density of states and partition function laplace transform
TRANSCRIPT
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