Density of States and Partition function: Laplace transform

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: September 08, 2018)

1. Laplace transformation

The partition function of a system is given by

0

)( )( EedEZ E (Laplace transformation)

The density of states satisfies the conditions

0)( , 0)(lim

E

EeE ( 0 )

The inverse Laplace transform of the partition function is the number of energy levels per unit

energy interval (density of states)

s

EseZ )(

The inverse Laplace transform )( is, by definition,

i

i

EZedi

E

'

'

)( 2

1)(

(Bromwich-Wagner integral)

Then we have

s

i

i

EE

i

i

E

sedi

Zedi

E

"

'

)(

"

'

2

1

)( 2

1)(

For iy ' with y to ,

idyd , ( 0' )

Im

Re

' i

' i

'0

complex plane

s

s

s

yEEiEEedyE s )(

2

1)(

)(

in terms of the Dirac delta function. Thus )(E is the density of states.

2. Entropy

One writes the integrand in the form

])(exp[ln)( EEEe E

The exponent of the above expression is expanded in the neighborhood of *E (let *E be the

value of E which maximizes the integrand)

])([ln !2

1

])([ln!1

1)(ln)(ln

**

2*

22*

**

*

***

EEE

EE

EEE

EEEEDEE

Noting that

)(ln *

*E

E, 0)(ln *

2*

2

E

E,

we get

)(ln2

1)(ln)(ln *

2*

22*** EE

EEEEEE

.

where the linear term becomes zero.

] 2

1exp[)()(

2*

2

*

*

*

EEeEeE

E

EE

(Gaussian distribution ; mean *E and standard deviation *E

)

Here we define

)(ln1 *

2*

2

2*

EEE

The partition function )(Z can be written as

] 2

1exp[)(

)( )(

2*

2

*

0

*

*

EEdEeE

EedEZ

E

E

C

or

*

* )(2)( * E

EC eEZ

The logarithm of )(Z can be written in the form

** - )](2ln[)(ln * EEZ

E

The Helmholtz free energy is

*

*

* *

*

ln ( )

ln[ 2 ( )]

ln[ 2 ( )]

B

B E

B E

F U ST

k T Z

k T E E

U k T E

where U = E. Thus the entropy S is given by

)](2ln[ ** EkSEB

((Note))

2

1)(ln)(ln

2*

2

**

*

EEEEEE

E

or

] 2

1exp[)()(

2*

2

)(*

*

*

EEeEE

E

EE

or

] 2

1exp[

)(

)( 2*

2*

*

*

*

*

EEeE

eE

E

E

E

which is a Gaussian distribution function with average energy *E and fluctuation *E

Fig. *

*

)(

)(*

*

E

E

eE

eE

vs E. Gaussian distribution function with mean *E and standard deviation

E

3. Convolution (Faltung) of the density of states

We consider the two kinds of partition functions,

1

1 1 1 1

0

( ) ( )Z d e

,

E

E E

2

2 2 2 2

0

( ) ( )Z d e

We calculate the product of these partition functions

1 2

1 2

1 1 1 1 1 2 2 2

0 0

( )

1 2 1 1 2 2

0 0

( ) ( ) ( ) ( )

( ) ( )

Z Z d e d e

d d e

We introduce new variables;

21 , '1

0 , '0

1 2 1 2 1 2

0 0 0

( ) ( ) ' ( ') ( ') *Z Z d e d d e

Thus )()( 21 ZZ is the Laplace transform of the convolution of the density of states

1 2 1 2

0

* ' ( ') ( ')d

.

or

1 1 1 2

0 0

1 2

0 0

1 2

0

( ) ( ) ' ( ') ( ')

' ( ') ( ')

*

Z Z d e d

d e d

d e

((Note))

E

EEEdEE0

212112 )'()'('*)( (convolution)

0

12111

0 0

12111

0

1221

1

)()(

)()(

)(

E

E

E

E

E

EEEdEedE

EEEdEdEe

dEEeZ

1

2

1 2

Fig. Region of integration. (a) EE 10 for the integration over 1E . (b) EE1

Here we put 21 EEE

)()()()( 21

0 0

2221112121

ZZEedEdEEeZE

4. The method of steepest decent (saddle point method)

In calculating )(E , one writes the integrand in the form

])(exp[ln)( EZZe E

and chooses as integration path a straight line (from i* to i* in the -complex plane,

with * determined by

EZ - )(ln *

*

The exponent of the above expression is expanded in the neighborhood of * ,

...)](ln[2

1)(ln

...)](ln[2

1

])(ln[)(ln)(ln

*

2*

22***

*

2*

22*

*

*

***

ZEZ

Z

EZEZEZ

Then the integral is approximated as

i

i

i

i

E

ZEZdi

Zedi

E

*

*

*

2*

22***

*

*

)](ln[2

1exp[])(exp[ln

2

1

)( 2

1)(

For "* i , " idd

Im

Re

i

i

0

complex plane

])(exp[ln)(ln

2

1

])(ln

"2

1exp["])(exp[ln

2

1

)](ln[2

1exp[])(exp[ln

2

1)(

**

2*

*2

2*

*22**

*

*

*

2*

22***

EZZ

ZdEZ

ZEZdi

E

i

i

Therefore, the logarithm of this expression gives

])(ln

2ln[2

1)(ln)(ln

2*

*2**

ZEZE .

The second term may be neglected in comparison with the first term. Thus one gets

EZE **)(ln)(ln

where

EZ - )(ln *

*

.

6. Canonical ensemble

),,( VNZC

0

0 }{

),,()exp(

)]((exp[),,(

N

C

N Ni

iG

VNZN

NENVZ

}{

)](exp[),,(Ni

iC NEVNZ (quantum mechanics)

)](exp[),,( NHdVNZ N

C (classical mechanics)

7. Ground canonical ensemble

**

**

**

**

),,(2

),,(22),(

**

**

NE

NE

NE

NEG

eeVEN

eeVENZ

),,(ln1 **

2*

2

2*

VENEE

, ),,(ln

1 **

2*

2

2*

VENNN

),,(ln **

*VEN

E,

),,(ln **

*VEN

N

**** - ),,(ln)2ln(),(ln ** NEVENZ

NEG

] 2

1exp[

)(

)( 2*

2**

*NN

eN

eN

N

N

N

Fig. *

)(

)(* N

N

eN

eN

vs N. Gaussian distribution function with mean *N and standard deviation

N .

8. Application-I: Simple harmonics

Multiplicity of simple harmonics (N oscillator systems)

N

N N

*

*

1( ) ( )

2

i

E

N N

i

E d e Zi

where

21 (1 )

N

N N

NZ Z e e

ℏ

ℏ

with

1 2( )2

1

0 1

n

n

eZ e

e

ℏ

ℏ

ℏ

Thus we get

*

( )2

*

*

2

*

1( ) (1 )

2

1 (1 )

2

iE

N

N

i

i

E N

i

E d e ei

e d e ei

ℏ

ℏ

ℏ

ℏ

*

2

*

* 1( )

2

0*

* 1( )

2

0*

1( ) (1 )

2

1 ( 1)

2

1 ( 1)!

2 ( 1)! !

i

E N

N

i

iM

E M

Mi

iM

E

Mi

E e d e ei

Ne d e

Mi

N Me d e

i N M

ℏ

ℏ

ℏ

ℏ

where

( 1)!( 1)

( 1)! !

MN N M

M N M

This formula can be proved by using the Mathematica. Thus we get

* 1[ ( ) ]

2

0 *

0

( 1)! 1( )

( 1)! ! 2

( 1)! 1[ ( ) ]

( 1)! ! 2

iE M N

N

M i

M

N ME e d

N M i

N ME M

N M

ℏ

ℏ

where

So the number of arrangement for 1

( )2

E M ℏ is

( 1)!

( 1)! !

N M

N M

9. Application-II: ideal gas

Using expressions for the partition function of classical ideal gases, evaluate the density of

states ( )E by the inverse Laplace transform. For simplicity, the gas molecules are assumed to

be of one kind. Ignore the internal degrees of freedom.

The partition function for the ideal free gas is given by

3 /2

1 2

1( ) ( )

! ! 2

NNN

N

V mZ Z

N N

ℏ

We now evaluate the density of states

*

*

3 /2 *

2 3 /2

*

1( ) e ( )

2

1 1 e

! 2 2

i

E

N

i

N iNE

N

i

E d Zi

V md

N i

ℏ

(Bromwich-Wagner)

where * 0 . Using the Jordan lemma, the integral can be rewritten as

*

*

1 ( )

2

i

a

i

e d ai

*

3 /2

*

3 /2

1 1 e

2

1 1 e

2

i

E

N

i

E

N

C

I di

di

where the contour C is in the left-half plane of the complex plane (E>0).

The case of even N

The function 3 /2

1( )

Ng

has a pole of rank 3 / 2N (integer) at the origin, which are inside

the contour C. Using the Cauchy’s theorem

( )

1

1 ( ) 1( )

2 ( ) !

n

n

C

f z dzf a

i z a n

we have

(3 /2) 1

(3 / 2)

NEI

N

The case of odd N

When 2 1N n (n; integer)

13 /2 3 13 1

2

1 1 1 1( )

N nn

gs

Because of the factor s , the function ( )g is discontinuous on the negative real axis. In this

case, the contour C should be changed as the path 1 2 3 4 5 6C C C C C C . Inside the

path there is no pole. According to the Cauchy’s theorem,

*

*2 3

4 5 6

3 /2 3 /2 3 /2

3 /2 3 /2 3 /2

1 1 10 e e e

1 1 1 e e e

i

E E E

N N N

C Ci

E E E

N N N

C C C

d d d

d d d

According to the Jordan’s lemma, we have

3

3 /2

1 e 0E

N

C

d

,

6

3 /2

1 e 0E

N

C

d

(a) The path integrals along the path C3 and path C5

The path 3 2 : 0 ise

3

3 3 /2

03

23 /2

3

23 /2

0

3

23 /2

0

1 e E

N

C

N sEi

i

N

N sEi

i

N

N sEi

N

I d

ee e ds

s

ee e ds

s

ee ds

s

The path 5 4 : 0 ise

5

5 3 /2

3

23 /2

0

3

23 /2

0

1 e E

N

C

N sEi

i

N

N sEi

N

I d

ee e ds

s

ee ds

s

Then we have

3 5 3 /2

0

32 sin( )

2

sE

N

N eI I i ds

s

For 2 1N n

a

123

4 56

s

s

1

1

2

3 3 (2 1)sin( ) sin[ ]

2 2

3sin(3 )

2

3sin( )

2

3sin( )cos( )

2

( 1)

( 1)

n

N

N n

n

n

n

We calculate the integral

3 /2

0

sE

N

eJ ds

s

When 2 1N n

(3 1) 1/2 (3 1)

0 0

1( )

sE sE

n n

e eJ E ds ds

s s s

Thus we get

(3 1) 3

0 0

( ) ( ) 1( 1)

sE sE

n n

dJ E s e eds ds

dE s ss s

22

2 3 1

0

( ) 1( 1)

sE

n

d J E eds

dE s s

Similarly, we have

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

(3 1)

0

( )( ) ( 1) ( 1)

n sEn n n

n

d J E eJ E ds

dE Es

where

0

sEeds

Es

From the sequential integration of the above relation, we get

3 1 3

12 2

3 132 12

1( ) ( 1)

1 3 5 3. .........( 1)

2 2 2 2

( 1)

3( )

2

N N

NN

J E EN

EN

where

3 1 3 5 3( ) . .........( 1)

2 2 2 2 2

N N

Thus we have

3 1

1 32 12 2

3 5

31

22

( 1)2 ( 1)

3( )

2

2 ( 1)3

( )2

NN N

N

N

I I i EN

Ei

N

(b) C4: circle: ie ( changes from to -)

4

3 /2

1 e E

N

C

d

It seems that this integral may diverge. This integral may be included in the path integral on the

path C3 and C5. In fact, in Kubo’s book, this assumption is included in the path integral on the

path C3 and C5.

[R. Kubo, H. Ichimura, T. Usui, and N. Hashitsume, Statistical Mechanics An Advanced Course

with Problems and Solutions (North Holland, 1988)].

(c)

*

*

3 53 /2

31

22

31

2

1 1 1 e ( )

2 2

( 1)[ 2 ( 1) ]

32( )

2

]3

( )2

i

E

N

i

N

N

N

d I Ii i

Ei

Ni

E

N

In conclusion, for any integer (both for even and odd numbers), we have the density of states

given by

3 /2 (3 /2) 1

2( )

! 2 (3 / 2)

NN NV m EE

N N

ℏ

Partition function for the simple harmonics:

1

1

0

(1 )n

C

n

Z e e

ℏ ℏ

For the system with N-simple harmonics,

1( ) (1 )N N

CN CZ Z e ℏ .

The density of states

*

*

*

*

1( ) e ( )

2

1 e (1 )

2

i

E

N

i

i

E N

i

E d Zi

d ei

ℏ

According to the Jordan’s lemma,

1( , ) e (1 )

2

E N

C

E N d ei

ℏ

The integrant has a pole at 0 of the rank N. Using the Cauchy’s theorem, we get

( , ) Residue[e (1 ) ,{ ,0}]

1 ( 1 )!

( 1)! !

E NE N e

N M

N M

ℏ

ℏ

((Mathematica))

Clear "Global` " ;

f1 N1 :Exp E1

1 Exp N1

. E1 M1;

Residue f1 2 , , 0

1 M1

Residue f1 3 , , 0 Factor

1 M1 2 M1

2

Residue f1 4 , , 0 Factor

1 M1 2 M1 3 M1

6

Residue f1 5 , , 0 Factor

1 M1 2 M1 3 M1 4 M1

24

Residue f1 8 , , 0 Factor

1 M1 2 M1 3 M1 4 M1 5 M1 6 M1 7 M1

5040