Statistical Mechanics 2014 Lecture Notes by Amit Charkrabati

7/21/2019 Statistical Mechanics 2014 Lecture Notes by Amit Charkrabati

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1

Physics 971 Statistical Mechanics

Lecture 1

Important experimental fact: we can answer many questions concerning systems in or near

thermodynamic equilibrium without solving the equations of motion of the particles in detail.

Thermodynamic Equilibrium

1. A system in contact with a heat reservoir for a long time.2. A large isolated system.

Experimental Observation

Pressure of a fixed quantity of gas at constant temperature same as in 1914 and in 2014.

Measurements are done over time intervals covering millions of atomic collisions or vibrations --- i.e., a

time average. What is going on with the system during this time?

Consider Hamilton’s equation of motion --- position q and momentum p are conjugate variables. N atoms 3N coordinates and 3N momenta.

Evolution graphically shown as a single orbit in the 6N dimensional space of p’s and q’s. Such a space

is known as phase space or Γ space.

[p] 3N momenta[q] 3N coordinates

Experimental Measurement: time averages over a segment of the orbit in the phase space of the system.

Example of phase space: Linear Harmonic Oscilllator

Hamilton

221

2 2

p H T V kq

m= + = +



2

What is the phase space?

( )

( )

0

cos /

sin

H pq

p m

H

p kqq

k q q

m

q A t k m

p m A t

ω φ ω

ω ω φ

∂= + =

∂

∂= − = −

∂

⇒ + =

= + =

= − +

2 2

1q p

A m Aω

+ =

Energy of the System

E is a constant of motion as the Hamiltonian does not have explicit time dependence.

Hamilton's Eqn. of motion

0 if there's no explict time-dependence

dH H H H q p

dt t q p

H H H H H

t q p p q

H

t

∂ ∂ ∂= + +

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂= + + −

∂ ∂ ∂ ∂ ∂

∂

= ∂=

2 21 (can be calculated from conservation of energy)

2 E m Aω =

( )

2 2 2 2

21 1

22 /

q p q p

A m A mE E mω ω

+ = ⇒ + =

Phase space is an ellipse.

We need to compute time averages over a segment of the orbit in phase space to find any measuredquantities. But time averages are difficult to perform

( ) x t

6N dimensional vector (3N positions, 3N momenta)

φ : some observable ( )0

0

1 ( )

t T

t dt x t

T φ φ

+= ∫

T τ >> , the relaxation time → characteristic time required for fluctuations in the properties of the system

to die out.

Since we cannot solve the detailed dynamics for ( ) x t

we cannot compute the time average.



3

Gibbs: systems appear to randomize themselves between observations ( )T τ >> . Instead of computing

a time average, imagine a group of similar systems (satisfying all external requirements) but

suitably randomized, and take averages over this group at one time.

The group of similar systems is called an ensemble of systems and is to be viewed as an intellectual

construction to simulate and represent at one time the properties of the actual system as developed in thecourse of time.

The ensemble is randomized suitably in the sense that every configuration of q and p accessible to theactual system in the course of time is represented in the ensemble at one instant of time.

Problem of Demonstrating time average ≡ ensemble average is the subject of Ergodic theory, which wewill not go into in detail.

We need to specify an ensemble through the probability density0( ) x ρ

for a selected member of the

ensemble to occupy the phase space point x

. Then the ensemble average

( ) ( )0

normalized

dx x xφ ρ φ =

↓

∫

I will introduce here a simple system of a collection of spins to discuss the concept of ensemble.

Question: As q, p move via Hamilton’s Eqn. how does the ( , ; ) p q t ρ change?

Liouville’s Theorem:

3 3( , ; ) N N q p t d q d p ρ = # of systems in the volume element3 3 N N d q d p at any time t around the point

q, p

Physical quantity f (q,p) ;

3 3

3 3

( , ) ( , ; )

( , ; )

N N

N N

f q p q p t d q d p f

q p t d q d p

ρ

ρ = ∫

∫

If the total # of systems in the ensemble does not change with time then ρ must satisfy the continuity

equation.

3 3 N N d d q d pω =



4

Change

d t

ρ ω ∂

∂ ∫

“flow” out

( )

( )

( ) ( )

ˆv

v

v 0 ; v ,

space dependent density

n d

d

q pt

ρ σ

ρ ω

ρ ρ

⋅

≡ ∇ ⋅

∂⇒ + ∇ ⋅ = ≡

∂

↑

∫

∫

We want the density function to be stationary so that < f > does not change w/time (equilibrium)

0t

ρ ∂

⇒ =∂

Thus

( )

( ) ( )3

1

3 3

1 1

v 0

0 N

i i

i i i

N N i i

i i

i ii i i i

q pq p

q pq p

q p q p

ρ

ρ ρ

ρ ρ ρ

=

= =

∇ ⋅ =

∂ ∂+ =

∂ ∂

∂ ∂∂ ∂⇒ + + +

∂ ∂ ∂ ∂

∑

∑ ∑

0

;

Hamilton's equation of

i i

i i

H H q p

p q

=

⇓

∂ ∂= = −

∂ ∂

motion

(Note that v 0∇ ⋅ =

for an incompressible fluid; here it corresponds to the conservation of phase space

volume)

3

1

0

v 0

N

i i

i i i

q pq p

ρ ρ

ρ

=

∂ ∂⇒ + =

∂ ∂

⇒ ⋅∇ =

∑

Equivalently,

3

1

0Liouville's Theorem

[ , ] 0 Poisson's Bracket

N

i i i i i

H H

q p p q

H

ρ ρ

ρ

=

∂ ∂ ∂ ∂− =

∂ ∂ ∂ ∂ ⇒ =

∑



5

Choices of ( , )q p ρ

1. ( , )q p ρ =constant ( Microcanonical Ensemble)

Uniformly distributed over the region of phase space—uniformly distributed over all possible

microstates.

2. ( , ) [ ( , )]q p H q p ρ ρ = i.e. any function of the Hamiltonian. Then the Poisson bracket is satisfied

We will see later that a natural choice is

( , )/( , ) H q p kT q p e ρ −∝ (Canonical Ensemble)

The Microcanonical Ensemble

•

Number of molecules N, volume V, and energy E --- all fixed.

• Motion of the N particle system will be confined to a (6N-1) dimensional hyper surface given by

( , )q p E =H --- we will call this surface ( ) E Γ .

• Instead of fixing the energy exactly at E we consider it to be

1 1( , )

2 2 E E q p E E

− ∆ ≤ ≤ + ∆

H

Actually this particular form does not matter. One can equivalently consider

( ) ( , ) E E q p E − ∆ ≤ ≤H , or ( )( , ) E q p E E ≤ ≤ + ∆H

With a specified macrostate, a choice still remains for the system to be in any one of a large # ofmicrostates.

• Uncertainty or tolerance E ∆ must be small i.e., it should not play any role in the thermodynamic limit

, , const N

N V V

→ ∞ → ∞ →

. This would require . E E ∆ On the other hand, we

need to allow many microstates.

E

E N ∆ So we should have

E

E E N ∆ .

Probability density function

1 1constant C if ( , )

( , ) 2 2

= 0 otherwise

E E H q p E E q p ρ

= − ∆ ≤ ≤ + ∆



6

• Equal a priori probabilities for different accessible regions of equal volume in the phase space.

• During any significant time in interval τ the phase point ( ) x t

will spend equal time intervals in all

regions of ( ) E Γ --- Ergodic Hypothesis

• This hypothesis is sufficient to allow us to replace a time average by an unweighted average over the

surface ( ) E Γ . However this is not necessary. A hypothesis that is weaker than the ergodic hypothesis

but still leads to Statistical Mechanics as we know it is called the mixing hypothesis. Mixing refers to

the notion that an initial compact distribution of points on ( ) E Γ very quickly distorts into a convoluted

object which permeates the entire surface while still occupying the same volume as required byLiouville’s theorem. (For more on this see the book by Balescu).



1

Lecture 2

Density of States

Consider N distinguishable particles (such as a perfect crystal or a polymer chain --- particles that can be

identified by their position). Volume of the phase space accessible to the system is the integral3 3

1 1( , )

2 2

1( ) N N

E E q p E E

E d q d pC

− ∆ ≤ ≤ + ∆

Ω = ∫

H

'3 3

3

1Make ( ) dimensionless; ( ) N N

N E E d q d p

hΩ Ω = ∫ , where h is the unit of action and

'

∫ means

total # of states with the energy in the specified range.

Choice of h is to make contact with Bohr-Sommerfield semi classical theory pdq nh=∫ and uncertainty

principle p q h∆ ∆ ≈

If we are dealing with indistinguishable particles, we do not distinguish between states which differ only by the labeling of the particles occupying the various regions of phase space. In this case our measure ofthe number of available states in the ensemble becomes

3 3

3

'1( )

!

N N

N E d q d p

h N Ω = ∫

Historically Gibbs introduced the N! term to remove a spurious entropy of mixing (Gibbs paradox)

0( ) E Ω = Total # of states for H E ≤ .

Density of States: 0 ( )( ) E g E E

∂Ω=∂

If E ∆ is small, ( ) ( ) E g E E Ω = ∆

Entropy in Statistical Mechanics

Thermodynamics tells us that entropy is a state function in the sense that the value of the entropy does notdepend explicitly on the past history of the system but only on the actual state of the system. However, inthermodynamics we do not have the tools which would enable us to understand physically what entropy

means in terms of the condition of the system.

We define S k n= Ω (without any additive constant. This is absolute value.)

We need the following properties of entropy to be satisfied:

1. dS is an exact differentialQ

T

δ =



2

2. 1 2S S S = + (additive)

3. 0S ∆ ≥ for an isolated system (thermally closed)

1. Entropy as defined above has a definite value for an ensemble in statistical equilibrium; thus the

change in entropy is an exact differential. If we define ( ) E Ω as a measure of the imprecision of our

knowledge or as a measure of randomness, then entropy also is a measure of these things.

2. Non-interacting system made of two parts with 1 N particles and 2

N particles

1 2Ω = Ω Ω

Phase space of the combined system is the product spaces of the individual parts

1 4

1st system 2 2nd system 5

3 6

Total system will have a total 9 states (such as 1,4; 2,5; etc.)

Then 1 2S S S = + (Note that N! to be replaced by 1 2! ! N N )

3. We need to assume that the equilibrium condition of a system is given by the most probablecondition. Then the volume of phase space with equilibrium properties will be a maximum. And the

entropy of a closed system has its maximum when the system is in the equilibrium condition.

Setting up Thermodynamics

( , , )

# of i-th chemical species

external parameters like Volume V, Magnetization M

iS S E x N =

↓

For simplicity, consider x V ≡

i

i i

S S S dS dE dV dN

E V N

∂ ∂ ∂ = + + ∂ ∂ ∂

∑

1

i i

i

dE PdS dN

T T T µ = + − ∑

and Chemical Potential i

i

S

T N

µ ∂= −

∂



3

Consideri

N fixed and compare with the 2nd Law of Thermodynamics

2nd Law of thermodynamicsTdS dE PdV = +

Then , Temperature

1 S

T E

∂

= ∂ Pressure

P S

T V

∂

= ∂

Generalizing further, Chemical Potential i

i

S

T N

µ ∂= −

∂

Entropy of a perfect gas using the Microcanonical Ensemble N non-interacting point particles

32

1

1( , ) with constraint ( , )

2

N

i

i

q p p E E q p E m =

= − ∆ ≤ ≤∑H H

1 3 1 3 1 3 1 33 3 3

2

' '1( )

! ! !

with the constraint 2 ( ) 2

N N

N N N N N N N

i

i

V V E dq dq dp dp dp dp dp dp

h N h N h N

m E E p mE

Ω = = =

− ∆ ≤ ≤

∫ ∫ ∫ ∫

∑

The accessible volume is that of a 3N dimensional hyper shell bounded by two hyperspheres of radii

2 ( ) and 2m E E mE − ∆

Volume of a Hypersphere

Equation of hypersphere2 2 2 2

1 2 n x x x R+ + + =

Start with the Gaussian Integral

( )

1

2 2 21 2

2 /2

n

n n

n x x x

x n

I e dx dx

e dx π π

∞

−∞

∞

−∞

− + + +

−

=

= = =

∫

∫

But in terms of sphere radius R (“polar” co-ordinate)

2 1

n 10

0

vol …element dV V= n

R n

n

R

nC R dR

I e C R dR− −

−

∞

⇒

=

∫

∫

Make a change of variable2t R=

1 2

1/22 2

0 0

1

2 2

Gamma Function2 2

n n

t t nn

n

C I C e t t dt e t dt

C n

− −− − −∞ ∞

= =

= Γ ←

∫ ∫



4

/2 /2

= 1 ! if is an integer 2 2 2

2 2

1 !

2 2

n

n n

n

C n n

C n n

π π

−

⇒ = ≡

− Γ

Then

/2 /21

0( )

! 12 2

n nn n n

n

RV R C R dR R R

n n

π π −= = =

Γ +

∫

Just consider ( ) nV R CR=

Volume of a shell of thickness s

( ) ( )

( )

1 1

S

n n

n

n

V V R V R

CR C R

CR R

= − −

= − −

= − −

s

s

s

/

/1 1

1 Note that: lim 1

as long as .

n

n

n

n nS R x

n

n

n RCR

n

xCR e e

n

CR n R

− −→∞

= − −

≈ − − =

≈ ⋅

s

s

Practically the volume of the whole sphere

( ) ( )

23

3 ~ 10 2

2 2 2 1 1

12

2

n N R mE

E m E m E E mE

E

E E mE m

E E

= = ∆

= − − ∆ = − −

∆ ∆≈ ⋅ ≈ ⋅

s

Then the condition n R⋅ s leads to E

N E E

E N E E E

N

∆⋅

⇒ ⋅ ∆ ⇒ ∆

We have already discussed this before.

Then ( )3 /2

3 /2

3( ) 2

3!!

2

N N N

N

V E mE

N N h

π Ω =



1

Lecture 3

( ) ( )

233 ~ 10 2

2 2 2 1 1

12

2

n N R mE

E m E m E E mE E

E E mE m

E E

= =

∆ = − − ∆ = − −

∆ ∆≈ ⋅ ≈ ⋅

s

Then the condition n R⋅ s leads to E

N E E

E N E E E

N

∆⋅

⇒ ⋅ ∆ ⇒ ∆

We have already discussed this before.

Then ( )3 /2

3 /23

( ) 23!

!2

N N

N N

V E mE N N h

π Ω =

( )3/23/2

3

32 ( !) !

2

S k n

S V N N n mE n N n

k hπ

= Ω

= ⋅ − −

Stirling’s Formula !n N N nN N ≈ −

( )

( )

3/23/2

3

3/23/23/2

3/23

3 3 32

2 2 2

1 52

23

2

S V N N N N n mE N nN N n

k h

V E N N n m

N h N

π

π

= − + − +

= ⋅ ⋅ ⋅ +

3/2 3/2

3

4 1 5ln3 2

m V E S Nk Nk N N h

π

= × × + (1)

Now

,

1 3 3 1 3

2 2 2V N

S Nk nE Nk E NkT

T E E E

∂ ∂ = = = ⇒ =

∂ ∂

and

,

1

N V

P S Nk PV NkT

T V V

∂ = = ⇒ =

∂



2

Using3

2 E NkT = , one can write,

4S Nk n=

2

3

mπ 2

3

h⋅

2

( )

3/2

1/2

5

2

where the denominator is the "average thermal momentum" of a molecule2

V kT Nk

N

h

mkT λ

π

+

=

(Note: actually the r.m.s. velocity is given by8

v kT

mπ = )

3

/ 5

2

Ratio of volume per particle/volume associate with de Broglie wavelength

V N S Nk n Nk

λ

= +

1/3

(Thus the classical description is valid for where V R R N

λ ≈

is the mean n-n distance).

Chemical Potential of a Perfect gas:

,

5/ 2

We rewrite Eq. (1) as

5

2

E V

S

T N

S Nk n A N Nk

µ

−

∂ − = ∂

= ⋅ +

5/ 2 5[ ]

2

S k n AN k

N

−∂= −

∂

5

2

k + 5/ 2[ ]k n A N −=

3 N

kT nV

λ µ

= +

Note that PV NkT = 3Thus, ( )

PkT n kT nP f T

kT µ λ

= = +

, a very useful expression.

Other Examples of Non-Interacting Systems

2. Two level systems.

N magnetic moments --- Classical “spins”; distinguishable

B applied magnetic field

Each spin can be parallel or antiparallel to the field

0 0

0 0

AP

P

E B

E B

µ ε

µ ε

= =

= − = − Two energy levels.

Fix the energy to some value 0 E M ε = ⋅ Note that



3

all spin parallel all spin antiparallel

, 2, , M N N N ↑ ↑

= − − +

Now

0

0

0 0

# of spins in state

# of spins in - state

N

N E N N

M N N

ε

ε ε ε

+

−

+ −

+ −

= +

== −

⇒ = −

How many possible ways of choosing N − particles out of N so that the energy is still 0 M ε ? This

number is the # of microscopic states.

!

( )!( )!

!(Interchanging among or gives nothing new)

! !

N

N N N

N N N

N N

− −

− +

− +

Ω =−

=

Now

1 1( ) and ( )

2 2

1 1 1 1( ) ( ) ( ) ( )

2 2 2 2

N N M N N M

S k n

k N nN N M n N M N M n N M

− += − = +

= Ω

≈ − − − − + +

1 1 Note: Terms ( ) ( ) canel

2 2 N N M N M

− + − + +

, 0

0

1 1

1 1 1( )

2 2 2

N V

S S

T E M

k n N M

ε

ε

∂ ∂ = = = ∂ ∂

= − −

( N M −

1)

1

2( N M −

1 1 1 1( )

2 2 2 2)

n N M

− − + −

( N M +1

)1

2( N M+

0

0

1

2

1 1

2

k N M n

N M

k N M n

T N M

ε

ε

−=

+

−⇒ =+

Note for 0 ( 0) 0 M E T > > < (Negative temperature!)

The system is not normal in the statistical sense, but for 0( 0) M E < < the system is normal



4

Now 02 /kT N N M e

N N M

ε −

+

−= =

+

0

0 0

0

0 0

/

/ /

/

/ /

Probabilities of finding any

one particle in the states

and

kT

kT kT

kT

kT kT

N e

N e e N e

N e e

ε

ε ε

ε

ε ε

−

+

−

−

−

=

+⇒

= +

Canonical distribution.

Then

( ) ( )0 0 0tanh / E N N N kT ε ε ε + −= − = −

Specific Heat

2

20 0/ cosh

dE C

dT

Nk kT kT

ε ε

=

=

3. Generalization of the 2-State System

K-States particles in each statei inε

1 2

! !

! ! ! !K ii

N N

n n n nΩ = = Π

S k n

k N n N N

= Ω

= −

i in n n n− +∑

i

i i

i

k N n N n n n

= −

∑

∑

Schottky specific heat“anomaly” characteristic ofsystems with a gap excitation

energy02 E ε ∆ −



5

Constraints:

1 1

K K

i i i

i i

E n N nε

= =

= =∑ ∑

Maximize ( , are Lagrange's Multipliers)i i iS S n nα β ε α β = − −∑ ∑

0i

S

n

∂

=∂

1/

/

1 0

i

i

i i

k k i

k

k n n

n e e

C e

α βε

βε

α βε

− − −

−

⇒ − − − − =

⇒ =

= ⋅

But

/1i

i

k

n N

N N C

Qe βε −

=

⇒ = =

∑

∑

Note also,1

1 (1 ) 1 N

n C k n C k nk Q

α α = − − ⇒ = − + = − +

Then

[ ]

1

/1

1

/ /

1 1 1"

/

1

( / ) /

Note that the last term is as

i

i i

i

i i

k i

k k i

Q

k i i i

S k N n N n n n

N k N n N e n N Q k

Q

N N N k N n N n e e

Q Q Q k

N n n e

k Q

βε

βε βε

βε

βε

β ε

β ε

−

− −

−

= −

= − −

= − +

=

∑

∑

∑ ∑

∑

S k N n= N N n− 1

1

1

1

1

1

1 Note that ; 's are the only variables, so is ind of E

1

But we know,

1

i

N N nQ E k

Nk n Q E

S n Q

T E

E S Nk n Q

Nk E S n Q

E TS F

E TS NkT n QT

β

β

β

β

β β

β

+ +

= +

∂= =

∂

⇒ − = −

⇒ − = −

− =

= − = −

( )1 1We will learn later, ln ln ln N

N F kT Q kT Q NkT Q= − = − = −



1

Lectures 4 and 5 (Lecture 4 was partly HW discussion)

Harmonic Oscillators

Distinguishable

Energy of each one (i-th one) 1 0, 1,22

i in hv n + =

Total energy

1

1

2

N

i

i

n hv

=

+

∑

1Fix

2 E N hv Mhv= ⋅ + So that 1 2 and

2 N

E N M n n n M

hv= + + + = −

( ) # E Ω = of ways distributing M particles in N (labeled boxes) (a box may be empty)

This is same as M while balls + ( N-1) Black balls which work as the barriers.Running index 1,2... M+N-1.

( )( )1 !( )

! 1 ! M N E M N

+ −Ω = −

To see that this counting makes sense, consider a simple example of M = 6 and N=3.

The above counting says that there are( )

( )

6 3 1 !28states

6! 3 1 !

+ −=

−

What are these states?0 0 6 (3 states of these types; remember the harmonic oscillators are distinguishable)1 0 5 (6 states)2 0 4 (6 states)

3 0 3 (3 states)1 1 4 (3 states)

2 1 3 (6 states)2 2 2 (1 state)

Total 28 states

(Q: if the harmonic osc. are instead indistinguishable, how many states are there for general M and N?)

Now, going back to Stat Mech.



2

1 1

S k n

N M

= Ω

>> >>

( ) ( )

1

Use2

S k M N n M N M n M N n N

S S M

T E M E k M N E N

n M hv M hv

= + + − −

∂ ∂ ∂= =

∂ ∂ ∂+

= = −

2 2

2 2

E N N N E hv

k k hvn n E N N hv hv

E hvhv

− + +

= = − −

/

/

1 11 2 2

1 12 2

1

2 1

hv kT

hv kT

E hv E Nhv

k N n e E T hv hv E Nhv N

hv E N hv

e

+ + + +

= ⇒ = − −

⇒ = +

−

Canonical Ensemble

• Difficult to control total energy E ; fixed temperature is more practical.

• On the technical side—microcanonical ensemble is difficult to use in practice because of difficulty in

evaluating the volume of phase space or number of states accessible to the system.



3

Total system can be expressed by a microcanomical ensemble with total Energyt E

( )t t E Ω “volume” of the phase space accessible to the total system for Energy between andt t t E E E δ +

The probability that the total system is an element of volume t d Ω of the appropriate phase space is

t t d C d ω = ⋅ Ω if the energy between andt t t E E E δ +

=0 otherwise

What is the probability that the subsystem is in sd Ω without specifying the condition of the reservoir?

volume of the phase space of then reservoir(so that total energy is in about ).

s s r

t t

d C d

E E

ω

δ

= Ω Ω

↑

Our task is to evaluate r Ω --- if we know the subsystem is in sd Ω , how much phase space is accessible tothe reservoir? (I will discuss a spin-system example to elaborate)

Entropy of Reservoir/ r S k

r r r S k n e= Ω ⇒ Ω =

Now r t s E E E = −

We may take s t E E <<

Expand:

( )( ) ( ) ( ) r t

r r r t s r t st

S E S E S E E S E E

E

∂= − = − +

∂

Then

( ) ( )1exp expr t r t

r st

S E S E E

k k E

∂ Ω = −

∂

Now

/0

Reservoir

( )1

s

t r

r t

t

E kT r

E E

S E

T E

A e−

≈

∂=

∂

⇒ Ω = ⋅

↑

Then

/0

/

e

may be viewed as a normalization constant.

s B

s B

E k T s s

E k T s

d C d A e

A d

ω −

−

= Ω

= Ω

↑



4

For the subsystem then the probability density

/( ) s B E k T s E A e ρ

−=

Normalizing( ) =

where for discrete energy states

s s

s

s

E E s E

E

e e E Qe

Q e

β β

β

β

ρ − −−

−

=

=

∑

∑

Thermodynamic Functions

• Entropy is fixed by the energy independent of whether the system is isolated or in contact with a heat

bath.

Canonical ensemble entropy for mean energy E = Microcanonical ensemble entropy for energy E

0

0

( ) # of states ,

( ) # of states H

( ) density of states =

E E E E

E E

g E E

δ Ω − +

Ω − ≤

∂Ω−

∂

If E δ is small enough ( ) ( )g E E E δ = Ω

0

( )

( ) (1) E

S k n E

E E E δ

= Ω

∂Ω

Ω = ∂

What is E δ -- the range of probable values for canonical ensemble?

Let ( ) p E dE be the canonical ensemble probability distribution that the system will have energy in the

range dE at E . Then

probability density of states

density at E (

( ) ( ) ( )

degeneracy for discrete system)

p E E g E ρ = ⋅

↑ ↑

0 ( )( )

E E

E ρ

∂Ω= ⋅

∂

discuss( ) sharply peaked at and normalized

( ) 1

p E E

p E E δ

=



5

0( )( ) 1

E

E E E

E ρ δ

∂Ω ⇒ =

∂ (2)

Then from (1) and (2)

/11( )( )

B E k T E A e E ρ

−Ω = =

Mean energy E U ≡

/

( )

A+

A F kT

S k n E

E k n

T

F E TS kT n A e

= Ω

= −

= − = ⇒ =

Partition Function and Helmholtz’s Free-energy

/ 1 =i E kT

i

Q e A

F kT n Q

−=

= −

∑

dF SdT PdV dN µ = − − +

( ), ,

, ,

N V N V

N T N T

F S k T n Q

T T

F P kT n QV V

∂ ∂ = − = ∂ ∂

∂ ∂ = − = ∂ ∂

,

V T

F kT n Q

N N µ

∂ ∂ = − = − ∂ ∂

Energy

1Write =

kT β

s

s

E

s E E e E U

e

β

β

−

−≡ = ∑∑

( ) 2 ( ) ( )U F TS kT n Q kT T n Q kT n Q n QT T β

∂ ∂ ∂ = + = − + = = −

∂ ∂ ∂

Alternatively, from the definition of Q ,1

( )Q

E U n QQ β β

∂ ∂≡ = − = −

∂ ∂



6

2-Level Systems

Classical magnetic moments in an external magnetic field H; can only be up or down; energy of 2 states

0 0and H H µ µ + − or 0 where 1i i H µ σ σ = ± The classical statistics should work at high temperatures.

( ) ( )

00

0 0 0 0

0Hamiltonian

ii i

i i

i

i

i

H H

N

i

N H H H H H

i i

H

Q e e

e e e e e

βµ σ βµ σ

σ σ

βµ σ βµ βµ βµ βµ

σ

µ σ

− −

= −∑

= =

= = + = +

∑∑ ∑ ∏

∑∏ ∏

H

Make sure you follow the difference between

iσ

∑ which is the sum over all possible iσ in the

whole system and

iσ

∑ which is the sum over all possible values of the i-th spin iσ

Energy ε + and ε − for each moment

(Distinguishable)( )1

1

N N

N

Q

Q e e Q βε βε −= + =



1

Lecture 6

2-Level Systems (Continued)

Partition Function

Discrete States

( , ) This sum is over all possible states

This sum is over energy states with degeneracy

Ensemble average of any observable, can be written as

s

s

E N

s

E s s

s

s

Q V T e

g e distinct g

f

f e

f

β

β

β

−

−

−

=

≡

< >=

∑

∑

∑

( , )

s E

N Q V T

Partition Function of a Two Level System from Counting Degeneracy

( ) ( )

1 0 2 0

11 0 0 1 0 0 0

1 1

1 0 2 0

1 2

1 2

1 2

( 2 )

1 1 1 1

( )

!

! !

!

! !

! !

! ! ! !

s E N s

s

s

s

N N N

s

N N N N N

N N

Q g e

E N N

N N N

N g

N N

N Q e

N N

N N e e e e e

N N N N N N

β

βε βε

βε βε βε βε βε

ε ε

−

− +

− − −

=

= + −

= +

=

=

= =− −

∑

∑

∑ ∑

Binomial Expansion( )

!(1 )

! !

n N

n

N x x

n N n≡ +

−∑

( ) ( ) 0 0 0 0 0 02

1 1 N N N

N N Q e e e e e e

βε βε βε βε βε βε − − −= + = + = +

Continuous States( , )

( ) ;( , ) ( , )

q p

N N

f e d e E f

Q V T Q V T

β β

ρ

−− Ω= < > =

∫ H

H



2

( , )

3

1( , )

!

q p N N

Q V T e d N h

β −= Ω∫ H

Or in terms of Density of States,0

( , ) ( ) E N Q V T e g E dE β ∞ −= ∫

Perfect Gas2

( , )2

i

i

pq p

m= ∑H

22 3/23 32

3 3

1( , )

! !

ii

N N p p mkT m

N i i i N N i

V Q V T e d q d p e dp

N h N h

β − ∞ −

−∞

∑ = Π = ∫ ∫

( )

2

3/2

3 3

1

32

Gaussian Integral ( )

1

Here = Hence,2

1 1( , ) 2

! !

Thus for non-interacting systems one can write, ( , )!

( , , )2

x

N N

N

N

N

N

I e dx

mkT

V V Q V T mkT

N N h

QQ V T

N

N hF N V T kT n Q NkT n

V mkT

α π α

α

α

π λ

π

∞ −−∞

= =

= =

=

= − =

∫

/2

1

−

( )

3/22

,

,

3/2

2,

2

2 5

2

3 (comes from )

2

V T

N T

N V

F N hkT n

N V mkT

F NkT P

V V

F V mkT S Nk n

T N h

U n Q U F TS NkT

µ π

π

β

∂ = = ∂

∂ = − =

∂

∂ = − = + ∂

∂= − = + =

∂

Partition Function from Density of States

0( , ) ( ) E

N Q V T e g E dE β ∞ −= ∫



3

( )

( )

0

3 /23 /2

0 3

3 33 /2 1 10 2 2

3 0

( ) compute as we did before in Microcanonical

( ) 23! !2

2( ) 1( )

3!1 !

2

N N N

N

N N N N E

E

V E mE

N N h

m E V g E E A e E dE

N E N h

β

π

π − −∞ −

Ω ←

Ω =

∂Ω = = = ∂ −

∫

The integral for ( , ) N

Q V T is of the form0

E e E dE β α ∞ −∫ . Note that the Γ − function is defined as

0( 1) xe x dxα α

∞ −Γ + = ∫

Make a change of variable, E x β = .Then one finds,

0

E e E dE β α ∞ −

∫ =1 0

1 x

e x dxα

α β

∞ −+ ∫ =

1 1

( 1) !α α

α α

β β + +

Γ +=

31

230

2

31 !

2( , )

N

E N N

N

Q V T A e E dE A β

β

−∞ −

−

= = ⋅∫

3 /2

3

1 2( , )

!

N N

N

V mQ V T

N h

π

β

=

Same as obtained by integrating over phase space volume



1

Lecture 7

Single-Particle Density of States

We know that 1one can write, ( , )

!

N

N

QQ V T

N

= where 1Q is the single-particle partition function. In

terms of single-particle density of states ( )a ε , 1Q can be written as

10

( )Q e a d βε ε ε

∞−= ∫

Now, our task will be to evaluate this single particle density of states ( )a ε .

0 ( ) :PΣ # of microstates available to one particle confined to volume V ; its momentum p being P≤

3 3 2 30 3 3 30

1 4( ) 4

3

PV V P d q d p p dp P

h h h

p P

π π Σ = = =

≤

∫ ∫

Now use energy-momentum relation to find 0( )ε Σ . For a non-relativistic particle,

2

2

P

mε = and for an

extreme relativistic particle, Pcε = . For a non-relativistic particle,3/2

0 3

4( ) (2 )

3

V m

h

π ε ε Σ = by

using

2

2

P

mε = . We then find the single particle density of states

3/2 1/203

( )( ) 2 (2 )

d V a m

d h

ε ε π ε

ε

Σ= =

3/2

13 30

13

2( )

1

! !

N N

N

V m V Q e a d

h

Q V Q

N N

βε π ε ε

β λ

λ

∞ − ⇒ = = =

= =

∫

Other Examples:

Free orientable classical dipoles (Paramagnetism)

0

01 1

cos0

1 1

cos

exp cos i

i i

N N

i i

i i

N N H

N i

i i

H H

Q H e βµ θ

θ θ

µ µ θ

βµ θ

= =

= =

= − ⋅ = −

= =

∑ ∑

∑ ∑ ∑ ∏

H



2

0

0

cos cos

1 1

cos1 1

independent dipoles

where

i i

i i

N N H H

i i

H N

e e

Q Q e

βµ θ βµ θ

θ θ

βµ θ

θ

= =

= =

= =

∑ ∑∏ ∏

∑

( )0

20cos

1 000

4 sinh( ) sin (sum changed into integral)

H H Q e d d

H

π π βµ θ π βµ β θ θ φ

βµ = =∫ ∫

Magnetization = mean magnetic moment along the field direction M z

01

cos angular bracket stands for thermal average

comapre with

N

z i

i

M

F F P

H V

µ θ

=

=

∂ ∂ = − = − ∂ ∂

∑

1

0

1 N

N n Q n Q

H H

H N

β β

βµ

β

∂ ∂= + =

∂ ∂

=

4π 0

4

sinh( ) H

π

βµ 0 βµ 2

1

H − 0 0

1sinh( ) H

H βµ βµ + 0cosh( ) H βµ

0 0 0 00

0 0

1 1coth( ) coth( )

( )

N H N H H H

N L H

µ βµ µ βµ β βµ

µ βµ

= − + = −

=

Langevin Function 1( ) coth L x x x

= − where 0 H xkT

µ =

3High temp large 0 ( )or small field 0 3 45

T x x x L x

H

→ → ≈ − +

→

20 0

03 3

z

H N M N H

kT

βµ µ µ ≈ ⋅ = Susceptibility

20

0lim Curie's law

3

z

H

N M C

H kT T

µ χ

→

∂ = = = ∂

m0 0( for per unit vol calculation where is the Avogadro's #) N N N →

Paramagnetism (Quantum)

Magnetic Moment 0 Bg J µ µ = where J is the angular momentum

Applied field is in the z direction

magnetic quantum #

z Bg mµ µ =

↑

, 1, 1,m J J J J = − − + −



3

( )

1

/ / 2 2 /1

(2 1) /

/

( ) substitute

( )

1 1 1sinh 1 / sinh

2 21

B

J g Hm

B

m J

m J mx J x x x J x xJ x J x J

m J

J x J

x

x J

Q e x g JH

Q e e e e e e e e

ee x x

J J e

β µ β β µ

β

=−

=− − − −

=−

+−

= =

= = + + + +

− = = +

−

∑

∑

1

0

( )

( ) where ( ) is the Brillouin function given by

1 1 1( ) 1 coth 1 coth

2 2 2 2

z

z z B

z J J

J

F N M n Q

H H

M N x g JH

B x B x

x B x x

J J J J

β β

µ β µ

µ µ

∂ ∂= − =

∂ ∂

= =

=

= + + −

2 2 ( 1)1 11 ( ) 1

3 3

Even in the Quantum case, susceptibility , but the Curie constant is changed.

B J z

g J J x B x x H

J kT

C

T

µ µ

χ

+ ≈ + =

=

0: 0

1 cosh( / 2 )( ) coth( )

2 sinh( / 2 )

0

1 1 1 ( ) coth( ) coth( ) ( )

Langevin Function2

2

B B

J

J

Classical Limit J g g J

x J B x x

J x J

x

J

B x x x L x x J x

J

µ µ µ → ∞ → →

≈ −

→

≈ − ≅ − ≡

where

0

0 / B x g J H H kT

µ

µ β µ = = . So we recover the Classical Expression.



1

Molecule in an Ideal Gas (Maxwell’s Velocity Distribution)

• One molecule—remaining ones constitute the reservoir

221

2 2

p E mv

m

= =

Probability3 3( , )P r p d r d p

that the molecule has position lying in the range ( ),r r dr +

, momentum

( ), p p d p+

23 3 3 3 ( /2 )

does not depend on ; can beintegrated out

( , ) p m

r r

P r p d r d p d r d p e β −∝

( ) 3P p d p →

Probability that momentum is in the range ( ), p p d p+

and that r

could be anything

( ) ( )2

2

3 3 3 /2

integralover

/2 3

and is a constant.

p m

r

p m

P p d p d r d p e

V e d p V

β

β

−

−

∝

∝ ⋅

∫

Since v, p m=

Velocity Distribution is given by2

3 v /2 3(v) v vmP d e d β −∝

or,

2v /2

(v) normalization constant

m

P C e

β −= ⋅

↑

• Molecule in an Ideal gas in the presence of gravity

2

2

pmgz

m= +

H

We want to figure out pressure (z). As before,

( )

( )

2

2

2

3 3 3 3 ( /2 )

/23 3 ( ) 3

integrate out

/23 3

( , )

( )

( ) same as before

p m mgz

p mmgz

p m

P r p d r d p d r d p e

P p d p d r e d p e

P p d p C e d p

β

β β

β

− +

−−

−

∝

∝

= ⋅

∫



2

Probability of finding a molecule between , z z dz+

integral integral

3 3

,

( ) ( , )

mgz

x y p

P z dz P r p d r d p

C e dz β −

∝

= ⋅

∫ ∫

Then number density

( ) (0) mgzn z n e

β −=

As Pressure if Temp is constant

Pressure( ) Pressure(0) Isothermal Atmospheremgz

n

z e β −

∝

=

Grand Canonical Ensemble

s r t r r

s r t

E E E S k n

N N N

+ = = Ω

+ =

( )( , )/

( ) ( )

r t s t s

s s s s r t s

S E E N N k s

d N C d N N N

C e d

ω

− −

= Ω Ω −

= Ω

( ) ( ) ( ) ( )

( )

, ,, ,

,

t t t t r r r t s t s r t t s s

t t

sr t t s

E N E N S S S E E N N S E N E N

E N

E S E N N T T

µ

∂ ∂− − = − − +

∂ ∂

≈ − +

Recall

, ,

1,

N V E V

S S

T E T N

µ ∂ ∂ = − = ∂ ∂

( )/

/

Then

; Grand Potential

s s E N

kT kT s s

N E kT

kT

d A e d

A e

e

µ

µ

ω

ρ

− +

−

Φ

= Ω

=

Φ ≡

( )/( , ) N E kT E N e µ ρ Φ+ −= Grand Distribution Function

Grand Partition Function /1 kT

e A

−Φ≡ ≡



3

( )

( )

,

,

//

//

( , ) N

N s

N s

N E kT kT

N s

N E kT kT

N s

Q V T

e e

e e

µ

µ

−−Φ

−

= =

=

∑∑

∑ ∑

Q

Define/ fugacitykT

z eµ = =

( , ) N N

N

z Q V T = ∑Q

Connection with Thermodynamics

• Similar to what we did for canonical ensemble

( ) ( , )

/

S k n N E

k N E kT

E TS N

d dE TdS SdT dN Nd

ρ

µ

µ

µ µ

= −= − Φ + −

⇒ Φ = − −

Φ = − − − −

Recall ( , , )S S E V N

S S S dS dE dV dN

E V N

dE P

dV dN T T T

µ

≡

∂ ∂ ∂= + +

∂ ∂ ∂

= + −

So, dE TdS PdV dN µ = − +

, , ,

; ;T V V

d PdV SdT Nd

P S N V T µ µ µ

µ

µ

⇒ Φ = − − −

∂Φ ∂Φ ∂Φ ⇒ = − = − = − ∂ ∂ ∂

Perfect Gas

3

/3

/ / /3 3

1 but!

/ ! exp

N N kT

N N

N

N kT kT zV

N

V e Q Q N

V V e N e e

µ

µ µ λ

λ

λ λ

= =

= = =

∑

∑

Q

Q

/

3 3

kT V V kT n kT e kT z

µ

λ λ Φ − ⋅ = ⋅ Q =



4

/3 3

3

But

and

kT V zV N e

kT N

N

kT n V

µ

µ λ λ

λ µ

∂Φ= − = = ∂

⇒ Φ = −

=

/

3,

1 kT

T

N P kT e kT PV NkT

V V

µ

µ λ

∂Φ = − = = =

∂

Ideal Gas in Grand Canonical Ensemble (Continued)

Probability of finding a subsystem with n particles where the average # of particles in that

volume V is N

We do not care what the energy of the subsystem is, so

( ) ( )

energyintegrated out

/ //

( , )n

n kT n kT E kT n

P E n d

e e d e Qµ µ

ρ

Φ+ Φ+−

= Ω

= Ω =

∫

∫

( )/

3

1

!

nn kT

nV

P en

µ

λ

Φ+ =

( )

( )

/ /3 3

/

Now

1

!

nnkT n kT

nkT n

V V N e N e

P e N n

µ µ

λ λ

−

Φ

= ⇒ =

⇒ =

Since kT N Φ = − , ( )1

!

n N nP e N

n

−= Poisson Distribution

0

Normalized such that 1n

n

P∞

=

=∑

N PNow Compute

Note that 0 !

1 !!

N

N

n P

P

=

=

Show that

1n

n

nP N ∞

=

=∑



5

( )

( )

1 1

1

0

0

1 Substitute 1

!

1

1 !

!

N nn

n n

N m

m

m N

m

LHS nP e n N m nn

me N

m

N e N N

m

−

= =

∞− +

=

∞−

=

= = ⋅ = −

+=

+

= =

∑ ∑

∑

∑

Adsorption of a Perfect Gas onto a Surface

Surface has 0 N adsorption sites (distinguishable) --- n gas molecules adsorbed on it. Molecules do not

move around.

Pressure of the gas is changed; how does the coverage0

n N

θ = change?

0ε − is the adsorption energy per site

Consider first just the adsorbed phase (no interaction between adsorbed molecules) in Grand canonicalEnsemble

0/

0

s

s

N n kT

n

n

E n

s

E s

s

e Q

Q e

g e

µ

β

β

=

−

−

=

=

=

∑

∑

∑

Q

n molecules are adsorbed 0s E nε = −

Note that: Only one state. What is the degeneracy of this state?

0

0

0

0

0

!

!( )!

! e

!( )!

s

nn

N g

n N n

N Q

n N n

β ε

=−

=−

Another way of looking at it and computing the Canonical Partition function



6

( )

( )

( )

1 2

0

0

0

1 2 3

1 2

0 0 0

0

0

1molecules are indistinguishable

!1

!

1

!

1( 1) 1

!

!Same as bef

! !

i

i

n

i

i

i

E n

molecules

E

mol mol mol

E E E

mol mol mol n

n

n

E

E E

Q e

n

en

e e en

N N N n en

N e

n N n

β

β

β β β

βε

β ε

ε

−

−

−− −

+

= −

=

∑=

∑=

=

= − − +

=−

∑

∑

∑ ∑ ∑

∑ ∑ ∑

ore

( )

( )

00

00

00

/ 0

00

0

00

!Then

!( )!

!

!( )!

1

N nn kT

n

N n

n

N

N e e

n N n

N e

n N n

e

β ε µ

β ε µ

β ε µ

=

+

=

+

=−

= −

= +

∑

∑

Q

( )

( )( )

( )

( )

0

0

0 0

0

00

0

1

1

1 1

kT n kT N n e

N en kT N e

e e

β ε µ

β ε µ β ε µ

β ε µ β ε µ β

µ

+

++

+ +

Φ = − = − +

∂Φ= − = ⋅ =

∂ + +

Q

( )

( ) ( ) ( ) ( )

0

0 0 0

0 00

0 0

1 11

1

N N nn e nn

N n n N ne e e

β ε µ

β ε µ β ε µ β ε µ β ε µ

+

+ + +

−= ⇒ = + ⇒ = ⇒ + =

−+

( ) ( )0 0

00

/ /0

|Partition function of one single

1kT kT

nkT n

N n

nkT n kT n

N n e eε ε

µ ε

θ

θ +

→

= −−

= =

− −

adsorbate molecule.(Problem 4.10 of Pathria)



7

( )

( )

0

0

Equilibrium demands

( )1

Then ( ) 1

ads gas

ads gaskT n kT n P f T

kT n P f T kT n

µ µ

θ µ ε µ

θ

θ ε θ

=

= − = +

−

+ = − −

( )

( )

0

0

"( )

/

( )1

( ) , where ( ) (Problem 4.11)1 ( )

f T

f T kT

kT nP kT n kT n g T

PP g T or g T e

P g T

ε

ε

θ

θ

θ θ

θ

− −

− +

= +

−

⇒ = ⋅ = =− +

Approach the Problem from Canonical Partition Function

( )

( )

00

0 0

0 0 0

! ! !

! ! !

nn

n

N Q e N n n

F kT n Q

kT n N n N n n n n

kT N n N N

βε

βε

= −

= −

= − − − − +

= − −

( ) ( )0 0 0 N n n N n N − − − + n− n n n n− + ( ) ( )

0 0 0 0

0,

0

0

( )

same as before

adsV T

ads

n

kT N n N N n n N n n n n n

F kT n n n N n

n

N nkT nn

nkT n

N n

βε

βε

µ βε

ε

µ ε

+

= − − − − − +

∂ = = − − + + −

∂

−= − −

= −−



1

Chemical Reaction in the Gas Phase

• Mixture of different gases

• Gas molecules are non interacting (Perfect gas) but chemically reacting

•

m different kind of molecules

Chemical symbols of these molecules 1 2, m A A A

2 2 22 2 H O H O+

2 2 2

1

2 2 0

0m

j j

j

H O H O

a A

=

⇒ + − =

=∑

(1)

1 2 2 2 3 2

1 2 3

a 2 a 1 a 2

# of molecules in the system| can change as a result of chemical reaction

j j

A H A O A H O

N A

≡ ≡ ≡

= = = −

=

→

2 2 2Conservation of atoms: : : 2 : 2 : 1

Thus for all

H O H O

j j

dN dN dN

dN a j

= − −

∝ (2)

Consider an isolated system. Equilibrium dictates 0dS =

1 2( , , , )S S E V N N ≡ but and E V do not change

0 0 j j

jdS dN µ = ⇒ =∑ (3)

[Note: (a) If instead of an isolated system, we consider T and V to be constant, then dF=0 which

yields 0 j j

j

dN µ =∑ .

b) Similarly if we consider T and P to be constant, then dG=0 which again

yields 0 j j

j

dN µ =∑ . ]

From(2) and (3)

1

0m

j j j

a µ

=

=

∑ General condition for chemical equilibrium (4)

Need to know jµ --- can use Canonical partition function to compute F and then jµ

1Recall ( , )!

N

N

QQ V T

N =

Since there are many species here, we change the notation a bit and write q = Q1 so that qi is the single

particle partition function of the i-th type.



2

31 21,1 1,2 1,3 1,

1 2

( , )

! ! ! !

i i N N N N N

i i N

i ii i

Q Q Q Q qQ V T

N N N N = = =∏ ∏

, ,

ii

T V N

N

F

N

F P kT n Q

V V

µ ∂

= ∂

∂ ∂ = − = +

∂ ∂

( ) ( )

! N i i i i i i i i

i i

i i i

ii

i

F kT n Q kT N n q n N kT N n q N n N N

kT n q n N

qkT n

N

µ

µ

= − = − − = − − +

⇒ = − −

= −

∑ ∑

( )

1 1

Now, we know 0

0

i i

i i

i i i i

m ma ai i

i i

a

a n q a n N

n N n q

µ

= =

=⇒ − =

⇒ =

∑ ∑

∑ ∑

1 2 1 21 2 1 2 ( , ) .a a a a

N N N q q K T V ⇒ = = − − − Law of Mass Action

N K → Equilibrium constant of the reaction independent of the numbers of molecules i N present (as it

only depends on single particle partition functionsi

q ’s raised toi

a ’s.)

Equivalent way of writing 1

ia

i

ii

N

q

= Π

3For a Perfect Gas andi

i

V q PV NkT

λ = =

Note that, we are neglecting any binding energies here.

[ ]3

( ) where

ii

i

aaai i i

i i i ii

N P N N c P f T c

q NkT N

λ ⇒ = ⋅ = =

Then ( ) 1

( )( )

i

ii i

i

ai i

i

aa ai ai

ii

c P f T

Pc g T P

f T

− −

Π =

∑ ∑⇒ Π = =Π

(Note that where = Partial Pressure. Thus or

and the above expression can be written in terms of rather than in terms of )

i ii i i i i i

i i

N PP V N kT P c P Pc

N P

P c

= = = =



3

1.

[ ] [ ] [ ]

2 2

1 2 3

1 1 22 2

1 1 2

2

1 1 2

Instead of we are now using [ ] concentration

( ) ( ) Independent of Pressure

i

H Cl HCl

a a a

H Cl HCL c

P g T g T

−

+ −

+

= = = −

=

= =

2.

[ ][ ]

2

2

2

( )

H H H

H g T

H P

+

=

At low pressure the system is largely dissociated as atomic hydrogen, whereas at high pressure the

formation of molecular hydrogen is favored. Binding energy favors molecular hydrogen; entropy favors alarge # of particles.

3. Thermal Ionization of Hydrogen Atoms

T is large 2[ ] H is negligible

H p e+ −+

0ε =ionization potential (ground state of hydrogen atom has 0ε − energy)

p e p e

H H

N N q qK

N q= =

Classical limit , p e+ − concentrations are small, T large; Coulomb interaction between separated

protons and electrons neglected.

( )

( ) ( )0

spin degeneracy

3/2

3

/3/2 3/2

3 3

2

2 2 and 4 2 where we have used

2

M m M

e

kT

p H

V

h

V V

h h

q mkT

q MkT q MkT e

ε

π

π π + ≈

↓

=

= =

( )0

3

p e

H e

N N eV

N

βε

λ

−

⇒ = Saha Ionization Formula 1921

Examples



4

You must have noted that we have considered that the H-atom is its ground state. What is the

probability that the H-atom is in its 1st excited state even at a temperature of 410 kT ?

1st excited state energy 01

4ε −

00

0

1/ 3

4 /1 4/

0

4001

0

4 12 61

0

1We know,

40

When 10 6 10

room

kT

kT

kT

room

T

T

N e e N e

kT eV

N e

N

N T e

N

ε

ε

ε −

−

− −

= =

≈

⇒ ≈

= = = ×

10

Thus, st

N

N small even if 4~ 10T K



Molecules with Internal Motion

1) Intermolecular interactions are negligible

2) Classical gas ( )3 1nλ <<

1

(1) (1)1 trans int

(1) (1) (1) (1)(1)nuc rotint elec vib

( , )!

( , )

N

N

QQ V T

N

Q V T Q Q

Q Q Q Q Q

=

= ⋅

=

For example diatomic molecules:

6 degrees of freedom for the 2 nuclei

3 translations for center of mass1 vibration along the axis

2 rotations of the axis in space ( , )θ ϕ

Diatomic Molecules

dissociation

4 5

4 5

10 10

10 10

kT E

K

T K

<<

↓ ≈ −

<< −

Practically no molecule in the excited state as well (exception oxygen at 2000 6000T K ≈ − )

diss 55,000 E K ≈ but excited 11000 E K ∆ ≈

Vibrational States

310 K k

ω

Almost full contribution for410T >

No contribution for210T <

Temperature is not high enough to excite vibrational states of high energy— harmonic oscillation.

Partition Function:

1

2vib

0

n

n

Q eω β

∞ − +

=

= ∑



( )( )

v

v

2 /v

v vvib / 2 ;

1

T

T

eC Nk

T k e

ω Θ

Θ

Θ = Θ =

−

( )

( )

v v vib

v v vib

0 Vibrational degrees of freedom are frozen

T C Nk

T C

Θ

Θ

Very high temperature → anharmonic effects

Rotational States

1) Heteronuclear molecules AB

Rotational states & states of the nuclei do not interact

Rotational states of a quantum mechanical linear rigid rotator

( )

m.i. about a line perp to the axis and through the cm.

2rot

2 1 20

1 2equilibrium distance

1 / 2

reduced mass

I

m m I Mr M

m m

ε

↓

= +

= =+

( ) ( )

( ) ( )

2(1)rot

0

0

2

2 1 exp 12

2 1 exp 1

10 1002

r

r

Q IkT

T

K Ik

∞

=

∞

=

= + − +

Θ = + − +

Θ = −

∑

∑

(1)





Putting all of these together

1) Very low temperature: rotational and vibrational degrees of freedom are frozen

v3 3

(as for a perfect gas)

2 2

C Nk U NkT = =

2) v

v5

(extra rotational contribution of )2

r T

C Nk Nk

Θ Θ

=

3) v

v7

vibrational contribution2

T

C Nk

Θ

=

Log-scale



Quantum Mechanical Considerations

System of many non-interacting identical particles:

Single particle in a potential field ( )V r

Schrödinger equation

22 ( ) ( ) ( )

2V r r E r

mψ ψ

− ∇ + =

Two identical particles

2 22 21 1 2 21 1 12 2 2

22

1

( ) ( , ) ( ) ( , ) ( , )2 2

( )2

N

N j j

j

V r r r V r r r E r r m m

H V r m

ψ ψ ψ

=

− ∇ + + − ∇ + =

= − ∇ +

∑

To compute the N-particle energy eigenstates, it is only necessary to solve the single particle Schrödingerequation

single-particle eigen state #

22 ( ) ( ) ( )

2 n n nV r u r E u r

m ↑

− ∇ + =

Complete set of eigenfunctions for the N particle problem can be taken by arbitrary products of single- particle states

Example

1) 1 2 3 1 1 2 2 3 3( , , ) ( ) ( ) ( )

eigenstate 1 1st particle

r r r u r u r u r ψ =

↓ ↓

Eigenvalue 1 2 3 E E E E = + +

1 2 3 1 1 2 3 1 2 2 3 1 2 3 3 1 2 3( , , ) N H r r r E u u u u E u u u u E u Eu u uψ = + + ≡

2) 1 2 3 1 1 1 2 4 3

1 4

( , , ) ( ) ( ) ( )

2

r r r u r u r u r

E E E

ψ =

= +

In these examples, permutation of particles among the single particle states gives the same energyeigenvalue.In example (1) N! permutations.



Symmetry Properties

Quantum Field Theory:Spin integer particles (Bosons) --- symmetric for 2 particle interchange

Spin half-integer particles (Fermions) --- antisymmetric

BosonsSymmetry 1 2 3 2 1 3 1 3 2( , , ) ( , , ) ( , , ) etc.r r r r r r r r r ψ ψ ψ ⇒ = =

Symmetrize – Permutations

( ) ( ) ( ) ( ) ( ) ( )1 2 3 1 1 4 2 1 3 4 1 1 2 1 3 1 1 1 2 4 3

distinct permutations3 of themnormalization

1( , , ) ( ) ( ) ( )

3r r r u r u r u r u r u r u r u r u r u r ψ

↑

= + +

Eigenvalue 1 1 4 1 42 E E E E E + + = + Same as before

Generally, if we have 1 N bosons in state 1, 2 N bosons in state 2, …..

K

i

i

N N =∑

( ) 1 11 2 3 1 1 1 2 1 3 2 1 2 2

1 2

States

!

, , ( ) ( ) ( ) ( ) ( )! distinct particles particles

K

ii

N N N K

K

N

r r r r u r u r u r u r u r u N N N

ψ + +

−

Π

= ∑

permutations

This wave function is called “permanent”.

1 1 2 2 K K E N E N E N E = +

Fermions

Antisymmetric

( ) ( ) ( )1 2 3 2 1 3 1 3 2, , , , , , etc.r r r r r r r r r ψ ψ ψ = − = −

Note that ( ) ( ) ( )1 1 1 2 4 3u r u r u r not allowed

Interchange particles 1 and 2

( ) ( ) ( ) ( ) ( ) ( )1 1 1 2 4 3 1 1 1 2 4 3u r u r u r u r u r u r = − Clearly not allowed

Pauli Exclusion Principle: no two Fermions can have the same set of quantum numbers



2 Particles

( ) ( ) ( ) ( )1 2 1 1 2 2 2 1 1 2( , )r r u r u r u r u r ψ = −

Eigenvalues

1 2 1 1 1 ( ) ( ) .i i i H H H H u r E u r etc= + =

( )( )

( )( ) ( )

( )( )

1 2 1 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1

1 2 1 2 1 2 2 1

1 2 1 2 2 1

1 2

H H u u u u E u u E u u E u u E u u

E E u u E E u u

E E u u u u

E E E

+ − = − + −

= + − +

= + −

= +

N particles

Slater determinants

( )

1 1 1 2 1 3

2 1 2 2 2 31 2 3

3 1 3 2 3 3

( ) ( ) ( )

( ) ( ) ( )1, ,

( ) ( ) ( )!

u r u r u r

u r u r u r r r r

u r u r u r N ψ

− − −

− − − = − − − − − − − − − − − −

Property of determinants

1) Interchange 1r and 2r meaning interchange two columns—picks up a negative sign.

2) If 1 2u u=

then two rows are the same and the determinant vanishes.

⇒ So of all possible ! N states, Nature picks only 1 (symmetric for bosons antisymmetric for

fermions). This is the origin of the1

! N in the classical case.

⇒ If the density is high, then the question of having same quantum # arises—classical statistics breaks

down.



Complications for Homonuclear Molecules such as H 2

r T Θ

Classical partition function (directly) azimuthal angle φ runs from (0,π ) because of exchange symmetry.

Low temperature: Coupling between nuclear spin and rotational states

⇒ Total wave function must be symmetric or antisymmetric for an interchange of two identical particle.

Consider 2 H

1

2 I = for both nuclei

i) Total spin 0 (spins are anti-parallel—spin function is antisymmetric; spin degeneracy 1)

ii) Total spin 1 (spins are parallel—spin function is symmetric; spin degeneracy is 3)

In general if the nuclear spins are sn --- total # of states ( )( ) 22 1 2 1n ns s+ + = s

Symmetric ( )( )( 1) ( 1)

1 2 1 antisymmetric (2 1)2 2

n n n ns s s s+

= + + = +s s s s -

Figuring it out:

Symmetric (1) (2) (2) (1)i j i j f f f f + Spin wave functions part of the total wave function with

1, 2,3,.....i s=

# of ways doing it( -1)

2

s s

but 1 1(1) (2) f f like states are also allowed

So total number of symmetric states are

( -1) ( 1)

2 2

++ =

s s s s s

and total number of antisymmetric states( -1)

2

s s

2 H 1

2 I = 2 1 2 I = + =s

Symmetric 3 OrthoAntisymmetric 1 Para



2 D I =1 2 1 3 I = + =s

Symmmetric 6 OrthoAntisymemetric 3 Para

Ortho →larger statistical weight ↓

spin wave function symmetric

Back to Considering H2

Nuclei Fermion

Total wave function must be antisymmetric

Symmetric spin = antisymmetric rotational part Ym

( )

1 /(1)rotodd

odd

odd

(2 1) r T Q e

↓

− + Θ

−

= +∑

Antisymmetric spin ≡ symmetric rotational part

( )

1 /(1)roteven

even

even

(2 1) r T Q e

↓

− + Θ

−

= +∑

(1) (1) (1)rot rot rot

odd even

3Q Q Q= +

Ratio of Ortho to Para

(1)rotodd ortho

(1) para rot

even

3Qn

n Q=

r T Θ

(1) (1) (1)rot rot rotodd even classical

1

2Q Q Q≈ ≈

Then ortho

para

3:1n

T n

→ ∞



r T Θ

(1) /6roteven

(1) /2rot

odd

1 5

3

r

r

T

T

Q e

Q e

Θ−

Θ−

≈ + +

≈

2 /2 /ortho

6 / para

3 39 0 when 0

1 5

r r

r

T T

T

n ee T

n e

− Θ− Θ

− Θ

×≈ ≈ ≈ →

+ All Para

When r T ≈ Θ , one needs to compute the full partition function and compute specific heat. Specific heat

computed this way dose not compare well with experiments. But the conversion from Ortho to Para time-

scale is very large. So the gas behaves like a mixture of two gases of pre-assigned ratio. If that is the case

then ( ) ( )v v vortho para

3 1

4 4C C C = + and this compares well with experiments. (Dennison 1924)

-------------------------------------------------------------------------------------------------

2 1

Bosons

D I =

So Ortho ≡ symmetric spin → even rotational Para ≡ antisymmetric spin → odd rotational

(1)rotevenortho

(1) para rot

odd

6

2 :13

Q

n T n Q

= → ∞

0T → all Ortho



Ideal Quantum Gas

Review of Classical Partition Function (Distinguishable Particles)

2-State Systems

( )

1

i

i

i

i

i

i

H N

H

i

H

i

H H

i

N

Q e

e

e

e e

Q

βµ σ

σ

βµ σ

σ

βµ σ

σ

βµ βµ −

∑=

= Π

= Π

= Π +

=

∑

∑

∑

Now consider a different way of computing this – in terms of single particle states

( )1 1 2 2

1 2

1 2

1 2

1 1 2 2

1 2

1 2,

!'

! !

s E

N s

n n

n n

H H

n n

E n n

N n n

Q e

N e

n n

β

β ε ε

ε µ ε µ

ε ε

−

− +

= − =

= +

= +

=

=

∑

∑

( ) ( )1 11 2

1

1 1 1 2

1

2 1

1 1

1 1

( )

!

!( )!

!where and

!( )!

n N n

N

n

n N n

n

n N n

N Q e e

n N n

N x y x e y e

n N n

βε βε

βε βε

−− −

− − −

= −

=−

= = =

−

∑

∑

( )1

1 2

11 1

1

!1 ( )

!( )!

n N N

N N N

n

N

N x x y y x y e e

n N n y y

Q

βε βε − − = = ⋅ + = + = +

−

=

∑



How does one handle indistinguishability in classical Statistical Mechanics?

Well that came as an ad hoc solution by Gibbs:

1

!

N

N Q

Q N

=

but the !!i

N nΠ

degeneracy factor was still used in the partition function !

In general

( )'

degeneracy factor

!

! non-interacting

i i

i

n N

ini

N Q e

n

ε −∑=Π

∑

( ) ( )1 21 2

1 21 2,

can be shown explicitly(see, for example, 3-state systems)

!'

! !

n n

n n

N e e

n n

βε βε − −= ∑

31 2

1

i

N

N

N

e e e

e

Q

βε βε βε

βε

−− −

−

= + + +

=

=

∑

3 terms

( )( )

( )

31 2

1 2 3

1 2 1 2

1 2

1 2

1 2

1 2

1 2

1 2 3, ,

1 2 1 2,

1 2 1 2,

1 2 1 2

!

! ! !

!

! ! ( )!

!

! !

1 !

! ! !

nn n

n n n

n n N n n

n n

n n N

n n

n n N

n n

N x y z

n n n

N x y z

n n N n n

N x y z

z zn n N n n

x N y z

n z zn N n n

− −

− +

=

− +

=

− +

∑

∑

∑

∑ ∑



( )( )( )

( )

( )

1 2

1

11

1

1

1 1 2 1 2

1 1

1 1

!1 !

! ! ! !

1 !1

! !

!1

! !

n n N

n

N nn N

n

N N

N n x N y z

n z N n n N n n z

x N y z

n z N n z

y N x z

z N n n z

−

− = − − −

= + −

= + −

∑ ∑

∑

∑ z

( )

1

( ) 1

n

N N

N

y z

x z y

y z

x y z

+

= + +

+

= + +

Quantum Systems (indistinguishable)

Canonical Partition Function

' i i

i

n N

n

i i i

Q e

E n n N

β ε

ε

− ∑=

= =

∑

∑ ∑

No of particles in occupying eigenstates iε

Indistinguishable: System degeneracy factor is unity. This makes it very hard to compute the Canonical

Partition Function

Grand Partition Function

( )

0 0

0

' where z

ii

ii

i

n N

N N

N N

n

i N n

e Q e Q

ze e

βµ βµ

βε βµ

∞ ∞

= =

∞−

=

∑= =

= Π =

∑ ∑

∑ ∑

Q

Note the double sum with the constraint: the first sum is over in under a fixed value of N and then over

all possible values of N. When N → ∞ , this double sum is the same as summing over in independent ofone another i.e. without the constraint.

To show that in Grand Partition Function

1 2 1 20 , ,

'

N n n n n

∞

=

≡∑ ∑ ∑ ∑

Consider only 2 states 1 2,n n where 1 2n n N + =



Suppose N=6

With constraint 1 2 6n n+ = : these points are on the top line

with N=5 2nd line

N=4 3rd line etc.

But you would have gotten all these (not really!) points by allowing

1

2

0,6independently

0,6

n

n

=

=

Actually, this would work only in the line . N → ∞

Then,

( ) ( )

( )

1 21 2

1 2

i i

i

n n

n n

ni

i in

ze ze

e

βε βε

β µ ε

− −

−

=

= Π = Π

∑ ∑

∑ So the Grand Partition Function Factorizes!

Q

Q

( )( )0,1 1 ii FD

in e

β µ ε −= = Π +

Fermi Dirac Statistics

Q

( ) i1

0,1,2 ; < for convergence1 i

i BE i

ne

β µ ε µ ε

−

= = Π

−

Bose - Einstein statistics

Q



Lectures 15-17

Grand Potential

Fermi – Dirac( )

( ) 1 i

FD FD

i

kT n kT n e β µ ε −Φ = − = − +∑ Q

Bose-Einstein

( )( )1 i BE

i

kT n e β µ ε −Φ = −∑

Together, ( )( )1 i

i

kT n e β µ ε −Φ = ± ∑ (upper sign BE)

( )

Also,

but

1i

i

E TS N

F G G F PV

PV

PV kT n z e

βε

µ

−

Φ = − −

= − = +

= −

⇒ = −∑

1

i

i

i

E U n z e a

βε

ε

β −

∂= = − =

∂ +∑ Q

( )

( )

( )

1

1

1

1

11

1

i

i

i

i

i

i

ai

e N kT

e

e

a BE N

a FD z e βε

β µ ε

β µ ε

β µ ε

β

µ

−

−

− −

− +

∂Φ= − =

∂

=

= −== +

∑

∑

∑

How about in ?

Since the Grand partition function factorizes and i N n= ∑

( )1 /

1 1Occupation numbers

i ii kT

n z e a e a

βε ε µ − −= =

+ +

1 (all ) otherwise 0

i

i i i

FD n BE nµ ε ε

≤< <

When 0µ ε = , 0 divergesn → BE condensation



Classical Statistical mechanics (Maxwell-Boltzmann or MB) is recovered when a=0:

( )/i kT in e

ε µ − −=

Note than the effect of a is negligible when ( )/1i kT

e ε µ −

1in⇒ (Low density)

Now you can see why Gibbs’ fix works. If you treat them as distinguishable you get an extra

degeneracy factor of !

!ii

N

nΠin the partition function. But this degeneracy factor becomes just ! N

as ! 1ii

nΠ ≈ when 1in (recall that 0! 1= ).

Also since ( )/1i kT

e ε µ − for all values of

iε , it follows that

/ /1 1 1kT kT e e zµ µ − ⇒ ⇒

Consistency Check: Ideal Gas

3 3

or, N N

kT n nV kT V

λ µ λ µ

= =

3

3

Or, 1

1 density

N

V

N n n number

V

λ

λ = ≡

Another way of looking at it:

( )1/3

1/3

1 3

1

3

3

3/2

mean interparticle distance 3d

1

1

1

smallconditions for recovering Classical or Maxwell-Boltzmann Statistics.

large

V

N

n

n

n

n

nT

n

T

λ

λ

λ

λ λ

−

−

−

−

⇒

⇒



Ideal Bose Gas

( )( )1 i

i

iPV

kT n e β µ ε

ε

−

↓−

↑

Φ = −∑

( )( ) ( )1 1 ; zi i

i i

PV n e n z e

kT

β µ ε βε βµ + − −⇒ = − − = − − =∑ ∑

( ) 1

1 1

11

iii

i i i

i i

n z ee

N n E n

βε β µ ε

ε

−− −= =

−−

= =∑ ∑

Single Particle Density of States

For large V the single particle states iε of a system are almost continuous.

0( ) ( ) ( )

i

i f a f d

ε

ε ε ε ε ∞

⇒ →∑ ∫

Consider x y zV L L L= and L λ .

Periodic boundary conditions (to eliminate reflections) ( , , ) ( , , ) etc. x x L y z x y zψ ψ + =

Free particle ik r pe k ψ ⋅= =

2( ) 2 x x x x x x

x

k x L k x n k n L

π π + = + ⇒ =

# of states for to be in between xk ( , ) x x xk k k + ∆ ,2

x x x

Ln k

π ∆ = ∆

# of states ( ),k k dk +

( )( )

33

2

x y z x y z

L L La k d k k k k

π = ∆ ∆ ∆

If in is not negative 1 z ≤ i.e., 0µ < .

Also, 0 z ≥



( )( ) ( )

3 3 23 3

42 2

element of volume in k-space

V V a k d k d k k dk π

π π = =

↑

( )( )

1/2

3 2

2 2

2 2 14 d

22

non-relativistic2

dk

V m ma d

k

m

ε ε ε π ε ε

π

ε

−

↑

= ⋅ ⋅

=

( )3/2

1/22 3

(2 )d

4

V ma d ε ε ε ε

π

−⇒ =

(Recall, classically 3 3 30 3 3

1 4( )3

p P

V P d q d p Ph h

π

≤

= =∑ ∫

( )3/2

30

4( ) 2

3

V m

h

π ε ε =∑

( ) ( )

0 3a

ε ε

ε

∂= =

∂

∑2 3

4V

h

π 2

3

π

( )

( )

3/2 1/2

3/23/2 1/2 1/2

3 32

2

(2 )2 2

4

m

V V mm

h

ε

π ε ε π

= =

Equation of State

Sum over iε can be replaced by integral in the Equation of State

( )

( )

0

1

0

0 BE distribution function

( ) 1

1

1

1

( ) 1

" ( )

( ) ( )

ii

i

BE

BE

PV a d n ze

kT

N n z e

a d e

f

E a d f

βε

βε

β ε µ

ε ε

ε ε

ε

ε ε ε ε

∞−

+−

∞

−

∞

= − −

= =−

= ⋅−

=

∫

∑ ∑

∫

∫



( )( )

3/2 1/2

3 0

2 2

1

V m d N

h e β ε µ

π ε ε ∞

−=

−∫

This integral can be evaluated by substituting x βε =

1/23/2 3/23 10

2 ( ) (2 )1 x

x dx N V kT mh z e

π ∞−

=−∫

General Form

1

10

1( )

( ) 1

n

n x

x dxg z

n z e

∞ −

−=

Γ −∫

1Recursion Relation ( ) ( )n n z g z g z z

−∂

=∂

(We will need it later)

( )

3/2 3/23/23

31/2

3/2 3/23

3/23

2 3 3 1 1 1( ) (2 ) ( ) :

2 2 2 2 2

2( ) ( )

1( )

N V kT m g z Noteh

mkT V V g z g z

h

N g z

V

π π

π

λ

λ

= Γ Γ = Γ =

= =

⇒ =

Note that 1 z ≤ , so we can expand 3/2 ( )g z

( ) ( )

( )

1

0

1

00

1 1

1 10 0

both denominator and numerator by ,

1( )

( ) 1

1

( )

1 1

( ) ( )

x

xn

n x

n x x

n x n x

Multiplying ze

zeg z x dx

n ze

x dx ze zen

x dx ze z x e dxn n

−

∞ −−

−

∞ ∞− − −

=

∞ ∞∞ ∞− − − −

= =

=

Γ −

=Γ

= =Γ Γ

∫

∑∫

∑ ∑∫ ∫

1

11 1

0

1

0

2 3

(Substituting )

1( )

( )

since ( )

Finally then, ( )2 3

n y

n n n

n y

n n n

lx y

y e dy zg z z

n

n y e dy

z zg z z

∞ − −∞ ∞

−= =

∞− −

=

= =Γ

Γ =

= + + +

∑ ∑∫

∫



Maximum value of ( )ng z is when z=11 1

(1) 1 ( )2 3

n n ng nζ = + + + =

Converges for 1n >2

43/2(2) ; (4) / 90; (1) (3 / 2) 2.612

6g

π ζ ζ π ζ = = = =

How does this convergence affect the physics?

( )

3

3/2

3

experimentalquantity

12.612

2 2.612

N

V

mkT n

h

λ

π

↑

⇒ ≤

≤ ×

(When you vary T, µ changes and so does z. z becomes = 1 and then stays at 1).

So we can choose small enough T such that the inequality is broken!! What is going on?

Bose-Einstein Condensation

Note that →∑ ∫ is valid only if no state is macroscopically occupied. So when this inequality is

broken, some state is macroscopically occupied.

Lowest Temperature cT T = when the equality is still satisfied (with 1 z = ; actually 1 z = for

all cT T < )

2/3

3

2/32

1 2.612 2

2 2.612

c

c

h nT mk

h nT

mk

π

π

= ⋅

=

Which state is macroscopically occupied? Bosons can occupy any state macroscopically. Makes sense

for 0ε = state though.

Check:1

1

1iin

z e βε −

=−

When 0 1

10, as z 1

11i

z N

z zε

−= = = → ∞ →

−−

Also note that1/2

( ) ~a ε ε which means ( ) 0a ε = for 0ε = and we have not taken the 0ε = stateinto proper account. So what? What is one state worth in a continuum? Problem here is that this stateis macroscopically occupied.

So we better write 00( ) ( ) BE

ex

N a d f N

N

ε ε ε ∞

↓

= +∫ 03/23

1( )

N N g z

V λ

−⇒ =



When cT T = , 0 0 N = , and when cT T < , 0 0 N >

At cT ,3/2

3/2~ (1)c N

T gV

(1)

Below cT ,3/2

0 3/2~ (1) N N

T gV

− (2)

Divide Equation (2) by (1) to get

3/2

0 cc

N N T T T

N T

−= ≤

3/2

0 1c

c

T N N T T

T

⇒ = − ≤

Revisit Equation of State and Other Thermodynamic Quantities

( )

( )

3/2 1/23 0

3/2 1/2

3 0

2 (2 ) 1 (1 )

( 0 )

2 1(2 ) 1 (1 )

PV m V n ze d n zkT h

term

Pm n ze d n z

kT V h

βε

βε

π ε ε

ε

π ε ε

∞ −

∞ −

= − − − −

↑ =

⇒ = − − − −

∫

∫

Even when 0 N is a large fraction of N 0

001 1

N z N z

z N

= ⇒ =

− +

The last term

0

0

1 11

N nV N

− − +

0 00

0 0

1 1 1 0 for macroscopic system1

N n N n n N V N V V N

= − = = →+

0k = Condensation (momentum space) does not contribute to pressure.

Now substitute x βε =

( ) ( )

( )

3/21/2

3 0

3/2 3/2

00

3/2

5/210

2 2 1

Integrating by parts,

2 21

3 31

The first term vanishes, and the second term can be written as

2 2( ) (5 / 2)

3 31

2

3

x

x x

x

x

mkT P x n ze dx

kT h

zen ze x x dx

ze

x dxg z

z e

π ∞ −

∞ −∞−

−

∞

−

= − −

− −−

− = − ⋅ Γ−

= − ⋅

∫

∫

∫

5/23 1 3 3 1

( ) as (5 / 2) (3 / 2) (1/ 2) and (1/ 2)2 2 2 2 2

g z π π ⋅ Γ = Γ = ⋅ ⋅ Γ Γ =



5/23

1( )

Pg z

kT λ ⇒ =

High temp limit:/

3

1( ) as 1; and kT

nP

g z z z e

kT

µ

λ

≈ =

Classically

3 N

kT nV

λ µ

=

⇒

3

3

1P N N

kT V V

λ

λ = =

Below the critical point, 5/23 3

1 1.342(1)

Pg

kT λ λ = =

Isotherms (or P-V Characteristics)

Recall the definition of critical temperature: 3/23(1)

V N g

λ =

Instead of setting cT T = , we keep T fixed and set cV V =

Then 3/23(1)cV

N gλ

= (1)

When there is no condensation, 3/23

1( )

N g z

V λ =

5/23 3 3/2

5/2

3/2

1 1 1( ); Replace by

( )

( )

( )

Write v ,Then

P N g z

kT V g z

g zP N

kT V g z

V

N

λ λ ⇒ =

⇒ =

=

5/2

3/2

( )v

( )

g zP kT

g z⇒ = No condensation cv>v

When condensate starts forming c cv=v Note that v is a function of temp.

5/2

3/2

(1)v

(1)c

gP kT

g= (2)

Substitute T from Eq.(1)

25/3 5/2

05/33/2

(1) v , some constant

2 (1)c

ghP c

m gπ = =



When condensate forms cv<v , 03/23

1(1)

N N g

V λ

−= (3)

and 0 5/25/23

3/2

1 (1)(1)

(1)

P N N gg

kT V gλ

− = =

(4)

Divide Eq.(3) by (1)3/23

0

3/23

1 (1)1

1(1) c

c

g N N

VN V V g

λ

λ

−= = 0 (5)

c

N N N

V V

−⇒ =

( ) ( ) 5/2

3/2

(1)From 4 and 5 ,

(1)c

gP N

kT V g=

5/2

3/2

(1)

v (1)c

gkT P

g= , independent of v

BE Condensation Revisited in a General Way

For BE condensation, the most important thing to understand is the behavior of the integral

10

1( ) when 1

1 I a d z

z e βε

ε ε ∞

−= →

−∫ (This integral is actually N .)

If ( 1) I z = diverges, everything is fine— no condensation.

0

1( 1) ( )

1 I z a d

e βε ε ε

∞= =

−∫

Consider ( ) ~ ma ε ε

1

1 1

1

Now if ~ , ~

( ) ~ ~ ~

n n

d d d n

n

k d k dk

d a k dk k dk k k d

k

ε ε

ε ε

−

− − −−



( ) ~

d n

na d d ε ε ε ε

−

i.e.,d n

mn

−=

When ε is large, e βε − dominates hence there is no divergence.

Consider ε small and cε is some cut off where this approximation cease to work.

1

0 0 0

1( 1) ~

1 1

cC cm m m I z d d ε ε ε

ε ε ε ε ε βε

−= ≈ ≈+ −∫ ∫

( 1) I z = diverges if 0m ≤ ( 0m = logarithmic divergence.) No condensation.

≥

≥

For condensation m 0

d - ni.e., 0

nor d > n

For non-relativistic particles,2 2~ ~ p k ε i.e., 2n = . That means no BE condensation in d=2 but BE

condensation takes place at d=3.

For relativistic particles,1~ pc k ε = i.e. 1.n = That means BE condensation can take place in both d=2

and d=3.



Ideal Fermi Gas

( )

( )( )

/1

0

1

1 where and 01

( )

1

i

i

i

i

kT i

i

FD

PV n n ze

kT

N n z e z z e

f

a d N

e

βe

e

µ βe

e

β e µ

e

e e

−

−

∞

−

≡ = +

= = = ≥ < ∞+

=+

∑

∑ ∑

∫

Q

3/2 3/21/2 1/2

2 3 3

3/2

1/2 1/23

(2 ) (2 )( ) 2

4

(2 )( ) 2 ; spin degeneracy = 2 1s s

V m V ma

h

V ma g C g s

h

e e p e p

e p e e

= =

= ⋅ = ≡ +

( )

3/2 1/2

30

(2 ) 2 substitute

1

s N m d

g xV h e

β e µ

p e e βe

∞

−= =

+∫

( ) ( )( )

3/2 1/2-3/2

3 10

3/2 1/2

2 10

1

3/23 10

(2 ) 2

1

2 1

1

1 1where

1

s x

s x

n

s n x

m x dxg

h z e

m kT x dxg

h z e

x dxg f z f z

n z e

p β

p

p

λ

∞

−

∞

−

∞ −

−

= ⋅+

= ⋅

+

= ⋅ =Γ +

∫

∫

∫

( ) ( )

( )

( )

1/2

3/2 3/2

0

Equation of state 1

1

2 ( ) 2

13 31

PV a d n ze d

kT

C d n ze d

z

C n ze C e ze

βe

βe

βe βe

βe

e e e

e e e

β

e e

−

−

∞+ − −

−

= +

= ⋅ +

−= + −

+

∫

∫

∫

3/2

10

3/2

5/23/2 1 30

2

3 1

2 1 = ( )

3 1

s x

d C x

z e

g V x dxC f z

z e

βe

e e β βe

β β λ

∞

−

∞

−

= =+

= ⋅ ⋅+

∫

∫



Internal Energy1

,

( )1 z V

d U n a

z e βe

e e e

β −

∂= − ≡ ∂ +

∫ Q

5/23

3 3 3( )

2 2 2sg V P

kT f z kT V PV kT λ

= = ⋅ =

Density low, temperature very high ( )non -degenerate

( )

3

3/21

2s

nh

g mkT p <<

3

3/

Classical limit 0

; 1kT

N kT n

V

N z e z

V

µ

λ µ

λ

= <<

= = <<

2 3

Similar to the expansion of ( ) for the Bose gas, here also one can show that

( ) , for small2 3

n

n n n

g z

z z

f z z z z= − + ≈

Then

5/23

5/2

3/2

3 ( )

2

( )3 3; z<<1

2 ( ) 2

2PV=

3

V kT g f z

f z NkT NkT

f z

U NkT

λ

= =

⇒ =

U=

At very low temperature (degenerate)3

0s

nT

g

λ → → ∞

( )0

/0

0 0

1 for1

0 for1

0 where is the Chemical Potential at 0

kT n

e

T T

e µ

e µ

e µ

µ µ µ

−

< = →

> +

→ = =

i) 0 0 e µ e µ < − = −∆ /

11 as 0

1kT n T

e−∆

= → →+

ii)/

0 0 /

1 0

1

kT

kT n e

ee µ e µ −∆

∆> − = ∆ = ≈ →

+

Step Function at T=0



Definition of Fermi Energy

( )

0

10

0

0

3/2

3/2

2/3 2/32 3 2 2

3/2

( )

( )

( ) ( )

1 for and 0 otherwise

( )

2

3

3 1

2

3 4 6

2 2(2 )

F

F

FD

F

F

F

F s

a d N

a d N

e

f a d

a d

C N

N C

n N

g mgV m

e

β e µ

e

e e

e e

e e e

e e

e e

e

e

p p e

∞− +

∞

=

=

=

≤

=

=

= ⋅

⇒ = =

∫

∫

∫

∫

Ground state energy (average) (T=0)

( )0 0

3/20

5/2 3/2

0

( ) since 1

2 2 3 3

5 3 5 5

3 2/ 0

5 5

F

F

F F F

F F

E a d n

C d

C C N

N E N P

V

e

e

e e e

e e

e e e

e e

= =

= ⋅

= ⋅ = ⋅ ⋅ =

⇒ = = ≠

∫

∫



Approximate Calculation of the Specific Heat of a Fermi gas

As the temperature is raised form 0 to T each free electron does not gain energy by an amount kT becausemost of them are occupying states below Fermi level and by the Pauli principle they cannot be excited to

the states which are already fully occupied. Only a small fraction of electrons with energy close to the

Fermi level that can be excited to empty states lying in the range kT above 0µ .

# of excited electrons

ex N ≈ (Density of states near the Fermi level) ×width of energy

3/21/2

3

3/2 3/23

2 3

3/2 3 3/2

( )

(2 ) 2

1 2 (2 )

1 2 3 4 (2 ) 2 (2 )

3 3

2 2

F

s F

s F F

sF s

F F

a kT

V mg kT

h

g V m kT h

N g V m kT

V h g V m

N T kT N

T

e

p e

p e

e

p p

e

e

≈ ⋅

≈ ⋅ ⋅

=

=

= =

10,000 , 300 ~ 4 5%exF

N T K T K

N ≈ −−

Then ( )23

( ) ~2

exF

T U T Nk N kT

T ≡ ⋅

~ 3V

F

T C Nk

T

close to the correct result

Electron Gas in Metals

3

electron phonon contribution contribution

V C T T γ δ ↓

= +

2V C T

T γ δ = +



Magnetic Behavior of an Ideal Fermi Gas

Pauli Paramagnetism; Alkali metals Pauli 1927

T=0

*µ magnetic moment

B magnetic field

(1) * * ( )

(2) * * ( )|

total energy

B KE B Parallel

B KE B Anti parallel

µ e µ

µ e µ

= −

= + −

→

N + = # of electrons in group 1; *KE Be µ = +

N − = # of electrons in group 2; *KE Be µ = −

Spin degeneracy is lifted by magnetic field

3/2 1/2 1/20

3||

2( ) (2 )

no heres

V a d m C

h

g

p e e e e = =

→

Magnetic moment *( ) M N N µ + −= −

0

0

( * ) ( )

( * ) ( )

FD

FD

N a B f d

N a B f d

e µ e e

e µ e e

∞+

∞−

= +

= −

∫

∫

At T=0 1/2

00

3/20

( * ) 1

2( * )

3

F

F

N C B d

C B

e e µ e

e µ

+ = + ⋅ ⋅

= +

∫B is small so * Bµ the lower limit 0→ after a change of variable

Similarly

3/20

2( * )

3 F N C Be µ − = −

3/2 3/23/23/2 3/2

3

3/2

3

*( )

4 *(2 ) * *1 1

3

4 *(2 ) 3 * 3 *1 1 If B is small

2 23

F F F F

F

F F

M N N

V m B B

h

V m B B

h

µ

p µ µ µ e e

e e

p µ e µ µ

e e

+ −= −

= + − −

= + − +



2 3/2 1/230

21

4lim * (2 )

* 3F

F B

M V m

B h

N e

p χ µ e

µ

→= =

= ⋅

In contrast to classical Langein paramagnetism where 1~ (Curie's law)T

χ

At low temp 0 χ χ → constant Pauli Paramagnetism

Physical Picture:

Langevin result: (Probability that an atom parallel to the magnetic field – Probability that it is

antiparallel).

1 1

1 1

(if it is small)

H H

H H

e e H H

H H e e

H

βµ βµ

βµ βµ

βµ βµ

βµ βµ

βµ

−

−

− + − +

→ + + −+=

For N atoms this gives a net magnetic moment

2 2~ / ~ / N H N H kT N kT µ βµ µ χ µ ⋅ ⇒

But now at a temp F T T << only ex N of the electrons contribute. This means

223 3

~ ~ /

2 2

ex F

F

T H M N H N N H kT

T kT

µ µ βµ µ ≈ ⋅

χ ⇒ independent of temperature



Fermi Gas : Degenerate Case (Detailed Calculation)

( ) ( ) ( )

( ) ( )( )

1/23/2

3 /0

3/23/2

3 /0

0

22

1

22

1

F F F

s kT

C

s kT

C

T T kT

V d N g m T

h e

V d U T g m

h e

e µ

e µ

e

p e e µ µ

p e e

∞

−

∞

−

< << =

= ≡+

=+

∫

∫

((((

((((

Integrals of the form0

( ) ( )FD I f d f e e e ∞

= ∫ where ( )FD f e is a slowly varying function of e , except

in the region e µ ≈

In order to take advantage of this property of ( )FD f e ′ , define0

( ) ( ) ( ) dF

F d d

e e f e e f e

e = ⇒ =∫

and note that (0) 0F =

Then0

( ) ( )FD I F f d e e e ∞

′= ∫ Integrate by parts

( )FD f e =

00

0

( ) ( ) ( )

(0) 0

because( ) 0

( ) ( )

FD

FD

FD

F f F d

F

f

f F d

e e e e

e e e

∞∞

∞

′−

⇓=

∞ =

′= −

∫

∫



Expand ( )F e about µ

2

0 1 2

1( ) ( ) ( ) ( ) ( ) ( )

2

( ) ( ) ( )

F F F F

I I F I F I F

e µ e µ µ e µ µ

µ µ µ

′ ′′= + − + − +

′ ′′= + + +

( ) ( )

( ) ( )

0

0

1

0

2 22

0

( ) ( )

( ) ( )

1 1( ) ( )

2 2

FD FD

FD FD

FD FD

I f d f d

I f d f d

I f d f d

e e e e

e µ e e e µ e e

e µ e e e µ e e

∞ ∞

−∞

∞ ∞

−∞

∞ ∞

−∞

′ ′= − ≡ −

′ ′= − − ≡ − −

′ ′= − − ≡ − −

∫ ∫

∫ ∫

∫ ∫

Now( )/

1( )

1FD kT

f e

e µ e

−=

+

Write ( ) / kT ye µ − =

( )2

( ) ( ) ;

1

yFD

y f e d dy

e

β e β e ′ = − =

+

( )

( )

0 2

2

21

1

; 1,

1

11

y

y

y y y

y

dy I e

e

e dy x e dx e dy

e

y xdx

y x x

β

β

∞

−∞

∞

−∞

∞

= +

+

= = + =

+

= −∞ = = =

= ∞ = ∞

∫

∫

∫

( ) ( )

( )( )

2

1 2 2/2 /2

2 22

2 2 2

" /3

10 odd function

1

1 1

2 61

y

y y

y y

y

y e dy dy I y

e e e

y e d I kT

e

λ

p

β β

p

β

∞ ∞

−−∞ −∞

∞

−∞

= + = =

+ +

= + =

+

∫ ∫

∫((((



Then ( )0

22

( )

( ) ( ) ( )6

FD I f d

F kT F

f e e e

p µ µ

∞

=

′≈ +

∫

Where ( )

( )

( )0

( ) F

F d F

e f e

e f e e f e

′ =′ ′= ⇒

′′ ′=∫

( ) ( ) ( )2

2

0 0

( )6

FDd

f d d kT d

µ

e µ

p f f e e e f e e

e

∞

=

⇒ = +∫ ∫

1/2

01/2

( )

( )

FD N C f d e e e

f e e

∞

= ⋅

=

∫

Then

0

0

21/2 2 1/2

0

2 21/2 1/2

1/2

00small error by

1( )

6 2

( )

12

N C d kT

kT C d d

µ

µ µ

µ

p e e µ

p e e e e

µ

− = +

≈ + +

∫

∫ ∫

0 setting μ μ here≈

Note that at T=0

0 1/2

0 N C d

N

µ e e =

⇒

∫

N = ( )

( )

( ) ( )

0

0

221/2

1/20

221/2

1/20

221/20 0 1/2

0

12

12

mean value theorem

12

kT C d

kT

d

kT

µ

µ

µ

µ

p e e

µ

p e e µ

p µ µ µ

µ

+ +

⇒ = −

↓

≈ − = −

∫

∫



( )22

0012

kT p µ µ

µ ⇒ − = − (1)

22

0

0

112

kT p µ µ

µ

⇒ = −

(degenerate Fermi Gas) (2)

Internal energy

( )

( )0

0

3/2

0

3/2 1/2

223/2 1/2

0

23/2 3/2 2 1/200

3/0 0

( )

3( ) ( )

23

6 2

1

4

small error

FDU C f d

U C d kT

d d kT C

U C

µ

µ µ

µ

e e e

f e e f e e

p e e µ

e e e e p µ

µ

∞

↓

=

′= =

= +

+ + =

≈ +

∫

∫

∫ ∫

( ) ( )

( )

222 1/2

0 0

2 2 2 223/2 2 1/2 1/2

0 0 0 0

0

4

( )( )

12 4 6

kT

kT U U C kT C kT

p µ µ µ

p p p µ µ µ

µ

− +

⇒ − = ⋅ − + =

Now0 1/2 3/2

00

23/2

0 0

2

3

2

3 6

N C d C

U U C

µ e e µ

p µ

= = ⋅

⇒ − = ⋅

∫

2

3⋅

( ) ( )2 22

0 02 4

kT kT N

p

µ µ

=

Then

( )

( )

0 1/20 00

22

00

22 2

00 0

3

5

35 4

3 51

5 12 2V

U C d N

kT U N N

kT kT N C Nk

µ e e e µ

p µ µ

p p µ

µ µ

= =

⇒ = +

= + ⇒ =

∫



Equilibrium between Phases

Two-Phase Co-Existence Curves (single-component)

Sample Phase Diagram

Equilibrium between phases

1 2 1 2 P P T T = = 1 2µ µ =

For example, all along the condensation curve (i.e., liquid-gas coexistence)

( , ) ( , ) L GP T P T µ µ =

For point A 1 2( , ) ( , )T P T Pµ µ =

B 1 2( , ) ( , )T T P P T T P Pµ µ + ∆ + ∆ = + ∆ + ∆

1 2d d µ µ ⇒ = Tangential derivative along the curve

Now ( , ) ( , )G T P N T P

G U TS PV

dG d

µ =

= − +

= U T d − S SdT P d − + V VdP SdT VdP+ = − +

In terms of per particle or molar variables

( ) ( )

v v

v v

v

g g

g g

s dT dP s dT dP

dP s s dT

dP s

dT

− + = − +

⇒ − = −

∆⇒ =

∆

s∆ (at fixed temp) =( ) L T

T where L(T) is the Latent Heat

( )

v

dP L T

dT T =

∆ Clausius-Clapeyron Eqn.



Slope of the coexistence curve:

Solid to liquid transition:

Most case v 0∆ > but water is an important exception for which v 0∆ <

As the entropy of liquid > entropy of solid, this means the slope is negative for water as shown below:





Order parameter v vg −

(or density difference g ρ ρ −

).

V-T Plane:

Co-existence occupies a whole region of plane and not simply a curve as in the P-T plane. This is

because the volume V , unlike the pressure, is not the same for the two phases.

Total molar volume vo ov v vg g x x= + (due to coexistence)

x: mole fraction 1g x x+ =

( ) ( )ov = v vg g g x x x x⇒ ⋅ + +

o

o

v vLever Rule

v v

g

g

x

x

−⇒ =

−

Van der Waals Eqn. of State for the Liquid Gas Transition (1873 Ph.D. Thesis)

(v, )P P T = Equation of State

Ideal gas law fails—non interacting systems do not give phase transition in a classical system (Bosecondensation is the only exception in Quantum Systems).

[ ]2

N

P a V Nb NkT V

+ − =

or ( )

2v

v

aP b kT

+ − =

Works away from the critical point, but near critical point fails.

The first term in the equation of state originates from long-range, attractive interactions between gasmolecules (see below) while the second term arises from the “excluded volume” of the gas molecules asthey are not really point particles as assumed in ideal gas law calculation.

F P

V

∂= − ⇒

∂ Since the extra pressure (compared to the ideal gas case) arising from the 1st term in the

equation of state is negative, this would mean the presence of a new attractive free-energy term(compared to the ideal gas case).



2

~2

extraa N

F V

− for the extra attractive term1

2

a N N

V ≡ − ⋅ ⋅ ⋅

Every particle interacts with everyone else with the same strength a

V − . (You need the

1

V term for the

interaction strength to be able to take the thermodynamic limit. Note thatF

N should be finite;

1~

2

F N a

N V − which is finite in the thermodynamic limit). Infinite – Range Model → gives so-called

mean-field behavior

Rewriting Van der Waals Eqn. of State

3 2 vv v 0

kT a abb

P P P

− + + − =

(1)

For cT T < 3 solutions 1 2 3(v , v , v ) for fixed P, T

cT T = 3 solutions coalesce at v vc= i.e. for ,C C P P T T = = LHS of Eqn. (1) is ( )3

C v v−

But ( )3 3 2 2 3v - v v -3v v 3v v - vc c c c= + (2)

Comparing (1) and (2)

From (iii) and (ii) dividing

Then2

3 2

and3

3 ( ) 33

3 ( )27 27

8( )

27

c cc c

c

c cc

c cc

kT b i b v b

P

a ab av ii P

P b b

ab av iii T

P bk

n n

+ = = ⇒ =

= = =

= =

Define reduced variables / , / , /c c cP P P T T T n n n = = =

Universal from

( )1 28 3 1 3 (3)P T n n

− −= − − Law of Corresponding States





Critical Exponents

Start with Eq. (3), get rid of the tildes so that v 1c c cP T = = = in this set of units.

2

8 3

3v 1 v

T P = −

−

Let’s set v 1 vδ = + to expand near the critical volume which is 1 in this set of units.

2

8 3

(3 3 v 1) (1 v)

T P

δ δ = −

+ − +

( )( )

12

2

2 3 2 3

8 3 34 1 v 3 1 v

3 21 v2 1 v2

3 9 27 2 3 2 3 44 1 v v v 3 1 2 v v v

2 4 8 2 3 2

T T

T

δ δ δ δ

δ δ δ δ δ δ

−−

= − = + − + ++

⋅ ⋅ ⋅ = − + − − − + −

⋅

( )

( ) ( ) ( )

2 3 2 3

2 3

274 6 9 3 1 6 9 12

2

274 3 1 6 9 12

2

T v v v v v v

T T v v T v

δ δ δ δ δ δ

δ δ δ

= − + − − − + −

= − + − − + − −

Now 1 v 0T δ → → ; Let’s write 1 ( 0)T τ τ − = →

Then ( ) ( )2 331 4 6 9

2P v v vτ τ δ δ δ ≈ + + − + − (7a)

If we set v 1 vδ = −

( ) ( )2 331 4 6 v 9 v v

2P τ τ δ δ δ ≈ + + + + + (7b)

Subtracting30 12 v 3 vτ δ δ = ⋅ +

( )

2 2

4 v v 4 1 T

τ δ δ ⇒ − = ⇒ = −

( )1/2

1/2

1/2

2(1 ) symmetricshape of the co existence curve

v ~ 1 2(1 )

v ~ 1 2(1 )

g

v T

T

T

δ = ± − −

⇒ + −

− −



Order parameter1/2v v ~ 4(1 ) ; 1g T T − − →

Order –parameter behaves like OP t β hence the order parameter exponent i.e β -exponent for the vdW

gas is1

2 (parabolic)

( )( )

1/21/2

v v 4 11 1~ ~ (1 )

v v v v 1 4 1

gg

g g

T T

T ρ ρ

− −− − = = −

− −

Experimentally one finds,0.327

~g cT T ρ ρ − − so vdW theory does not work near the critical point.

Simple Statistical mechanical Calculation to obtain van der Waals Equation of StateOrnstein (1908)

Interaction between two particles separated by a distance r

( ) ( ) >0r q r V

α φ α = −

With0

0

( )0

r r q r

r r

∞ ≤=

> Hard-core repulsion + infinite-range attractive interaction

Can be solved exactly only in 1-d for particles can be ordered

( )2

12

ii j

i i j N

p x x

mφ

≤ < ≤

= + −∑ ∑H

( )2

21 1

1( , )

ii j

p x xm

N N N N Q L T e dp dp e dx dx

h

φ β β −∞ − −

−∞∑ ∑= ×∫ ∫ ∫ ∫

The N! term does not occur in ( , ) N Q L T as particles can be ordered in1d, they are distinguishable.

Then



( )

( )

1

( 1)

21

1( , )

1

i j

i j

hc

q x x L

N n N

N N q x x

Ln N

Q

Q L T e dx dx

e e dx dx

α β

βα β

λ

λ

− − + −

−− −

∑=

∑=

∫ ∫

∫ ∫

((((((((((((

Gas of hard rods of length 0 (Tonks 1936)

Positional marking on the LH corner of each rod. Maximum value of is . N

x L

0

1 2

-th __ ___

L

N

x x

((((((((((((

( ) ( )1 2 1 10 0 0 i j i jq x x q x x L L Lhc N N N Q dx dx dx e dx dx dx e β β

− − − −−∑ ∑= ≡∫ ∫ ∫ ∫ ∫

If they do not overlap 0ijq = and so the integrand is one.

If they do overlap ijq = ∞ and the integrand is zero.

We will construct conformations such that they do not overlap.

Limit on2 0

1 10

x x dx

−

∫

Limit on x2

3 0

02

0

particle 1 is already occupying 0 to space.

xdx

−

∫

Limit on4 0

03 32

x x dx

−

∫

……..

Limit on

( ) 01

L

N N

N

x dx

−∫

( )

2 0 3 0 4 0

0 0 00

1 2 3

2 1

L x x x

hc N

N

Q dx dx dx dx− − −

−

⇒ = ∫ ∫ ∫ ∫

Change of variable

01 1 2 2 0 3 3 02 ........ ( 1) N N u x u x u x u x N = = − = − = − −

2 0 3 0 4 0 0

0 0 0

2 3 ( 1)

1 2 3 0

x x x L N

hc N Q du du du du− − − − −

= ∫ ∫ ∫ ∫

but 2 0 2 3 0 3; 2 .. . x u x u etc− = − =



2 3 4 0

0 0 0

2 3 0

0 0

( 1)

1 2 3 10 0

( 1)

10

N u u u u L N

hc N N

u u L N

N

Q du du du du du

du du

− −−

− −

=

=

∫ ∫ ∫ ∫ ∫

∫ ∫ ∫

Change the order

0 4 3 2

2

23

34

-( -1) 0-1 3 2 10 0 0 0 0

"

"2

3.2

- ( -1)

!

N N L N u u u u

N N

u

u

u

L N du du du du du

N = … =∫ ∫ ∫ ∫ ∫

((

((((

((((((((

Then

( 1)02

( 1)1( , )

!

N N N

L N N

L N Q L T e

N

α β

λ

−− −

=



Magnetic Phase Transition

Isotherms

Spontaneous Magnetization 0 0( , 0) M M M H = →

Order Parameter 0 M

Coexistence curve

Magnetic Phase Transition

Curie-Weiss TheoryMean-Field TheoryMolecular Field Theory



B

Ising model

| unit of energy

i j i

ij i

J h

B

s s s

m < >

= − −

→

∑ ∑H

If 0 J = spins are independent

[ ]

1 2 3

1 2

1 2

1 2

1

1

0

( )

1 1

1 1

12cosh

ii

i

N

h

h

h h

N

N h

Q e

e

e e

e h

β s

s

β s s s s

s s

β s β s

s s

β s

s β

+

+ + +

=± =±

=± =±

=±

∑=

=

=

= =

∑

∑ ∑

∑ ∑

∑

0 2cosh( )F kT n Q NkT n h

F M N kT

h

β = − = −

∂= − = +

∂

1

22

cosh( )h β ⋅ ( )Sinh h β β ⋅ tanh( )

tanh( )

N h

M m h

N

β

β

= ⋅

= =

When 0 J ≠ , we will try to write the Hamiltonian in terms of an effective field due to all other spins.

Suppose we can write

i i

i

hs = −∑H

Then

[ ][ ] [ ]

1 1 2 2

1 21 1 1

1 22cosh( ) 2cosh( ) 2cosh( )

N N

N

hh h

N

Q e e e

h h h

β s β s β s

s s s

β β β

=± =± =±

=

=

∑ ∑ ∑

If 1 2 N h h h h= = = =

then tanh( )m h β = is recovered.

Now,

( )

i i

i

i j

j

j j j

j j

H h

h h J

h J J

s

s

s s s

= −

⇒ = +

= + + −

∑

∑

∑ ∑



= applied field + mean-field + fluctuations. Neglect fluctuations and note that im s =

2| co-ordination #

ih h J d m≈ + ⋅ ⋅

→ All ih are the same

2tanh

B

h dJmm

k T

+=

Set 0h = and look for spontaneous magnetization

tanh(2 / ) B

y

m dJm k T =((((((

For 2

1dJ

kT < , there is only one solution.

2c

B

dJ T

k =

Now look at the eqn. of state ( 0)h ≠ define /cT T τ =

( ) ( )( ) ( )

tanh

tanh / tanh

1 tanh / tanh

tanh( )tanh

1 tanh( )

B

B

B

B

hm m

k T

h k T m

h k T m

h m m

k T m m

τ

τ

τ

τ

τ

= +

+

=+ ⋅

−⇒ =

−

(Note that τ is not small!)

Then for small h and m we can expand the tanh-function as3

tanh3

x x x≈ − .

( )

( )

3

3

3

13

B

mm mh

k T m

m m

τ τ

τ τ

− +≈

− −

3

3 2(1 ) 1

3m m m

τ τ τ

≈ − + +

( ) 33 2

31

B

hm m

k T

τ τ τ τ ≈ − + − + Equation of State 1)



Set 0h = to study spontaneous magnetization

( ) 3 2

2

2

30 1

3

11 (note

13

m m

m

τ τ τ τ

τ τ τ

τ τ τ

− + − +

−⇒ = − → →

− +

0) cT

T

τ =

2 (1 )3 1 3 3 3 (2)

11 1 1

3

define 1

c c c c c

c

c

c c

T T T T T T T T m

T T T T T

T T T t

T T

τ − − −− ≈ − ≈ − − ≈ − = = − +

−= = −

12 2Then, 3 1 3 (1 ) 3 3 3 as 0m t t t t t t t t

−≈ − = + ≈ + ≈ →

Thus, we see that the order-parameter exponent is 12

β = --- the same as in a vdW gas!

The concept of Universality arises in terms of critical exponents.

Free Energy

( )

2 2cosh

22cosh 0

B

B

h dJmF NkT n

k T

F dJm f kT n h

N k T

+ = −

= = − =

Write, ( ) and look at the expansion of x A T m= 2 coshn x

2 4

2 4 2

2 4 4

2nd term1st term

2 4

2 cosh 2 12 24

2 1 careful! (1 )2 24 2

22 24 2 4

1 2

2 12

x x x xn x n e e n

x x xn n n x x

x x xn

xn x

−

= + = + + +

= + + + + + ≈ − +

≈ + + −⋅

≈ + −

!

!

((

( )

22 4 4

2 40 2 4

2 ( ) ( )2 12

( ) ( )

m kT f kT n kT A T A T m

a t a T m a T m

= − − +

= + + Landau expansion



Classification of Phase Transitions: Ehrenfest’s Scheme

Note that when we derived Clausius-Clapeyron Eq. we moved along the co-existence curve and for each

point we have used 1 2µ µ =

But we have not put any restrictions on orG G

P T

∂ ∂

∂ ∂

;T P

G GV S

P P

∂ ∂ = = −

∂ ∂ “normal derivatives” of G

If these derivates are discontinuous – First Order Phase Transition

If these derivates are continuous, but the higher order derivates are discontinuous – n-th order transition.

Unfortunately, Nature has not co-operated with Ehrenfest’s scheme except for the important case of First

Order Phase Transition.

For example, Sp. Heat → 2nd derivative

Diverges at the critical point of a magnetic phase transition

Modern Usage – 1st order and Continuous Phase Transition

Unambiguous Definition of Phase Transition

In single phase regions of the P-T plane, G (T, P) is an analytic function bounded by co-existence curvesalong which the function is not analytic.

Then, Taylor expansion would not converge as one crosses co-existence curves.

Landau’s Phenomenological Theory

• Near the critical point, free-energy can be expanded in terms of the order parameter m.

• Minimize the free-energy in terms of m.

Landau Free-Energy ( ) order parameter LG ψ ψ →

Not quite the Gibbs free-energy but will not get into that. In the absence of any external field they behavequite similarly.

2 32 30 1( ) ( ) ( ) ( ) ( )

2 3 L

a aG a T a T T T ψ ψ ψ ψ = + + + +

The expansion is motivated by the expansion of the mean-field equation of state for m small i.e., near cT

Symmetry: in the absence of applied field ( ) LG ψ is symmetric under ψ ψ → −

( ) 2 42 40

( )( ) ( )

2 4 L

a T a T G a T ψ ψ ψ = + + +



1. Continuous Transition

4a is positive

2 2 0( ) ( )a T A T T = − a2 changes sign when 0T T =

0 Ls

Gψ ψ

ψ

∂= ⇒ =

∂ 0 2, aT T < is negative

32 4 0s sa aψ ψ + = ⇒ 0sψ = or ( )2 2 2

04 4

s

a AT T

a a

ψ = − = − −

( )1/22

04

s

AT T

aψ = − ψ is continuous at cT T = . No Latent Heat

1

2 β = 0cT T =

0ψ = is maximum.

2. First order phase transition

(i) Symmetry ψ ψ → − survives

(Ferroelectric crystals like Potassium dihydrogen phosphate or KH2PO4 )

2 2 0( ) ( )a T A T T = − 4a is negative.

We need 6a term (positive) in the expansion to “produce” a first-order phase transition.

( ) 2 4 60 2 0 4 6

1 1 1( ) ( ) ( ) ( )

2 4 6 LG a T A T T a T a T ψ ψ ψ ψ = + − − +



First Order Phase Transition: Discontinuous jump in the order parameter which is related to

Latent Heat (via the discontinuity in entropy).

Conditions:

1) 0 LG

ψ

∂=

∂

minimum

2) At , ( 0, ) ( , )c L c L s cT G T T G T T ψ ψ ψ = = = = =

The “weak” secondary minimum seen above Tc becomes equal strength at Tc and then

becomes deeper i.e. global minimum below Tc .

From 0 LG

ψ

∂=

∂

3 52 0 4 6( ) 0s s s A T T a aψ ψ ψ − − + =

2 42 0 4 60 or ( ) 0 (1)s s s A T T a aψ ψ ψ ⇒ = − − + =

22 4 4 6 2 0

6

4 ( )

2s

a a a A T T

aψ

± − −⇒ =

24 6 2 04 ( )a a A T T ≥ − for the second minimum to develop.

Let 1T T = when2

4 1 6 1 2 1 0( ) 4 ( ) ( )a T a T A T T = − . Then 1T T < for the double well shape to form.



Now at cT From (1) ( ) ( )2 42 0 4 6( ) 0c c s c s A T T a T a T ψ ψ − − + = (2)

Also recall the second condition: (0, ) ( , ) L c L s cG T G T ψ =

2 4 60 0 2 0 4 6

2 42 0 4 6

1 1 1( ) ( ) ( ) ( ) ( )

2 4 6

1 1 1( ) 0 (3)2 4 6

c c c s c s c s

c s s

a T a T A T T a T a T

A T T a a

ψ ψ ψ

ψ ψ

⇒ = + − − +

⇒ − − + =

From (2) and (3)2 4

6

( )3

4

c

s

a T

aψ = (4)

Substituting2sψ in eq.(2)

( ) 42 0 4

6

3

4c

a A T T a a

a− = ⋅ −

24

6 26

9

16

a

a

⋅24 6

0

3/ 0

16

c

a a

T T

= >

⇒ >

When both 2a and 4a vanish at the same temp. → Tri-critical points.

(ii) Asymmetric case 3,ψ ψ terms may appear (but ψ term can be eliminated)

Nematic Liquid Crystals

( ) 2 3 3 40 2 0 4

1 1 1( ) ( ) ( ) ( )

2 3 4 LG a T A T T a T a T ψ ψ ψ ψ = + − + +



Magnetic Phase Transition in the Infinite-range model and its connection to

the Mean-Field Model

1

i j

i j N

J

N σ σ

≤ < ≤

= − ∑H

1 N − term necessary for thermodynamic limit to exist

Note:

( )

( ) ( )

2 2 21 2 1 2 1 2

2 2 2 21 2 2 3 1 3 1 2 3 1 2 3

2

2

.etc

σ σ σ σ σ σ

σ σ σ σ σ σ σ σ σ σ σ σ

= + − −

+ + = + + − − −

Hence,2

2

1 12

2

1

2 2

but 1, so

2 2

N N

i i

i i

i

N

i

i

J J

N N

J J N

N N

σ σ σ

σ

σ σ

= =

=

= − + ⋅

=

= − + ⋅

∑ ∑

∑

H

H

i

i

N Q e β σ

σ

−= ∑ H define J K β =

( )2

22

i

i

K K

N e eσ

σ

− ∑= ∑

Use the identity

22

221

2

xa ax

e e dxπ

− + ∞

−∞= ∫ This is just a Gaussian integral but in this

context it is called the Hubbard-Stratanovich Transformation

1/2

Choose iK

a N

σ

=

∑

22

1exp

22i

K

N i x K

Q e x dx N

σ

σ π

∞−

−∞

= − +

∑ ∑∫

Spins are now decoupled and we know that 1iσ = ±



22

1exp 2cosh

22

N K

N x K

Q e x dx N π

∞−

−∞

= −

∫

Note:

i

i

e

α σ

σ

∑

∑

( )

( )

1 2

1 2

31 2

1 2

1

2cosh

N

N

N

i

i

N

N

e

e e e

e

α σ σ σ

σ σ σ

ασ ασ ασ

σ σ σ

ασ

σ

α

+ +

=±

=

= ⋅ ⋅

= =

∑ ∑ ∑

∑ ∑ ∑

∑

New variable 1/2( )m x KN −=

2

2cosh( )1/222

2

KmK N n Km

N

KN Q e e dm

π

− −∞ −

−∞

=

∫

Note: N cb e c N n b= ⇒ =

Saddle Point or Method of Steepest Descent to evaluate the integral:

write2

( ) 2cosh( )

2

m f m K n Km= −

Minimum value of f(m) contributes the most to the integral when N is very large.

0( ) 0, f m m m

K

′ = =

⇒ 01

2m =

0

2cosh( )Km

K 0sinh( )Km

0 0tanh( )m Km=

1K < only one solution

1 1cc

J K

kT = ⇒ = and we find that

20

02cosh( )1/222

2

KmK N n Km

N KN

Q e eπ

− − −

=



How do we know 0m is the magnetization?

Need to consider the presence of an applied field

20

02cosh( )

1/2 220 0tanh( );

2

mK N K n Km B

N KN m Km B Q e e

h B

kT

π

− − +

− = + =

=

20

0

00

0

0 0

1 2cosh( )

2 2 2 2

1 12sinh( )2cosh( )

tanh( )

tanh( )

N F kT n Q

mKN K kT n Nk N n Km B

F M kT N Km Bh Km B kT

N Km B

M Km B m m

N

π

= −

= − − − + +

∂= − = + ⋅ + ⋅ ∂ +

= +

= + ⇒ ≡

Documents

Statistical Mechanics 2014 Lecture Notes by Amit Charkrabati