Notes:
• Reminder, Exam: Tues, Oct 13.
§
§
§
In class.
Covers through ch. 5
Open book = Kittel only.
• HW 4: Can turn in next Thursday, Oct. 8. (I will have solutions ready.)
Quantized Modes, shown last time:
− β!ω
⎡1
e k ,ν
E = ∑ !ωk ,ν ⎢ +
− β!ω k ,ν
2
1
−
e
k ,ν
⎣
⎤
⎡1
⎤
=
!
ω
+
n
⎥ ∑ k ,ν ⎢
kν ⎥
2
⎣
⎦
⎦ k ,ν
nk ,ν =
1
e
β!ω k ,ν
−1
• Planck formula, sum over lattice normal modes
kT << !ω :
E ≈ !ωk ,ν e
kT >> !ω :
E ≈ kT
− β!ω k ,ν
For each mode; except doesn’t
work for lowest acoustic modes;
Also omitted zero-point term.
for each mode: Classical result.
Quantized Modes, shown last time:
− β!ω
⎡1
e k ,ν
E = ∑ !ωk ,ν ⎢ +
− β!ω k ,ν
2
1
−
e
k ,ν
⎣
⎤
⎡1
⎤
=
!
ω
+
n
⎥ ∑ k ,ν ⎢
kν ⎥
2
⎣
⎦
⎦ k ,ν
nk ,ν =
1
e
β!ω k ,ν
−1
• Planck formula, sum over lattice normal modes
kT << !ω :
E ≈ !ωk ,ν e
− β!ω k ,ν
For each mode; except doesn’t
work for lowest acoustic modes;
Also omitted zero-point term.
Debye model works very well at low T. ω ≅ kc
Specific heat ~ T3
kT >> !ω :
E ≈ kT
for each mode: Classical result.
Dulong-Petit law (as expected by correspondence
principle). 3N total modes:
CV = 3Nk B / V
includes basis
Specific Heat (Debye model):
1
ω
⎡
⎤
E = ∑ ω k,ν ⎢ β ω k,ν
=
D(
ω
)d
ω
β ω
⎥⎦ ∫
e
−1
e
−1
⎣
k,ν
D(ω)/V
∂
C=
∂T
ωD
∫
0
Debye model showed last time. ω D = c 3 6π 2 n ≡ c 3 6π 2 N / V
3ω 2 ω dω
3kB4T 3
= 2 3 3
2 3 β ω
2π c e −1 2π c 
kT << !ω :
∫
0
T
x 4 dx e x
T3
= 9kB n 3
x
2
(e −1)
ΘD
ΘD
∫
0
T
x 4 dx e x
(e x −1)2
∞
x 4 dxe x
4π 4
1 / T → ∞; ∫ x
=
2
(e − 1)
15
0
kT >> !ω :
ΘD
CV = 3Nk B / V
C = (12 / 5)π 4 nk BT 3 / Θ3D ≡ βT 3
classical limit, also easy to show.
Specific Heat, Phonon notes:
3
T
C = 9k B n 3
ΘD
ΘD
(Debye)
∫
0
T
x 4 dxe x
(e x − 1) 2
• Normally at low temperatures two terms observed,
phonons + metallic electrons (for non-magnets).
C ≡ βT 3 + γT
• Phonons “elementary excitations in crystal vacuum ”.
• Zero-point contribution small; detectable.
• Einstein vs. Debye: Atoms as independent SHO’s
(Einstein) applies in some cases (optical modes)
• Scattering by phonons: often is the most important
electrical resistivity contribution.
Crystal Momentum:
o “Crystal momentum” is k-vector we have been using
!
!
(K-vector in text). (Momentum: p = "k )
o Crystal potential mixes states with k, k ± G .
o Result: k can only be specified ± G. “Folding” process
brings k ± G states together. Thus we only need to
define k in a single Brillouin zone.
Free-space translation
QM
result
[Tr! , H ] = 0
[T , H ] = 0
!
R
Discrete translation
!
pr! = constant of motion
!
crystal k = constant of motion
Crystal Momentum:
o Crystal momentum: k-vector folded with states k ± G.
! ! ! !
Laue condition k − k ′ ≡ q = G
!
!
!
!
3
A ∝ ∫ d r n(r ) exp i[k − k ′] ⋅ r
(
)
vol
Lattice symmetry; Fourier
components have wave-vectors G
!
crystal k = conserved
o Crystal potential mixes states with k, k ± G .
!
! !
U (r ) = U (r + R)
! ! !
!
ψ k (r ) = ∑G! CG ei ( k +G )⋅r
Bloch theorem – later chapters.
Anharmonic lattice effects:
► Real vibrational potential not
precisely quadratic; most cases can
be treated as perturbation.
Results:
• Thermal expansion as higher vibrational states become
populated.
• Phonon states not stationary states; finite lifetime.
• Leads to “phonon-phonon scattering”:
energy and crystal momentum conserved.
Anharmonic lattice effects:
Neutron scattering:
•
Neutron energy vs. k comparable to phonon (& magnetic) excitations
•
Used to observe both elastic and inelastic scattering.
Neutron scattering:
Atomic displacements
Elastic scattering attenuation
measured this way – This
vs. T – appendix A;
example ReO3
Debye-Waller Factor.
(Rodriguez et al., J Appl Phys 2009)