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