 aBT

4

78g2

π2

153c3kBT( )4 = 78

g2aBT4

u = gbosons +78gfermions

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∑ aB2T4


Is the pressure the force per unit area? A last task: Show that the pressure we compute is indeed the

average force per unit area cf. Kittel and Kroemer Chapter 14 p. 391

Describe the particles by their individual density in momentum space (ideal gas)

If the particles have specular reflection by the wall, the momentum transfer for a particle arriving at angle θ is

Integration on angles gives

that we would like to compare with the energy density

θ

�

non relativistic⇒ pv = 2ε ⇒ pressure P = 23u (energy density)

u = 32NVτ ⇒ P = N

Vτ = same pressure as thermodynamic definition = τ∂σ

∂V U,Nultra relativistic ⇒ pv = ε ⇒ P =

13u

2 pcosθ

�

23× 2π pv n p( )p 2dp

0

∞∫

!

�

u = ε n p( )d 3 p0

∞∫ = 4π ε n p( )p 2dp0

∞∫

density in d 3p = n p( ) p2dpdΩ

vΔt

dA

P =ForcedA

=d ΔpΔtdA

=1

dAΔt2pcosθ

Momentum transfer vΔtdAcosθ

Volume of cylinder n p( ) p2dpdΩ

density in cylinder ∫

= dϕ0

2π

∫ d cosθ 2pvcos2θ n p( ) p2dp0

∞

∫0

1

∫

Documents

Fermi-Dirac and Bose-Einstein - University of …cosmology.berkeley.edu/Classes/S2012/Physics_112/...Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein 1 B.Sadoulet Fermi-Dirac and Bose-Einstein