1

Chapter 13

Classical and Quantum Statistics

2

So far, with the exception of the previous chapter, we have dealt with the 1st and 2nd laws of thermodynamics. In using these laws to make numerical calculations it is usually necessary to appeal to experimental measurements. We would like to calculate all thermodynamic properties of a system from a microscopic model of that system. We have made a start in the last chapter and we will make significant progress in the present chapter.

3

Lagrange Undetermined MultipliersI would like to give you a trivial example of the use of

Lagrange Undetermined Multipliers. The term is somewhat misleading because the multipliers can, in fact, be determined. In the following example it is not necessary to use this sophisticated method and you should solve the problem in a simple fashion.

Consider the equation ax+by=0……………..(1)If y=y(x) then y=-(a/b)x However this is not true if both

x and y are independent. The only solution is then a=b=0Suppose, however, that x and y are not completely independent

but satisfy a constraint condition such as x+2y=0………(2)What can we say about the coefficients a and b?

The procedure, using Lagrange Multipliers, is to multiply each constraint condition by a multiplier, which is initially unknown. In the present case we have one constraint condition and let multiplier.

0)2()1(0)2( yx giving

)3.....(0)2()( ybxa

4

asoa 0)(

20)2(

bsob

Now that the constraints have been implicitly taken into consideration we treat x and y as two independent variables. We must have

This yields ab 2

Placing this in equation (1) shows that the constraint condition is satisfied.

5

EXAMPLE: Lagrange Undetermined MultiplierA cylindrical nuclear reactor has radius R and height H

We wish to minimize the volume of the reactor.There is a constraint supplied by neutron diffusion theory

constraint

HRV 2

constantHR

4048.222

)2(0)4048.2(

02)4048.2(2

3

2

3

2

3

2

3

2

dHH

dRR

dHH

dRR

0)2()1( 0

)4048.2(2

3

2

3

22 dH

HdR

RdHRRHdR

dHRRHdRdV 22 For an extremum dV=0

)1(02 2 dHRRHdR differentiating the constraint

0)4048.2(

23

22

3

2

dH

HRdR

RRH

6

)4.....(0)3.....(0)4048.2(

23

22

3

2

H

RR

RH

Now we consider R and H to be independent

From equation (4) 321HR

Substituting into equation (3)

32

3

2)4048.2(2

HR

RRH

)4048.2(2

)4048.2(2

2

222 R

HR

H

(multiplier is determined)

7

Boltzmann Statistics.We consider N distinguishable particles and we can place any

number into a particular state. We wish to determine the thermodynamic probability for a particular macrostate.

To guide us we consider a simple example: N=3 (A B C) and take the macrostate in which 32 11 gN

We begin with a box labeled and we wish to throw 2 particles into the box.

1st particle: 3 choices ( N)2nd particle 2 choices (N-1)

The total number of choices is (3)(2) (N)(N-1)

These choices are shown on the next slide

1N

8

Choices

1st 2nd

A BA CB AB CC AC B

However we can permute the particles in the box without changing the contents of the box. The number of permutations is 2! ( !)1N

Therefore the number of distinct choices is:

!)1(

!223

1N

NN

They are AB AC BC

Now we arrange these particles into the 3 degenerate states.1st particle has 3 possibilities2nd particle has 3 possibilities

1g

1g

The total number of possibilities is 1

123 Ng

We will list these possibilities for the case where the box has AB

9

STATES

1 2 3 AB AB ABA BA B A BB AB A B A

We see that there are 9 possibilities.We also have the AC and BC possibilities, each with 9 possible distributions among the states

23

The total number of possibilities is therefore

1

11

2

!)1(

3!223 Ng

N

NN

Now we concentrate on the general case of N particles and consider placing particles in energy level which has a degeneracy

1N

1g1

As above selecting the particles for the box, we have 1N 1

!)1()2)(1(

1

1

N

NNNNN possibilities

We can write this as

1)1)((!1)1)()(1()2)(1(

111

111

NNNNN

NNNNNNNNN

10

and this is evidently )!(!!

11 NNN

N

Now we consider the possibilities for distributing these particles into the degenerate states of 1The total number of possibilities is then

)!(!!

11

11

NNN

gN N

Now we go to the level. The procedure is obviously the same except that we no longer have N particles. The number of available particles is

2

)( 1NN

As above we have, for this level, possibilities)!(!

)!(

212

212

NNNN

gNN N

For the level3)!(!

)!(

3213

3213

NNNNN

gNNN N

11

Hence the thermodynamic probability is

)!(!

)!()!(!

)!()!(!

)!(),,(

3213

321

212

21

11

121

321

NNNNN

gNNN

NNNN

gNN

NNN

gNNNNw

NNN

n

!!!

)!(),,(

3

3

21

2121

321

N

g

NN

ggNNNNw

NNN

n

n

i i

Ni

n N

gNNNNw

i

121 !

!),,( Boltzmann statistics(distinguishable particles)

This is the total number of accessible microstates for a particular macrostate. The value of w will be different for each particular macrostate. The greater the value of w, the greater the probabilityof occurrence. Remember:

The equilibrium macrostate is the one for which w is a maximum.

12

Example: Before continuing with our discussion of Boltzmann statistics, let us consider a simple situation.

Consider 6 distinguishable objects (A,B,C,D,E,F) which can be placed in 6 boxes (1,2,3,4,5,6). We will calculate w for several macrostates. The degeneracy will be unity for the boxes.

6

1i i

6

1i i

Nn

1i i

N

in21 !N

1!6

!N

1!6

!N

g!N)N,N,N(w

ii

720)1,1,1,1,1,1(w!6!1!1!1!1!1!1

1!6)1,1,1,1,1,1(w

1)6,0,0,0,0,0(w1!6!0!0!0!0!0

1!6)6,0,0,0,0,0(w

360)0,2,1,1,1,1(w!2

!6

!0!2!1!1!1!1

1!6)0,2,1,1,1,1(w

6),0,0,0,0,1,5(w!1

6

!0!0!0!0!1!5

1!6),0,0,0,0,1,5(w

(most disorderd)

13

We vary the to find the maximum value for w. This will give the in terms of and

iN

ig iiN

For convenience we will work with ln(w) instead of w. (The range of values is much smaller.) This will also permit us to use Stirling’s fromula, valid for large x: ln(x!)=xln(x)-x

)!ln()ln()!ln(!

!ln)ln(111

n

iii

n

ii

n

i i

Ni NgNN

N

gNw

i

Apply Stirling’s formula to the last term :

n

ii

n

iiii

n

ii NNNgNNw

111

)ln()ln()!ln()ln(

Ng

NNNw

n

i i

ii

1

ln)!ln()ln(

Now we maximize to obtain the equilibrium distribution.

n

ii

ii

iii

n

i i

i dNgN

gNdN

g

Nwd

11

1ln))(ln(

constants

Now continuing with Boltzmann Statistics:

14

n

iii

n

i i

i NddNg

Nwd

11

ln))(ln(

)1(ln))(ln(1

i

n

i i

i dNN

gwd

We have the constraints:

constantUNconstantNN i

n

1ii

n

1ii

Differentiating 0011

ii

n

i

n

ii dNdN

We multiply the constraint conditions by the Lagrange multipliers

)(,)ln( We also have the condition for a maximum that d(ln(w))=0

0)ln(ln111

i

n

ii

n

iii

n

i i

i dNdNdNN

g

(This form of the multipliers is for convenience.)

15

The are now taken as independent, so the coefficient of each must vanish. This gives

iN

idN)2(0)ln(ln

i

i

i

N

g

Solving for yields iN

)3(iegZ

NN ii

Hence varies with the degeneracy and with the energy of the level. Now we need to determine the Lagrange multipliers.

iN

n

ii

n

ii

iegNN11

We write this in the form

Z

N with

n

ii

iegZ1

Z comes from zustandsumme, which means a sum over states. It is usually referred to as a partition function. As we shall see, partition functions are very useful. It turns out that Z depends only on T and the parameters that determine the energy levels.

iegN ii

16

Once Z is known it is straightforward to calculate many thermodynamic properties, such as U, S and P. Of course there are other ways of calculating these quantities, but the simplest way is to first calculate the appropriate ln(Z).

We still need to determine β.

17

Now we bring T into statistical mechanics. For a process

taking place between two equilibrium states:

đQ = dU+ PdV and dS = đQ/T so

Considering S as a function of U and V,

and comparing we see that

The state variables S and U may be calculated by

statistical methods and so this equation brings the

macroscopic concept of T into statistical thermodynamics.

V1 ddUdS TP

T

VVVddUdS

US

US

V

1US

T (reciprocity relation)

18

Now we are in a position to discuss the Lagrange multiplierWe have the following two equations (equations 1 and 2 above)

i

n

i i

i dNN

gwd

1

ln))(ln( 0)ln(ln

i

i

i

N

g

Substituting the second equation into the first one

n

iii

n

iii

n

ii

n

ii NdNddNdNwd

1111

)ln()ln())(ln(

We are dealing with a closed system with constant V so 0

1

n

iiNd UN i

n

ii

1

and

dUwd ))(ln(This gives

Using the Boltzmann relationship S=kln(w) dS=k d(ln(w))

kdUdSk

VUSor1

19

From the reciprocity relationship:

Tk

1ork

T

1

The temperature turns out to be a Lagrange multiplier!Now that we have determined the Lagrange multipliers,

we write down the following two fundamental equations (see equation (3)):

kT

i

ii

i

eZ

N

g

Nf

Boltzmann distribution

The Boltzmann distribution gives the probability of occupation of a single state belonging to the ith energy level.

n

i

kTi

i

egZ1

The partition function, an explicit function of T and an implicit function of V (through the energy levels) contains the statistical information about the system.

(See note on next slide)

20

NOTE: There is some confusion regarding terminology. The term Boltzmann distribution for the equation kT

i

ii

i

eZ

N

g

Nf

is unfortunate because, among other things, it is then easilyconfused with the Maxwell-Boltzmann Distribution. The Boltzmanndistribution applies to systems which have distinguishable particlesand N, V and U are fixed. The Maxwell-Boltzmann Distribution is applicable only to dilute gases.

21

Consider the case in which the energy levels are very closely spaced. Then, instead of considering discrete levels with degeneracies , we consider a continuum and replace the by , the number of states in the energy range between

igig dg )(

dand The Boltzmann distribution is then written in the form:

deg

Nef

kT

kT

)(

)(Distinguishable particles

22

Example of Boltzmann distribution.

Suppose that we have 1000 particles and T=10000K. The available states and their multiplicities are shown below. The partition function and the distribution of particles amongst the states are calculated using

(eV)

3

2

1

0

gi

4

3

2

1

n

i

kTi

i

egZ1

kT

ii

i

egZ

NN

Z=2.045

Ε(eV) Ni

0 489 1 307 2 144 3 60

23

24

Canonical ensemble

This consists of a set of systems in contact with a large thermal reservoir. For the Boltzmann distribution V, N and U are fixed.We now consider a system in which V, N and T are fixed. The energy of a system is not fixed, but will fluctuate about some average value due to the continuous interchange of energy between the system and reservoir. For large N these fluctuations are small. The formula giving the probability distribution is one of the most important in statistical mechanics. It applies to any system with a fixed number of particles in thermal equilibrium (V, N, T fixed). We will not provide a derivation of this probability formula, which is essentially the same as the Boltzmann formula {See section 13.9 of the textbook.}

kTiii

eZ

g

N

N

25

kTi

e

is called the Boltzmann Factor

We see that at low temperatures only low energy states are populated. As the temperature increases, the population shifts to higher energy states.

26

Example: An atmosphere at a temperature of 6000K has, as one constituent, neutral atoms with the first excited state 1.20eV above the ground state. The ground state (taken as the reference energy level of 0eV) is doubly degenerate while the 1st excited state is 6-fold degenerate. Assuming that there is negligible population of higher excited states, what fraction of these molecules are in the 1st excited state? (Use Canonical Distribution.)

1.2eV 6

0eV 2

)1(egZ

NN kT

ii

i

n

i

kTi

i

egZ1

59.2e6e2Z))K106(

K

eV1062.8(

eV20.1

kT

eV0 35

Placing in (1) kTi

kTi

iii

eg59.2

1eg

Z

1

N

N

))K106(K

eV1062.8(

eV20.1

235

e)6(59.2

1

N

N

227.0N

N 2

27

Now we are going to discuss two other important distributions. In preparation we review some aspects of fundamental particles. These particles are divided into two classes, depending on their quantum number called the intrinsic angular momentum or “spin”Spin quantum number an integer: Bosons (examples are photons, gravitons, pi mesons,……..)Spin quantum number an odd number of (1/2):

Fermions (examples are electrons, quarks, muons,……….)

Fermions obey the Pauli Exclusion Principle: In an isolated system, no two fermions can occupy the same state.

28

Fermi-Dirac Distribution.This distribution is for indistinguishable fermions. There can

be no more than one particle in any state. This places the following restriction on any macrostate: ii gN

Consider the ith energy level. We wish to place particles into the states.

iN

ig

For the 1st particle there are possibilities.ig

For the 2nd particle there are possibilities.)1( ig

For the ith particle there are possibilities.)1( ii Ng

The total number of possibilities is then )1()1)(( iiii Nggg

Since the particles are indistinguishable, we can permute them in a particular distribution without obtaining a different distribution.

29

The total number of distinct possibilities is then

!)1()1)((

i

iiii

N

Nggg or )!(!

!

iii

i

NgN

g

The total thermodynamic probability for a particular macrostate is

n

i iii

in NgN

gNNNw

121 )!(!

!),(

Again we will work with ln(w)

n

iii

n

iii

n

i

NgNgw111

))!ln(()!ln()!ln()ln(

The first term is constant. We use Stirling’s approximation for the other terms.

n

iii

n

iiiii

n

ii

n

iiii

n

i

NgNgNg

NNNgw

11

111

)()ln()(

)ln()!ln()ln(

(gi constant!)

{What are the Ni

at equilibrium?}

30

n

ii

ii

ii

n

iiii

n

ii

i

in

iii

dNNg

Ng

NgdNdNN

NNdNwd

1

111

)()(

)(

)ln()()ln())(ln(

n

iiii

n

iii NgdNNdNwd

11

)ln()()ln())(ln(

n

ii

i

ii dNN

Ngwd

1

)(ln))(ln(

As before, we have the constraints:

constantUNconstantNN i

n

1ii

n

1ii

We introduce the Lagrange multipliers and

0)()()(ln111

n

iii

n

ii

n

ii

i

ii dNdNdNN

Ng

31

0)(ln1

n

iii

i

ii dNN

Ng

With the constraints included in the equation, the are independentiNThe coefficient of each 0idN

ii

iii

i

ii

N

Ng

N

Ng

ln0ln

1

11

i

i

eg

Ne

N

g

i

i

i

i

It is not trivial to determine the Lagrange multipliers. It turns out, as before, that

Tk

1

The other multiplier is related to the chemical potential

kT

With these assignments we have

32

1

1/)(

kT

i

ii

ieg

Nf Fermi-Dirac distribution

For a continuous energy distribution, this becomes

1

1)(

/)(

kTef

Later in the course we will use the Fermi-Dirac Distribution.

33

Bose-Einstein StatisticsThe particles are again indistinguishable, but now any number

of particles can be in a particular energy state. (The Pauli Exclusion Principle does not apply.)

Example: 13 particles in the ith energy level which has a degeneracy of 8

xxx x xxx x xx xx x

xx xx x xx x x x xx x

1 2 3 4 5 6 7 8 State

We can obtain different distributions by moving particles and partitions around. How many ways can we arrange particles and partitions to form different distributions given that the particles and partitions are identical? The answer is

iN)1( ig

Number of ways=

!1!!1

ii

ii

gN

gN

partition

(Students: convince yourselves.)

34

Example: 3 particles in 4 degenerate states 20!3!3!6

Students: show these 20 distributions (BEEx.ppt)

The thermodynamic probability for a macrostate is then:

n

i ii

iin gN

gNNNNw

121 !1!

!1),,(

We proceed as before:

n

ii

n

iiii

n

i

gNgNw111

))!1ln(()!ln())!1(ln()ln(

Using Stirling’s formula

n

ii

n

iii

n

ii

n

iii

n

iiiii

n

iii

gggN

NNgNgNgNw

111

111

)1()1ln()1(

)ln()1()1ln()1()ln(

35

n

iii

n

iiiii

n

iii ggNNgNgNw

111

)1ln()1()ln()1ln()1()ln(

n

ii

n

iii

n

iiii

n

ii dNNdNdNgNdNwd

1111

)ln()()1ln()())(ln(

n

iiiii

n

ii NdNgNdNwd

11

)ln()()1ln()())(ln(

i

n

i i

ii dNN

gNwd

1

1ln))(ln(

Introducing Lagrange multipliers with the constraint conditions:

01

ln111

n

iii

n

ii

n

ii

i

ii dNdNdNN

gN

Proceeding as before:i

i

ii

N

gN

1ln

Neglecting the 1 (gi large!) ii

ii

N

gN

ln

36

ieN

g

N

g

i

ii

i

i

11ln

1

11

i

i

eg

Ne

N

g

i

i

i

i

Again, a more detailed analysis gives the same values as before for the Lagrange multipliers

1

1/)(

kT

i

ii

ieg

Nf

Bose-Einstein Distribution

This distribution will be used later in the course.

37

Maxwell-Boltzmann Statistics and ReviewLet us consider an assembly (system) of N indistinguishable

particles. A macrostate is a given distribution of particles in the various energy levels. A microstate is a given distribution of particles in the energy states.

Basic postulate of statistical mechanics: All accessible microstates of an isolated system are equally probable of occurring.

We have considered some general assembly with:

1N particles in any of the states of 1g 1particles in any of the states of

particles in any of the states of

2N

iN

2g

ig

2

i

We now impose the restriction that for all i. This gives the Maxwell-Boltzmann Statistics. This condition holds for all real gases except at very low temperatures. At low temperatures one must use either Bose-Einstein or Fermi-Dirac Statistics depending on the

ii Ng

38

spin of the molecules. This restriction means that it is very unlikely that more than one particle will be in a given state. Subject to this restriction we first consider the number of ways that distinguishable particles can be distributed among the states.The first particle can be placed in any one of the states. The second particle can then be placed in any of the remaining states, and so forth. The total number of different ways is then:

iN

ig

ig)1( ig

)1()2)(1( iiiii Ngggg Since this is approximately ii Ng iN

igAt present we are interested in indistinguishable particles and so many of these distributions will be the same when the condition of distinguishability is removed. We can start with one particular distribution and then obtain identical distributions by permuting the indistinguishable particles among themselves. The number of such permutations is . Hence the number of ways, subject to the restriction, that the indistinguishable particles can be distributed among the states is

!iN

iN)1()!N/g( i

N

ii

39

A fundamental problem in statistical mechanics is to determine the particular macrostate of a system when at equilibrium.The U of the isolated system is fixed and so each microstate has this value. The laws of mechanics do not lead one to expect that the system will be found in one microstate rather than in any other. This is consistent with with our postulate that all microstates are equally probable. Of course, all such postulates must be verified by comparing the calculations based on the postulates with experimental results.

Now let us return to a consideration of a particular macrostate, that is, a particular set As we stated earlier, the number of ways this particular macrostate can be achieved is called the thermodynamic probability w. We have calculated this probability (unnormalized) above for one energy level. The total thermodynamic probability is the product of the individual probabilities for all the accessible levels. From equation (1):

),( ii N

n

i i

Ni

n N

gNNNw

i

121 !

),,(

40

As shown in the textbook, the Fermi-Dirac and Bose-Einstein probabilities reduce to this equation when one makes the approximation ii Ng

The thermodynamic probability for Boltzmann Statistics and Maxwell-Boltzmann Statistics is

)Boltzmann-Maxwell(!)Boltzmann( wNw

These two probabilities differ by a constant and hence their derivatives (dw) will be the same. In addition the constraint conditions (constant N and U) are the same. Hence, in using the method of Lagrange Undetermined Multipliers, we will obtain the same result.

kT

i

ii

i

eZ

N

g

Nf

n

i

kTi

i

egZ1

Maxwell-Boltzmann

Distinguishable Indistinguishable

ii Ng so fi<<1

41

Statistical Interpretation of Heat and WorkFor concreteness we return to the quantum mechanical formula

for the energy levels available for molecules of a gas in a container of volume V. 23/2

22223/2

2

8)(

8nV

m

hnnnV

m

hzyxi

For a given quantum level (given n) the energy depends only upon the volume.

dVdV

ddV i

iii

)(

The internal energy of a gas is i

n

iiNU

1

dVdV

dNdNdNdNdU i

n

ii

n

iiii

n

ii

n

iii

1111

dVdV

dNdNdU i

n

ii

n

iii

11

Macroscopically dU= Q-PdVđ

42

Now suppose that the volume of the gas does not change. (No work is done on the gas.) Comparison of the two equations yields

n

1iii dNQ

When energy is added by the heating process, the energies do not change, but the distribution of particles among the energy levels changes. (See Fig. 13.4). In heating the gas you are promoting molecules from lower levels to higher levels.Now suppose that, instead of heating the gas, one does work on the gas, so that its volume decreases. In this case, the above formulae shows that:

i

n

1ii

in

1ii dNdVdV

dNW

In this case the distribution of particles among the energy levels does not change, but the energy of the levels change. Doing work on the gas raises the energy levels. (See Fig. 13.5).

đ

đ

43

Q

W

44

The entropy, Helmholtz function and the chemical potential in terms of the partition function.

We obtain formulae using the Maxwell-Boltzmann distribution.

n

i i

Ni

n N

gNNNw

i

121 !

),,( kT

i

ii

eZ

N

g

N

n

1i

n

1iiii )!Nln(glnNkwlnkS Using the Stirling approx.

i

in

ii

n

i

n

iii

n

iiii g

NNNkNNNgNkS lnlnln

11 11

But kT

ZNg

N i

i

i lnlnln

U

kTNZNNNkS

1))((ln))((ln

kT

ZlnNlnNNkS in

1ii

45

1lnln NZNkT

US

Since we can write the Helmholtz function in terms of U and S we can easily obtain an expression for F in terms of Z

F=U-TS

1lnln NZNkT

UTUF

1lnln NZNkTF

We have already determined the chemical potential in terms of F:

VTN

F

,

Differentiating F with respect to N

NNkTNZkT

11lnln

Z

NkT ln

{Note: Z=Z(T,V) so F(T,V)}

(M-B Dist.)

(M-B Dist.)

46

Finally we write the Maxwell-Boltzmann distribution in terms of the chemical potential. This will permit a comparison of the three distributions.

kT

i

ii

i

eZ

N

g

Nf

Using the above expression for the chemical potential: kTeZ

N

Substituting into the distribution: kTkT

i

ii

i

eeg

Nf

kTi

ii

ieg

Nf

/)(

1

47

Comparison of the distributionsThe chemical potential, which enters these distributions, will

be discussed when we use the distributions. At this stage we will simply plot the distributions, which can all be represented by

aef

kTii

/)(

1

+1 Fermi-Dirac a= -1 Bose-Einstein 0 Maxwell-Boltzman

We plot these distributions as a function of (MAPLE plot distrib.mws)

kTix )(

000 xxx iii

48

limit of validity (M-B)

49

Consider the FD distribution.For x=0 2/1if

For 1 ifx

The low-energy levels are very nearly uniformly populated with one particle per state.

Consider the MB distributionThis distribution is only valid for

In this limit this distribution is an approximation for the BE and FD distributions.

1if

Consider the BE distribution:For the distribution is infinite and for

the distribution is negative and hence meaningless. The particles tend to condense into the lower energy states.

i i

50

We have introduced four distributions. These give the distributionof particles in the accessible states.Boltzmann: Distinguishable particles. Any number can go into

an energy state.Fermi-Dirac: Indistinguishable particles which obey the PEP.Bose-Einstein: Indistinguishable particles. Any number can go

into an energy state.Maxwell-Boltzmann: Indistinguishable particles. Any number

can go into an energy state. Valid when ii Ng

An extremely useful function was introduced: The partition function.

n

1i

kTi

i

eg)V,T(Z

Many thermodynamic variables and functions can be written in terms of this function. This function contains all the statistical information about the system.

{Boltzmann Factor}

51

kTii

eZ

1

N

N

We also introduced the canonical distribution, a probability distribution,

which has wide applicability.