5. systems in contact with a thermal...

Phys 112 (S2009) 05 Boltzmann-Gibbs 1

5. Systems in contact with a thermal bath

So far, isolated systems (micro-canonical methods)

5.1 Constant number of particles:Kittel&Kroemer Chap. 3Boltzmann factorPartition function (canonical methods)Ideal gas (again!)

Entropy = Sackur-Tetrode formulaChemical potential

5.2 Exchange of particles: Kittel&Kroemer Chap. 5Gibbs FactorGrand Partition Function or Gibbs sum

(grand canonical methods)

5.3 Fluctuations


Ideal Gas Thermal BathConsider 1 particle in thermal contact with a

thermal bath made of an ideal gas with a large number M of particles

We assume that the overall system is isolated and in equilibrium at temperature τ. We call U the total energy (which is constant).

We would like to compute the probability of having our particle in a state of energy ε. Because in equilibrium, all states of the overall system are equiprobable, this is equivalent to computing the number of states such that our particle has energy ε and the thermal bath E=U-ε.

Using the fact that for an ideal gas of energy E and particle number Np, the number of states is

and that

we obtain for large Np

gRes

ε

exp −ετ

⎛⎝⎜

⎞⎠⎟

U =32Npτ

g ∝ E 3/2Np

prob ε( )∝ U − ε( )32Np ∝ 1− ε

32Npτ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

32Np

Np→∞⎯ →⎯⎯ exp −ετ

⎛⎝⎜

⎞⎠⎟

Phys 112 (S2009) 05 Boltzmann-Gibbs

# States in configurationFundamental postulate (or H theorem): at equilibrium

Notes: • =Boltzmann factor • Z=partition function • Probability of each state not identical!

<= interaction with bath3

Boltzmann factor

R , Uo-ε S,ε

Total system =Our system S + reservoir R isolated

Configuration: 1 state of S + energy of reservoir

1× gR Uo − ε( )

Prob configuration( )∝ gR Uo − ε( )

with Z = exp −εsτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

all states s∑

⇔ Prob state ε s( ) = 1Z

exp −ε sτ

⎛⎝⎜

⎞⎠⎟=

1Z

exp −ε skBT

⎛⎝⎜

⎞⎠⎟

⇒Prob state ε1( )Prob state ε2( ) ≈ exp ∂σR

∂Uo

ε2 − ε1( )⎛⎝⎜

⎞⎠⎟= exp 1

τε2 − ε1( )⎛

⎝⎜⎞⎠⎟

⇒Prob state ε1( )Prob state ε2( ) =

gR Uo − ε1( )gR Uo − ε2( ) = exp σR Uo − ε1( ) − σR Uo − ε2( )( )

exp −ε sτ

⎛⎝⎜

⎞⎠⎟


Examples2-state system

State 1 has energy 0State 2 has energy ε

4

Pr ob 2( ) =

exp −ε

τ( )

1 + exp −ε

τ( )

=1

1 + expε

τ( )

Pr ob 1( ) =1

1 + exp −ε

τ( )

ε

0

Z = 1+ exp −ετ

⎛⎝⎜

⎞⎠⎟

U = ε =ε exp − ε

τ⎛⎝⎜

⎞⎠⎟

1+ exp − ετ

⎛⎝⎜

⎞⎠⎟=

ε

exp ετ

⎛⎝⎜

⎞⎠⎟ +1

CV =dQdT

⎛⎝⎜

⎞⎠⎟V (,N )

=∂U∂T

⎛⎝⎜

⎞⎠⎟V (,N )

= kb∂U∂τ

⎛⎝⎜

⎞⎠⎟V (,N )

= kbεkbT

⎛⎝⎜

⎞⎠⎟

2 exp εkbT

⎛⎝⎜

⎞⎠⎟

exp εkbT

⎛⎝⎜

⎞⎠⎟+1

⎛

⎝⎜⎞

⎠⎟

2

CV

T

Prob(0)

Prob(ε)

1


ExamplesFree particle at temperature τ

Assumption: No additional degrees of freedome.g. no spin, not a multi-atom molecule

Maxwell-Boltzmann distribution KK p 392Normalizing to N particles, the probability to find one particle per

unit volume in the velocity interval dv and solid angle dΩ is

5

Prob dVvolumeelement

, p, p + dp[ ]Momentum interval , dΩ

solid angleelement

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=

1Z

exp −p2

2Mτ⎛⎝⎜

⎞⎠⎟

Boltzmann factordoes not vary appreciably over dp

dVp2dpdΩh3

number of spatialquantum states

in phase space element

dΩ = d cosθdφ

= Prob dVvolumeelement

, ε,ε + dε[ ]energyy interval , dΩ

solid angleelement

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=

2M 3/2

Zexp −

ετ

⎛⎝⎜

⎞⎠⎟dV εdεdΩ

h3

= Prob dVvolumeelement

, v,v + dv[ ]velocity interval , dΩ

solid angleelement

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=M 3

Zexp −

12Mv2

τ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

dVv2dvdΩh3

f (v)dvdΩ = n M

2πτ⎛⎝⎜

⎞⎠⎟

32exp −

Mv2

2τ⎛⎝⎜

⎞⎠⎟v2dvdΩ with n = N

V


Another interpretation of theBoltzmann Distribution

We derived it from number of states in the Reservoir

But each excitation in the reservoir will have a typical energy

Thermal fluctuations of the order ofDifficult to get

6

≈ τ

τ

ε >> τ

Prob ∝ exp −ετ

⎛⎝⎜

⎞⎠⎟


Definition

If states are degenerate in energy, each one has to be counted separately

Relation with thermodynamic functionsEnergy

Entropy

Free Energy

Chemical potential

7

Partition FunctionZ = exp −

εsτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

all states s∑

T

F =U −τσ = −τ log Z

U = ε =1Z

εs exp −εsτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

s∑ = −

∂ logZ

∂1τ

= τ2∂ logZ∂τ

σ = − pss∑ logps = −

e−ε sτ

Zs∑ log e

−ε sτ

Z

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = logZ +

Uτ

=∂ τ log Z( )

∂τ

⇒ σ =−∂F∂τ

U = τ 2 ∂ log Z∂τ

= −τ 2∂ F

τ⎛ ⎝

⎞ ⎠

∂τas implied by the thermodynamic identity

µ =∂F∂N τ ,V

= −τ ∂ logZ∂N


ExamplesEnergy of a two state system

Free particle at temperature τAssumption: No additional degrees of freedom

e.g. no spin, not a multi-atom molecule

Method 1: density in phase space Energy:

Method 2: quantum state in a box: see Kittel-Krömer (chapter 3 p 73) Same result, of course!

Z = 1+ exp −ετ

⎛⎝⎜

⎞⎠⎟

U = ε =τ 2∂ logZ

∂τ=

ε exp − ετ

⎛⎝⎜

⎞⎠⎟

1+ exp − ετ

⎛⎝⎜

⎞⎠⎟=

ε

exp ετ

⎛⎝⎜

⎞⎠⎟ +1

ε =px2 + py2 + pz2

2M

Z1 =d3q d3p

h3∫ exp −ετ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

Vh3

dpxe− px

2

2Mτ−∞∞∫

= 2πMτ

dpye−

py2

2Mτ−∞∞∫ dpze

− pz2

2Mτ−∞∞∫

⇒ Z1 = V 2πMτ

h2⎛ ⎝ ⎜ ⎞

⎠ ⎟ 32 = V Mτ

2π2⎛ ⎝ ⎜ ⎞

⎠ ⎟ 32 = VnQ

with nQ =

Mτ2π2

⎛⎝⎜

⎞⎠⎟

32 ⇒ ε = τ 2 ∂ logZ1

∂τ=

32τ


Partition FunctionPower comes from summation methods

• Direct e.g. geometric series

• Few term approximation useful if f decreases fast

• Integral approximation

Other example: Harmonic oscillator

rii=0

∞∑ = lim

m→∞ri

i=0

m∑ = lim

m→∞

1− rm+1

1− r=

11− r

f i( ) = f 0( )i=0

∞∑ + f 1( ) + f 2( ) + ....

f i( ) ≈ f (x)dx0

∞

∫i=0

∞∑

Consider an harmonic oscillator ωThe probability of mode of energy sω , s integer ≥ 0 is

Prob s( ) = 1Zexp −

sωτ

⎛⎝⎜

⎞⎠⎟

Z = exp −sωτ

⎛⎝⎜

⎞⎠⎟s=0

∞

∑ =1

1− exp − ωτ

⎛⎝⎜

⎞⎠⎟⇒U = sω = τ 2

∂ log Z( )∂τ

=ω

exp ωτ

⎛⎝⎜

⎞⎠⎟ −1


Applies to classical gases in equilibrium with no potentialStart from Boltzmann distribution Prob (1 particle to be in state of energy ε)=

with

=>

Change of variable to velocities:Probability of finding particle in v, v+dv, dΩ

Maxwell distributionParticle density distribution in velocity space Note: 1) Could get the factor in front directly as normalization

2) Note that no h factor! 3) One can easily compute Equipartition!

Maxwell Distribution K&K p. 392

1Z1exp −

ετ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Z1 =

d3q d3ph3∫ exp −

ετ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = V Mτ

2π2⎛ ⎝ ⎜ ⎞

⎠ ⎟ 32 = VnQ

For N particles, the volume density between p and p + dp = NV1nQexp −

ετ

⎛⎝⎜

⎞⎠⎟d 3ph3

=nnQexp −

ετ

⎛⎝⎜

⎞⎠⎟d 3ph3

= n 12πMτ

⎛⎝⎜

⎞⎠⎟3/2

exp −ετ

⎛⎝⎜

⎞⎠⎟d 3p

f (v)dvdΩ = n M2πτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ 32 exp −

Mv2

2τ⎛

⎝ ⎜

⎞

⎠ ⎟ v2dvdΩ

p = Mv

Mv2

2=32τ =

32kBT

Probability of finding this particle in volume dV and between p and p + dp = 1VnQ

exp −ετ

⎛⎝⎜

⎞⎠⎟

dVd 3ph3

Number of states


Density of States

11

εε

D ε( )dε

ε + dε Individual States of probability ps

It is convenient to replace expressions such as ps fss∑ (fs any function of the state s)

by an integral over the energy ε : ps fss∑ ≈ ps ε( ) fs0

∞

∫ ε( )D ε( )dε

D ε( )dε is the number of states between ε and ε + dε and is called the density of states. We replace ps

sε<εs <ε+dε

∑ by ps ε( )D ε( )dε since for small enough dε, ps does not vary

Note that if you ask what is the probability p ε( )dε for the system to be betweenenergy between ε and ε + dε p ε( )dε = ps ε( )D ε( )dε i.e., p ε( ) ≠ ps ε( )For example, for Boltzmann distribution and a non relativistic ideal gas (monoatomic,spinless)

ps ε( ) = 1Zexp −

ετ

⎛⎝⎜

⎞⎠⎟ and using p = 2Mε d 3p = p2dpd cosθdϕ = p2dpdΩ

D ε( )dε =d 3x

V∫ dΩ p2dpangles∫h3

=V 4πh3

p2 dpdε

dε =V 4πh3

122M( )3/2 εdε = V

4π 2

2M

⎛⎝⎜

⎞⎠⎟3/2

εdε


Weak interaction limit2 sub systems in weak interaction:

<=>States of subsystems are not disturbed by interaction (just their population)<=>

=> Partition functions multiply for distinguishable systems

Distinguishable systemsSuppose first that we can distinguish each of the individual

components of the system: e.g. adsorption sites on a solid, spins in a lattice

For 2 subsystems

and for N systems

Partition Function for N systems

Ψ1+2 n1,n2( ) =Ψ1 n1( )Ψ2 n2( ) with n1, n2 still meaningfull

Z = exp −ε s1

+ε s2

τ⎛

⎝ ⎜

⎞

⎠ ⎟ ⇒ Z 1+ 2( )

s2

∑s1

∑ = Z 1( )Z 2( )

Z N( ) = Z 1( )( )N


ExamplesFree energy of a paramagnetic system

In a magnetic field B, ε = −mB if m is the magnetic moment.

For 1 spin Z1 = exp −−mBτ

⎛⎝⎜

⎞⎠⎟+ exp −

mBτ

⎛⎝⎜

⎞⎠⎟= 2cosh mB

τ⎛⎝⎜

⎞⎠⎟

For N spins Z=Z1N since they are distinguishable

Z = 2N coshN mBτ

⎛⎝⎜

⎞⎠⎟

⇒U = τ 2 ∂ logZ∂τ

= −NmB tanh mBτ

⎛⎝⎜

⎞⎠⎟⇒ Magnetization M = −

UVB

= nm tanh mBτ

⎛⎝⎜

⎞⎠⎟

B

M

τ

M

mBτm


Indistinguishable SystemsIndistinguishable:in quantum mechanics, free particles are

indistinguishableWe should count permutations of states among particles only once

Therefore if we have N indistinguishable systems

This is a low occupation number approximation due to Gibbs ( no significant probability to have 2 particles in the same state: we will

see how to treat the case of degenerate quantum gases)

e.g. for 2 particles Z 1( )Z 2( ) = e−εs1τ

s1∑ e

−εs2τ

s2

∑ overcounts s1, s2 ≡ s2 , s1for s1 ≠ s2we have to divide by 2

More generally if we have N identical systems, the Nth power of Z1can be expanded as

(a1 + ...+ as + ....am )N =N !

n1!...ns !...nm !ni∑ a1

n1 ..asns ...am

nm

If we have a large number of possible states, most of the terms are of the form N !aiaj ...akN terms

corresponding to N different states which are singly occupied. If the multiple occupation ofstates is negligible we are indeed overcounting by N !

Z N( ) ≈ 1N!

Z 1( )( )N


Ideal GasN atoms

Indistinguishable

Interpretation: nQ as quantum concentrationde Broglie wave length

λ = hMv

≈ h3Mτ

⇐ 12Mv2 ≈ 3

2τ

1

λ3≈3Mτ

h2⎛⎝⎜

⎞⎠⎟

32 ≈

Mτ

2π2⎛

⎝⎜⎞

⎠⎟

32 = nQ

ZN =1N!

Z1( )N =1N!

nQV( )N

logZN = N lognQn

⎛⎝⎜

⎞⎠⎟+ N with n = N

V= concentration

nQ =Mτ2π2

⎛⎝⎜

⎞⎠⎟

32

logN!≈ N logN − N⇒ log ZN = N log nQVN

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + N


Ideal gas lawsEnergy

equipartition of energy 1/2τ per degree of freedom

Free energy

Pressure

Entropy: Sackur Tetrode identical to our expression counting states

if extra degrees of freedom, additional terms!Well verified by experimentInvolves quantum mechanics

Chemical potential

F = −τ log Z = −τN lognQn

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −τN

U = τ2∂log Z∂τ

=32Nτ =

32NkBT

σ =SkB

= −∂F∂τ

= N lognQn

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +

52

⎡ ⎣ ⎢

⎤ ⎦ ⎥

S =dQT0

T∫

µ =∂F∂N

=∂ −τN log

nQVN

⎛⎝⎜

⎞⎠⎟− τN

⎛⎝⎜

⎞⎠⎟

∂N= τ log n

nQ

⎛

⎝⎜⎞

⎠⎟

p = −∂F∂V

=τNV

⇒ pV = Nτ = NkBT with nQ =

Mτ2π2

⎛⎝⎜

⎞⎠⎟

32


Barometric Equation: Three different waysµ=constant

Thermodynamic equilibrium between slabs at various z

Single particle occupancy of levels at different altitude

Hydrodynamic equilibrium

Generalizes to case when the temperature is not constant

At constant temperature µtotal=C ⇔τ logn z( )nQ

= −Φ z( ) + C ⇔ n z( ) = n0 exp −mgzτ

⎛⎝⎜

⎞⎠⎟

n z( )∝ prob εK , z( )∝ exp −εK +Φ z( )

τ⎛⎝⎜

⎞⎠⎟⇒ n z( ) = n0 exp −

mgzτ

⎛⎝⎜

⎞⎠⎟

− p + dp( )dA

pdA−mgn z( )dAdz

pdA − p + dp( )dA − mgn z( )dAdz = 0⇒ 1n z( )

∂p∂z

= −mg

pV = Nτ ⇔ p z( ) = n z( )τ ⇒1

n z( )∂n∂z

= −mgτ


Internal Degrees of Freedom (Optional)In general, we need to add internal degrees of freedom

e.g. for multi-atomic molecules Rotation ≈ 10-2 eV

Vibration ≈ 2 10-1 eV similar to harmonic oscillator

Spin in a magnetic field etc..

Rotation energy ε j = j j +1( )ε0 with multiplicity 2 j+1

Zint = 2 j+1( )j∑ exp −

j j +1( )ε0

τ⎛⎝⎜

⎞⎠⎟≈ 2 j+1( )

−1/2

∞

∫ exp −j j +1( )ε0

τ⎛⎝⎜

⎞⎠⎟dj

= τεo

exp −x( )x0

∞

∫ dx with x = j j +1( )ε0

τ dx =

2 j +1( )ε0

τdj, x0 = −

ε0

4τ

⇒ Zint ≈τεo

exp εo4τ

⎛⎝⎜

⎞⎠⎟≈τεo

for τ >>ε0 ⇒ for τ >>ε0 U =τ 2∂ logZ

∂τ= τ Crot = kB

For τ < < ε0, keep the first terms of the discrete sum Zint ≈ 1+ 3exp −2 εoτ

⎛⎝⎜

⎞⎠⎟+ ...

logZint ≈ 3exp −2 εoτ

⎛⎝⎜

⎞⎠⎟ U =

τ 2∂ logZ∂τ

≈ 6ε0 exp −2 εoτ

⎛⎝⎜

⎞⎠⎟ Crot ≈ 12kb

εoτ

⎛⎝⎜

⎞⎠⎟

2

exp −2 εoτ

⎛⎝⎜

⎞⎠⎟≈ 0

Zint = exp −nωτ

⎛⎝⎜

⎞⎠⎟n=0

∞

∑ =1

1− exp − ωτ

⎛⎝⎜

⎞⎠⎟≈

τω

for τ >>ω ⇒ Cvib = kB

-1/2 0 1 2 3 4 5 j

In most cases, the kinetic and internal degrees of freedomare independent, for one particle Z = Zkin

kinetic Zint

internal

Diatomic moleculescf. KK fig 3.9 kB

log τ

CV

kBRotation

Vibration

Slightly betterapproximationthan 0


Exchanging Particles with an Ideal gasLet us consider again a system with an energy level ε , which can now be occupied by a variable number N of particles

We put in contact with a thermal reservoir constituting on a large number Np-N of free particles. Energy and particles can be exchanged with the reservoir.

We consider a single state with N particles and an energy NεWe can compute the number of states of the overall system as a

function of Ν and ε .

gRes

N

exp −ε − µ( )N

τ⎛⎝⎜

⎞⎠⎟

g = gRes ×1 = e5 /2

M2π2

2 U − Nε( )3(Np − N )

⎛

⎝⎜⎞

⎠⎟

3/2

Np − NV

⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Np −N

<=Direct Counting

or Seckur Tetrode

= e5 /2 1+ NNp − N

⎛

⎝⎜⎞

⎠⎟

5 /2⎡

⎣⎢⎢

⎤

⎦⎥⎥

Np −NM2π2

2U3Np

⎛

⎝⎜⎞

⎠⎟

3/2

Np

V

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Np −N

1− NεU

⎛⎝⎜

⎞⎠⎟3/2⎡

⎣⎢⎢

⎤

⎦⎥⎥

Np −N

=

Np→∞

U =32Npτ

⎯ →⎯⎯⎯ e5 /2nQn

⎡⎣⎢

⎤⎦⎥

Np

exp −N lognQn

⎛⎝⎜

⎞⎠⎟exp −

Nετ

⎛⎝⎜

⎞⎠⎟∝ exp −

N ε − µ( )τ

⎛⎝⎜

⎞⎠⎟⇐

lim 1+ aN

⎛⎝⎜

⎞⎠⎟N

N→∞⎯ →⎯⎯ exp a( )

bN ≡ exp −N logb( )


S,ε

20

Gibbs factorSystem in contact with thermal bath (reservoir)

Exchange of particles

# States in configuration

At equilibrium

Notes: • Z = grand partition function

Total system =Our system + reservoirConfiguration: 1 state s of Sof energy εs and # of particles Nis+ energy of reservoir

R , Uo-ε

1× gR Uo − ε s ,Noi − Nis( ) Prob configuration( )∝ gR Uo − ε s ,Noi − Nis( )

⇒

Pr ob state ε1

, Ni 1

( )

Pr ob state ε2

, Ni 2

( )=

gR

Uo− ε

1, N

oi− N

i1( )

gR

Uo− ε

2, N

oi− N

i 2( )

= exp −∂σ

R

∂Uε 1 − ε 2( ) −

∂σR

∂Ni

Ni1 − Ni 2( )

i

∑⎛ ⎝ ⎜ ⎞

⎠ ⎟

Absolute activity λi = exp µi

τ⎛ ⎝

⎞ ⎠

⇔ Prob state s with ε,Nis( ) = 1Z

exp −ε s,Nis( ) − µiNis

i∑

τ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ =

1Z

exp −ε s,Nis( ) − µiNis

i∑

kBT

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

with Z = exp −ε s,Nis( ) − µiNis

i∑

τ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟all states s,Nis

∑


Grand Partition FunctionDefinition

Note: ε(s,Ν) depends on Ni both through through increase of N, and the interactions

In weak interaction limit (ideal gas) and only 1 species ,

Note If the Nis =Ni are constant,

the probabilities probabilities are the same as with Z

This formalism is useful only if the Nis change.

Z = expµiNi

i∑

τ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ Z Ni( )

Z = expµiNis −

i∑ ε s,Nis( )

τ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ all states s

and occupation Nis

∑

ε s,Ns( ) = Ns ε senergy of each level

⇒ Z = exp −ε s − µ( )Ns

τ⎛⎝⎜

⎞⎠⎟s,Ns

∑

Prob Ns particles in state s( ) = 1Z

exp −ε s − µ( )Ns

τ⎛⎝⎜

⎞⎠⎟


Examples few state systems Kittel-Kroemer 140-145

Gas adsorption (heme in myoglobin, hemoglobin, walls)Assume that each potential adsorption site can adsorb at most one gas molecule. What is the fraction

of sites which are occupied?State 1= 0 molecule => energy 0State 2= 1 molecule => energy ε

If ideal (applies equally well to gas and dissolved gas and any low concentration species)

Langmuir adsorption isotherm

Z = 1+ exp µ − ε

τ⎛⎝⎜

⎞⎠⎟

Fraction occupied = f =exp µ − ε

τ( )1 + exp µ − ε

τ( )=

1

exp ε − µτ( ) +1

= same form as Fermi- Dirac

f = p

τ nQ expετ( ) + p

µgas = τ log nnQ

⎛

⎝ ⎜ ⎞

⎠ ⎟ = τ log p

τnQ

⎛

⎝ ⎜ ⎞

⎠ ⎟ ⇐ pV = Nτ n =

NV

=Pτ


Ionized impurities in a semiconductor cf Kittel and Kroemer p370

Donor= tends to have 1 more electron (e.g. P in Si)State 1= ionized D+ : no electron => energy= 0State 2=neutral 1 extra electron spin up => energy =εd

State 3= neutral 1 extra electron spin down => energy =εd

Acceptor= tends to have 1 fewer electron (e.g. Al in Si)State 1= ionized A- : one electron completing the bond=> energy= εa

State 2= 1 neutral: missing a bond electron spin up => energy =0 State 3= 1 neutral: missing a bond electron spin down => energy =0

µ determined by electron concentration in semiconductor = Fermi level

Z=1+ 2exp −εd −µτ

⎛⎝⎜

⎞⎠⎟

Fraction ionized f = 1

1+ 22 spin stateswhen donoris neutral

exp − εd − µτ

⎛⎝⎜

⎞⎠⎟

εd = energy of electron on donor site K&K use ionization energy I=-ε d

Fraction ionized f =exp − εa − µ

τ⎛⎝⎜

⎞⎠⎟

exp − εa − µτ

⎛⎝⎜

⎞⎠⎟ + 2

2 spin stateswhen acceptor

is neutral

=1

1+ 2exp − µ − εaτ

⎛⎝⎜

⎞⎠⎟

εa = energy of electron on acceptor site


Grand Partition FunctionRelation to thermodynamic functions

• Mean number of particles

• Entropy

• Energy

• Free Energy

• Grand Potential Ω defined as

σ = − ps,NisNis∑

s∑ log ps,Nis( ) = logZ−

µi

τNis

i∑ +

ετ

=∂ τ logZ( )

∂τ V ,µi

U = ε = τ 2 ∂ logZ

∂τ+ µi Ni

i∑ = τ 2 ∂

∂τ+ τµi

∂∂µii

∑⎛⎝⎜

⎞⎠⎟logZ

F =U − τσ = −τ log Z+ µi Ni

i∑ G = F + pV

Ω ≡ F − µi Ni

i∑ = −τ logZ

Ni = Nis ps,NisNis∑

s∑ =

Nis

Zexp

µiNis −i∑ ε s,Nis( )

τ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟Nis

∑s∑ = τ ∂ logZ

∂µi τ ,V


FluctuationsMicroscopic exchanges

In a system in equilibrium, exchanges still go on at the microscopic level! They are just balanced!

Macroscopic quantities= sum of microscopic quantities ≈ N x mean

But with a finite system, relative fluctuations on such sum is of the order 1/√N

e.g., fluctuation on total energy

Computation with partition function; By substitution we can see that

Similarly when there is exchange of particles

U = ε = ε s pss∑ = ε s

e−ε sτ

Zs∑

σE2 = E − E( )2 = E2 − E 2 = εs

2 e−εsτ

Zs∑ − εs

e−εsτ

Zs∑⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2U = E =

τ 2∂ log Z∂τ

σE2 = E2 − E 2 =

τ2∂U∂τ

=τ 2∂τ

2∂log Z∂τ

∂τ

Variancenot entropy!

N =

τ∂ logZ

∂µ τ ,V

and σ N2 = N 2 − N 2 =

τ∂τ∂ logZ

∂µ∂µ

=τ∂ N∂µ


ExamplesFermi Dirac

Fermions= particles with half integer spinPauli exclusion principle: each state contains at most one particle

26

For a state of energy ε

Z = 1+ exp −ε − µτ

⎛⎝⎜

⎞⎠⎟

N =τ∂ logZ

∂µ τ ,V

=exp − ε − µ

τ⎛⎝⎜

⎞⎠⎟

1+ exp − ε − µτ

⎛⎝⎜

⎞⎠⎟=

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ +1

= Prob 1 particle( )

σ N2 =

τ∂ N∂µ

=exp ε − µ

τ⎛⎝⎜

⎞⎠⎟

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ +1

⎡⎣⎢

⎤⎦⎥

2 = N 1− N( ) ≤ N Less fluctuation than a Poisson distribution


ExamplesBose Einstein

Bosons= particles with integer spin

27

For a state of energy ε

Z = exp −ε − µ[ ]N

τ⎛⎝⎜

⎞⎠⎟N

∑ =1

1− exp − ε − µτ

⎛⎝⎜

⎞⎠⎟

N =τ∂ logZ

∂µ τ ,V

=exp − ε − µ

τ⎛⎝⎜

⎞⎠⎟

1− exp − ε − µτ

⎛⎝⎜

⎞⎠⎟=

1

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

σ N2 =

τ∂ N∂µ

=exp ε − µ

τ⎛⎝⎜

⎞⎠⎟

exp ε − µτ

⎛⎝⎜

⎞⎠⎟ −1

⎡⎣⎢

⎤⎦⎥

2 = N 1+ N( ) ≤ N More fluctuation than a Poisson distribution


Minimization of the Free Energy Consider a system in interaction with reservoir imposing

temperatureConstant exchange of energy (+ in some case volume, particles)=> Fluctuations, probability distributionContrary to a system which is able to loose energy to vacuum, system does

not evolve to state of minimum energy (constantly kicked)!Contrary to an isolated system, does not evolves to a configuration of

maximum entropy! (The entropy of the combination of the reservoir and the system is maximized)

It evolves to a configuration of minimum Free Energy“Balance between tendency to loose energy and to maximize entropy.”

More rigorouslyWe cannot define the temperature of the system out of equilibrium. So we

introduce

They are minimized for the most probable configuration(≈equilibrium)! Can be generalized to any variable! (Will be useful for phase transitions)

FL US( ) ≡US − τ Rσ S US( ) Landau Free Energy (constant volume)GL US ,Vs( ) ≡US − τ Rσ S US ,VS( ) + pRVS Landau Free Enthalpy (constant pressure)


Landau Free EnergyLet us consider a system S exchanging energy with a large reservoir R.(We are no more considering a single state for system S).

S,USR , U-Us

FL

US

Constant volumeσ tot US( ) = σ R U −US( ) +σ S US( )

σ R U( ) − ∂σ R

∂UUS +σ S US( )

= σ R U( ) − 1τ RUS +σ S US( )

= σ R U( ) − 1τ R

FL US( ) with

FL US( ) ≡US − τ Rσ S US( ) ≡ Landau Free EnergyMost probable configuration maximizes σ tot i.e. minimizes FL US( )Note that this implies

∂FL US( )∂Us

= 0⇔∂σ S US( )∂US

=1τ R

e.g. for a perfect gas

σ S US( ) = N log

MUS

3π2N⎛⎝⎜

⎞⎠⎟3/2

NV

⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

+52

⎧

⎨⎪⎪

⎩⎪⎪

⎫

⎬⎪⎪

⎭⎪⎪

(Sackur Tetrode with τS =2Us

3N)

⇒US =32Nτ R


Exchange of ParticlesIf we exchange particles, what is minimized is

in a way similar to the previous slide

30

ΩL US ,NS( ) ≡US − τ Rσ S US ,NS( ) −µRNS Landau Grand Potential

σ tot US ,NS( ) = σ R U −US ,N − NS( ) +σ S US ,NS( ) σ R U( ) − ∂σ R

∂UUS −

∂σ R

∂NNS +σ S US ,NS( )

= σ R U( ) − 1τ RUS +

µR

τ RNS +σ S US ,NS( )

= σ R U( ) − 1τ R

ΩL US ,NS( ) with

Most probable configuration maximizes σ tot i.e. minimizes ΩL US ,NS( )Note that this implies

∂ΩL US ,NS( )

∂Us

= 0⇔∂σ S US ,NS( )

∂US

=1τ R

∂ΩL US ,NS( )∂Ns

= 0⇔∂σ S US ,NS( )

∂NS

= −µR

τ R

5. systems in contact with a thermal...

Documents