Chapter 7 Bose and Fermi statistics. §7-1 The statistical expressions of thermodynamic quantities 1...

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Chapter 7 Bose and Fermi statistics

§7-1 The statistical expressions of thermodynamic quantities

1 、 Bose systems:

Define a macrocanonical partition function:

1 lnYy

ln ln 1 ll

Yy e y

1 lnpV

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

ln lndU Ydy d dyy

1 lnYy

ln ln lnlnd d d dyy

Because: ln ln 1 ll

ln lndU Ydy d dyy

ln ln lnlnd d d dyy

ddddyy

lnlnlnln

ddddYdydU

lnlnlnln

lnlnlnlnln

dYdydU

So that,

lnlnln ddd

NdYdydU

lnlnln dddk

lnlnlnkS

NUk ln

lnlnlnkS

ln ln 1 ll

llEB !1!

Boltzmann relation

2 、 Fermi system

ln ln 1 ll

e (Bose)

ln ln 1 ll

1 lnYy

1 lnpV

ln ln lnS k

Boltzmann relation

Addition:

Define a new thermodynamic quantity –grand potential

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:

NTSUJ Because of Euler’s equation:

NpVTSU

The grand potential is identical with -pV

lnJ kT

NUk ln

lnlnlnkS

NUkTTS ln

Because：

§7-2 Bose and Fermi weak degeneracy ideal gas

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.

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

3 2 1 23

2( ) (2 )VD d g m dh

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

laN Photon g=2

Electron g=2

1/ 23 2

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

3 23 0

2 (2 )1x

V x dxU g mkT kTh e

1 23 2

2 (2 )1x

V x dxN g mkTh e

1/ 23 2

2 (2 )1

V dN g mh e

2/32/3

1 11 (1 )x x xe e e

3 23 2

2 (2 )1x

V x dxU g mkT kTh e

1 23 2

2 (2 )1x

V x dxN g mkTh e

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

xxxx n

1 23 2

2 (2 )1x

V x dxN g mkTh e

3 22 3 2

2 1( ) [1 ]2

mkTN g Ve eh

2/12/33 22 dxxedxxemkT

hVgN xx

2/122/32/33 22222 xdxeeemkT

hVgN x

1 11 (1 )x x xe e e

222 2/32/3

3 eemkT

3 22 5 2

3 2 1( ) [1 ]2 2

mkTU g VkTe eh

3 23 2

2 (2 )1x

V x dxU g mkT kTh e

1 (1 )1

x xx e e

2/32/33 22 dxxedxxekTmkT

hVgU xx

2/322/52/33 222

4322 xdxeekTemkT

hVgU x

4322 2/52/3

3 ekTemkTh

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.

lle 1lnln

Vg 2/1

2/33 1ln22ln

3 2 1 23

2( ) (2 )VD d g m dh

ex 1ln

321ln1ln 2/3

2/32/1

2/33 22 m

eeeA13

2321lnln 2/3

132ln 2/3

1 (1 )1

x xx e e

1 11 (1 )x x xe e e

132ln 2/3

dxeeA xx

132ln 2/3

dxexexA xx 222/32/3

dxexedxexeA xx

2/32/3

2/32/52/3 2132 dxexeeA x

2/32/52/3 2132ln dxexeeA x

2/3 dxex x

2/52/3 2121ln eeA

2/33 22 m

2/52/3

2 212ln

2/52/3

2 212ln

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.

( ) / 1l

ll kTe

Because the can not be negative, So that la

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,

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 .

2 21kT

N nV e V

（ /）

If we substitute integration for the summation, there is

2 2V m dh d —V

2 21kT

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.

2 21ckT

Because:

2 21ckT

2 21c x

x dxmkT nh e

02.612

2.612cT n

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

2 21kT

cT T 0

But, when cT T

There is a contradiction in it, because :

2 21kT

N nV e V

（ /）

2 2V m dh

cT T ckTkT ee

2 2V m dh

2 21kT

dn T m nh

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

0 cT T 0

dxxmkTh x

2 21ckT

2 21c x

x dxmkT nh e

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.

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.

2 21kT

V dU mh e

3 5 22 2

V x dxU m kTh e

0.770c

TU NkTT

1.925VV c

U TC NkT T

0.770c

TU NkTT

CT T 3/ 2VC T

CT T 1.925VC Nk

32VC NkcTT

2.612cT n

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.

Bose statistics:

.constNal

Here,r

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

Substitute for the equation

lladTU ,

The equation above is called Planck’s formulation.

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

VUkT 1

VU x 1

We can also use another solution:

lle 1lnln

l e 1ln

h rl 2

V 1lnln

1ln1ln

exexdxex x

x x 11

2 13145

It is the same with what we have obtained before.

1 lnpV

lnlnkS VTc

§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.

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

V — d 1322

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

4 2V m dh

We want to rewrite the sums in terms of integrals.

4 21kT

V m Nh

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

When T=0K 1

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

Is called Fermi energy level.

0 3 NpV

Is called Fermi momentum, which is the largest momentum.

33 022

4 32 05

VU m d Nh

For example: Cu (cuprum)

18 38.5 10N m

120 1.1 10 J

kT 1 kTee

T ＞ 0

4 21kT

V dN mh

4 21kT

V dU mh

kTzI kTdz

0 01 1kT

kTz kTzkT dz kT dz

ekTzkTdz

ekTzkT z

0 01 1kT

kTz kTzkT dz kT dz

0 0 1z

kTz kTzI d kT dz

In the first term

000 11

kTzkTdze

kTzkTdzkTzkTI zkT

zI d kT dze

0 6d kT

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

4 21kT

V dN mh

2/12/33 6

24 kTdm

2/12/33 6

24 kTdm

2/12/33 2

24 kTdm

222/32/3

3 8124

22 2 333 1

2 8N kTc

4 21kT

V dU mh

0 6d kT

2/32/33 6

24 kTdm

2/32/33 2

24 kTdm

222/52/3

3 85124

We can also use another method:

lle 1lnln

4 2V m dh

V 1ln24ln 2/12/33

2/33 24 m

deA 1lnln 2/1

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

132 2/3

0 6d kT

132 2/3

0 6d kT

32 kTA

32 kTkTkTA

2/33 24 m

3 8512

1516ln

Exercise:

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

Chapter 7 Bose and Fermi statistics. §7-1 The statistical expressions of thermodynamic quantities 1...

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