10.4 sommerfeld formula - binghamton university
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1
Sommerfeld formula
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(November 01, 2018)
Arnold Johannes Wilhelm Sommerfeld: 5 December 1868 – 26 April 1951) was a
German theoretical physicist who pioneered developments in atomic and quantum physics,
and also educated and mentored a large number of students for the new era of theoretical
physics. He served as doctoral supervisor for many Nobel Prize winners in physics and
chemistry (only J. J. Thomson's record of mentorship is comparable to his). He introduced
the 2nd quantum number (azimuthal quantum number) and the 4th quantum number (spin
quantum number). He also introduced the fine-structure constant and pioneered X-ray
wave theory.
https://en.wikipedia.org/wiki/Arnold_Sommerfeld
A Sommerfeld formula is an approximation method developed by Arnold Sommerfeld for
the integrals represent statistical average using the Fermi-Dirac distribution.
2
2 4
(1) (3)
2 4
0 0
6 8(5) (7)
6 8
10 12(9) (11)
10 12
7( ) ( ) ( ) ( ) ( )
6 360
31 127( ) ( )
15120 604800
73 1414477( ) ( ) ...
3421440 653837184000
f g d g d g g
g g
g g
where ( )
1( )
1f
e
is the Fermi-Dirac distribution function. is the chemical
potential. When 1 (the condition of strong degeneracy), the derivative ( )f
becomes a Dirac delta function, which takes a very sharp peak around . The chemical
potential is dependent on temperature.
1. Derivation of Sommerfeld formula
We assume that
0
)()( dxxgG
where ( )g is a slowly varying function around
)(')( Gg
We now calculate the integral
0
0
0
0
0
( ) ( )
( ) '( )
( )( ) ( ) ( )
( )[ ] ( )
I d f g
d f G
ff G d G
fd G
3
Note that )()(
f
which is the Dirac delta function having a very sharp peak at
. We expand )(G by using Taylor expansion around
....)()(!3
1)()(
!2
1)()(
!1
1)()( )3(3)2(2)1( GGGGG
Thus we get
(1) 2 (2)
0 0
3 (3)
(1) (2) 2
0 0
(3) 3
0
( ) 1 1( ) ( ) [ ][ ( ) ( ) ( ) ( ) ( )
1! 2!
1( ) ( ) .......]
3!
1 ( ) 1 ( )( ) ( ) [ ]( ) ( ) [ ]( )
1! 2!
1 ( )( ) [ ]( ) ...
3!
fd f g d G G G
G
f fG G d G d
fG d
We put )( x with 1 .
0
)()( dgG
)()()1( gG , (2) (1)( ) ( )G g , (3) (2)( ) ( )G g ,…..
Then we have
4
0 0
2 3(1) (2)
2 3
4 5(3) (4)
4 5
0
( )( ) ( ) ( )
( 1)( 1)
1 1( ) ( )
2 ( 1)( 1) 6 ( 1)( 1)
1 1( ) ( ) ...
24 ( 1)( 1) 120 ( 1)( 1)
1( )
x x
x x x x
x x x x
g xd f g g d dx
e e
x xg dx g dx
e e e e
x xg dx g dx
e e e e
g d
2
(1)
2
4(3)
4
( )2 ( 1)( 1)
1( ) ...
24 ( 1)( 1)
x x
x x
xg dx
e e
xg dx
e e
We have the Sommerfeld formula
2 4
(1) (3)
2 4
0 0
6 8(5) (7)
6 8
10 12(9) (11)
10 12
7( ) ( ) ( ) ( ) ( )
6 360
31 127( ) ( )
15120 604800
73 1414477( ) ( ) ...
3421440 653837184000
d f g g d g g
g g
g g
where
2 2
( 1)( 1) 3x x
xdxe e
,
4 47
( 1)( 1) 15x x
xdxe e
6 6
31
( 1)( 1) 21x x
xdxe e
,
8 8127
( 1)( 1) 15x x
xdxe e
2. T dependence of the chemical potential
We start with
0
( ) ( )N f D d
where
5
3/2
2 2
2( )
2
V mD a
ℏ
and
3/2
2 2
2
2
V ma
ℏ
We get
0
22 2
0
23/2 2 2
( ) ( )
( ') ' '( )6
2
3 6 2
B
B
N f D d
D d k T D
a ak T
.
But we also have )0( TF . Then we have
2/3
03
2)( F
adDN
F
.
Thus the chemical potential is given by
2
3/2 3/2 2 22 2
3 3 6 2F B
a a ak T
,
or
2/12
22/3
81
FF
B
F
Tk
.
which is valid for the order of
2
F
BTk
in the above expansion formula.
3/2 2 1/2 2
2 2
1 18 8
B B
F F F F
k T k T
6
or
2 2
2 22/3[1 ] 1
8 12
B B
F F F
k T k T
The chemical potential is approximated by the forms
2
2
[1 ]12
BF
F
k T
(3D case)
This result confirms that the chemical potential (Fermi level) remains close to the Fermi
energy as long as B Fk T . As the temperature increases, the chemical potential falls
below the Fermi energy by a margin that grows quadratically with the temperature.
For the 1D case, similarly we have
2
2
[1 ]12
BF
F
k T
(1D case)
3. Total energy and specific heat
Using the Sommerfeld’s formula, the total energy U of the electrons is approximated
by
0
2 2
0
2 2
0
( ) ( )
1( ) ( ) [ ( )] '( )]
6
1( ) ( ) ( ) ( ) [ ( ) '( )]
6
F
B F F F
F F F B F F F
U f D d
D d k T D D
D d D k T D D
.
The total number of electrons is also approximated by
0
2 2
0
2 2
0
( ) ( )
1( ) ( ) '( )
6
1( ) ( ) ( ) ( ) '( )
6
F
B F
F F B F
N f D d
D d k T D
D d D k T D
.
Since 0/ TN (N is independent of T), we have
7
2 21' ( ) '( ) 0
3F B FD k TD ,
or
2 2 '( )1'
3 ( )
FB
F
Dk T
D
.
The internal energy U is
2 2
0
1( ) ( ) ( ) ( ) [ ( ) '( )]
6
F
F F F B F F FU D d D k T D D
The specific heat Cel is defined by
2 2
2 2 2 2
1[ ( ) '( )] ' ( )
3
1 1( ) [ '( ) ' ( )]
3 3
el
B F F F F F
B F F B F F
dUC
dT
k T D D D
k TD k TD D
.
The second term is equal to zero. So we have the final form of the specific heat
2 21( )
3el B FC k D T .
In the above expression of Cel, we assume that there are N electrons inside volume V (= L3).
The specific heat per mol is given by
TkNDTkNN
DN
N
CBAF
A
BAF
Ael 2222 )(
3
1)(
3
1
.
where NA is the Avogadro number and )( F
AD [1/(eV at)] is the density of states per unit
energy per unit atom. Note that
22
3
1BAkN =2.35715 mJ eV/K2.
The entropy S is obtained as follows.
8
2 21( )
3el B F
SC T k D T
T
or
2 21( )
3B F
Sk D
T
Thus we have
2 21( )
3B FS k D T
which is the same as the heat capacity.
________________________________________________________________________
((Note)) The heat capacity of free electron
FBFF
FF
Tka
a
N
D
2
3
2
3
3
2
)(
2/3
The electronic heat capacity per mol is
F
B
FB
BAF
T
TRTRk
TkTkN
N
D
22
3
3
1)(
3
12
222
Then is related to )( F
AD as
)(3
1 22
F
A
BA DkN ,
or
(mJ/mol K2) = 2.35715 )( F
AD . (2)
We now give the physical interpretation for Eq.(1). When we heat the system from 0
K, not every electron gains an energy kBT, but only those electrons in orbitals within an energy range kBT of the Fermi level are excited thermally. These electrons gain an energy
of kBT. Only a fraction of the order of ( )B Fk TD can be excited thermally. The total
electronic thermal kinetic energy E is of the order of 2
( ) ( )B Fk T D . The specific heat Cel
is on the order of 2
( )B Fk TD .
((Note))
9
For Pb, = 2.98, )( F
AD =1.26/(eV at)
For Al = 1.35, )( F
AD =0.57/(eV at)
For Cu = 0.695, )( F
AD =0.29/(eV at)
Table 2(mJ/mol K ) (H.P. Myers)
Na: 1.38 Ti: 3.35
K: 2.08 V: 9.26 Mg: 1.3 Cr: 1.4
Al: 1.35 Mn: 9.2 Pb: 2.98 Fe: 4.98
Cu: 0.70 Co 4.73 Ag: 0.65 Ni 7.02
Au: 0.73 Pt: 7.0
4. Grand potential G for free electrons with the use of Sommerfeld expansion
Using the formula for fermions,
3/2 3/2
( )2 3
0
2
3 12G
gVmd
e
ℏ
and
3/2
2 3
02
gVmN d
ℏ
(at T = 0 K)
with 2g for spin 1/2, we get the ratio
10
3/2
( )
0
0
5/222
3/2
5/2 22
2 22 2
5/2
22 2
12( )
3
2
2 55( ) [1 ]23 8
3
2 5[1 ]
5 8
2 5[1 ] [1 ]
5 12 8
2 5 5[1 ][1
5 24 8
F
G
B
F
BF
F
B BF
F F
B BF
F F
de
Nd
k T
k T
k T k T
k T k T
2
]
or
2
22 5
[1 ]5 12
BG F
F
k TN
where we use the Sommerfeld expansion
2 2(1)
( )
0 0
( )( )( ) ( )
1 6
Bk Tgd g d ge
We note that
2ln
3G B G
UPV k T Z
The internal energy is
2
22 3 5
[1 ]3 5 12
BG F
F
k TU
The entropy S is obtained by
11
2 2
, 3
G B
V F
k TS N
T
5. Summary
(i) Internal energy U
0
0
22
0
( ) ( )
( ) ( )
( ) ( ) '( )]6
B
U D f d
u f d
u d k T u
or
22
0
( ) ( ) [ ( ) '( )]6
B F F FU u d k T D D
where
( ) ( )u D , '( ) ( ) '( )u D D
(ii) Number N:
0
22
0
( ) ( )
( ) ( ) '( )6
B
N D f d
D d k T D
Since N is independent of T,
2
20 ' ( ) '( )
3B FD k TD
or
12
22
0 ' ( ) '( )3
BD k TD
(iii) Heat capacity C:
22
22
2 22 2
22
' ( ) '( )3
' ( ) [ ( ) '( )]3
[ ' ( ) '( )] ( )]3 3
( )]3
B
B
B B
B
UC
T
u k Tu
D k T D D
D k TD k TD
k TD
or
C T
with
2
2( )]
3B Fk D
Note that
( ) ( )u D '( ) ( ) '( )u D D
(vi) Grand potential G
0
0
0
ln ( ) ln(1 )
1( )
11
( ) ( )
GZ D ze d
d
ez
f d
13
where
'( ) ( )D
Thus we have
0
22
0
ln
( ) ( )
[ ( ) ( ) '( )]6
G B G
B
k T Z
f d
d k T
When
( )D a
we get
3/22( ) ( )
3D d a
3/2 3( ) ( ) ( )
2u D a
The internal energy is
2 2
0
2 2
0
1( ) ( ) '( )]
6
3 1[ ( ) ( ) '( )]
2 6
3
2
B
B
G
U u d k T u
d k T
or
2
3G PV U
14
6. Derivation of entropy S (by R. Kubo)
Here we show the method used by Kubo. This method is very instructive to students.
The chemical potential can be derived as follows. We start with
0 0
( ) '( ) ( )F F
FN D d d
(1)
at T = 0 K, where ( ) '( )D .
2 2
0 0
1( ) ( ) ( ) ( ) '( )
6B FN f D d D d k T D
(2)
Subtracting Eq.(1) from Eq.(2), we get
2 21( ) ( ) '( ) 0
6F
B FD d k T D
Noting that
( ) ( )( )
F
F FD d D
we have
2 21( )( ) ( ) '( ) 0
6F F B FD k T D
Then the chemical potential is obtained as
2
2
22
22
'( )( )
6 ( )
ln ( )( ) [ ]
6
"( )( )
6 '( )
F
FF B
F
B
FB
F
Dk T
D
d Dk T
d
k T
15
or
2
2 "( )( )
6 '( )
FF B
F
k T
where
'( ) ( )D
The internal energy:
0
22
0
22
0
( ) ( )
( ) ( ) [ ( )] |6
( ) ( ) [ ( ) '( )]6
FB
B F F F
U D f d
dD d k T D
d
D d k T D D
Here we use the Taylor expansion
0 0
( ) ( ) ( ) ( )F
F F FD d D d D
Thus the internal energy can be rewritten as
2
2
0
2 22 2
0
2 22 2
0
22
0
( ) ( ) ( ) ( ) [ ( ) '( )]6
'( )( ) ( ) ( ) ( ) [ ( ) '( )]
6 ( ) 6
( ) ( ) '( ) ( ) [ ( ) '( )]6 6
( ) ( ) (6
F
F
F
F
F F F B F F F
FB F F B F F F
F
B F F B F F F
B
U D d D k T D D
DD d k T D k T D D
D
D d k T D k T D D
D d k T D
)F
or
16
22
0
22
0
( ) ( ) '( )6
( ) ( ) ( ) '( )6
F
F
B F
F F B F
U D d k T
d k T
where
0 0 0
( ) '( ) ( ) ( )F F F
F FD d d d
The heat capacity:
2 2
2 2'( ) ( )
3 3B F B F
dUC k T k D T
dT
Note that ( )F BD k T is the number of particles which are in the softened edge in the
vicinity of Fermi energy.
The grand potential:
0
0
0
22
0
ln ( ) ln(1 )
1( )
11
( ) ( )
[ ( ) ( ) '( )]6
G
B
Z D ze d
d
ez
f d
d k T
Using the Taylor expansion,
0 0
( ) ( ) ( ) ( )F
F Fd d
17
22
0
2 2
0
22 2
2
0
ln
( ) ( ) ( ) ( ) '( )6
"( )1( ) ( ) [ ( ) '( )]
6 '( )
"( ) ( ) [ '( )]1( ) ( ) '( ){ }
6 [ '( )]
F
F
F
G B G
F F B F
FB F F
F
F F FB F
F
k T Z
d k T
d k T
d k T
where
2 2 2 2'( ) "( )1 1( ) ( )
6 ( ) 6 '( )
F FF B B
F F
Dk T k T
D
Thus we have
22 2
2
0
"( ) ( ) [ '( )]1( ) ( ) '( ){ }
6 [ '( )]
F
G
F F FB F
F
PV
d k T
using these expressions, the entropy can be obtained as follows.
22 2
2
0
22
0
2 2
2 2
2 2
"( ) ( ) [ '( )]1( ) ( ) '( ){ }
6 [ '( )]
( ) ( ) ( ) '( )6
"( )1[ ( ) ] ( )
6 '( )
1( ) '( )
3
1( ) ( )
3
F
F
G
F F FB F
F
F F B F
FF B F
F
B F
B F
ST U N
d k T
d k T
k T
k T
k T D
or
2 21( )
3B FS k TD
7. Grand potential
18
The Gibbs free energy:
G N F PV U ST PV
The grand potential is given by
G PV U ST N
Since
( )Gd d U ST N
TdS PdV dN SdT TdS dN Nd
SdT PdV Nd
we have
,
G
T V
N
, ,
G
V
ST
We now calculate the grand potential using the Sommerfeld formula,
0
0
0
ln
( ) ln(1 )
1( )
11
( ) ( )
G
B G
B
PV
k T Z
k T D ze d
d
ez
f d
with
'( ) ( )D , ( )
1 1( )
1 11
fe
ez
8. The derivation of S from the grand potential
The entropy S is given by
19
, 0
(( )G
V
fS d
T T
Noting that
(f f
T T
ST can be rewritten as
0
0
( )( )
( ) ( )
fTS d
h f d
where
( ) [( ) ( )] ( ) '( ) ( )h
and
'( ) ( ) "( ) 2 '( )h
Using the Sommerfeld formula, we get
2
2
0
22
0
( ) ( ) '( )6
( ) ( ) '( )3
B
B
TS h d k T h
h d k T
Here we note that
0 0
0
( ) ( )
[ ( )( )]
0
h d h d
Thus ST can be approximated as
20
2 22 2
( ) '( ) ( ) ( )3 3
B F B FTS k T k T D
The entropy S is
2
2( )
3B FS k TD
9. The derivation of the number N from the grand potential
The number N is expressed by
,
G
T V
N
Note that
0,
0
0
0
(( )
( )( )
'( ) ( )
( ) ( )
G
T V T
fd
fd
f d
D f d
where
( ) ( )
T
f f
This leads to the results which is familiar to us,
0
( ) ( )N D f d
10. Derivation of the entropy using the Maxwell’s relation
From the thermodynamics, dG van be expressed as
21
( )dG d U ST PV
TdS PdV dN SdT TdS PdV VdP
SdT VdP dN
leading to the relations
,P N
GS
T
, ,T N
GV
P
, ,T P
G
N
From these relations, the Maxwell’s relation can be expressed by
, ,P N P T
S
T N
since
2
, , ,P N P N T P P
G G
T T N T N
2
, , ,P T P T P N P
S G G
N N T N T
Suppose that N depends only T and . We get
( , )
( , )
( , )
( , )
( , )
( , )
( , )
( , )
( , )
( , )
N
T
N
T T N
N
T
T N
T
N
T
N T
T
N
T
N
22
and
( , )
( , )
( , )
( , )
( , )
( , )
T
T
T
S S T
N N T
S T
T
N T
T
S
N
Thus we have the relation
T
S N
T
Here we note that
2 2
0
1( ) ( ) '( )
6BN D d k T D
Then we get
2 21( ) ( ) "( ) ( )
6B F
T
ND k T D D
2 2
2 2'( ) '( )
3 3B B F
Nk TD k TD
T
(a) Chemical potential
Using the above relation, we have
22 '( )
3 ( )
FB
N F
T
N
T Dk T
T DN
or
23
2
2 2 '( )
6 ( )
FF B
F
Dk T
D
(b) Entropy S
Using the above relation, we can calculate the entropy S as
2
2'( )
3B
T
S Nk TD
T
The integration of this with respect to leads to
2 2
2 2( ) ( )
3 3B B FS k TD k TD
11. Sommerfeld formula: 1D case
We consider the case of one-dimensional system.
1 1
0 0
( ) ( ) ( )F
N d f D d D d
where the density of the state for the 1D system is
1/2
1/2
1 2
2( )
L mD
ℏ
since
1/2
1 2
2( ) 2 2
2
L L m dD d dk
ℏ
The factor 2 of 2dk comes from the even function of the 1D energy dispersion relation.
We use the Sommerfeld formula
2
2 2
0 0
( ) ( ) ( ) '( )6
BN d f d d k T
where
1/2
1/2 1/2
1 12
2( ) ( )
L mD a
ℏ
24
1/2
3/2 3/2
12
( ) 2 1 1( )
2 2
d L ma
d
ℏ
Thus we have
2
1/2 2 2 3/2
1 1
0
21/2 2 2 3/2
1 1
1)( )
6 2
212
B
B
N a d d k T a
a k T a
(1)
But we also have
1/2 1/2
1 1 1
0 0
( ) 2F F
FN d D a d a
(2)
Combining Eqs.(1) and (2), we get
2
1/2 1/2 2 2 3/2
1 1 12 212
F Ba a k T a
or
2
1/2 1/2 2 2 3/2
24F Bk T
Then the chemical potential is
1/2
2 223/2
1/2
2 22
2
124
124
B
F F
B
F
k T
k T
or
2 2 2 22 2
2
2 2(1 ) 1
24 12
B B
F F F
k T k T
We make a plot of the number distribution as a function of / Fx
25
22
1 1( ) ( )
1exp[ (1 )] 1
12
D fxx
where /B Fk T
is changed as a parameter. We choose 0.3 .
Fig. The number distribution vs / Fx
. / 0.3B Fk T
for the 1D case. The
chemical potential for the temperature with / 0.3B Fk T
is denoted by the
dashed line. It shifts to the high energy side from the Fermi energy at T = 0 K (x =
1). The area for 1x (shaded region denoted by green) is the same as the area
below 1x .
26
27
28
12. Sommerfeld formula for the 3D system
Next we consider the case of three-dimensional system.
3 3
0 0
( ) ( ) ( )F
N d f D d D d
where the density of states for the 3D system is
29
3/2
1/2
3 2 2
2( )
2
V mD
ℏ
We use the Sommerfeld formula
2
2 2
0 0
( ) ( ) ( ) '( )6
BN d f d k T
Thus we have
2
1/2 2 2 1/2
3 3
0
23/2 2 2 1/2
3 3
1
6 2
2
3 12
B
B
N a d k T a
a k T a
Combining Eqs.(1) and (2), we get
2
3/2 3/2 2 2 1/2
3 3 3
2 2
3 3 12F Ba a k T a
or
2
3/2 3/2 2 2 1/2
8F Bk T
Then the chemical potential is
3/2
2 1/22 2
3/2
2 22
2
18
18
B
F F
B
F
k T
k T
or
2 2 2 22 2
2/3
2 2(1 ) 1
8 12
B B
F F F
k T k T
or
30
2 22
2(1 )
12
BF
F
k T
Fig. Chemical potential as a function of temperature for the ideal 3D Fermi gas and the
1D Fermi gas.
31
32
33
Fig. The number distribution vs / Fx
. /B Fk T
0.1 – 0.6 for the 3D case.
The chemical potential for the temperature with /B Fk T
is denoted by the
dashed line. It shifts to the low energy side from the Fermi energy at T = 0 K (x =
1) with increasing temperature. The area for 1x (shaded region denoted by green)
is the same as the area below 1x .
13. Exact solution for the chemical potential for the 2D case
The density of stares for the 2D system is
2 2
2 2 2 2
2 2 2 1( ) 2 2
(2 ) (2 ) 2
L L mD d kdk d
ℏ
or
2
2 2( )
mLD d d
ℏ
, 2
2 2( )
mLD
ℏ
The density of states for the 2D system is independent of . The chemical potential can be
evaluated exactly as follows. In other words, we do not have to use the Sommerfeld
approximation.
2
2 2 2
0
( )F
D F
mLN D d
ℏ
34
2
22 ( ) 2 ( )
0 0
( )
1 1D
D d mL dN
e e
ℏ
Then we have
2
2F Dnm
ℏ
with
22 2
DD
Nn
L
We get
( )
0 011
1F
d d
eez
We put
x e
dx e d xd
0 1
1
1
1
1 11 1
1 1 1( )
1[ln( )]
1ln(1 )
d dx
e x xz z
zdx
x x z
dxx x z
x
x z
z
ln(1 )F z
or
35
1Fe z
Thus we have
ln( 1)F
Bk T e
Fig. Chemical potential of the 2D system. Plot of / Fy
as a function of
/B Fy k T
x kBT F
y F
2D system
0.5 1.0 1.5 2.0 2.5 3.0
2
1
1
36
Fig. Chemical potential of the 1D, 2D, and 3D systems. Plot of / Fy
as a
function of /B Fy k T
REFERENCES
J. Rau, Statistical Physics and Thermodynamics, An Introduction to Key Concepts
(Oxford, 2017).
R. Kubo, Statistical Mechanics An Advanced Course with Problems and Solutions
(North-Holland, 1965).
M.D. Sturge, Statistical and Thermal Physics, Fundamentals and Applications (A.K.
Peters, 2003).
H.P. Myers, Introductory Solid State Physics, second edition (Taylor and Francis, 1997).
2D
3D
1D
xkB T
F
yF
0.5 1.0 1.5 2.0
2
1
1
2
3
4