Fermi liquid theory

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: January 13, 2019)

1. Introduction

Here we consider the normal state of interacting Fermi particles without any long range order. The normal state at low temperatures is called the Fermi liquid and is considered to be the system of free quasi-

particles that is continuously connected with free Fermi gas. The concept of the Fermi liquid was introduced

and developed by L. D. Landau. Landau’s Fermi liquid theory, which concentrates rich contents.

The electrons are fermion with spin 1/2. The system of many electrons is a collection of Fermi

particles. The behavior of electrons in solids can be well explained in terms of quantum mechanics;

the nature of duality of wave and particle. The electron has a negative charger. There is a repulsive

Coulomb interaction between electrons. Nevertheless, the electron in metal is well described by a

free fermion without any interactions. Why is a simple free electron model so useful in spite of

complicated nature of electrons in metals? What is the role of interactions between electrons? Here

we discuss the concept of the Fermi liquid theory which is essential to understanding the

effectiveness of free electron model.

Here we discuss the relaxation time in the Fermi liquid theory. Several methods for the

derivation of the relaxation time will be introduced below.

2. Relaxation time due to the collisions of Fermi particles

The interaction between electron in metal is relatively large. It seems that there are many

collisions between electrons repeatedly. This is not the case. Because of the Pauli’s exclusion

principle, the probability for collisions is suppressed as low. The reciprocal of the relaxation time

for the scattering of electrons is given by

2( )B

F

k T

ℏ≃

When B Fk T ,

ℏ

becomes sufficiently small. The scattering effect is neglected. The wave

vector k of electron can be regarded as good quantum number. This is the fundamental of the

Landau Fermi liquid theory. Typical examples of Fermi liquids are (i) electrons in metals and (ii)

liquid 3He.

According to the Fermi golden rule, the relaxation time due to the scattering of electrons is

given by

2 1 2

1 2 1 2 2 1 2

21 2 1 22

, ', '1

, ' ' ' '

1 2 1( ' ')

( )

(1 )(1 )

A

UN

f f f

p p p

p p p p p p p

p ℏ

where NA is the number of particles.

Fig. Four states of electrons in the vicinity of the Fermi surface. See the detail in the text.

3. Derivation of the relaxation time by Anderson

It is no doubt well known that the resulting mean free time varies as the square of the energy

increment,

2 2

2 2[( ) ]2 F F FE k a k k a

m m

ℏ ℏ

22

3consts.

F

V E

ℏ

This may be seen by observing that for any final state with the hole in state k’ energy conservation

requires:

2 2 2 2( ) ( ' ) ' k k q k q k

or

2 2 2 2 2 2( 2 ) ( ' 2 ' ) ' k k q k q k q k q k

or

2( ') k k q q

or

( ' ) 0 k k q q

Fig. Four states: k (A, outside the Fermi sphere), k q (B, outside the Fermi surface), 'k

(D, inside the Fermi sphere), 'k q (C, outside the Fermi sphere). H’ is the center of the

circle (denoted by green line) where the points A, B, C, and D lie on. The diameter of the

circle is 'k k .

This is the equation of a sphere whose diameter is 'k k : thus we know that 'k q must lie on

the sphere whose diameter is the vector from k to 'k . When a is small, this sphere can be

approximated by planes through k and 'k , since 'k , k , k q , and 'k qmust all lie near the

Fermi surface (this is true except for ' k k , which is a very small fraction of the available states).

Now clearly, if 'k lies within ( cos )a of the surface, two types of scattering process are possible;

if we pick a ( '). q k k and small ( ' cosFq k k , cosq a ), we may have

A

B

C

D

HH'

O

E qq

k'k' qk

k q

kF

a

k k q , ' ' k k q

or

' k k q , ' k k q

In either case, the possible ranges both of 'k and of q for a given are proportional to

a E :thus the total scattering probability at this is 2E . I leave it as an exercise to perform

the integration over and get the full expression for the /ℏ of the electron of a given E .

F

ax

k

[( (1 )cos , (1 )sin ]FOA k x x ����

outside the Fermi surface

[( (1 ) cos , (1 )sin ]FOB k x x ����

(0, 2 sin )Fk x q

[0,sin ]FOH k ����

[cos ,sin ]FOG k ����

2[cos ,sin ] (0, sin )

[cos , (1 )sin ]F F

F

OD OG

k k x

k x

q���� ����

on the Fermi surface

2[cos ,sin ] (0, sin )

[cos , (1 )sin ]F F

F

OC OG

k k x

k x

q���� ����

inside the Fermi surface

(a)

We note that

2 2 2 2

2

(1 ) cos (1 ) sin

1 2 cos(2 )

F

F

OB k x x

k x x

����

When the point B is on the Fermi sphere with 1OB ����

, we have

2cos(2 )x ,

or

2cos(2 )x 1

arccos( / 2)2

x

When x = 0, 4

(b)

We note that

2 2 2

2

cos (1 ) sin

1 (2 )sin

F

F

OC k x

k x x

����

2 2 2

2

cos (1 ) sin

1 (2 )sin

F

F

OD k x

k x x

����

indicating that

FOC k����

, FOD k����

FAE a k x ����

, FOE k����

cos cosFEF a k x ����

cos cosFBG a k x ����

, 2 sin (2 sin )FAB a k x ����

cosFDJ KG BG k x ���� ���� ����

______________________________________________________________________________

2. Approach by Abrikosov and Khalatnikov

Fig.

Let us define the probability for a weakly interacting gas. If there is a particle 1 outside the

Fermi sphere, then the process of first order in the interaction will be as follows. Particle 1 interacts

with particle 2 inside the Fermi sphere, following which the two particles pass over to states 1’

and 2’ outside the Fermi sphere. Because of the Pauli principle, this is the only possibility. The

law of momentum conservation requires that

1 2 1 2' ' p p p p

and, in accordance with what has been seen said above

1 Fp p , 2 Fp p , 1 ' Fp p , 2 ' Fp p ,

The planes 1 2( , )p p and 1 2( ', ')p p do not coincide, generally speaking, and in Fig.1 they are

simply superposed by rotation. The scattering probability is given to within a constant by

3 3

1 2 1 2 2 1( ' ') 'W d d p p

The integration is carried out only over 2p and 1 'p , since 2 'p is determined by the law of

momentum conservation. The angle between the vectors 1 'p and 2 'p is actually specified by the

law of energy conservation. The integration over this angle eliminates the -function. It now

remains only to integrate the absolute values of the vectors.

Suppose that p1 is close to pF. Then, all the remaining momenta will also be close to pF in

absolute value, and, consequently, in Fig.1 they will make nearly equal angles with the horizontal

line (with the sum 1 2p p ). Hence, from the relationship between the projections on this axis we

can write the relation between the absolute values:

1 1 2 2' 'p p p p .

Since 2 ' Fp p , it follows that we have

1 1 2' Fp p p p

But at the same time 1 ' Fp p , from which it follows that

1 1 2'F Fp p p p p or 1 22 Fp p p

The upper limit for 2p is Fp . Thus, we have

1 2 0F Fp p p p

1 1 20 ' ( ) ( )F F Fp p p p p p

On integration we obtain

22 1 1

1' ( )

2 Fdp dp p p

Hence

2

1( )Fp p

The complete formula for can be obtained from dimensionality considerations. It must be

proportional to the square of the interaction constant and, according to the above calculation, to

the quantity 21( )Fp p . Following this, we have to introduce a further factor made up of

Fp , m

and ℏ in such a manner that the result has the dimension of energy.

1 ' Fy p p , 2 Fx p p , 0 1 0Fx p p

0 0x x , 00 y x x

Fig. Area shaded in green. 20

1

2A x .

Fig.

Fig. The conservation of momentum and energy. 1OA p����

. 2OC p����

. 1 'OB p����

. 2OC p����

.

1 'OB p����

. 2 'OD p����

. 2AP p����

. 2BP p����

. 1 2 1 2' 'OP p p p p����

. H is the center of mass.

Fig. Point B and point D can rotate around the axis OP by the same angle. The point H is the

center of mass between OA����

and OB����

. Triangle OAP is fixed. In other words, the vector

OP����

is fixed. The vector AB����

is always parallel to the vector C D����

from the momentum and

energy conservation laws: A B C D ���� ����

. 1 2 1 2' ' OP p p p p����

or 1 1 2 2' ( ') p p p p

Note that is the angle between 1p and 2p ;

1 0 (cos ,sin )2 2

p

p , 2 0 (cos , sin )

2 2p

p ,

1 0( ') sin2

r p

p

where GBJ , and 2

BOJ AJO

. The vector f is in the plane where 1p and

2p lie;

( , )z rf ff

A

B

D

C

H

O

Eqq

p2 k'

p2 k'

k' q

p2 ' k' q

p1 k

p1' k q

p1 p2 p1' p2 '

kF

G

J

where zf is the component of the vector f along the axis parallel to

1 2p p and rf is the

perpendicular component. We have the relation

1 1' p p f

or

22 21 1 1

2 21 1

2

1

1

' 2

2 ( cos sin )2 2

1[1 2 ( cos sin )]

2 2

z r

z r

p p f

p f p f f

p f fp

p f

or

1 1' cos sin2 2

z rp p f f

We also have the relation

2 2' p p f

or

2 2 22 2 2

2 22 2

2

' 2

[1 2 ]

p p f

pp

p f

p f

or

2 2' (1 cos sin )2 2

z rp p f f

Jacobian:

1 21 2

1 2

( , ) ' '' '

( ', ') 2 sin cos2 2

z rz z

f f dp dpdf df dp dp

p p

with

1 1

1 2

2 2

' '

( ', ')sin

' '( , )z r

z r

z r

p p

f fp p

p pf f

f f

For 2p , we use the polar co-ordinate

2 2( , , )p around the vector 1p .

3 2

2 2 2 2sind p dp d d p

3

1 1

1 20

1 20

' ( ')

' 'sin

2 2sin cos2 2

' '

2cos2

r z rd df df d

dp dpp d

dp dpp d

p p

and

3 3 21 21 2 0 0 2 2

30 2 1 2 2

' '' sin

2cos2

sin ' '2

dp dpd d p d p dp d d

p d d d dp dp dp

p p

We put

2 20

0 0

00

( )

( )2

( )( )2

( )

x

p p

m

p p p pm

pp p

m

p

0pdx dp

m

2 2 02 2 0

0

mpmp dp p dx dx

p

1 1

0

' 'm

dp dxp

, 2 2

0

' 'm

dp dxp

3 3 31 2 1 2 1 2 0 2 1 2 2

1 2 1 2

30 2

3

1 2 1 21 2 2

0

2

32

1 2 2 1 2 1 2

' ( ' ') sin ' '2

( ' ')

sin2

' '' ' ( )

1sin

2

' ' ( ' ')

d d p d d d dp dp dp

p d d d

x x x xmdx dx dx

p

m d d d

dx dx dx x x x x

p p

The relaxation time is obtained as

3 22

1 2 1 2 2 1 2 2 1 2

3 2

1 2 1 2 2 1 2 2 1 2

1( ) sin

2

( ' ') ' ' ( )[1 ( ')][1 ( ')]

2 ( ) sin2

( ' ') ' ' ( )[1 ( ')][1 ( ')]

B

B

m k T d d d

x x x x dx dx dx f x f x f x

m k T d d

x x x x dx dx dx f x f x f x

with

2 1 22 1 2 ' '

1 1 1( )[1 ( ')][1 ( ')]

1 1 1x x xf f f

e e e

where

1 2 1 2 1 2 1 2

1 2 1 2

1( ' ') [ ( ' ')]

1( ' ')

x x x x

x x x x

Note that the integral given by

0 0

1 1 1( )

1 1 1x y a x yF a dx dy

e e e

has a finite value when 0 1a . When a = 0,

2

(0)24

F

0.41123

((See APPENDIX)) Thus the relaxation time is

2( )1 B

F

k TA

ℏ

where A is a dimensionless number and is of the order of unity. ((Heisenberg’s principle of uncertainty))

This equation can be explained using the Heisenberg’s principle of uncertainty,

2( )1 1 1

( ) B BB

F F

k T k Tk T

ℏ ℏ

where B

F

k T

is the fraction of electrons in the energy interval Bk T (thermal energy) in the

vicinity of F .

((Note)) The density of states for free electrons

1/2( )D a (a, constant)

The total number of electrons;

1/2 3/2

0 0

2( ) ( )

3

F

F

aN d D f a d

The fraction of number of electrons in the vicinity F nis

1/2( ) ( )

F B F BN D k T a k T

Thus we have the fraction

1/2

3/2

( ) 32 23

F B B B

F FF

a k T k T k TN

aN

3. Approach by Kittel

Fig. The point H is the wave vector of the center of mass of 1 and 2. All pairs of states 1’ and

2’ conserve momentum and energy if they lie at opposite ends of a diameter of the small

sphere. The small sphere is drawn from the center of mass of 1 and 2. But not all pairs of

points 1’ and 2’ are allowed by the exclusion principle, for both 1’, 2’ must lie outside the

Fermi sphere; the fraction allowed is / F . (Kittel, ISSP 8-th edition)

It is an astonishing property of metals that conduction electron, although crowded together only 2 Å apart, travel long distances between collisions with each other. The mean free paths for electron-electron collisions are longer than 104 Å at room temperature and longer than 10 cm at 1K.

Two factors are responsible for these long mean free paths, without the free-electron model of

metal would have little value. The most powerful factor is the exclusion principle, and the second factor is the screening of Coulomb interaction between electrons.

We show how the Pauli’s exclusion principle reduces the collision frequency of an electron that has a low excitation energy

1 outside a filled Fermi sphere. We estimate the effect of the

exclusion principle on the two collision, 1 2 1' 2' between an electron in the excited state (1)

and an electron in the filled state (2) in the Fermi surface. It is an astonishing property of metals that conduction electron, although crowded together

only 2 Å apart, travel long distances between collisions with each other. The mean free paths for electron-electron collisions are longer than 104 Å at room temperature and longer than 10 cm at 1K.

Two factors are responsible for these long mean free paths, without the free-electron model of metal would have little value. The most powerful factor is the exclusion principle, and the second factor is the screening of Coulomb interaction between electrons.

We show how the Pauli’s exclusion principle reduces the collision frequency of an electron that the energy

1 is outside a filled Fermi sphere. We estimate the effect of the exclusion principle

on the two collisions, 1 2 1' 2' between an electron in the excited state (1’) and an electron in

the filled state (2’) in the Fermi surface. Because of the exclusion principle the states 1’ and 2’ of the electrons after collision must lie outside the Fermi sphere, all states within the sphere being already occupied; thus both states are outside the Fermi sphere.

The conservation of energy (1 2 1 2' ' ) requires that 2 1F F . This means that

collisions are possible only if the state 2 lies within a shell of thickness a within the Fermi

surface. Thus the fraction / F of the electrons in filled states provides a suitable target for

electron 1. But even if the target electron 2 is in the suitable energy shell, only a small fraction of the final states compatible with conservation of energy and momentum are allowed by the exclusion principle. This gives a second factor of / F . In Fig. we show a small sphere on which

all pairs of states 1’, 2’ at opposite ends of a diameter satisfy the conservation laws, but collisions can occur only if both 1’ and 2’ lie outside the Fermi sphere. The product of the two fractions is

2/ F .

Fig.

1 p k , 2 'p k , 1 ' p k q , 2 ' ' p k q

4. Heat capacity and entropy for Fermi gas

The internal energy FGU of the Fermi gas

A

B

D

C

H

O

Eqq

k'

k'

k' q

k' q

k

k q

kF

a

22 22 5/2

22 22

22

3 5[1 ( ) ] [1 ]

5 12 8

3 5 5[1 ( ) ][1 ]

5 24 8

3 5[1 ( ) ]

5 12

FG B B

F F F

B B

F F

B

F

U k T k T

N

k T k T

k T

or

2

23 5[1 ( ) ]

5 12B

FG F

F

k TU N

The heat capacity:

2 22

23 5( ) 2

5 12 2FG B B

FG F

F F

dU k N kC N T T

dT

or

The heat capacity FGC :

2

2FG B

B F

C k T

Nk

.

The entropy FGS :

2 2

2FG B

FG

F

C N kdS dT dT

T

or

2 2

2B

FG

F

N kS T

2 2

2 2FG B

B F F

S k TT

Nk T

((Note))

In general case,

2

2( )3FG F B

C D k T

where ( )F

D is the density of states at the Fermi energy F Using the relation

3( )

2F

F

ND

, we

have the specific heat for free electron Fermi gas

2 22

23

3 2 2 2B B B

FG B

F F F

Nk k T NkN TC k T

T

or

2

0.998948 (K)2

FG

B F

C TT

Nk T

where 4.94 KF

T for liquid 3He.

5. Pauli paramagnetism for Fermi gas model

The Pauli paramagnetism is given by

2 2

2 3 3( )

2 2FG F

F B F

N ND

k T

with

3( )

2F

F

ND

The magnetic moment of liquid 3He is

1

2I ℏ ℏ

where 2.1227N

and I = ½. The nuclear magneton is defined by

2N

p

e

m c

ℏ

(mp is the mass of proton). The gyromagnetic ratio (the ratio of the magnetic moment to the angular

momentum) is given by

2.1227

4.2454/ 2

N N

I

ℏ ℏ ℏ

= 4.2454 x 4789.43=-2.033 x 104 (rad/s G)

with

2N

p

e

m c

ℏ=4789.43

where

2 2

Bf

=32.361 (MHz/T)

The Pauli susceptibility can be rewritten as

2

2 2 233 32

2 2 8FG

F B F B F

n n nk T k T

ℏ

ℏ

6. Fermi energy of liquid 3He

Liquid 3He as a Fermi gas

spin 2

1I (fermion)

Density 081.0 g/cm3

The Fermi energy:

3/22

0

2

)3(2

nm

F ℏ

where m0 is the mass of 3He atom,

gg

m2423

230 100.5105.010022.6

016.3

The number density:

322

240

/1062.1100.5

081.0cm

mV

M

M

N

V

Nn

=1.62 x 1028/m3

Then the Fermi energy is

416 10254.4erg10815.6 F eV.

The Fermi temperature

B

FF

kT

4.94 K.

That is only a little higher than the boiling point, 3.2 K.

7. Heat capacity of fermi liquid

As predicted for the fermi gas mode, the heat capacity of liquid 3He should be

2

0.99892

FG

B F

C

Nk T T

K-1

So although the linear temperature dependence agrees with experiment, the predicted coefficient

is too small by almost a factor of 3. According to the Fermi liquid theory, the heat capacity of the

liquid 3He is predicted as

( )2 2

1(1 ) 2.802 3 2

s

FL

B F F

C F

Nk T T T

where ( )

1 5.39sF for P (pressure) = 0. Similarly, the entropy S is predicted as

( )2 2

1(1 ) 2.802 3 2

s

FL

B F F

S F

Nk T T T

The entropy of the solid, meanwhile, should be

since each nucleus has two possible spin orientations. This constant value should apply down to

very low temperatures, when the nuclear spins finally align and the entropy freezes out. Here is a

sketch of both entropy functions. We make a plot of the entropy of the fermi gas and Fermi

liquid as a function of T as well as the entropy of solid 3He.

For the Fermi gas, we have

2

2ln 2 0.69F

FG

TT

K

where 2

ln 22

FG

FG

B F

ST

Nk T

.

For the Fermi liquid, we have

2

2 ln 20.25

2.80F

FL

TT

K.

where

ln 2 ln 2N

B Bk k N

S NkB

T K0.25 K 0.69 K

Fermi liquid Fermi gas

ln 2solid

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

2

2.80 ln 22

FLFL

B F

ST

Nk T

.

The temperature FLT (=0.25 K) is much lower than the temperature FG

T (=0.69 K).

According to the Clausius-Clapeyron relation, the slope of the solid-liquid phase boundary on

a graph of P vs T should be proportional to the entropy difference, liquid solidS S . The above analysis

predicts that the slope should be positive for T>0.25 K, and negative at lower temperatures. The

experimental phase diagram shows just this behavior, with the transition from positive to negative

slope at about 0.3 K, just slightly higher than the prediction 0.25 K. This discrepancy could be

because of lattice vibrations giving the solid some additional entropy, and/or the entropy of the

liquid no longer being quite linear at relatively high temperatures. At very low temperatures, where

the entropy of the solid also goes to zero, the phase boundary becomes horizontal.

8. Entropy

The entropy is mainly due to the nuclear spin of the solid 3He around T = 0.3 K. The 3He atoms

form a lattice. Below 10 mK the nuclear spins starts to order because of the antiferromagnetic

exchange interaction. The nuclear spins antiferromagnetically ordered below the Neel temperature

TN = 2 mK. The antiferromagnetic exchange interaction is -0.85 mK.

Solidliq VV

For example, at T = 0.3 K, it is determined experimentally that

atomcmN

VV Solidliq /101.2 324

9. Pomeranchuk cooling

In the P-V phase diagram, the solid-liquid boundary below 0.3 K has a negative slope. This is

very unusual. Note that the solid-liquid boundary for most materials (except water) has a positive

slope. What are the entropies for the liquid phase and solid phase? The 3He atom is a fermion. The

heat capacity is proportional to T at low temperatures

2

2liquid A B

F

TC N k

T

2.80

The entropy is evaluated as

2

2.802

liquid

liquid A B

F

C TS dT N k

T T

where FT is the Fermi temperature. The entropy is proportional to T.

The entropy of solid 3He is dominated by the much larger contribution of the disordered nuclear

spins (spin 1/2). Each the nuclear spin (spin 1/2) of 3He atom has a magnetic moment, just like a

paramagnetic salts. In the paramagnetic state where the directions of spins are random, the solid

entropy is given by

2lnBAsoilid kNS .

Note that these spins are antiferromagnetically ordered with a Neel temperature NT 1 mK and

the entropy drops rapidly to zero. The liquid entropy coincides with the solid entropy at a

characteristic temperature 1T (=0.32 K). Below

1T , the solid entropy is higher (more disordered)

than the liquid entropy. Above 1T the liquid entropy (more disordered) is higher than the solid

entropy. Such a cross-over of the liquid and solid entropies of 3He at about 0.32 K produces a

pronounced minimum in the melting curve at about 2.93 MPa, followed by a rise in the melting

pressure to 3.45 MPa at T = 0 K. This negative slope is used to produce adiabatic compressional

cooling along the melting curve, a technique called Pomeranchuk cooling after its proposer, and

used by Osheroff et al. (1972a) in their discovery of the superfluid phases of 3He.

Fig. Entropy per atom in the coexisting solid and liquid phases of 3He. The entropy of the liquid

phase is less than that of the solid phase below a characteristic temperature 1 2

2 ln 2FT T

.

The dotted line from a to b corresponds to slow adiabatic compression from pure liquid to

pure solid.

T

S kB

Solid

Liquid

ln2

ab

T1

The phase boundary is expressed by the Clausius-Clapeyron equation

solidliq

solidliq

VV

SS

dT

dP

In the phase diagram of 3He around 0.3 K,

0dT

dPfor T>0.3 K corresponding to solidliq SS

0dT

dPfor T<0.3 K corresponding to solidliq SS

since Solidliq VV in the vicinity of 0.3 K.

Consider the liquid-solid boundary (denoted by red open circle) below 0.32 K, where solidliq SS ,

When temperature is increased, the entropy increases. The only way this can happen here is to

solid which has higher entropy. So the liquid freezes when the temperature rises. This explains the

negative slope below 0.32 K.

In order for entropy to increase, heat must be absorbed from the surrounding (like when a

normal solid melts). This gives rise to a cooling effect. Higher pressure forces the liquid to become

solid. This gives a cooling effect. This compressional cooling" method was proposed by Issak

Pomeranchuk in 1950.

Pomenranchuk predicted that adiabatic compression of coexisting solid and liquid would cool 3He. If the initial temperature is below 0.32 K, then an increase in the external pressure moves the

helium along the melting curve up and to the left: toward higher pressure and lower temperatures.

Typically, the temperature decreases from 0.3 K to 1 mK. The solid-liquid boundary (arrow) below

0.3 K has a negative slope.

APPENDIX-I Fermi liquid theory

The life time a quasi-particle excitation dressed by interactions is rather long owing to the

surrounding degenerate Fermi sea. Thus the physical properties at FTT may be well described

by a theory based on quasi-particle excitations, the so-called Landau Fermi liquid theory. Since

the critical temperature Tc of this superfluid is of the order of 1000/FT , similar to that for

superconductivity in a metal, one can naturally study the system on the basis of this theory.

Fermi liquid relations between the Landau parameters and experimentally measured quantities

(a) Effective mass:

)(

13

11

* sF

m

m

(b) Specific heat:

)(1

3

11

* s

g

N Fm

m

C

C

where

F

BgT

TNkC 2

2

1

(c) Spin suceptibility:

)(

0

)(

1

)(

0 13

11

1

1*a

s

a

g

N

F

F

Fm

m

where

FB

B

F

BgTk

NN 1

2

3

2

3 22

(d) Compressibility:

)(

0

)(

1

)(

0 13

11

1

1*s

s

s

g

N

F

F

Fm

m

where

F

gn

1

2

3

(e) The sound velocity:

)(

1

)(0

2

3

11

1

s

s

g F

F

c

c

where

)3(2

22

nmk

TB

F ℏ

, )3(2

2

*

2*

nkm

TB

F ℏ

3/22 )3( npF ℏ

____________________________________________________________________________

REFERENCES

P.W. Anderson, Concepts in Solids (World Scientific, 1997).

A.A. Abrikosov, Fundamentals of the theory of Metals (North-Holland, 1988).

A.A. Abrikosov and I.M. Khalatonikov, Soviet USPEKHI 66, 68 (1958), “Theory of the Fermi

Liquid.”

K. Yamada, Electron Correlations in Metals (Cambridge, 2004).

H. Shiba, Electron Theory of Solids (Maruzen, 1996). [in Japanese].

N.W. Ashcroft and N.D. Mermin, Solid State Physics (Holt, Rinehart and Winston, 1976).

______________________________________________________________________________ APPENDIX

The integral

0 0

1 1 1( )

1 1 1x y a x yF a dx dy

e e e

When a = 0,

2

(0)24

F

0.41123

When a increases, the function ( )F a increases as shown in Figure below.

Fig. Plot of ( )F a as a function of a.

a

F a

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8