Fermi-Dirac and Bose-Einstein - University of...
Transcript of Fermi-Dirac and Bose-Einstein - University of...
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
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
Fermi Dirac and Bose Einstein
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet4
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet5
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Mean number of particles in state of energy
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Mean number of particles in state of energy
How do we determine µ?B: We compute the mean number of particles
A1:
B1
C1
How to determine µ?
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Mean number of particles in state of energy
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
How to determine 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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet9
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet10
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet11
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet12
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)
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet13
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
What is the effect of temperature
15
<s(ε)>
µ ε
1
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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!
What is the effect of temperature
16
<s(ε)>
µ ε
1
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A hole is a missing electron in the “filled sea” of the Fermi Dirac distribution at zero temperature
What apparent charge? A: negative B: positive C: no charge
Holes
18
<s(ε
)>
εF
ε1
Holes
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A hole is a missing electron in the “filled sea” of the Fermi Dirac distribution at zero temperature
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A hole is a missing electron in the “filled sea” of the Fermi Dirac distribution at zero temperature
What energy? A: 1 has energy greater than 2 B: 2 has energy greater than 1
Holes
20
<s(ε
)>
εF
ε1
12
Holes
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A hole is a missing electron in the “filled sea” of the Fermi Dirac distribution at zero temperature
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet23
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet24
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet25
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Why does the heat capacity increase linearly with temperature?
Why ?
26
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Why does the heat capacity increase linearly with temperature?
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Why does the heat capacity increase linearly with temperature?
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet29
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( )
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet
Example Ge,Si,GaAsfrom Sze “Physics of Semiconductor devices” p13
Wiley-InterScience 1981
30
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet31
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet32
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet
from Sze “Physics of Semiconductor devices” p850Wiley-InterScience 1981
33
Example Ge,Si,GaAs
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
How does the number of holes in the valence band depend on µ?
Dependence on µ?
34
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
How does the number of holes in the valence band depend on µ?
A: Increases with increasing µB: Decreases with increasing µ
Dependence on µ?
35
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
How does the number of holes in the valence band depend on µ?
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class How do we determine µ?
37
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
C: Impose charge neutrality
How do we determine µ?
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet40
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
logne(µ)lognh µ( )
1
1/2
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet41
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet42
Donors
negative carriers (n type)
Acceptors
positive carriers (p type)
Semiconductors cf. K&K fig 13.6
εv εcµ
µ
logne(µ)lognh 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
+
-
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet43
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− + .....
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet44
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet45
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!
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
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
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
Fermi Dirac and Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Mean number of particles in state of energy
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
Mean number of particles in state of energy
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A:
B:
How do we determine µ?
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A:
not accurate enough!B is correct: too many particles can land in the ground state
How do we determine µ?
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet52
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet53
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A:
B:
How do we determine µ?
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
In class
A:
not accurate enough!B is correct: too many particles can land in the ground state
How do we determine µ?
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet56
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet57
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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
B.SadouletPhys 112 (S12) 7 Fermi Dirac/Bose Einstein
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)
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet61
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet62
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.
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet63
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.
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet64
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet65
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
Phys 112 (F2012) 7 Fermi Dirac/Bose Einstein B.Sadoulet66
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
∫