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VibrationalVibrational PropertiesProperties
The physical properties of a crystal can be roughly divided into those that are determined by the electrons and those that relate to the movement of the atoms about their equilibrium positions.
This division is qualitatively easy to justify: the motions of atomic nuclei are, due to their high mass, much slower than the motions of the electrons.
If the atoms are displaced from their equilibrium positions, then the electronsadopt a new distribution (with higher total energy). If the initial positions of the nuclei are restored, then the energy is recovered.
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VibrationalVibrational PropertiesProperties
∑∑∑∑∑≠≠
−+
−−
−++=
jiji jiij ji
j
jjjj jj
jj
j j
j
i i
i
rre
RreZ
RR
eZZ
MP
mpH
,
2
,
2
, '
222
21
21
22'
'
'
The Hamiltonian describing a perfect crystal can be written as
ri denotes the position of the ith electronRj is the position of the jth nucleusZj is the atomic number of the nucleus pi and Pj are the momentum operators of the electrons and nuclei.
Obviously the many-particle Hamiltonian can not be solved without a large number of simplifications:1) separate the electrons into two groups: valence and core electrons. The core electrons are assumed to move rigitly with the nuclei.2) adiabatic approximation (Born-Oppenheimer). Since the frequencies of the ionic vibrations are typically < 1013 s-1 while the frequencies of the electronic motion are ≈ 1015
s-1 the electrons can respond to ionic motion almost instantaneously or to the electronsthe ions are stationary. On the other hand, ions can not follow the motion of the electrons and they see a time-averaged adiabatic electronic potential.
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BornBorn--OppenheimerOppenheimer approximationapproximation
),(),()( 0 jiionejiejions RrHRrHRHH δ−++=
Hion is the Hamiltonian describing the ionic motion under the influence of ionicpotentials plus the time-averaged adiabatic electronic potentials.
He is the Hamiltonian for the electrons with the ions frozen in their equilibrium positions (Rj0).
He-ion describes the change in the electronic energy as a result of the displacements (δRj) of the ions from their equilibrium positions. He-ion is knownas the electron-phonon interaction.
By using the Born-Oppenheimer approximation the Hamiltonian can be expressed in three terms:
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HarmonicHarmonic approximationapproximationIn this approach an equation of motion for the ions is obtained by expanding Hionas a function of their displacement (δRj) from their equilibrium positions Rj0:
),....,(),......,( 010'
010 nnoion RRHRRHH δδ+=
H0 is the Hamiltonian of the crystal with all nuclei at their equilibrium positionsH’ is the change in Hion due to displacements of the nuclei by small amount (δR) from the equilibrium positions.
We expand Hion around R10, ..,Rn0. Since Rj0 are the equilibrium positions of the ions, the first-order tems in δRj vanish. In addition, when δRj are identical, the crystal is uniformly displaced and not distorted. Thus, the lowest order terms in the expansion relevant to the vibration of the crystal are the second order termsin δ(Rj-Rk).
If we keep only the terms quadratic in H’, the motion of the nuclei are describedby a collection of simple harmonic oscillators (harmonic approximation).
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Force Force constantsconstants
ukl is the displacement of the ion k in the unit cell l from equilibrium and the matrixΦ(kl,k’l’) contains the force constants describing the interaction between the ionsdenoted (kl) and (k’l’). The force constants contain two parts: the first part is the direct ion-ion interaction due to their Coulomb repulsion, the second is an indirect interaction mediated by the valence electrons. Infact, the motion of one ion causes the electron distribution to change and this rearrangementof the electrons produces a force on the neighboring ions.
The determination of the lattice dynamics described by the above Hamiltonian can be carried out in two steps.First we treat the Hamiltonian classically and solve the equation motion. In thisclassical approximation H describes the energy of a collection of particles executingsmall-amplitude oscillations (normal modes).
''''
2
)'',(21
21)(' lk
lkkl
klkkl ulkkludtdu
MuH ∑ Φ+
=
Within this approximation H’ can be expressed as:
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NormalNormal modesmodes and and phononsphononsIn the second step we quantized the energies of the normal modes. Each quantum of lattice vibration is called phonon.
Instead now we want to determine the vibrations of the normal modes described by the H’.Since the matrix Φ(kl,k’l’) possesses translational symmetry, we expect that the atomic displacements (ukl) which diagonalize the H’(ukl) can be expressed in terms of planewaves similar to the Bloch functions for electrons in a crystal:
)exp(),( 0 tRqiuqu lkkl ωω −⋅=rrrr
ukl is the displacement of the kth ion in the lth unit cell and it can be related to uk0 of a corresponding ion in the unit cell located at the origin by a Bloch wave, where q and ωare, respectively, the wave vector and the frequency of the wave. There is however an important difference between electrons and phonon Bloch waves. Electrons can be anywhere in the crystal, ion positions are discrete and follow the periodicy of the crystal.
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DynamicalDynamical matrixmatrix
[ ]
[ ] 0)(Das written becan eq. the
exp))('0,()(D
as of ansformFourier tr modified-mass a gIntroducin
)exp()0',(
)exp()0',(
0'k'
'2
kk'
2/1'kk'
0'''
02
0'''
=−
⋅−Φ=
Φ
⋅−Φ=
⋅−Φ=
∑
∑
∑
∑
−
kkk
mm
kk
kmlk
kk
kmlk
kok
uq
RqiMMkmkq
uRqikkmuM
uRqikkmuM
δω
ω
r
rrr
rrr
r&&
By substituting ukl in to the Hamiltonian H’ we obtain the equation for uk0(Newton law):
Dkk’(q) is know as the dynamical matrix. The vibrational frequencies ω(q) are found bysolving the secular equation:
The vibrational amplitudes uk0 are obtained by substituting the solutions of the secularequation into the above equation.The degrees of freedom or the number of indipendentwaves is equal to three times the number of atoms N in the crystal (3N).
0)(D '2
kk' =− kkq δωr
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DiatomicDiatomic LinearLinear ChainChainWe are going to discuss the special case of a diatomic linear chain (1D). This case has little in common with a real crystal but it is frequently discussed because ot itssimple mathemathics.We consider a linear chain in which all nearest neighbors are connnected by ideal springs with force constant f. The unit cell contains two atoms of masses M1 and M2.
k and k’ are 1,2 and we are in 1D and m can take only the values n+1, n, and n-1. One thus obtains the following equations:
ff
uuuuM
uuuuM
nn
nn
nn
nn
nn
nn
nnnn
nnn
nnn
nnnn
nnn
nnn
2
0
0
11
22
1,12
12
21
2,11
1,11,1
22221
1222
2211
112,1
2,1111
=Φ=Φ
−=Φ=Φ=Φ=Φ
=Φ+Φ+Φ+
=Φ+Φ+Φ+
+−
++
−−
&&
&&
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DiatomicDiatomic LinearLinear ChainChain
By inserting this plane wave solution in the system of equations we have:
( )
( ) 0211
0112
22
11
1
21
21
211
2
1
=
−++−
=+−
− −
uMfue
MMf
ueMM
fuMf
qa
qa
ω
ω
[ ] 2,1 )(exp)(),( =−⋅= atnaqiququn ωω αα
0)2(0)2(
1,11222
2,12111
=−++
=−−+
+
−
nnnn
nnnn
uuufuMuuufuM
&&
&&
We use the plane wave solution:
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DiatomicDiatomic LinearLinear ChainChainThe dynamical matrix Dkk’(q) is therefore
2 )1(
)1( 2
221
211
+−
+−
−
−
Mfe
MMf
eMMf
Mf
iqa
iqa
Setting the determinant of the system equal to zero, leads to the dispersion relation (ω versus q):
2/1
2
21
2
2121
2
2sin411 11
−
+±
+=
qaMMMM
fMM
fω
This dispersion relation is clearly periodic in q witha period given by:
aqqa ππ 2 ,
2==
The solution must satisfy the condition:m2rG and )()(
have weefurthermor and )()(
πωω
ωω
=⋅=−
+=rrrr
rrr
Gqq
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DiatomicDiatomic LinearLinear ChainChain
For example we show the twodispersion relations (branches) for a mass ratio ofM1/M2=5 The branch that goes to zero at small q is know as a the acoustic branch(for small q, we have ω ∝q).The upper branch is called opticalbranch. 1/µ= 1/M1+1/M2For the diatomic linear chain we onlyallowed displacements ot the atomsalong the direction of the chain(longitudinal waves). For a 3D crystalthere are also transverse waves
acoustic branch
optical branch
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Taking these facts together we can represent ω(q) in the region2
0 Gq ≤≤
WaveWave solutionssolutions at at q=0q=0 )(exp),( )(exp),(
22
11
tnaqiAqutnaqiAqu
ωωωω
−⋅=−⋅=
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For q<<π/a ω∝qa A1=A2 (acoustic mode)ω=(2f/µ)1/2 A1=-A2 (optical mode)
The first belongs to the acoustic mode and describes a motion in which the two ions in the cell move in phase with one another. The minus belongs to the high-frequencyoptical mode, and describes a motion in which the two atoms in the cell are 1800 out of phase.
ω=(2f/M1)1/2 A1=-A2ω=(2f/M2)1/2 A1=A2
When q=π/a, the motions in neighboring cells are 1800 out of phase. In each case, only one type of spring is stretched.
WaveWave solutionssolutions at q=at q=ππ/a/a
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3D 3D phononphonon dispersiondispersion curvescurves forfor SiSiFor Si ( 2 atoms per primitive cell) in 3D there are six phonon branches (three acousticbranches and three optical branches). Along high-symmetry directions the branches can beclassified as transverse and longitudinal branches.A clear separation of the vibrational modes into longitudinal and transverse is only possible in certain symmetry directions of the crystal.
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3D 3D phononphonon dispersiondispersion curvescurves forfor GaAsGaAsFig. 3.2
For GaAs and other zincblende-type semiconductors, the LO phonon has higher energy than TO phonons. At exactly at the zone center, the TO and LO phonons in the zinc-blend crystal are notdegenerate because of the cubic symmetry of the zinc-blend structure. The difference from Si comes into the partially ionic nature of the bonding in zinc-blend crystals.
TA
LA
L0
TO
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PhononPhonon SpectroscopySpectroscopyThe momentum and energy conservation laws that govern the interaction of light with the crystal can be used to advantage for the experimental determination of phonon dispersioncurves.
The two most commonly employed electromagnetic techniques, each with its limitations, are the inelastic scattering of X rays and of visible light. Inelastic scattering of thermal neutronsand atoms can also be used for phonon determination.
X-Ray photons can be inelastically scattered with emission and/or absorption of one or more phonons. However the change in energy is extremely difficult to measure:⇒X ray energy is several 103eV while phonon energy is several 10-3eV.
Usually one can only measure the total scattered radiation of all frequencies, as a function of scattering angle, and determining the angles away from those satisfying the Bragg condition.
On the contrary, if photons of visible light are scattered with the emission or absorption of photons, the energy shifts are still very small but they can be measured.
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PhononPhonon SpectroscopySpectroscopyThe inelastic scattering of light in and around the visible region is known either asRaman scattering, or, when the interaction is with the acoustic waves, as Brillouinscattering.
The scattering is produced by the polarization of the atoms in the radiation field and the subsequent emission of the dipole radiation.In the frequency range of visible light the maximum wave vector transfer is:
i.e., ∝1/1000 of a reciprocal lattice vector (G≈Å-1). Thus the Raman scattering can only be used to study lattice vibrations near to the center of the Brillouin zone (q=0).
This is not the case for inelastic scattering of X-rays (E=104 eV, λ=1 Å) were the momentum transfert is ∝1/λ≈G= Å-1 (reciprocal vector).
130 10242 −− Α≈= &xk
λπ
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Scattering Scattering AmplitudesAmplitudesThe solution of the equations of motion of the lattice have the form of plane waves. In analogy tothe wave-particle dualism of quantum mechanics, one might ask whether lattice waves can alsobe interpreted as particles.Any such particle aspect would manifest itself, for example, in the interactions with otherparticles, i.e. electrons, atoms, neutrons, and photons (light).
In the diffraction experiment from a crystal one can make use of X-ray, electrons, neutrons and atoms. The scattering amplitude (from diffraction theory) scattered from a rigid lattice ρ(r) integrating over the entire scattering region is:
To simplify the matemathics we consider a primitive lattice and assume the atoms to be point-likescattering centers situated at the time-dependent positions rn(t). Thus we write:
∫ −−∝ dreretA rKKitiB
o )( 0)()( ρω
∑∝
−∝
−
∑
n
trKKitiB
nn
neeA
trrtr
)()( 00
))(())((
ω
δρ
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InelasticInelastic scatteringscatteringWe separate the rn(t) into the lattice vector rn and a displacement from the lattice site sn(t)
with this one obtains
for small displacement s(t) we can make the expansion
we obtain the elastic scattering espression:
Thus for scattered waves, ω differs from that of the primary waves (ω0) by exactly the frequency of the lattice waves.
)()( tsrtr nnn +=
ti
n
tsKKirKKi eeeA nn 000 )()()( ω∑ −−∝
[ ][ ]tqrqi
n
ti
nn
rKKi
n
n
uets
etsKKieA
)(
0)(
)(
)()(1 00
ω
ω
−⋅±
−
∝
⋅−+∝ ∑plane waves
[ ] tqi
n
rqKKi euKKieA n ))((0
)( 00 )( ωω ±±− −∝ ∑
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MomentumMomentum and and energyenergy conservationconservationThe scattered waves must obey a further conditions relating to the wave vector:
by multipying both equation by
one sees that the second of these classical equations can be interpreted quantum mechanically as the conservation of energy.
The plus sign corresponds to the excitation of a lattice wave by the primary wave; the minus signapply to the process in which a lattice wave loses energy to the primary wave (this occurs onlywhen the lattice wave has sufficient energy).
The first condition can be interpreted as the conservation of momentum if it is assumed that the q is the momentum of the lattice wave.In this sense we can regard the lattice waves as a particle (phonon) with a momentum (defined within an arbitrary reciprocal lattice vector).
)(0
0
qGqKK
ωωω ±==±−
0)(0
0
0
=±−=−±−
qGqKK
ωωω hhh
hhhh
h
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RamanRaman SpectroscopySpectroscopyThe interaction with visible light with solid occurs via the polarizability of the valence electrons. The electric field E0 of the incident light wave induces, via the susceptibility tensor χ a polarization P. P=ε0χE0 and E0=E0cosωot
The susceptibility tensor depends from the collective excitations that they can be regarded as a perturbation. By just considering the first two terms:χ=χ0+(δ χ /δX)X If the susceptibility itself oscillates with the frequency ω(q) of an elementary excitation (e.g. phonon) then the oscillation of polarization induced by the incident radiation (ω0) is modulated by ω(q) .
The periodic modulation of P leads tothe emission of a wave (scattered)
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RamanRaman SpectroscopySpectroscopy
Elastic process
Stoke and anti-stoke lines.
The modulation in the polarization leads to contributions in the scattered light at ω0 ±ω(q). The plus and minus corresponds to scattered light that have, respectively, absorbed the energy of, and lost energy to, the relevant elementary excitation [ω(q),q].
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RamanRaman spectrumspectrum of of GaAsGaAs at 5Kat 5KElastic contribution
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1~ −= λν
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