thermal properties of crystal lattices remember the concept ofcapacity heat amount of heat...
TRANSCRIPT
Thermal Properties of Crystal Lattices
dTvCQ
Remember the concept of capacity heat
Amount of heat Temperature increase
Heat capacity at constant volume
pdvQdU
Internal energy
constant volumedV=0
0
vT
UvC
How to calculate the vibrational energy of a crystal ?
Classical approach
2
2
1xDEpot
x
222
1
2
2rm
m
pE
•In the qm descriptionapproach of independentoscillators with singlefrequency is calledEinstein model
•In classical approachdetails of Epot irrelevant
dTvCdUQ
Click for details about differentials
Let us calculate the thermal average of the vibrational energy NEU
Classical: )r,p(EE
defines state (point in phase space)
Energy can change continuously to arbitrary value
rdpd)r,p(E
e
rdpd)r,p(E
e)r,p(E
E
33
33
Average thermal energyof one oscillator
Boltzman factor, whereTBk
:1
rdpd
rmm
p
e
rdpd
rmm
p
ermm
p
E
3321
2
3321
22221
2
2
22
2
22
2
pdm
p
e
pdm
p
em
p
32
322
2
2
2
rdrm
e
rdrm
erm
321
321
2221
22
22
TBk2
3 TBk
2
3 TBk3
zdpydpxdpm
ppp
epdm
p
e
zyx
232
2222
3
2
2
dpm
p
e
32 2
dxxem
pm
x2
3
2
m
zdpydpm
pp
exdpmp
emxp
pdm
p
em
pzyx
222
2332
2
2222
2
2
223
22
dpmp
edpmp
e
mm 22
3
mm 232
3
pdm
p
e
pdm
p
em
p
32
322
2
2
2
mm 232
3
32
m
343
m
m
TBk2
31
2
3
The same applies for the second integral
rdrm
e
rdrm
erm
32
1
32
122
2
1
22
22
TBk2
3
TBkE 3 TBkNU 3
Classical value of the thermal average of the vibrational energy TBkNU 3
Understanding in the framework of: Theorem of equipartition of energy
every degree of freedom TBk2
1
Example: diatomic molecule
TBkTBkTBkU
2
1
2
1
2
3
Only rotations relevant where moment of inertia 0J
Vibration involveskinetic+pot. energy
TBk
vT
UvC
with nRBkNvC 33
# of moles
Gas constant R=kB NA = 8.3145J/(mol K)
= n 24.94J/(mol K)
Classical limit
Solid: N atoms 3N vibrational modes TBkNU 3
Requires quantum mechanics
1D Quantum mechanical harmonic oscillator
)x(E)x(xmdx
)x(d
m
222
12
2
2
2Schrödinger equation:
Solution: Quantized energyn
1E (n )
2
E
n nE (x)
x
2
n nE (x)
x
0
1E
2
1
1E (1 )
2
5
1E (5 )
2
CORRESPONDENCE PRINCIPLE
Large quantum numbers: correspondence between qm an classical system
2
20(x)
x Classical point of reversial
Classical probability density
Einstein model: N independent 3D harmonic oscillators
x y zn n nE(3D) E E E 3 E
n
n
En
n 0
E
n 0
E eE
e
Energies labeled by discrete quantum number n
Boltzman factor weighting everyenergy value TBk
:1
n
n
E
nE
n 0
ep
e
Probability to find oscillator in state n
nn 0
p 1
Quantum mechanical thermal average of the vibrational energy N)D3(EU
1(n )
2
n 0n 1
(n )2
n 0
1(n ) e
2E
e
Let us introduce
1(n )
2
n 0
Z e
partition function
n
1 ZE
Z
therefore calculate Z
n/ 2
n 0
e e
1
(n )2
n 0
Z e
/ 2 1
e1 e
/ 2 1Z e
1 e
/ 2 / 2Z 1 1e e
1 e 1 e
/ 2 2/ 2e
e 1 e e2 1 e
1 ZE
Z
e
2 1 e
1 1E
2 e 1
1E n
2
where1
ne 1
Bose-Einsteindistribution
With NEU 3 and 1E n
2
1n
e 1
In the Einstein model where E for all oscillators
2
3
1
3 ETBk/
E N
e
NU
zero point energy
vT
UvC
Heat capacity:
2
2
13
TBk/E
TBk/EBE
Bve
eTk/kNC
Note: typing error inEq.(2-57) in J.S. Blakemore, p123
Classical limit
2
2
13
TBk/E
TBk/EBE
Bve
eTk/kNC
1 for
EBTk TBk/E
B
E eTk
2
for
EBTk
•good news: Einstein model explains decrease of Cv for T->0
•bad news: Experiments show
3TCv for T->0
0 1 2 30,0
0,5
1,0
CV
/3N
k B
T/TE
Assumption that all modes have the same frequency E unrealistic
refinement
Debye Model
We know already: )k(
wave vector k labels particular phonon mode
1)
2)
3) total # of modes = # of translational degrees of freedom
3Nmodes in 3 dimensions N modes in 1 dimension
Let us remind to dispersion relation of monatomic linear chain
N atoms N phonon modes
labeled by equidistant k values within the1st Brillouin zone of width
a
2
distance between adjacent k-values
LaNdk
221
Ldk
2
.constE
A more detailed look to the origin of k-quantization
D a
Quantization is always the result of the boundary conditions
Let’s consider periodic boundary conditions
Atom position n characterized by
After N lattice constants a we end up again at atom n
( ( ) )i k n N a tn Nu A e
( )i kna tnu A e
( )i kna tnu A e
1ikNae 2kNa n
2k n
Na
2 2k n and k
L L
In 3D we have: ,L
dkx
x
2
yy L
dk
2
andz
z Ldk
2
One phonon mode occupies k-space volume
VLLL
dkdkdkkdzyx
zyx
33 2222
Volume of the crystal
How to calculate the # of modes in a given frequency interval , d ?
11 d
22 d
k
D( ) (k) Density of states
3
3k
Vd k
2
Blakemore callsit g(), I prefer D()
)k( 1 )k( 2
))k(( 1
11 d))k((
k
vL
vT,1=vT,2=vT
Let us consider dispersion of elastic isotropic medium
d)(Dmax
0
total # of phonon modes In a 3D crystal Nd)(Dmax
30
Particular branch i: kv i
kd)()(
V)(D k
332
dkkkd 23 4here
kv)k()k( i
22
i
k
vk
ki
dv
dk 1
k
ii
kk d
vv)(
)(
V)(D
14
2
2
3 3
2
22 iv
V
Taking into account all 3 acoustic branches
332
2
21
2 TL vv
V)(D
What is the density of states D(ω) good for ?Calculate the internal energy U
00
Ud)T,(n)(DUmax
# of modes in , d
Energy of a mode # of excited phonons )T,(n
temperature independentzero point energy
= phonon energy
D(ω)
00
Ud)T,(n)(DUmax
00
2
332 1
21
2Ud
evv
VU
max
TL
How to determine the cut off frequency max ?also called Debye frequency DDensity of states of Cu
determined from neutron scattering
2)(D
Nd)(DD
30
choose D such that both curves enclose the same area
Nd)(DD
30
332
2
21
2 TL vv
V)(D
3332
921
2 DTL
N
vv
V
00
2
3 1
9Ud
e
NU
max
D
withvT
UvC
de
e
Tk
NC
max
TBk/
TBk/
BDv
02
2
231
9
de
e
Tk
NC
max
TBk/
TBk/
BDv
02
2
231
9
D
energy
BD k/
temperature
D:
Tkx
B
Substitution:
dxTk
d B
Debye temperature
dxe
exTNkC
T/D
x
x
DBv
0
2
43
19
0T dxe
exdx
e
exx
xT/D
x
x
02
4
02
4
11
34
5
12
DBv
TNkC
Click for a table of Debye temperatures