Flashcards

Week 5 LNs

R S C ICMP Year 4 English 34 cards

Write mathematically the relations for the DOS in terms of N w and K

Cards in this deck (34)

1Why is the term isotropic used for systems when relating density of states to phonon dispersion relations? How does the wave vectors in a sphere relate to the phonon frequency?consider the simple case where the dispersion
relations depend only on the magnitude of the wave vector, but not its direction.
In other words, we restrict this discussion to isotropic systems
All wave vectors
on a sphere around the origin in reciprocal space correspond then to the same
phonon frequency
2From DOS, how is dN/dK and dK/dw found?The second term can be evaluated from the dispersion
relation 𝜔(𝐾), and the first term can be evaluated taking into account that the
allowed wave vectors form a dense, regular grid in reciprocal space
3Write mathematically the relations for the DOS in terms of N w and K
4Define DOS
5Describe dN/dK and dK/dw for a 1D crystal of length L
6Yield a general expression for DOS that depends on the derivative of the dispersion relationIn three dimensions, we use again that the allowed wave vectors form a regular
grid and can then proceed with two slightly different approaches. In the first
approach, we calculate the number 𝑁sph(𝐾) of wave vectors inside a sphere of
radius 𝐾 in reciprocal space. As in the 1D case, this is achieved by dividing the
volume 𝑉sph by the reciprocal-space volume per allowed wave vector. d𝑁/d𝐾 is
then obtained by differentiation. Overall, this yields a general expression for the
density of states that depends only on the derivative of the dispersion relation,
[d𝜔(𝐾)/d𝐾]^−1 = 𝑑𝐾(𝜔)/d𝜔
7Find a 2nd way to yield a general expression for DOS that depends on the derivative of the dispersion relation
8Explain why some features in the DOS are not explained from the ph dispersion relation plotsThe reason is that a dispersion plot
samples only selected paths through the Brillouin zone and may thus be missing
phonon dispersion extrema at wave vectors away from the high-symmetry lines
that are usually sampled in a dispersion plot.
9Write the total number of states in terms of the DOS and a 2nd one for the no of atoms.
10Describe evaluating the vibrational energy of a crystal from integrals over the phonon DOSWe can now evaluate the vibrational energy of a crystal from an integral over the
phonon density of states rather than a sum over all wave vectors and phonon
branches. The lattice (or phonon) heat capacity, 𝐶lat ≡ 𝐶𝑣
, is then the temperature
derivative of 𝑈lat, which can be calculated with the chain rule. The phonon density
of states 𝐷(𝜔) (or 𝐷𝑝(𝜔) for each phonon branch 𝑝 individually) contains all the
necessary information about the crystal.
11Reproduce (explain) how to get the Einstein model value for the DOSThe density of states is a delta function centred at 𝜔0 and normalised with the
total number of states (or oscillators), 3𝑁, as a prefactor. The integral over the
density of states yields the total number of states as required.
Using the general expression for 𝐶lat from the previous slide, we reproduce the known result for the Einstein model.
12Explain the Debye model and compare it to the Einstein modelThe Debye model is an analytical model that describes the phonon heat capacity
of most materials fairly well while still using only a single material parameter.
In contrast to the Einstein model, it describes the variation of the heat at low
temperatures correctly. At the same time, it is not necessary to know the complete
phonon density of states.
13Describe the phonon modes that dominate the heat capacity a low temperaturesThe phonon modes that dominate the heat capacity at low temperatures are
the low-frequency acoustic modes near the Brillouin zone centre (as the higherfrequency modes are not thermally populated at low temperatures). They are
characterised by a nearly linear dispersion relation, and this is the starting point
for the Debye model.
14In the Debye model, why is the cut off frequency independent of crystal size at constant pressurethe
number of states/oscillators is proportional to the crystal volume 𝑉
15For the Debye model, why is the DOS set to 0 above a cut off frequency?The linear dispersion relation yields a quadratic expression for the phonon density
of states, 𝐷(𝜔) ∝ 𝜔^2
. However, an integral over this phonon density of states
would diverge and yield an infinite number of states. Therefore, the density of
states is set to zero above a cut-off frequency 𝜔𝐷, which is chosen such that
the correct number 𝑁 of oscillators and states is obtained.
16State the phonon internal energy in the Debye model (describe) How is the heat capacity then found?
17How does the heat capacity lead to the Debye temperature?The heat capacity (and also the lattice energy) can be expressed either with an
integral over the phonon frequencies up to the cut-off frequency 𝜔𝐷 or with a
dimensionless integral up to a dimensionless parameter 𝜃𝐷/𝑇 , where 𝜃𝐷 is the
so-called Debye temperature. We will see below that the Debye temperature is a
characteristic temperature: for 𝑇 ≪ 𝜃𝐷, quantum behaviour will be observed in
the form of 𝐶 → 0 for 𝑇 → 0, whereas classical behaviour is restored for 𝑇 ≫ 𝜃𝐷.
18Explain the following experimental results by comparing the Einstein and Debye modelThe Debye heat capacity has a universal temperature dependence when plotted
against the reduced temperature 𝑇 /𝜃𝐷, where 𝜃𝐷 is the only material-specific
parameter. This reproduces the experimental observations discussed above. The
resulting 𝐶(𝑇 ) is shown by the solid (blue) line in the graph, and one can see that
the heat capacity at low temperatures is indeed somewhat higher than in the
Einstein model. In the high-temperature limit, the classical value of 3𝑁 𝑘𝐵 for 𝑁
atoms (or 3𝑅 for one mole) is obtained — the cut-off frequency 𝜔𝐷 was chosen
so as to ensure this. The low-temperature limit can be obtained analytically
19Explain Debyes T^3 law, with illustrations from Einstein and Debyethe compromise in the Debye model lies in describing the low-frequency part accurately while
neglecting the details of the higher-frequency part. The latter is less important —
even the very simple Einstein model managed to describe the high-temperature
behaviour of the heat capacity fairly well
20Explain the two equations of motion for a diatomic linear chain
21Obtain the solutions to the EoM from the diatomic linear chain
22Explain the solutions to the diatomic linear chain (frequency) producing the optical and acoustic branches by mentioning the BZ centre and boundaryThe solutions corresponding to the plus and the minus sign in the ± expression
produce the higher- and the lower-frequency branch.
For the upper (optical) branch at the zone centre, we evaluate the phonon frequency at 𝐾 = 0, which removes the second term in the square root. The resulting frequency depends on both atomic masses. When expressed in terms of the
reduced mass 𝑀red, we obtain a familiar expression, 𝜔 ∝ √𝐶/𝑀red.
The phonon frequency of the lower (acoustic) branch goes to zero as 𝐾 → 0, just
like in the monatomic linear chain. If we evaluate 𝜔(𝐾) to leading order in 𝐾,
we recover the previous result of a nearly linear dispersion, 𝜔(𝐾) ∝ 𝐾, near the
zone centre, 𝐾 ≪ 𝜋/𝑎. In the long-wavelength limit, 𝜔(𝐾) = 𝑣𝑠𝐾 gives the sound
velocity 𝑣𝑠
, which we can now express in terms of the two atomic masses, spring
constant and lattice periodicity. Note that the lattice parameter 𝑎 is here twice
the interatomic distance. With that taken into account, one can recover the result
for the monatomic linear chain, where 𝑀1 = 𝑀2
23For the diatomic linear chain explain the zone boundary phonon frequencies giving the optical and acoustic branch frequencies
24Describe how neighbouring atoms are displaced relative to each other in acoustic and optical modes Describe the ionic crystals caseWith the same approach you will determine how neighbouring atoms are displaced relative to each other in acoustic and optical modes. The result is that
neighbouring atoms move in parallel (in-phase) in long-wavelength acoustic
phonons, as expected for sound waves. In optical phonon modes, neighbouring
atoms move in opposite directions (anti-phase). In the case of an ionic crystal,
this will produce a fluctuating electric dipole moment, which can couple to electromagnetic radiation. Optical phonons in polar materials can be measured with
optical (far-infrared) absorption spectroscopy. For example, a phonon frequency
of 𝜈 = 10 THz corresponds to a far-infrared wavelength of 𝜆 = 𝑐/𝜈 = 30 µm. Note
that the distinction between acoustic and optical branches is meaningful only
near the Brillouin zone centre.
25Describe longitudinal and transverse modesIn the one-dimensional chain, atoms can only be displaced along the one dimension of space. However, in real, three-dimensional crystals, atoms can be
displaced along the propagation direction of the phonon (i.e., parallel to the wave
vector) or perpendicular to that direction. The corresponding modes are said to
be longitudinal and transverse modes, respectively. As there are two independent
directions that are orthogonal to the wave vector (and to each other), the are in
general two transverse branches but only one longitudinal branch. As for the
acoustic and optical branches, the distinction between longitudinal and transverse
branches is meaningful only near the Brillouin zone centre.
26Describe the properties of an acoustic phonon branchIn general, an acoustic phonon branch is characterised by two properties: (i) neighbouring atoms move in phase in the limit of long wavelengths and (ii) the phonon
frequencies become zero in the limit 𝐾 → 0, i.e., in the long-wavelength limit.
This nature of the acoustic modes permits only three acoustic phonon branches
in any three-dimensional crystal, all other phonon branches are optical branches.
For a three-dimensional crystal with 𝑝 atoms in the primitive unit cell, one has 3𝑝
oscillator modes in the Einstein picture of independent oscillators. The number
of modes is preserved when the oscillator modes are coupled to form phonon
modes. Therefore, there are 3𝑝 phonon frequencies (possibly degenerate to some
extent) for every wave vector 𝑲 in the Brillouin zone. Correspondingly, there are
3𝑝 phonon branches in total, of which three are acoustic branches.
27Diamond has only one type of atom, but there are two atoms per primitive unit cell, what can be expected of this?we expect to observe a total of six phonon
branches, three acoustic branches (as always) and the others are optical branches
(three). This can indeed be seen along Γ–K–X. Along X–Γ–L, the two transverse
acoustic branches are degenerate, as are the two transverse optical branches. We
can also see once more how flat regions of the dispersion relations relate to kinks
and peaks in the density of states
28Why cant vibrational amplitude explain why most materials expand when heated?In a harmonic system, it requires the same force and energy to expand or compress
a bond (or spring) by a certain amount. In other words, a harmonic system is
perfectly symmetric with respect to compression and expansion. Therefore, an
increase in vibrational amplitude with increasing temperature cannot result in
thermal expansion.
29Explain in detail how materials expand when heatedWe start with an anharmonic potential 𝑈(𝑥) in terms of the displacement 𝑥 of the atoms from their equilibrium separation. Note that the minimum of this 𝑈(𝑥) at
𝑥 = 0 is only local minimum, not a global minimum. But this is not a problem as
long as we consider only sufficiently small displacements. From the discussion
above, we expect the asymmetric, cubic term (−𝑔𝑥3
) to play the crucial role, but
we include also the symmetric quartic term (−𝑓 𝑥4
) to illustrate that it does not
contribute to thermal expansion.
30Define the thermal expansion coefficientdefined as the relative change in sample
length with temperature
31What results in negative thermal expansionThe selective population of low-frequency phonon modes
32Explain Ficks 1st Law
33In regards to thermal conductivity in phonon gases; What limits mean free path of phonons between collisions? Is there the same limit with a crystal with purely harmonic potential? For quantum mechanics, what needs to be added to achieve coupling between phonons? What is the dominant scattering at room temp and above?The mean free path of phonons between collisions is limited by two types of
scattering: geometrical scattering due to crystal imperfections and phonon-phonon
scattering. In a hypothetical crystal with a purely harmonic potential, all phonon
modes are completely independent and don’t interact with each other — there
is no phonon-phonon scattering. In the language of quantum mechanics, the
phonons are stable solutions or eigenmodes of the Schrödinger equation. As such,
there is no coupling between them until one adds a perturbation in the form of an
anharmonic contribution to the oscillator potential. Therefore, it requires once
more consideration of the anharmonic terms of the crystal potential of a real
crystal. Phonon-phonon scattering occurs even in perfect, defect-free crystals,
and is usually the dominant scattering process at room temperature and above.
34Explain why geometrical scattering is an elastic processthe phonon energy and magnitude of the phonon wave
vector do not change, only the propagation direction. This scattering process
cannot redistribute energy between different phonon modes, but that is required
to establish a local thermal equilibrium, where the phonon population is given
by the Planck distribution.