固体物理ppt 4

Chapter 4
Phonons I. crystal vibration
Key points: Crystal vibrations Phonons Phonon momentum
Phonon is one kind of elementary excitations in solids
Name Field Electromagnetic wave Elastic wave Collective electron wave Magnetization wave Electron + elastic deformation Polarization wave
The homogenous linear equations have a solution only if the determinant of the coefficients of the unknown u, v vanishes.
2C M 1ω 2 C[1 + exp(iKa )] C[1 + exp(iKa )] 2C M 2ω
For K<<1/a (i.e. λ>>a), ω ∝ K At the boundary of first Brillouin zone (K = ±π/a), dω/dK = 0
First Brillouin zone Only K in the first Brillouin zone is physically significant for the elastic waves.
If there are p atoms in the primitive cell, there are 3p branches to the dispersion relation: 3 acoustical branches (1 LA and 2 TA) and 3p3 optical branches (p1 LO and 2p2 LO ).
Two atoms per primitive basis
The dispersion relation shows new features in crystals with two or more atoms per primitive basis. acoustical and optical branches (声学支和光学支)
v g = dω / dK in one dimension
4C With the particular dispersion relation ω = M 2 1/ 2 Ca Ka cos vg = 2 M
1/ 2
Ka sin 2
Long wavelength limit When λ >> a i.e. Ka << 1
With p atoms in the primitive cell and N primitive cells, there are pN atoms, 3pN degrees of freedom, N LA branches, 2N TA branches, (p1)N LO branches and (2p2)N LO branches.
The range of independent values of K is specified by:
π < Ka ≤ π , or
π
a
<K≤
π
a
We may treat a value of K outside the 1st Brillouin zone by subtracting a appropriate reciprocal lattice vector in order to obtain an equivalent wavevector in the 1st Brillouin zone. r r r K'≡ K G (in one dimension, K ' = K 2πn / a) r where K '∈1st Brilouin zone. At the Brillouin zone boundaries, K = ±π/a, us = u exp[i(± sπ ωt)] = (1)s u exp(iωt) (standing wave) The critical value, K = ±π/a, satisfies the Bragg condition. Note: for x-ray, it is posible to have other n which cause the wavevector K outside 1st Brilouin zone.
For the acoustical branch
u=v
The atoms and their center of mass move together.
at the 1st Brillouin zone boundary (Ka = ±π)
j =1 p >0
The equation of motion is p >0 d 2u s M = 2∑ C j (u s + j u s ) 2 dt j The dispersion relation 2 p >0 ω 2 = ∑ C j (1 cos jKa ) M j =1
Ma π / a 2 Cp = dKω K cos pKa 2π ∫π / a
2 4C 2 Ka Ca ω 2 = sin ≈ K2 2 M M
C i.e. ω ≈ a K M
C v g = dω / dK ≈ a M
1/ 2
1/ 2
Derivation of force constants from experiment In metals the effective forces may be of quite long range, carried from ion to ion through the conduction electron sea. Considering p Байду номын сангаасearest planes, the force on s plane F = 2∑ C j (u s + j u s )
longitudinal mode
transverse mode
Note: for N atoms there are 3N mode, N L-modes and 2N T-modes.
Assume that the elastic response of the crystal in linear function of the force. (Hooke’s law) i.e. the elastic energy is a quadratic function of the relative displacement of any two points in the crystal. Neglect cubic and higher order terms. From Hooke’s law, the force on the s plane caused by the displacement of the plane s+p is proportional to the difference of their displacement, i.e. Fsp = Csp (us+p us) Consider only the nearest neighbor interactions. The total force on s comes from the plane s±1 Fs = C (us+1 us) + C (us-1 us) = C (us+1 + us-1 2us)
Electron Photon Phonon Plasmon Magnon Polaron Exciton
Important elementary excitations in solids
Vibrations of crystals with monoatomic basis
Consider the elastic vibrations of a crystal with one atom in the primitive cell. Describe with a single coordinate us the displacement of the plane s from its equilibrium position. Three mode for each wavevector, one of longitudinal polarization and two of transverse polarization.
Assume that each plane interacts only with the nearestneighbor planes. The equations of motion is d 2u s M 1 2 = C (vs + vs 1 2u s ) dt d 2 vs M = C (u s +1 + u s 2vs ) 2 dt 2 the solution in a form of a traveling wave:
2
=0
or M 1M 2ω 4 2C ( M 1 + M 2 )ω 2 + 2C 2 (1 cos Ka ) = 0 the dispersion relation 2C ( M 1 + M 2 ) ± [2C ( M 1 + M 2 )]2 8C 2 M 1M 2 (1 cos Ka ) ω2 = 2 M 1M 2
ω 2 M = C[exp(iKa ) + exp(iKa ) 2] = 2C (1 cos Ka )