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64

CHAPTER 3. FERMIONS

3.4Landau-Fermi-Liquid

First we considered the non interacting FERMI gas where we have the filled F ERMI sphere as ground state and we were capable of describing all properties for T eF causing excitations (particle transitions) near eF. If we now switch on the interaction adiabaticly the behavior of the system remains similar to the previous one. In the interacting system a particle disturbs locally the surrounding particles. If we consider the particle together with the disturbance as a new particle (quasiparticle) we can transfer our previous discussion of the non interacting case to the interacting case. This procedure is called LANDAU conjecture and it can be justified by using a more complicated approach based on GREEN functions technique.

We are not interested in the general E p dependency but only near the FERMI surface as we have seen that the most important physics takes place there. We describe the ground state of an interacting (normal) FERMI system as a FERMI sphere filled with quasiparticles. The number of the quasiparticles is the same as the number of particles by the above conjecture, thus

p3

g F = n = nqp (3.141)

6p2~3

This means the FERMI momentum also remains the same.

This FERMI liquid approach is experimentally favorably because as – we will show now – the description boils down to a few, experimentally accessible parameters.

If we briefly assume we have only one type of F ERMIon we can relate the change in energy of the system to a change in the distribution function as

dE = åe(~p)dn~p

(3.142)

~p

 

which provides the definition of e(~p) for quasiparticles. LANDAU introduced the f function which determines the change of the energy of the quasiparticle e(~p)

due to the change in the distribution function:

 

de(~p) = å f~p~p 0dn~p 0

(3.143)

~p 0

 

The LANDAU f -function is a direct consequence of interparticle interaction and, as we will see, it actually governs collective behavior of the system.

We can now expand the energy-momentum relationship around the most interest-

3.4. LANDAU-FERMI-LIQUID

65

ing point, the FERMI momentum:

 

e(~p) = m +~vF(~p ~pF) with

(3.144)

 

 

 

p2

 

m =

F

+ interaction

(3.145)

 

 

 

 

2m

 

v

F

=

pF

 

(3.146)

 

 

m

 

 

 

 

m = m + interaction

(3.147)

Thus the complicated energy-momentum relation is substantially simplified since only momenta p pF are important. As expected, the energy depends only on the absolute value of the momentum.

If we compare this result with our previous discussions ((3.117) and (3.120)) in the weakly interacting range

E = E0

+

4p~2a

 

1

å ån~p jn~p 0

 

 

 

 

j

0

 

 

m 2 j= j0

~p~p 0

 

 

 

 

 

 

6

 

 

 

we have

p2 ej(~p) = 2m

which means that

m =

and therefore

 

E0 = åå

p2

n~p j

 

 

 

 

 

 

 

2m

 

 

 

 

 

 

 

 

 

 

~p

j

 

 

 

 

 

 

 

 

 

 

 

 

 

4p~2a

å n~p 0 j0 =

 

p2

4p~2a

 

å n j0

+

 

 

 

 

 

 

 

+

 

 

 

m

~p

2m

m

 

 

 

 

 

 

0 j6= j0

 

 

 

 

 

j6= j0

e(pF) =

pF2

+

4p~2a

 

 

å n j0

 

m = m

 

 

 

 

 

 

 

 

 

 

2m

 

 

m

j= j0

 

 

 

 

 

 

 

 

 

= f~p j~p 0 j0 = (

6

 

 

 

 

 

 

n~p j

4p~2a

j00

= j

 

¶ej

(p)

 

 

 

 

 

 

 

 

0

 

j

= j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m

 

6

 

 

(3.148)

(3.149)

(3.150)

(3.151)

(3.152)

in first order in the interaction. If we also consider second order terms we get rather complicated terms with non trivial f , lengthy expressions and m 6= m. The effective mass m can be measured experimentally e.g. in the specific heat (cf. (3.35) and (3.13)).

 

p2

 

 

c =

 

n(m)T

with

(3.153)

3

 

 

 

 

 

n(m) =

m pF

g

 

(3.154)

 

 

 

2p2~3

 

 

66

CHAPTER 3. FERMIONS

In our regime of low temperatures cV = cp and we thus do not have to distinguish between them.

We now want to calculate the effective mass m via the f -function for general FERMI liquid. The liquid moves with the velocity ~v in the laboratory system thus generates the current

~

= mn~v = å~pn˜~p j

(3.155)

j

~p j

The distribution function in the frame of the moving liquid relates to the distribution function in the laboratory n˜ as follows

n˜~p j = n j (e(~p) ~p ~v) = n(e(~p)) + dn~p j

(3.156)

dn~p j =

n~p j

de

de = ~p ~v + interaction

(3.157)

¶e

The distribution function of FERMIons is a step function at T = 0 thus its derivative is a d-function. The current now reads

~j = å~p j ~p n j(e(~p)) + dn~p j

n~p j

 

 

= å~p j ~p

 

de(~p)

(3.158)

¶e

The first term in the summation vanishes because the filled F ERMI sphere is rotational invariant and the summation runs over all momenta.

n~p j

1

 

 

= d(e(p) m) =

 

d(p pF)

¶e

vF

~

= å~p

 

1

d(p pF) d(e(~p))

 

 

j

 

v

F

 

 

 

~p j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

p2

 

 

 

 

 

dΩ

 

 

 

= vF

 

(2p~)3 pFg4p Z

4p ~ede(pF~e)

 

 

 

 

 

 

 

F

 

 

 

 

 

 

 

 

 

 

 

 

 

g

 

p3

 

 

dΩ

3

 

dΩ

~ede(pF~e)

 

= vF

 

2p2~3

Z 4p ~ede(pF~e) = vF n Z

4p

 

 

 

 

F

 

 

 

 

 

 

 

 

 

 

 

(3.159)

(3.160)

(3.161)

(3.162)

In (3.162) we used (3.141). Looking at the shift in energy assuming pF( j) pF8j we get

de(~p) = ~p ~v + å f~p j~p 0

j0dn~p 0 j0

 

(3.163)

~p 0

j0

 

n~p 0 j0

 

 

= ~p ~v + å f~p j~p 0

 

 

 

j0

 

de(~p

0)

(3.164)

¶e

~p 0

j0

 

 

 

 

 

3.4. LANDAU-FERMI-LIQUID

(2p~)0

 

 

 

 

 

= ~p ~v vF åj0

Z

3 f~p j~p 0 j0d(p pF)de(~p 0)

 

1

 

 

 

 

d3 p

 

 

 

 

 

1

 

 

pF2

 

 

åj0

Z

dΩ0

= ~p ~v

 

 

 

4p

 

f~p j pF~e 0 j0de(pF~e 0)

vF

(2p~)3

4p

|{z }

(3.146) and (3.154)

= ~p ~v Z dΩ0 ånj0(m) f~p j p ~e 0 j0de(pF~e 0)

4p j0 F

This is interesting only for j~pj = j~pFj and we can write using ~pF = pF~e

de(pF~e) = pF~e ~v Z

dΩ

 

 

 

 

 

 

 

0

F(~e~e

0)de(pF~e 0)

 

with

4p

 

F(~e ~e 0) = ånj0(m) fpF~e j pF~e 0 j0

 

 

 

 

 

 

 

 

j0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= å(2l + 1)Fl Pl (~e ~e 0)

 

 

 

 

 

 

 

 

l=0

 

 

 

 

 

 

 

 

 

 

 

 

 

Here Pl are the LEGENDRE-Polynomials.

 

 

 

 

 

 

To solve this we make the ansatz

 

 

 

 

 

 

 

 

 

 

de(pF~e) = ApF~e ~v

 

A = const

 

 

 

 

 

 

 

 

 

 

 

 

dΩ0

 

 

 

 

 

 

A~e

 

~v = ~e

~v

 

A

å(2l + 1)Fl Pl

(~e

~e

0) ~e 0

 

~v

4p

 

 

 

Z

l=0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P|{z}1(~e 0 ~v)

67

(3.165)

(3.166)

(3.167)

(3.168)

(3.169)

(3.170)

(3.171)

(3.172)

= ~e ~v AF1~e ~v

 

 

 

 

 

 

 

 

 

 

 

 

 

A =

 

 

1

 

 

 

de

(p ~e) =

 

pF~e ~v

 

 

 

 

 

1

+ F1

 

 

1 + F1

 

 

 

 

 

 

 

F

 

 

 

 

 

 

Thus the current becomes

 

1 +FF1 Z

 

4p

 

~e(~e ~v)

 

 

 

 

 

~j = vF n

 

 

 

 

 

 

 

 

 

3

 

 

 

 

p

dΩ

 

 

 

 

 

 

=

3

n

 

pF

 

 

1

~v = n

 

m

 

 

~v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

vF 1

+ F1 3

1

+ F1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z

dΩ

 

1

) m = m(1 + F1)

with

 

 

 

 

 

~ei~e j

=

 

 

di j

 

 

 

 

4p

3

(3.173)

(3.174)

(3.175)

(3.176)

(3.177)

where F1 is usually positive. The value of F1 has to be derived from an appropriate model or from measurement. Only for very few systems – e.g. F ERMI gases – direct calculation is possible.

We now want to discuss some non trivial examples.

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