Phonon Attenuation and Acoustic Velocity Modulation by Two-Level Systems

Phonon Attenuation and Acoustic Velocity Modulation by Two-Level
Systems
Aviram Steinbok
It has been shown by experiments that disordered and amorphous solids have universal
properties at low temperatures. A model that explains this phenomenon was proposed
by Anderson, Halperin, and Varma (AHV) [1]. According to this model the resonant
phonon attenuation is governed by localized two-level systems. A second contribution
to the phonon attenuation arises from the relaxation of the localized excitations.
In this project I have investigated these attenuations. In addition I investigated the
variation of velocity, a property that is related to the phonon attenuation. With this
knowledge I constructed the attenuations region of influence.
Most of the following derivations can be found in the literature [2,3,4].
I first describe the ”Tunneling model”, proposed by AHV. Their first hypothesis is that in a glass, a
certain number of atoms have two equilibrium positions corresponding to the minima of asymmetric
double-well potentials. The motion of such atoms can be approximately described as an oscillation
around either of the two potential minima. At low temperatures the localized oscillators are in their
ground states.
Below I use the following notation:
∆ = the energy difference between the localized ground states.
~Ω/ = the zero-point energy.
2
V = the height of the potential barrier.
d = the distance d between the two potential minima.
Now I can write single tunneling system (TS) Hamiltonian as:
1
∆
−∆0
H0 =
,
(1)
2 −∆0 −∆
21
where ∆0 = ~Ω exp [−λ], and λ = d 2mV
.
2
~
In the diagonal form the Hamiltonian is (Appendix I):
1 E
0
H0 =
,
(2)
0 −E
2
p
where E = ∆2 + ∆0 2 .
TS couple to their environment via strain field. The interaction can be accompanied by simultaneous
transition of the tunneling particle from one well to the other, leading to variation of ∆0 . If the
position of the particle is not changed during interaction, then the asymmetry ∆ is varied. These
two processes can formally be described by the introduction of the deformation potentials γ∆0 and
γ∆ :
2γ∆ = δ∆/δe,
(3)
2γ∆0 = δ∆0 /δe,
(4)
and
Where e is the strain field.
The interaction Hamiltonian can be written as:
γ∆
−γ∆0
H1 =
· e,
−γ∆0 −γ∆
(5)
1
in the first base (1), or
1
D 2M
H1 =
· e,
2 2M −D
(6)
in the second base (2) (Appendix I), where
D = 2γ∆
∆
∆0
+ 2γ∆0
,
E
E
(7)
M = γ ∆0
∆
∆0
− γ∆
.
E
E
(8)
and
The total Hamiltonian is:
1
1 E
0
D
H = H0 + H1 =
+
0 −E
2
2 2M
2M
−D
· e.
(9)
This Hamiltonian is identical with the Hamiltonian of a spin S = 1/2 in a magnetic field B,
Hs = −~γB · S,
while S = 12 σ =
By writing
1
2
(10)
(σx , σy , σz ), σi is the appropriate standard Pauli matrix, and B = B0 + B0 .
−~γB0 = (0, 0, E) ,
(11)
−~γB0 = (2M, 0, D) e · S,
(12)
and
I can now treat the TS as spin, precession around Z axis according to magnetic field B0 . There is
another magnetic field, that is oscillating, according to B0 .
The internal energy is related to the expectation values of the spin component hSi i in the magnetic
field as follows:
u = −N ~γB · hS (t)i ,
(13)
where N is the number of spin systems per unit volume.
The dynamical contribution to the stress is given by the strain derivative of the internal energy.
Therefore, according to (11), (12) and (13),
δσ = N (2M hSx i + D hSz i) .
(14)
The spin components are related to the perturbing field through the dynamical susceptibility:
Si = χij (ω) ~γB 0j .
(15)
We will see later that the off-diagonal components are negligible compared to the diagonal components, so I will neglect them for now.
With expressions (12), (14) and (15) I can now obtain the contribution of the TS’s to the dynamical
elastic constant:
δC (ω) = δσ/e = −N 4M 2 χx (ω) + D2 χz (ω) ,
(16)
where the elastic constant is C = C0 + δC and χi ≡ χii .
2
Finally, I can obtain the sound dispersion and absorption (Appendix II):
1
Re [δC (ω)] ,
δv (ω) =
2ρv0
l
−1
(ω) = −
ω
ρv0 3
(17)
Im [δC (ω)] ,
(18)
where ρ is the mass density of the solid and v0 is the sound velocity without interaction of the sound
wave with the TS’s. As we can see, I now need to find the dynamical susceptibility and the spins
expectation values. For this purpose I need to solve the Bloch equation [5], since they describe the
dynamical behavior of the spins.
Bloch equation:
d hSx i
hSx i
,
= γ (hSy i Bz − hSz i By ) −
dt
τ2
(19)
d hSy i
hSy i
= γ (hSz i Bx − hSx i Bz ) −
,
dt
τ2
(20)
hSz i − Szl [Bz (t)]
d hSz i
= γ (hSx i By − hSy i Bx ) −
,
dt
τ1
(21)
where τ1 and τ2 are the longitudinal and the transverse relaxation times respectively and ~γSzl [Bz (t)]
represents the magnetization in the Z direction for instantaneous equilibrium. The quantity ~γSzl [Bz (t)]
expresses the difference in population of the two levels and is given by:
~γBz (t)
1
l
.
(22)
Sz [Bz (t)] = tanh
2
2kT
I will use (22) only till first order in B 0 ,
Szl [Bz (t)] = Sz0 [B0 ] + B 0z (t)
dSz0
.
dt
(23)
Without loss of generality, I will assume magnetic field of the form
B0 (t) = 2B00 cos (ωt) = B00 eiωt + e−iωt .
(24)
I will present hS (t)i as a series:
m=∞
X
hS (t)i = S0 +
Sm exp [−iωm t],
(25)
m=−∞
where ωm = ω · m and S0 = 0, 0, Sz0 (B0 ) .
In order to simplify the writing I introduce the usual combination of the tranverse components of
the spin and the susceptibility:
S± = Sx ± iSy ,
χ± = χx ± iχy .
(26)
Inserting all of the above into Eq. (19)-(21) and with some simplifications (Appendix III), I get:
ωm ± ωr + iτ2 −1 S±,m =
±γBz 00 (S±,m+1 + S±,m−1 ) ∓ γBx 00 Sz0 δm,±1 + (Sz,m+1 + Sz,m−1 ) ,
(27)
3
−iωm + τ1 −1 Sz,m =
1
iγBx 00 {(S+,m+1 + S+,m−1 ) − (S−,m+1 + S−,m−1 )} + Bz 00 τ1 −1
2
dSz0
dB0
δm,±1 ,
(28)
where ωr = E/~ = −γB0 . From those equations it follows that near resonance, i.e., for ω ∼
= ωr ,
the components S+,+1 , S+,−1 , S−,+1 , S−,−1 , and Sz,0 are the most important ones, provided that
ωτ2 >> 1. Neglecting all other component, I get:
±ω + ωr + iτ2−1 S
= −γBx 00 Sz0 + Sz,0 ,
(29)
+, +1
+, −1
±ω + ωr + iτ2−1 S
+, +1
+, −1
= −γBx 00 Sz0 + Sz,0 ,
1
τ1−1 Sz,0 = iγBx 00 (S+,+1 − S−,−1 + S+,−1 − S−,+1 ) .
2
From these equations I obtain the following expression for the susceptibilities:
− Sz0 + Sz,0 ~
− Sz0 + Sz,0 ~
,
.
χ+ (ω) =
χ− (ω) =
ωr + ω + iτ2−1
ωr − ω + iτ2−1
(30)
(31)
(32)
For simplicity and because (S+,−1 − S−,+1 ) are usually much larger than (S+,+1 − S−,−1 ), I will
neglect the latter in the further discussion. For Sz,0 I get:
2
Sz,0 =
(γBx 00 ) τ1 τ2 Sz0
2
1 + (γBx 00 ) τ1 τ2 + (ω − ωr )2 τ2 2
.
(33)
As can be seen in Eq. (27)-(28), as well as in Eq. (29)-(31) and (33), the first order components
of S± is not lineary dependent on Bz 00 and is lineary dependent on Bx 00 . This is the reason of
neglecting the off diagonal components of the susceptibility. In Eq. (33) there is square dependence
on Bx 00 , and no dependence on Bz 00 . So, to find χzz (ω) I need the next component of Sz :
dS 0
−iω + τ1−1 Sz,1 = Bz 00 τ1−1 z .
dB0
(34)
Now I can write χz :
χz (ω) =
dSz0 1 + iωτ1
.
d (~γB0 ) 1 + ω 2 τ12
(35)
Translating all I have found to the strain field ”language” using Eq. (11)-(12), and inserting into
Eq. (16) and (17)-(18), I get:
"
#
2N M 2 ωτ2 0
1
1
−1
lres = −
Sz + Sz,0
−
,
(36)
~ρv03
1 + (ω − ωr )2 τ22 1 + (ω + ωr )2 τ22
δvres
"
#
N M 2 τ2 2 0
ω − ωr
ω + ωr
=−
Sz + Sz,0
−
,
~ρv0
1 + (ω − ωr )2 τ22 1 + (ω + ωr )2 τ22
−1
lrel
=
N D2 dSz0
ω 2 τ1
,
ρv03 d (~γB0 ) 1 + ω 2 τ12
(37)
(38)
4
δvrel = −
N D2 dSz0
1
.
2ρv0 d (~γB0 ) 1 + ω 2 τ12
(39)
Here:
Sz0
1
= − tanh
2
E
2kT
,
(40)
dSz0
exp [E/2kT ]
=
,
d (~γB0 )
kT (1 + exp [E/2kT ])2
Sz,0
(41)
−M 2 e2 τ1 τ2 πSz0 ~2
0
= q
f τ2 ,
τ2 1 + M 2 e2 τ1 τ2 π ~2
0)
(42)
τ2 0
1
π 1+τ2 02 (ω−ωr )2
0−1
−1
q
1 + M 2 e2 τ1 τ2 ~2 , is
= τ2
where f (τ2 =
is the line shape function, and τ2
the strain dependent relaxation time.
I split the sound dispersion and absorption into two separate equations each, according to their
nature. One of them, determined by χx , is due to resonant interaction between the sound wave and
the TS. The other one, determined by χz , has the typical form of a relaxation process.
In order to obtain expressions that can be compared with the experimental Rresults I have to integrate
Eq. (36)-(39) over the distribution of the energy splitting of TS: N ⇒ n0 dE. I will make the
calculation with the assumption of constant distribution as first suggested. At the end of the paper
I will point out some differences that were suggested in a later period.
I will start with Eq. (36). Thanks to the resonant nature of the line shape function, if ω ≈ ωr I can
treat it as a delta function, and the integration will give:
πn0 M 2
~ω
ω
−1
p
lres =
tanh
,
(43)
2kT
ρv03
1 + J/Jc
~2 ρv 3
where J ≡ 12 ρv03 e2 is the definition of acoustic intensity and Jc ≡ 2M 2 τ10τ2 .
The resonant absorption of a sound wave is proportional to the difference in population of the
lower and the upper levels. In thermal equilibrium the population difference is proportional to
~ω
, as in Eq. (43). As the acoustic intensity increases, the upper level becomes more and
tanh 2kT
more populated. The reason for this is that as the more phonons interact with the TS, the TS
doesn’t have sufficient time for relaxation. In Eq. (43) this fact is taken into account through the
factor √ 1
. When the intensity of the sound wave is high above Jc the population difference
1+J/Jc
equals zero, and the attenuation approaches zero. This is called saturation and reflects the dynamic
equilibrium between excitation and recombination processes. The attenuation is also proportional
to M 2 as could been expected from FGR and to n0 the density of states.
The integration of Eq. (38) is much more complicated. I need to integrate over all energies and for
this I need the explicit structure of τ1−1 [6]:
!
2
2
M
2M
E3
E
t
−1
l
τ1 =
coth
,
(44)
5 + v5
2πρ~4
2kT
v0,l
0,t
where the sub-labels l and t stand for longitudinal and transverse respectively.
In general, the integration can’t be done analytically, but it can be done in some limits. In the limit
of ωτ1 >> 1, the integration will give (Appendix IV):
!
2
3 T 3 3!ζ (4, 0.5)
2
M
k
2M
t
−1
l
lrel
=
n0 D 2
,
(45)
5 + v5
32πρ2 ~4 v03
v0,l
0,t
5
where ζ (4, 0.5) is the Hurwitz zeta function and 3!ζ (4, 0.5) ∼
= 97.38.
The reason for this absorption is that the sound wave modulates the level splitting of the TS.
Because of the finite value of τ1 the response of the TS is delayed with respect to the strain of
the sound wave, resulting in energy dissipation and hence attenuation. That for the relaxation
absorption proportional to the square of the energy splitting modulation, D, and to M 2 through
τ1 .
As can be seen in Eq. (45) there is no saturation in this process. Thus the absorption at higher
power levels is due to this process and not due to the resonant absorption. Two more differences
between the two processes are the dependence on ω and on T . From the dependence on T , and if
we recall that we are dealing with low temperatures, we can say that the resonant process is the
dominant one, while it is not saturated.
I will discuss the other limit, i.e. the limit ωτ1 << 1, after presenting the Tunneling Model (TM)
distribution.
The influence of the resonant absorption on the velocity of sound is described by Eq. (37). In
contrast to (36), (37) does not show the resonant behavior and therefore the integration will be
over all E. For simplicity I will calculate the integral for small acoustic intensities, where Sz,0 can
be neglected. I get for ~ω << kT (Appendix V):
∆v v (T ) − v (T0 ) n0 M 2
T
=
=
ln
,
(46)
2
v res
v (T0 )
T
ρv
0
0
res
where T0 is an arbitrary reference temperature.
As can be seen in Eq. (46), the resonant variation of velocity is independent of ω. This is surprising
because of the strong dependence of the resonant absorption, (43), on ω.
For ~ω ≥ kT the resonant variation of velocity becomes negligible, leading to a constant velocity at
T = 0.
As in the integration of Eq. (38), the integration of Eq.(39) can be calculated
analytically only
∝ T 26 . Because of the
in certain limits. In the limit of ωτ1 >> 1, the integration will give δv
v rel
ω
fact that this limit means very low Temperature, the resonant variation of velocity will mask the
relaxation variation of velocity. For the other limit, i.e. the limit ωτ1 << 1, I will present the TM
distribution.
In the TM the quantities ∆ and λ assumed to be independent of each other and to have a constant
distribution:
P (∆, λ) d∆dλ = P̄ d∆dλ,
(47)
where P̄ .
Another assumption is that γ∆ >> γ∆0 , so I can write
D = 2γ
∆
,
E
(48)
and
M = −γ
∆0
,
E
(49)
where γ ≡ γ∆ .
I can rewrite Eq. (47) using new variables E and u =
P̄
P (E, u) dEdu = √
dEdu.
u 1 − u2
∆0
E :
(50)
6
The main difference between the TM picture to the old one lies in the fact that in the first the
relaxation time τ1 shows a distribution even if the energy splitting E is kept constant. For ∆0 = E
the relaxation time has a minimum value τ1,m and with the decreasing value of ∆E0 the relaxation
time increases and tend to infinity.
Now I need to address to the changes if I take into account this distribution. A minor alteration is
required for the resonant interaction: in Eqs. (43) and (46) I replace n0 by P̄ and M by γ.
In the relaxation interaction the situation is not that simple. As long as ωτ1,m ≥ 1 holds, only a
slight change is necessary: the second integration due to the u distribution, leads to factor 3 in the
denominator of Eq. (45). But in the limit of ωτ1,m << 1, after some complicated calculation I get
(Appendix VI):
P̄ γ 2 πω
l−1 rel =
,
2ρv 3
(51)
δv 3 P̄ γ 2
T
=−
ln
.
2
v rel
2 ρv
T0
(52)
and
Recalling Eq. (46), the total variation of velocity in the limit of ωτ1,m << 1 is:
δv δv 1 P̄ γ 2
T
δv =
+
=−
ln
.
2
v total
v res
v rel
2 ρv
T0
(53)
The only problem is that almost all attempts to verify Ex. (53) experimentally failed. My next
goal is to try and apply a new model [7] on the calculation of the relaxation variation of velocity.
This calculation is beyond the scope of this project.
7
Appendix I
The following derivations can be found in the literature [8].
I wish to get the Hamiltonian to diagonal form, and I will do this with the so called ”Bloch sphere
picture”. The Hamiltonian:
1
1
∆
−∆0
H0 =
= (∆ · σz − ∆0 · σx ) = Ω · S,
(1.1)
2 −∆0 −∆
2
where S = 12 σ, and σ are the standard Pauli matrixes.
The evolution operator is:
U (t) = e−itH0 = e−itΩ·S = R (Φ (t)) ,
(1.2)
where Φ (t) = Ωt. As can be seen,
p the evolution is just a rotation around some axis. It is clear that
the oscillation frequency is Ω = ∆2 + ∆0 2 and hence the energy splitting equal to this frequency,
i.e., the diagonal Hamiltonian is:
1 E
0
,
(1.3)
H0 =
0 −E
2
p
where E = Ω = ∆2 + ∆0 2 .
0
The precession axis is tilted relative to the z axis with the angle θ = arctan −∆
∆ . Accordingly the
eigenstates can be obtained by rotating the old states at an angle θ around the y axis. I get:
 q
 q


E+∆
E−∆
θ
θ
−
cos 2 − sin 2
=  q 2E  ,
=  q 2E  .
|+i =
|−i =
(1.4)
E−∆
E+∆
sin 2θ
cos 2θ
2E
2E
With the eigenstates I can rotate every matrix from the old base to the new one:
 q
q

q

 q
E+∆
E−∆
E+∆
E−∆
−
γ∆
−γ∆0  q 2E
q 2E 
q 2E  e
H1 =  q 2E
E−∆
E+∆
E−∆
E+∆
−γ
−γ
∆0
∆
−
.
2E
2E
2E
2E
∆0
∆0
∆
∆
2γ
+
2γ
2γ
−
2γ
D
2M
∆E
∆0 E
∆0 E
∆ E
= 12
e = 12
e
∆0
∆0
∆
2M −D
2γ∆0 ∆
−
2γ
−2γ
−
2γ
∆
∆
∆
0 E
E
E
E
8
(1.5)
Appendix II
The dynamical elastic constant is
C = C0 + δC,
(2.1)
whre δC
hqis ithe contribution of the two level systems, and C0 >> δC. The sound velocity is
C
v = Re
ρ , and to first order:
v0 +
hq i
C
q ρ
ρ
1
2ρ
C0 ·
v = Re
∼
= Re
hq
C0
ρ
+
Re [δC] = v0 +
q
√ 1 d C
ρ dC δC=0
i q
i
h q
· δC = Cρ0 + Re 12 ρC1 0 · δC =
1
2ρv0 Re [δC]
.
(2.2)
From (2.2) the sound dispersion is:
δv = v − v0 =
1
Re [δC] .
2ρv0
(2.3)
The sound absorption is defined to be where the intensity or power of the field decays to e−1 of its
surface value. The intensity of the sound wave is proportional to the square of the strain field. If I
will present the strain field as an exponent series I get:
I ∝ e2 ∝ exp [2i (kx − ωt)] = exp [2i (Re [k] · x − ωt) − 2Im [k] · x] ,
(2.4)
and so:
l−1 = 2Im [k] .
(2.5)
Again, to first order:
q
h q i
√ dC − 12 ρ
ρ
= 2Im [k] = 2Im v = 2Im ω C = 2Im ω C0 + ω ρ dC · δC =
δC=0
.
3
i
h
ρ 2
− 23
1 √
ω
ω
2Im − 2 ω ρC0 · δC = − ρ C0
· Im [δC] = − v0 3 ρ · Im [δC]
l−1
ω
9
(2.6)
Appendix III
I will start with Bloch equation:
d hSx i
hSx i
,
= γ (hSy i Bz − hSz i By ) −
dt
τ2
(3.1)
hSy i
d hSy i
= γ (hSz i Bx − hSx i Bz ) −
,
dt
τ2
(3.2)
d hSz i
hSz i − Szl [Bz (t)]
.
= γ (hSx i By − hSy i Bx ) −
dt
τ1
(3.3)
Recall the combination of the transverse component of the spin: S± = Sx ± iSy , I combine Eq.
(3.1) and (3.2) the same way and get:
d hS± i
hS± i
= γ (hS∓ i Bz − hSz i (By ∓ iBx )) −
.
dt
τ2
(3.4)
Inserting to (3.3) the combination of the transverse component of the spin i get:
l
dhSz i
hS+ i+hS− i
hS+ i−hS− i
=
γ
B
−
Bx − hSz i−Sτz1[Bz (t)] =
y
dt
2
2i
.
iBx −By
By +iBx
hSz i−Szl [Bz (t)]
hS
i
−
hS
i
−
γ
+
−
2
2
τ1
(3.5)
Inserting
B = (0, 0, B0 ) + B 0x (t) , 0, B 0z (t) = (0, 0, B0 ) + B 00x , 0, B 00z eiωt + e−iωt ,
(3.6)
m=∞
m=∞
X
X
hS (t)i = 0, 0, Sz0 (B0 ) +
Sm exp [−imω · t] = S0 +
Sm exp [−iωm t],
(3.7)
and
m=−∞
m=−∞
to Eq. (3.4) and (3.5) I get:
m=∞
P
−i
ωm S±,m exp [−iωm t] = γ
m=−∞
m=−∞
∓iγ Sz0 +
−i
m=∞
P
m=∞
P
m=−∞
S∓,m exp [−iωm t] · B0 + B 00z eiωt + e−iωt
m=∞
P
m=∞
P
Sz,m exp [−iωm t] · B 00x eiωt + e−iωt −
ωm Sz,m exp [−iωm t] =
S+,m exp [−iωm t] −
iγ 00
2B x
m=∞
P
,(3.8)
S∓,m exp[−iωm t]
m=−∞
τ2
m=−∞
m=−∞
m=∞
P
eiωt + e−iωt ·
S−,m exp [−iωm t] −
m=∞
P
m=−∞
m=−∞
dS 0
Sz,m exp[−iωm t]−B 0z (t) dBz
τ1
By comparing the exponent coefficient I finally get:
00
−1 S
ωm ∓ γB
0 + iτ2
±,m = ±γBz (S
±,m+1 + S±,m−1 )
,
00
0
∓γBx Sz δm,±1 + (Sz,m+1 + Sz,m−1 )
00
1
−1 S
−iωm + τ
1
z,m = 2 iγBx {(S+,m+1 + S+,m−1 ) − (S−,m+1 + S−,m−1 )} .
0
dSz
+Bz 00 τ1 −1 dB
δm,±1
0
10
. (3.9)
0
(3.10)
(3.11)
Appendix IV
Starting with
N D2 dSz0
ω 2 τ1
,
3
ρv0 d (~γB0 ) 1 + ω 2 τ12
−1
lrel
=
(4.1)
Setting the limit ωτ1 >> 1 I get:
−1
lrel
E
exp kT
ω 2 τ1 ∼ N D2 dSz0
1
N D2
N D2 dSz0
−1
=
=
=
τ1 .
ρv03 d (~γB0 ) 1 + ω 2 τ12
ρv03 d (~γB0 ) τ1
ρv03 kT 1 + exp E 2
kT
(4.2)
Recall
τ1−1
Ml2 2Mt2
5 + v5
v0,l
0,t
=
!
E3
coth
2πρ~4
E
2kT
,
(4.3)
I get:
−1
lrel
E
exp[ kT
]
N D2
3
ρv0 kT (1+exp[ E ])2
kT
=
N D2
ρv03
N D2
ρv03
Ml2
5
v0,l
Ml2
5
v0,l
+
+
2Mt2
5
v0,t
2Mt2
5
v0,t
Ml2
5
v0,l
+
2Mt2
5
v0,t
E3
2πρ~4
coth
E
2kT
=
E3
1
2πρ~4 kT (exp[ E ]+exp[− E ])(exp[ E ]−exp[− E ])
2kT
2kT
2kT
2kT
2
1 (kT ) 2E 3
kT
2πρ~4 8
exp[−
0,l
(4.4)
1
E
kT
2E
](exp[ kT
]−1)
R
Now I can integrate by N → n0 dE:
2E 3
2 R
Ml2
( kT
)
2Mt2
−1
n0 D2
1 (kT )
lrel = ρv3
+
dE =
5
5
4
2E
E
8
2πρ~
v0,l
v0,t
exp[− kT ](exp[ kT
]−1)
0
x
3 R
Ml2
2Mt2
n0 D2
1 (kT )
x3 e 2
x −1) dx =
5 + v5
4
16
(e
2πρ~
ρv03
v0,l
0,t
2
R x3 e x2
2
k3 T 3
2 Ml + 2Mt
n
D
0
3
5
5
2
4
(ex −1) dx
32πρ ~ v
v
v
0
= .
(4.5)
0,t
R 3 e x2
where 3!1 (exx −1)
dx = ζ (4, 0.5) is the Hurwitz zeta function.
Finally I get:
!
2
2
3 T 3 3!ζ (4, 0.5)
M
k
2M
t
−1
l
n0 D 2
.
lrel
=
5 + v5
32πρ2 ~4 v03
v0,l
0,t
11
(4.6)
Appendix V
For small acoustic intensities Sz,0 can be neglected in Eq. (32) and I get:
"
#
Sz0
1
1
+
.
χx = −
2~ ωr + ω + iτ2−1
ωr − ω + iτ2−1
(5.1)
Using a constant density of states n0 up to certain maximum value Emax of the energy splitting I
can now integrate:
ER
ER
E
max
max
tanh( 2kT
)
1
1
χ̄x =
χx dE =
+ E − ω+iτ −1 dE =
E
4~
+ ω+iτ2−1 )
(
2 )
~ (
0
" 0
#~
ER
E
max
tanh( 2kT
)
1
1
1
E
+
d
=
−1
−1
2
2
2kT
~(ω+iτ2 )
~(ω+iτ2 )
E
E
,
(5.2)
0
+
−
2kT
2kT
2kT
2kT
Emax
h 2kT
i
R
~τ2−1
Emax
1
x
~ω
!2 dx = − 1 Ψ 1 +
tanh (x)
+
−
ln
2
2
2
2πkT
2πikT
2kT
~(ω+iτ2−1 )
0
x2 −
2kT
where Ψ represents the digamma function.
Recall
Z
1
2n0 M 2
δvres =
Re
δC(ω)res dE = −
Re [χ̄x (ω)] .
2ρv0
ρv0
With the notion that
δvres
1
2
(5.3)
~τ −1
2
>> 2πkT
for all experiments, I get:
n0 M 2
1
~ω
Emax
=
Re Ψ
+
− ln
,
ρv0
2 2πikT
2kT
(5.4)
and for the limit ~ω << kT :
v (T ) − v (T0 ) =
n0 M 2
1
~ω
Re Ψ
− ln
,
ρv0
2
kT
or for arbitrary temperature T0 ,
v (T ) − v (T0 ) n0 M 2
T
= ρv 2 ln T0 .
v0
0
res
(5.5)
(5.6)
12
Appendix VI
I will calculate the relaxation attenuation and variation of velocity together. For this I need the
relaxation dynamical elastic constant:
E
N D2 sec h 21 kT
δC = −
.
(6.1)
4kT 1 − iωτ1
I need to integrate over E and u. Because of the limit ωτ1,m << 1, the confines of the u integration
must have a cut off at some umin .
E
E
R∞ R1 P̄ D2 sec h( 12 kT
R∞ R1 P̄ γ 2 (1−u2 ) sec h( 12 kT
) √1
) √1
δC = −
dEdu
=
−
4kT
1−iωτ1 u 1−u2
kT
1−iωτ1 u 1−u2 dEdu
0 umin
0 umin
√
∞
1
R
R
u2 ≡ r
1−r
E
E
=
= −P̄ γ 2 sec h 21 kT
= [τ1,m = r · τ1 ] =
drd 21 kT
.
r(1−iωτ
)
1
2u · du = dr
rmin
0
R1 √1−r
R1 √1−r
R∞
R∞
E
1 E
E
1 E
2
−P̄ γ 2 sec h 12 kT
sec h 21 kT
r−iωτ1,m drd 2 kT = [ωτ1,m ≡ µ] = −P̄ γ
r−iµ drd 2 kT
rmin
0
rmin
0
I can develop the numerator of the first integration for better understanding:
h
i
R1 1
R1 r
R1 rn
R1 √1−r
1
r
dr
=
dr
−
dr
+
...
+
a
dr
+
...
=
s
≡
n
r−iµ
r−iµ
2
r−iµ
r−iµ
µ =
rmin
rmin
1
µ
R
rmin
µ
rmin
rmin
1
µ
1
s−i ds
− 12 µ
1
µ
s
s−i ds
R
rmin
µ
+ ... + an µn
R
rmin
µ
.
sn
s−i ds
+ ...
Recall that µ = ωτ1,m << 1, so µ1 → ∞. Furthermore, I assume that rmin is so small that even
rmin
when µ << 1 I still have rmin
µ << 1, or, µ → 0. I get:
1
In = an
Rµ
µn
rmin
µ
sn
s−i ds
∼
= a n µn
R∞
0
sn
s−i ds
Now I can neglect all terms except for the first, and to get:
R1 1
R1 √1−r
1
∼
∼
dr
=
r−iµ
r−iµ dr = ln (r − iµ)|rmin = ln (1 − iµ) − ln (rmin − iµ) =
rmin
rmin
ln (1) − ln (−iµ) = − ln (µ) − ln (−i) = − ln (µ) +
And to the original integral:
R∞
E
E
− ln (ωτ1,m ) + i π2 d 12 kT
=
δC = −P̄ γ 2 sec h 12 kT
0



R∞
4
 − 3 ln (T ) − i π +
P̄ γ 2 ln (ω) + ln  Mπρ~
sec h
2
2
2M 2
4k3
=
P̄ γ 2
.
i π2
t
l +
2
v2
v0.t
0.l
ln (a) − 3 ln (T ) −
i π2
+
0
R∞
sec h (x) ln
x−3
−1
· coth (x)

1 E
2 kT
ln
1 E
2 kT
−3
· coth
1 E
2 kT
−1 d
dx
0
Inserting the above to Eqs. (17) and (18) I finely got:
P̄ γ 2 πω
l−1 rel =
,
2ρv 3
δv v
=
rel
1 P̄ γ 2
2 ρv 2

ln (a) − 3 ln (T ) − i π +
2
(6.2)
Z∞

sec h (x) ln x−3 · coth (x)−1 dx .
(6.3)
0
For arbitrary reference temperature T0 I can get:
δv 3 P̄ γ 2
T
=−
ln
.
v rel
2 ρv 2
T0
13
(6.4)
1 E
2 kT

References
[1] Anderson, P.W., Halperin, B.I. and Varma, C.M. (1972). Phil. Mag. (1).
[2] Hunklinger, S. and Arnold, W. (1976).Phys. Acou. (Ed. Mason, W.P. and Thurston, R.N.),Vol.
XII, p. 155. New York: Academic Press.
[3] Hunklinger, S. and Raychaudhuri, A.K. (1986).Prog. Low. Temp. Phys. 9, 265.
[4] Doussineau, P., Frnois, C., Leisure, R.G., Levelut, A. and Prieur, J.Y. (1980). J. Phys., 41.
[5] Kittel, C. (1996). Introduction to Solid State Physics, 7th edition, p. 471-493. New York: John
Wiley and sons.
[6] Jackle, J. (1972). Z. Phys., 257, 212.
[7] Schechte, M., and Stamp, P.C.E. arXiv:0910.1283.
[8] Cohen, D. arXiv:quant-ph/0605180.
14