Statistical Thermodynamics - PBworksStatistical Thermodynamics 1. Introduction 2. Coin-Tossing Experiment 3. Assembly of Distinguishable Particles 4. Thermodynamic Probability and

Statistical Thermodynamics1. Introduction

2. Coin-Tossing Experiment

3. Assembly of Distinguishable Particles

4. Thermodynamic Probability and Entrpy

5. Quantum States and Energy Levels

6. Density of Quantum States

Introduction

• Energy ---- Entropy

the great of success of T.H.D. , how.

• But why?

• Needs particle theory or microscopic interpretation of equilibrium properties of macroscopic systems.

• Fundation is based on Q.M..

quantum states, energy levels and inter-particle interactions.

• The basic postulate of S.T.H.DEqual probable of allowed micro-states

• Ideas are centered around probability density function applies to a collection of many identical particles.

• Macrostate or configuration of a system :denoted by a set {Nj ,j=1,2,…,n; all energy levels} and Σ Nj = N .

• Microstate : number of particles in each quantum state. Degeneracy is possble for certain energy levels.

• Probability of a macrostate is given by number of corresponding microstates ωk, this is called the thermodynamic probability which is from 0 to 1 but true probibiltiy can be given by #microstates/total microstates or

pk = ωk / Ω because Ω = Σ ωk

Summary:

1. system can be in several macrostates

2. several microstates corresponding to every macrostate

3. each microstate is equally probable

Coin-tossing experiment• 4 coins: coin1, coin2, coin3, coin4

2 states of coin: H and T• 4 macrostates:

M1 : 4H M2: 3H1T M3: 2H2T M4: 1H3T M5: 4T

• Corresponding microstates of M1 to M4:M1: HHHH ( 1 microstate)M2: HHHT,HHTH,HTHH,THHH(4 microstates)M3: HHTT, TTHH,THTH,HTHT,HTTH,THHT(6 microstates)M4: HTTT,THTT,TTHT,TTTH (4 microstates)M5: TTTT (1 microstate)

probability of each macrostateM1: 1/16 M2: 4/16 M3: 6/16 M4:4/16 M5: 1/16

Result:M3(2H2T) is the most probable

Any number of coins

• N coins• Macrostates:

N1 H N2 T (N1 + N2 = N)# of microstates

w= N!/ N1!(N- N1)! Example:

(1) N=8, the most probable macrostate is N1=4wmax = 8!/4!4! = 70(2) N=1000 , wmax = 1000!/500!500! = ????

A quick calculation of N! when N is very large

• Stirling’s formula:㏑ n! ~ n ㏑ n –n

• ㏑ wmax = ㏑ 1000!/500!500! = ㏑ 1000! – 2 ㏑ 500!~ 1000 ㏑1000-1000 -2(500 ㏑500-500)~1000 ㏑1000 - 1000 ㏑500~ 1000㏑2 ~ 693

• And, ㏑10 x = ㏑10 2 ㏑ x = 0.4343 ㏑ x so ㏑10 wmax = 0.4343 ㏑ wmax = 0.4343*693=300

• Then wmax = 10300 !!!

•When the system is large…….(if N=1023=NA !!!)only the most probable macrostate stands out!

• This is the thermal equilibrium state of the system.

12.3 Assembly of distinguishable particles

• Macrostates specified by (N,V,U) or (N or n,V,T)

where T is a measure of the internal energy of the system.

We are talking about thermodynamics!

• Weakly interacting and energy levels of each particle (εj), microstates of the system must respect 2 conditions:

Σ Nj = N (conservation of particles)

Σ Nj εj = U (conservation of energy)

Example 1

3 particles: A,B and C

4 energy levels: εj,=j ε, j=0,1,2,3

ε0 = 0, ε1 = 1ε, ε2 = 2ε, ε3 = 3ε

2 conditions:

N=3, U = 3 ε

Possible macrostes (3, 3 ε ):

N0 = N1= N2 = N3 =

k=1 2 0 0 1

k=2 1 1 1 0

k=3 0 3 0 0

Microstates:K= 1 = 2

C B A ● --- --- --- --- --- --- ------ --- --- --- A A B B C C ●

--- --- --- --- B C A C A B ●AB AC BC ● ● C B C A B A ●

K = 3--- ------ ---ABC ● ● ● wk : w1 = 3 , w2 = 6 , w3 = 1 --- --- pk : p1 = 0.3 , p2 = 0.6 , p3 =0. 1

the most probable macrostate is k=2Note: ● and --- represents the occupation number of the levels

What if the number of particles N and the system’s internal energy U getting larger and larger??

• The most probable macrostat will stand out and it will be the equilibrium state or the most disordered state.

12.4 Thermodynamic probability and Entropy

• Boltzmann:S = k ㏑ w , k = 1.38x10-23 J K-1

Which satisfies Stotal = SA + SB ( entropy is an extensive quantity)

andwAB = wA wB (Law of combined probability)

Proof:

Stotal = k ㏑ wAB = k ㏑(wA wB) = k ㏑wA + k ㏑wB

= SA + SB

12.5 Quantum States and Energy Levels

• Energy level , quantum state, stationary state

,degenerate states(degeneracy)

• Ground states and Excited states

• Example:

a particle of mass m in an one-dimensional box with infinitely high walls(confined within the region 0 ≦ x ≦ L ).

• Wavefunction

• Ψ(x) = A Sin(kx) , 0 ≦x ≦L

k = n π/L , n= 1,2,3,….

p = h k , h = 1.054x10-34 JS

ε = ½ m v2 = p2/2m = (h k)2/ 2m

= (πh )2 n2 /2mL2

= ε 0 n2, n=1, 2, 3,..

where ε 0 = (πh )2 /2mL2 (ground state energy).

Extend to three-dimensional box with sides Lx , Ly , Lz

• Energy levels:

ε = (πh )2 /2m( n x2/ L x

2 + n y2/ L y

2 + n z2/ L z

2 ),

nx,y,z = 1,2,3,…..

• Hence, εj α ( n x2+ n y

2+ n z2) = nj

2 when Lx = Ly = Lz = L

energy levels only depend on nj2 , not individual

numbers n x , n y , n z .

• V = L3,

ε j = n j2 V -2/3 (πh )2 /2m , can apply to a container of

any shape of large size.

Degeneracy: ( n x2+ n y

2+ n z2) = nj

2

• Degeneracy gj↗ as nj ↗ .

• Gap or spacing of energy levels Δ ε j↗ as V ↘

• macrostates:

occupation numbers {Nj }

• 2 conditions:

Σ Nj = N

Σ Nj ε j = U

= nj2

•Helium atoms in V = 1 liter at room temperature(293k):

1 unit of energy of the quantum states ~ 5x10 -21 eV

while k.e. ~ 1/40 eV at room temperature

nj ~ 2x109

• closely spaced levels , occupied up to very high levels

energy levels can be considered continuous not discrete.

12.6 Density of Quantum States

1. Number of quantum states within ε + Δ ε and εg(ε)d ε = N(ε + Δ ε)-N(ε)

~ dN(ε )/dε ． Δ εwhere

N(ε ) = 1/8．4/3．π n3 = π/6 V(8m/h2)3/2 ε3/2

g(ε) = dN(ε )/dε= γ s 4 √2 π V/h3 m3/2 ε1/2

Spin factor: γ s = 1 (spin zero particles)or

γ s = 2(spin ½ particles)

ε = n 2 V -2/3 (πh )2 /2m