Fermi liquid: Liquid 3He

Masatsugu Sei Suzuki

Department of Physics

(Date: December 23, 2018)

1. Heat capacity and entropy for Fermi gas

The internal energy FGU of the Fermi gas

2

2 22 5/2

22 2

2

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 12

BFG F

F

k TU N

The heat capacity:

2 22

23 5( ) 2

5 12 2

FG B BFG F

F F

dU k N kC N T T

dT

or

The heat capacity FGC :

2

2

FG B

B F

C k T

Nk

.

The entropy FGS :

2 2

2

FG BFG

F

C N kdS dT dT

T

or

2 2

2

BFG

F

N kS T

2 2

2 2

FG B

B F F

S k TT

Nk T

((Note))

In general case,

2

2( )

3FG F BC D k T

where ( )FD is the density of states at the Fermi energy F Using the relation 3

( )2

F

F

ND

, we

have the specific heat for free electron Fermi gas

2 22

23

3 2 2 2

B B BFG 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 KFT for liquid 3He.

2. 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 2.1227 N 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

2.12274.2454

/ 2

N N

I

ℏ ℏ ℏ

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

with

2

N

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

3. 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

m 2423

230 100.5105.010022.6

016.3

The number density:

322

24

0

/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.

4. 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 5.39s

F 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

ln 2 ln 2N

B Bk k N

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

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

ln 22

FGFG

B F

ST

Nk T

.

For the Fermi liquid, we have

2

2 ln 20.25

2.80

FFL

TT

K.

where

2

2.80 ln 22

FLFL

B F

ST

Nk T

.

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

5. 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

6. 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

2ln 2F

T 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

dP for T>0.3 K corresponding to solidliq SS

0dT

dP for 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.

7. Problem: Clausius-Clapeyron equation

Ralph Baierlein

12-9

((Solution))

Fig. Phase diagram. P (mmHg=torr) vs T (K)of liquid He3.

0.20 0.25 0.30 0.35 0.40 0.45 0.50

10 5

10 4

0.001

0.010

0.100

2( )

v v v v

g l g g B

l l Pl PldP

dT T v v Tv TPv k T

or

2

1 v

B

ldP dT

P k T

or

0ln ln v

B

lP P

k T

We make a plot of ln P vs 1/ T for 3He.

Fig. Least squares fit to the straight line; Plot of ln P vs 1/T.

(a) The least-squares fit of the data (lnP vs 1/T) to a straight line yields

ln 4.34001 3.14569 1/P T

3.14569v Bl k (latent heat of evaporation per particle)

or

2.5 3.0 3.5 4.0 4.5 5.0

10

8

6

4

2

42.7108 10vl

(eV/particle)

(b) T = 0.1 K.

ln 4.34001 3.14569 1/ 0.1 27.1169P

121.672 10P torr.

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8. Problem: Phonon

Ralph Baierlein

12-10

((Solution))

3

412

5V

D

TC R

C

The entropy:

3

4212 1

5

V

D

CdS dT R T

T

3

44

5solid

D

TS R

with

16K .

Nuclear spin of 3He

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

ln 2nuclearS R

Fig. Entropy of lattice vibration (red) and nuclear spin (blue) for Helium three.

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:

T K

S R

0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

)(

13

11

* sF

m

m

(b) Specific heat:

)(

13

11

* s

g

N Fm

m

C

C

where

F

BgT

TNkC 2

2

1

(c) Spin suceptibility:

)(

0

)(

1

)(

0 1

3

11

1

1*a

s

a

g

N

F

F

Fm

m

where

FB

B

F

BgTk

NN 1

2

3

2

32

2

(d) Compressibility:

)(

0

)(

1

)(

0 1

3

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

______________________________________________________________________________

APPENDIX II Proton; magnetic moment

Nuclear magnetic moment 2.792 N and I = 1/2 for proton.

where

2N

p

e

m c

ℏ

The gyromagnetic ratio:

2.7925.584

/ 2

N N

I

ℏ ℏ ℏ

= 5.584 x 4789.43=2.67442 x 104 (rad/s G)

2

N

p

e

m c

ℏ=4789.43

2 2

Bf

=42.5647 (MHz/T)

___________________________________________________________________________