Chapter 7 Bose and Fermi statistics

§7-1 The statistical expressions of thermodynamic quantities

1 、 Bose systems:

1 le

ll

1l

lll l

Ne

Define a macrocanonical partition function:

1 lnYy

ln ln 1 ll

l

e

1l

l

l ll

l l

Yy e y

1 lnpV

Now, let’s see the entropy and Lagrange Variable factor

ln lndU Ydy d dyy

lnU

1 lnYy

ln ln lnlnd d d dyy

Because: ln ln 1 ll

l

e

ln lndU Ydy d dyy

ln ln lnlnd d d dyy

ddddyy

lnlnlnln

ddddYdydU

lnlnlnln

lnlnlnlnln

ln

ddddd

dYdydU

So that,

lnlnln ddd

NdYdydU

kT

,

lnlnln dddk

dS

lnlnlnkS

NUk ln

lnlnlnkS

ln ln 1 ll

l

e

l ll

llEB !1!

!1.

lnS k

Boltzmann relation

2 、 Fermi system

1 ll

ll l

e

1 le

ll

1 ll

ll l

e

ln ln 1 ll

l

e (Bose)

ln ln 1 ll

l

e

lnN

1 le

aN l

lll

ln ln 1 ll

l

e

l

l

ll ll

l

eee

111ln

lnU

lnN

lnU

1 lnYy

1 lnpV

ln ln lnS k

lnS k

Boltzmann relation

Addition:

Define a new thermodynamic quantity –grand potential

NFJ

NTSUJ It is of great importance for the statistical treatment of thermodynamic problems.The total differential reads

NdPdVSdTNddNSdTTdSdUdJ

The remaining thermodynamic quantities can be calculated by differentiating the grand potential:

VTTV

JNVJp

TJS

,,,

NTSUJ Because of Euler’s equation:

NpVTSU

The grand potential is identical with -pV

pVJ

lnJ kT

NTSUJ

NUk ln

lnlnlnkS

NUkTTS ln

NUkTTS ln

Because：

§7-2 Bose and Fermi weak degeneracy ideal gas

1 e

non-degeneracy condition:

13 nNow, we consider a condition, just as and is small ,but can not be neglected. We called this weak degeneracy condition.

e 3n

Under this condition, we want to deal with the Bose ideal gas and Fermi ideal gas respectively.

2 2 21 ( )2 x y zp p pm

V d

3 2 1 23

2( ) (2 )VD d g m dh

rl

h

Here g is the degeneracy caused by the particle’s spin.

l

laN Photon g=2

Electron g=2

1 le

ll

1 le

ll

Bose

Fermi

1/ 23 2

3 0

2 (2 )1

V dN g mh e

3 2 1 23

2( ) (2 )VD d g m dh

This equation can be used to determine the Lagrange factor .

3 2 1 23

2( ) (2 )VD d g m dh

ll

lU 1

le

ll

x 3 2

3 23 0

2 (2 )1x

V x dxU g mkT kTh e

1 23 2

3 0

2 (2 )1x

V x dxN g mkTh e

1/ 23 2

3 0

2 (2 )1

V dN g mh e

0

2/32/3

3 122

e

dmh

VgU

1 11 (1 )x x xe e e

3 23 2

3 0

2 (2 )1x

V x dxU g mkT kTh e

1 23 2

3 0

2 (2 )1x

V x dxN g mkTh e

1 e

1 xe

So that, we expand the into: 11 xe xe 1

xxxx n

111 2

1 23 2

3 0

2 (2 )1x

V x dxN g mkTh e

3 22 3 2

2 1( ) [1 ]2

mkTN g Ve eh

0

2/122

0

2/12/33 22 dxxedxxemkT

hVgN xx

0

2/122/32/33 22222 xdxeeemkT

hVgN x

1 11 (1 )x x xe e e

22

222 2/32/3

3 eemkT

hVgN

3 22 5 2

3 2 1( ) [1 ]2 2

mkTU g VkTe eh

3 23 2

3 0

2 (2 )1x

V x dxU g mkT kTh e

1 (1 )1

x xx e e

e

0

2/322

0

2/32/33 22 dxxedxxekTmkT

hVgU xx

0

2/322/52/33 222

4322 xdxeekTemkT

hVgU x

432

4322 2/52/3

3 ekTemkTh

VgU

3 1[1 ]2 4 2

U NkT e

3 22 5 2

3 2 1( ) [1 ]2 2

mkTU g VkTe eh

3 22 3 2

2 1( ) [1 ]2

mkTN g Ve eh

In the equation, the first term is the energy calculated according to the Boltzmann distribution, and the second term is the correlation energy caused by the quantum statistics.

l

lle 1lnln

demh

Vg 2/1

0

2/33 1ln22ln

3 2 1 23

2( ) (2 )VD d g m dh

ex 1ln

2/3

32 y

d

eeede

132

321ln1ln 2/3

00

2/32/1

0

2/33 22 m

hVgA

d

eeeA13

2321lnln 2/3

00

2/3

0

de

eA

132ln 2/3

0

de

A1

132ln 2/3

0

de

A1

132ln 2/3

0

1 (1 )1

x xx e e

e

1 11 (1 )x x xe e e

deeA

132ln 2/3

0

x

deeA

132ln 2/3

0

dxeeA xx

132ln 2/3

0

x

dxexexA xx 222/32/3

0

2/3

32

dxexedxexeA xx

0

22/3

0

2/32/3

32

0

2/32/52/3 2132 dxexeeA x

0

2/32/52/3 2132ln dxexeeA x

43

0

2/3 dxex x

2/52/3 2121ln eeA

2/33 22 m

hVgA

2/52/3

2 212ln

ee

hmgV

2/52/3

2 212ln

ee

hmgV

1 lnpV

lnlnlnkS

§7-3 Bose-Einstein conglomeration

From this section, we can see that when equals or is more than 2.612, the unique Bose-Einstein conglomeration phenomenon will appear.

3n

( ) / 1l

ll kTe

1 le

ll

1l

kTe

Because the can not be negative, So that la

0 1

l

kTe

That is ,here is the lowest energy level of Bose particles.0

0 If we assume the lowest energy of the Bose particle is 0, then,

11l

lkT

l

N nV e V

（ /）From the equation

We can see that, the chemical potential is the function of temperature T and the density of particle number n .

1

3 22

3 0

2 21kT

dm nh

e

11l

lkT

l

N nV e V

（ /）

If we substitute integration for the summation, there is

1322

3

2 2V m dh d —V

1

3 22

3 0

2 21kT

dm nh

e

If the n is fixed, according to the equation above, we can see that the higher temperature, the smaller chemical potential. When the temperature reduces to a critical value Tc ,the chemical potential will reach its highest value 0.

0

1

3 22

3 0

2 21ckT

dm nh

e

Because:

c

xkT

1

3 22

3 0

2 21ckT

dm nh

e

1

3 22

3 0

2 21c x

x dxmkT nh e

12

02.612

1 2x

x dxe

2 23

23

2

2.612cT n

mk

Then, if the temperature keep on reduces from Tc, what will happened?

1

3 22

3 0

2 21kT

dm nh

e

0

cT T 0

But, when cT T

Tne

dmh

kT

0

21

23

3

122

c

kT

Tn

e

dmh

c

0

21

23

3

1

22

There is a contradiction in it, because :

1

3 22

3 0

2 21kT

dm nh

e

11l

lkT

l

N nV e V

（ /）

1322

3

2 2V m dh

Tne

dmh

kT

0

21

23

3

122

c

kT

Tn

e

dmh

c

0

21

23

3

1

22

cT T ckTkT ee

<

1322

3

2 2V m dh

0 0l

1

3 22

3 0

2 21kT

dm nh

e

cT T

13 22

0 3 0

2 21kT

dn T m nh

e

Here n0(T) is the density of particle number on the energy level , when temperature is T( ), .

0 cT T 0

Tne

dmh

kT

0

21

23

3

122

kTx

Tne

dxxmkTh x

0

21

23

3 122

1

3 22

3 0

2 21ckT

dm nh

e

c

xkT

1

3 22

3 0

2 21c x

x dxmkT nh e

32

0 1c

Tn T nT

32

0 1c

Tn T nT

When ,Bose particles will accumulate on the energy level rapidly, and the density of particle number reach the same order of magnitude with the total particle number density n .This phenomenon is just the Bose-Einstein conglomeration.

cT T0

0n

0 Tc is called conglomeration temperature, on the energy level of ,we can see that the momentums of Bose particles are also 0, so, Bose-Einstein conglomeration is also called momentum conglomeration.

cT T

3

3 22

3 /0

2 21kT

V dU mh e

xkT

3

3 5 22 2

3 0

2 21x

V x dxU m kTh e

32

0.770c

TU NkTT

32

1.925VV c

U TC NkT T

32

0.770c

TU NkTT

CT T 3/ 2VC T

CT T 1.925VC Nk

32VC NkcTT

2 23

23

2

2.612cT n

mk

3

3 2.6122 c

hn nmkT

3 2.612n

§7-4 Photon gas

T: N(t) const.

According to the idea of particles, we can regard the photon field in the cavity as a photon gas.

h

khp

2

k

h

khp

cp

Bose statistics:

1 le

ll

.constNal

l

1

lel

l

1

lel

l

Here,r

ll h

dpppV

dpph

V 23

4

Because the spin degeneracy of photon is 2 (1.-1),

dpph

Vh r

l 23

8

cp

cp d

cdp

dpph

Vh r

l 23

8

Substitute for the equation

dV

dc

V

dcch

Vh r

l

232

2

3

8

1

lel

l

dc

Vh r

l 232

1

232

kT

l

e

dc

V

UdV :

lladTU ,

d

ec

V

kT 1

3

32

The equation above is called Planck’s formulation.

de

cVdTU

kT 1,

3

32

Integrating the equation above, we can obtain the total energy of the cavity.

d

ec

VUkT 1

3

032

dxe

xkTc

VU x 1

3

0

4

32

xkT

15

44

32

kTc

V 433

42

15VT

ck

We can also use another solution:

l

lle 1lnln

llel

1

l

l e 1ln

d

cV

h rl 2

32

dec

V 1lnln

0

232

dec

V 1lnln

0

232

x

dxexc

V x

1ln1ln

0

23

32

dxe

exexdxex x

xxx

13

1ln3

1ln0

3

0

3

0

2

dxe

x x 11

31

0

3

45

4

451ln

43

32

c

V

3

3

2 145

cV

lnU

4

3

3

2 13145

c

V

433

24

15T

cVk

It is the same with what we have obtained before.

1 lnpV

3

3

2 145

ln

cV

433

24

45T

ck

4

33

2 145

c

433

24

15T

cVkU

VUp

31

lnlnkS VTc

k 433

24

454

§7-5 The free-electron gas in the metal

A further, very useful model system is that of a noninteracting non relativistic gas of Fermi particles. Nucleons in atoms, as well as electrons in metals, can be regard as an ideal Fermi gas to first approcimation. The case T= 0 has here a special importance.

1 le

ll

1

1kT

fe

It stands for the mean particle number on the each quantum state.

V — d 1322

3

2 2V m dh

However, since the particle possess 2s+1 different spin orientations which are energetically degenerate in the interaction free case, Equation above must be multiplied by an additional degeneracy factor g=2

1322

3

4 2V m dh

l

laN1

kT

ll l

e

We want to rewrite the sums in terms of integrals.

1

3 22

3 0

4 21kT

V m Nh

e

d

We can also see that the is the function of T and n.

When T=0K 1

1kT

fe

10

ff

0

0

Pauli principle requires an energetically higher state for each new particle, and the is the highest energy level of electrons.

0

Ndmh

V 0

021

23

3 24

10

ff

0

0

22 32

0 32

Nm V

Is called Fermi energy level.

20

0 2pm

132

0 3 NpV

Is called Fermi momentum, which is the largest momentum.

33 022

3 0

4 32 05

VU m d Nh

Ndmh

V 0

021

23

3 24

For example: Cu (cuprum)

22 32

0 32

Nm V

18 38.5 10N m

V

120 1.1 10 J

300K0

1260

kT 1 kTee

T ＞ 0

1

1kT

fe

,21

,21

,21

f

f

f

1

3 22

3 0

4 21kT

V dN mh

e

3

3 22

3 0

4 21kT

V dU mh

e

0

1kT

I de

kTz

1z

kT

kTzI kTdz

e

－

0 01 1kT

z z

kTz kTzkT dz kT dz

e e

0

0

11dz

ekTzkTdz

ekTzkT z

kTz

0 01 1kT

z z

kTz kTzkT dz kT dz

e e

11

11

z

zz

z eee

e

0 0 1z

kTz kTzI d kT dz

e

In the first term

000 11

dze

kTzkTdze

kTzkTdzkTzkTI zkT

zkT

kTz

2

0 02

1z

zI d kT dze

2

2

0 6d kT

Because the integration comes from ,especially when the z is small. So, we can expand the numerator into power series.

ze

1

3 22

3 0

4 21kT

V dN mh

e

01kT

I de

2/12

2

0

2/12/33 6

24 kTdm

hVN

2/12

2

0

2/12/33 6

24 kTdm

hVN

2/12

2

0

2/12/33 2

16

24 kTdm

hV

222/32/3

3 8124

32

kTm

hV

22 2 333 1

2 8N kTc

3

3 22

3 0

4 21kT

V dU mh

e

0

1kT

I de

2

2

0 6d kT

2/32

2

0

2/32/33 6

24 kTdm

hV

2/12

2

0

2/32/33 2

36

24 kTdm

hV

222/52/3

3 85124

52

kTm

hV

We can also use another method:

llel

1

l

lle 1lnln

1322

3

4 2V m dh

edmh

V 1ln24ln 2/12/33

2/33 24 m

hVA

l

deA 1lnln 2/1

If we integrate the term in the equation above by parts, it follows that:2/1

de

eA

132 2/3

0

de

AkT

132 2/3

0

01kT

I de

2

2

0 6d kT

de

AkT

132 2/3

0

2

2

0 6d kT

0

1kT

I de

2

12

2

0

2/3

23

632

kTdA

2

12

225

452

32 kTA

2

12

225

452

32 kTA

kTkT

2

12

2

25

452

32 kTkTkTA

2

2

25

851

52

32

kTA

2/33 24 m

hVA

2

2

25

851

52

32ln

kTA

2/33 24 m

hVA

2

2

25

2/3

3 8512

1516ln

mhV

Exercise:

8.3 Request the entropy and pressure of weak degeneracy Bose ideal gas and Fermi ideal gas.