68 CHAPTER 3. FERMIONS
3.4.1 Zero Sound
Using a semi-classical approach we can describe small density fluctuations |
dn j |
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n around the ground state (equilibrium) as |
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n j(~p;~r;t) = n j(~p) + dn j(~p;~r;t): |
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(3.178) |
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The kinetic equation for n j(~p;~r;t) reads |
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¶n j |
+~vÑrn j + ( Ñrej)Ñ~pn j |
= Icoll |
! 0 |
(3.179) |
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¶t |
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where the collisional term vanishes for T ! 0 because t T 2 ! ¥ which corresponds to the so called collisionless regime, wt 1, where w is a characteristic frequency. Therefore the kinetic equation in this regime is
¶n j |
+~vÑrn j ÑrejÑ~pn j = 0: |
(3.180) |
¶t |
The first two terms describe the ballistic motion of particles while the last term acts as a non trivial collective force. This equation has to be solved self-consistently because small fluctuations of density generates the force which in turn changes the density and so on. But first we want to rewrite the equation
Ñrej |
= Ñr |
0e(p) + å f~p j~p 0 j0dn~p |
0 j0 |
1 |
= å f~p j~p 0 j0Ñrdn~p 0 j0 |
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~p 0 j0 |
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A |
~p 0 j0 |
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¶dn |
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j |
+~vÑrdn j å f~p j~p 0 j0( d(p pF))(~eÑr)dn~p 0 j0 = 0 |
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¶t |
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¶dn |
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~p 0 j0 |
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~p j |
+~vÑrdn~p j + d(p pF) å fpF~e j~p 0 j0 (~eÑr)dn~p 0 j0 = 0: |
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¶t |
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~p 0 j0 |
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Here we used |
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¶n(p) |
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¶n ¶e |
1 |
d |
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= ~ed(p pF): |
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= |
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= |
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(p pF)~vF |
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¶~p |
¶e |
¶~p |
vF |
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To solve this, we choose the ansatz
(~ )
dn~p j = d(p pF)c(~e)ei k ~r wt :
(3.181)
(3.182)
(3.183)
(3.184)
(3.185)
3.4. LANDAU-FERMI-LIQUID
Inserting this into (3.183) we get
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~ |
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0 = ( iw + ik ~v)c(~e) + |
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~på0 j0 |
fpF~e j~p 0 j0i ~e ~k d(p pF)c(~e 0) |
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(w kvF)c(~e) =~k ~e å fpF~e j~p 0 j0d(p pF)c(~e 0) |
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~p 0 j0 |
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v j0(m)vF Z |
dΩ |
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=~k ~e åj0 |
0 |
c(~e 0) fpF~e j pF~e 0 j0 |
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4p |
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69
(3.186)
(3.187)
(3.188)
=~k ~e Z |
dΩ |
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0 |
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F(~e ~e 0)c(~e 0)vF |
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4p |
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c(~e) = w |
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4p0 |
F(~e ~e 0)c(~e 0): |
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FvF~k ~e Z |
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k |
~e |
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dΩ |
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This is difficult to solve. We assume the simplest case, i.e.
F(~e ~e 0) = F0
and consider all further deviations as higher order terms. This leads to
(3.189)
(3.190)
(3.191)
c(~e) = w |
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4p0 |
c(~e 0) |
(3.192) |
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FvF~k ~e F0 Z |
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k |
~e |
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dΩ |
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~ ~
= ~c vFk e : (3.193)
~ ~ w vFk e
If we define (3.192) over
w = vFks with s =const and z = cos(~k ~e 0) and integrate both sides of |
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dΩ |
we get |
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4p |
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1 = F0 Z |
dΩ |
~ |
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~e |
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v k |
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F |
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~e |
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(3.194) |
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4p |
w vF~k |
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1 |
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1 |
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dz |
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z |
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= |
F |
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s |
(3.195) |
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= F0 Z 1 |
2 |
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s z |
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Z 1 dz |
1 z s |
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0 |
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0 |
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s + 1 |
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= F |
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1 |
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s |
ln |
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(s |
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1) |
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(3.196) |
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1 |
+ 1 = |
s |
1 |
+ s |
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, |
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ln |
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(3.197) |
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F0 |
2 |
j1 |
sj |
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If 0 < F0 1 the left side became huge therefore s has to be close to 1:
s = 1 + e e 1 |
(3.198) |