Post on 03-Apr-2020
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Phonons and Magnons
Phonons
Natom atoms in crystal3Natom normal modesp atoms in the basisNatom/p unit cellsNatom/p translational symmetriesNatom/p k-vectors3p modes for every k vector3 acoustic branches and 3p-3 optical branches
fcc phonons
3N degrees of freedom
fcc phonons
energy spectral density
internal energy density specific heat
http://lamp.tu-graz.ac.at/~hadley/ss1/phonons/phonontable.html
NaCl
2 atoms/unit cell
6 equations
3 acoustic and 3 optical branches
1 2 3 1 2 3exp expx x x xnml nmlk ku u i k a k a k a t v v i k a k a k a t
NaCl
Two atoms per primitive unit cell
Si
CsClM1 = M2
M1 = 5M2M1 = 2M2
M1 = 1.1 M2
Hannes Brandner
GaAs Hannes Brandner
Phonon quasiparticle lifetime
Phonons are the eigenstates of the linearized equations, not the full equations.
Phonons have a finite lifetime that can be calculated by Fermi's golden rule.
00 1 0 0
0 01 0 1 1 1 1
1
0 1
i Ni
i Ni
NN N N iN
i N
dPdt PdP Pdt
PdPdt
Occupation is determined by a master equation (not the Bose-Einstein function).
22i f ph ph f if H i E E
Acoustic attenuation
The amplitude of a monocromatic sound wave decreases as the wave propagates through the crystal as the phonon quasiparticles decay into phonons with other frequencies and directions.
Magnons
Magnons are excitations of the ordered ferromagnetic state
Magnons
1 12 p p pJS S S
Energy of the Heisenberg term involving spin p
The magnetic moment of spin p is
p B pg S
1 12
p p pB
J S Sg
This has the form -p.Bp where Bp is
1 12
p p pB
JB S Sg
Magnons
The rate of change of angular momentum is the torque
1 12
p p pB
JB S Sg
p B pg S
1 12pp p p p p p
dSB J S S S S
dt
If the amplitude of the deviations from perfect alignment along the z-axis are small:
1 1
1 1
2 2
2 2
0
xp y y y
p p p
yp x x x
p p p
zp
dSJ S S S S
dtdS
J S S S Sdt
dSdt
Magnons
1 1
1 1
2 2
2 2
0
xp y y y
p p p
yp x x x
p p p
zp
dSJ S S S S
dtdS
J S S S Sdt
dSdt
These are coupled linear differential equations. The solutions have the form:
expx xp ky yp k
S ui kpa t
S u
( 1) ( 1)
( 1) ( 1)
2 2
2 2
x ikpa ik p a ikpa ik p a yk k
y ikpa ik p a ikpa ik p a xk k
i u e J S e e e u
i u e J S e e e u
Cancel a factor of eikpa.
Magnons
4 1 cos( )0
4 1 cos( )i J S ka
J S ka i
These equations will have solutions when,
The dispersion relation is:
4 1 cos( )J S ka
2 2
2 2
x ika ika yk k
y ika ika xk k
i u J S e e u
i u J S e e u
Magnon dispersion relation
4 1 cos( )JS ka
A phonon dispersion relationwould be linear at the origin
Magnon density of states
4 1 cos( )JS ka
Mathematically this is the same problem as the tight binding model for electrons on a one-dimensional chain.
Ferromagnetic magnons - simple cubic
The dispersion relation in one dimension:
4 1 cos( )J S ka
The dispersion relation for a cubic lattice in three dimensions:
2 cos( )J S z k
The magnon contribution to thermodynamic properties can be calculated similar to the phonon contribution to the thermodynamic properties.