Karteikarten
Week 4 LNs
Karten in diesem Stapel (34)
- 1Why is the understanding of atomic vibrations crucial for crystals?crucial for understanding
the electrical and thermal conductivity of solids, superconductivity and other
properties. There is evidence from heat capacity measurements and many other
experiments that atoms in a crystal are never truly at rest, not even at 0 K - 2Define heat capacity in terms of internal energy State its internal U formula and its formulae for atoms and moles and the Dulong Petit law explaining the relationsThe heat capacity 𝐶 quantifies the change of the internal energy 𝑈 of a thermodynamic system when the temperature 𝑇 is changed. For gases, one distinguishes between the heat capacities at constant volume (𝐶𝑉 )
and constant pressure (𝐶𝑃 ). For solids, the difference is very small and usually
not relevant. In the classical limit, the heat capacity is 1
2
𝑘𝐵 per degree of freedom,
giving 3
2
𝑘𝐵 per atom of an ideal gas with its three translational degrees of freedom.
𝑘𝐵 and 𝑅 are the Boltzmann and ideal gas constant, respectively.
In a solid, we have six degrees of freedom, 3 × potential and 3 × kinetic. However, a
heat capacity of 3𝑘𝐵 per atom (or 3𝑅 per mole) is only observed at sufficiently high
temperatures. At low temperatures, the heat capacity decreases with decreasing
temperature, and 𝐶 → 0 for 𝑇 → 0. This motivated Einstein’s model for the heat
capacity of a an insulating solid, a quantum model, which was discussed in detail
in Physics of Matter and Statistical Mechanics. - 3For an atom is displaced from the equilibrium position experiencing a restoring force, what potential describes this?
- 4How is the heat capacity found from the internal energy of oscillators in 3D? What is a disadvantage of the Einstien model for the reduction of heat capacity?The calculation of the heat capacity is now straight forward. The internal energy
𝑈 is differentiated using the chain rule. Note that the constant 1
2
in the internal
energy is not physically relevant here and that it disappears after differentiating.
The Einstein model reproduces the reduction of the heat capacity at low temperature. However, the heat capacity is actually suppressed too much at low 𝑇 compared
to experimental results.
- 5State the key assumption and approximation in the Einstein model for the heat capacityThe key assumption and approximation in the Einstein model was that the oscillators are completely independent and that only a single oscillator frequency 𝜔𝐸
needs to be considered. The displacement of one atom is assumed not to affect the
neighbouring atoms. This cannot be strictly true, and it is probably the reason for
the slight discrepancy between the results of the Einstein model and experiments. - 6For the lattice vibrations of a 1D linear chain connected by springs where one atom is displaced and springs back; What is expected of each atom? What can finally be obtained from the coupling of sinusoidal waves?
- 7For the lattice vibrations of a 1D linear chain connected by springs where one atom is displaced and springs back; State the equation of motion for all atoms in the monatomic linear chain and its wave like solutions
- 8For the lattice vibrations of a 1D linear chain connected by springs where one atom is displaced and springs back; Use the ansatz solution to derive the dispersion relation(1) refers to the EoM
The dispersion relation shows that there are many (an infinite number) of wave
solutions, and the frequency 𝜔 depends on the wave vector/wavelength 𝐾 = 2𝜋/𝜆
of the excitation. Using standard trigonometric relations, the dispersion relation
can be expressed in various ways.
- 9Describe the dispersion relation in the 1st BZThe dispersion relation is periodic in the
wave vector or reciprocal space. The period is 2𝜋/𝑎, which is the length of the
one-dimensional reciprocal unit cell. The one-dimensional reciprocal unit cell
that is symmetric about the origin is called the first Brillouin zone; it covers wave
vectors from −𝜋/𝑎 to +𝜋/𝑎.
Each of the normal modes describes a wave motion of all atoms for a single
wave vector 𝐾. The overall motion of every atom is given by a superposition of all normal modes, with weights or population numbers given by the Planck
distribution. - 10Describe a wave with wave vector K at the BZ boundary (Linear Chain)A wave with a wave vector 𝐾 = 𝜋/𝑎 at the Brillouin zone boundary has a
wavelength of 𝜆 = 2𝜋/𝐾 = 2𝑎, so that neighbouring atoms oscillate anti-phase
(top row∗
). This can be regarded as a standing wave. The long-wavelength modes
with 𝐾 ≈ 0 near the Brillouin zone centre show the propagating waves most
clearly. - 11For any wave solution with a wave vector outside the 1st BZ what can be said of atomic motion?In general, for
any wave solution with a wave vector 𝐾
′ outside the 1st Brillouin zone, we can
find an equivalent solution with wave vector 𝐾 inside the 1st Brillouin zone that
describes exactly the same atomic motion. 𝐾 can obtained from 𝐾
′ by adding
or subtracting an integer multiple of 2𝜋/𝑎. Therefore, we need to consider only
the solutions with the shorter wave vectors (or longer wavelengths) inside the
Brillouin zone. - 12Define and Describe the Wigner Seitz CellsThe Wigner-Seitz construction is a recipe for the construction of a uniquely defined
and primitive unit cell for a crystal lattice, in either direct or reciprocal space. The
so-constructed unit cells can often no longer be represented by three vectors,
but they still serve the purpose of a unit cell in that they fill the lattice space (or
plane in 2D) by periodic repetition and without gaps. Wigner-Seitz cells are not
commonly used in real space to describe crystal lattice or structures, but they
are used routinely for reciprocal space lattices to define Brillouin zones in two or
three dimensions unambiguously.
- 13How does a Wigner Seitz cell relate to the BZ? How does an fcc lattice relate to a bcc lattice?The BZ is the WS cell of a reciprocal lattice.
The rec lattic of an fcc lattice is a bcc lattice - 14Describe the dispersion relations similarities and differences between 3D and 1D phononsThe treatment of the coupled atomic oscillations in a three-dimensional crystal
is mathematically more involved, but yields fundamentally the same solutions.
The reason is that the 3D travelling wave solutions describe a situation where
whole crystal layers move relative to each other. Instead of a single chain we have
now many chains moving in parallel, and we obtain exactly the same dispersion
relations. The only difference is that the force constant 𝐶 will in general depend
on the direction of the wave vector 𝑲. - 15Describe calculating phonon dispersion relations from first principlesWithout going into any details, the calculation of phonon dispersion relations
from first-principles electronic structure calculations have become almost routine.
For these calculations, one needs to know only the crystal structure (hence “firstprinciples” or ab initio, if you prefer Latin), and many properties of the crystal
can be calculated in the framework of density functional theory (DFT), which
allows to solve the quantum-mechanical many-electron problem to a very good
approximation. In particular, the forces on displaced atoms can be calculated and
hence the force constants and their direction dependence. - 16Illustrate with sketches the square crystal and rec lattice with special points and directions.
- 17From phonon dispersion graphs; Describe where the transverse and longitudinal waves come fromThe phonon dispersion graph shows up to three frequencies for any wave vector
𝑲. In three dimensions, atoms can oscillate with displacements parallel to the
propagation direction of the wave or perpendicular, giving longitudinal and
transverse waves, respectively. In 3D, there are two directions perpendicular to
the propagation direction, and hence there can be two different transverse phonon
branches. For a crystal structure with one atom in the primitive unit cell, there are
three phonon modes per wave vector (along Γ–K–X), but they can be degenerate
(along Γ–X, due to symmetry). In practice, the longitudinal branch has usually
the highest frequency at any wave vector. The distinction between longitudinal
and transverse displacement is well defined only in the long-wavelength limit,
near the Brillouin zone. This explains why the low-frequency transverse branch
along Γ − 𝑋 can turn continuously into the high-frequency longitudinal branch
along K–X. - 18For dispersion relations, describe the wave vectors and their role in fcc crystals with mention of [110] and [111] directions
![For dispersion relations, describe the wave vectors and their role in fcc crystals with mention of [110] and [111] directions](https://examtimeassets.s3.amazonaws.com/uploads/media/image/51517542/desktop_9c35abc1-afff-400a-a738-78bab80dba84.JPG)
- 19Describe the quantised energy from coupled quantum harmonic oscillatorsWe recall from quantum mechanics that the energy levels of the quantum harmonic oscillator are quantised. This means that, unlike in the classical case, the
energy of the system can only be changed in increments of ℏ𝜔0, where 𝜔0 is
the natural angular frequency of the oscillator. In a system of coupled quantum
harmonic oscillators, there is now a distribution of frequencies instead of a single
oscillator frequency 𝜔0. As we saw in the last lecture, each wave vector 𝐾, which
describes the wavelength of the lattice wave, is related to a specific oscillator
frequency 𝜔(𝐾), as described by the dispersion relation. The energy quantisation
applies also to the system of coupled oscillators — the energy of the system can
only be changed in increments of 𝜔(𝐾) - 20Describe scattering experiments measuring phonon dispersion relations in terms of conservationThe measurement of phonon dispersion relations relies on scattering experiments
with the conservation of energy and momentum. It is obvious that the overall
energy of the involved scattered particle (neutron or photon) and phonon needs
to be conserved. If a phonon of energy ℏ𝜔 is created in the scattering process,
its energy must come from the scattered neutron or photon. Conversely, if a
phonon is annihilated in the scattering process, the neutron or photon will gain
the phonon’s energy.
Phonons do not carry a linear momentum in the mechanical
sense. Every movement of an atom with a momentum 𝒑 is compensated by
the momentum −𝒑 of another atom half a wavelength along the propagation
direction.
- 21First; how does a crystal recoil with elastic scattering state relations for inelastic scattering for phonon, incident and final wave vectors Describe the physics here
- 22Are phonon frequency dispersion relations periodic or non periodic in reciprocal spacePeriodic
- 23What does the measurement of phonon dispersion relations in elastic scattering experiments rely on? What type of method is preferred for phonon dispersion relations? Define the triple axis spectrometer(i) the quantisation of the phonon excitation energies, (ii) energy conservation during the scattering process as well as (iii) the wave vector conservation
during scattering. Inelastic neutron scattering (INS) has long been the method of
choice for the measurement of phonon dispersion relations. A so-called tripleaxis spectrometer is used to define the neutron wave vector before and after the
scattering process, and to measure the energy transfer between neutrons and
phonons. - 24For inelastic neutron scattering; How is the energy transfer between a neutron and crystal measured during scattering? Explain; method in determining the scattering rate
- 25Define the scattering vectorthe scattering vector is the vector of the created or annihilated phonon
- 26In regards to the phonon spectra (neutron scattering rate as a function of energy transfer) for scattering vectors; What can be said of the peak position?The
peak position in each spectrum gives the frequency (or energy) of the phonons created with a particular wave vector. If one plots these phonon frequencies
versus wave vector along selected lines in the Brillouin zone, one obtains the
phonon dispersion relations. - 27Define and describe the group velocity in terms of the BZ boundary How is the sound velocity found from the phonon dispersion?
- 28What do surfaces of real crystals give rise to?Boundary conditions
- 29For a linear chain of atoms where atoms at the ends are fixed; Describe; the phonon wave solution, longer wave vectorsA simple illustration of the effect of boundary conditions is given by a linear
chain of 𝑁 + 1 atoms with a length of 𝐿 = 𝑁 𝑎, where 𝑎 is the lattice constant.
Let’s assume that the atoms at either end are fixed, i.e., they must be nodes of
the wave solutions. It follows that the phonon wave solution can only have
discrete wavelengths. There are 𝑁 − 1 solutions for this case of 𝑁 − 1 atoms
that are free to oscillate. Longer wave vectors are outside the first Brillouin zone
and therefore not relevant. This illustrates and explains in principle why it is
appropriate to consider a discrete set of allowed wave vectors. Considering a
particular crystal shape is, however, impractical in practice, and we consider a
more general approach in the following. - 30For a 1D linear chain of atoms where atoms at the ends are fixed and it is bent to form a ring; Describe: displacement of any atom Phase of the wave Two standing wave solutionsWe can argue that we could bend a linear crystal to form a closed ring. For a
macroscopic crystal, this should not change its properties. We can then consider
the travelling wave solutions. The displacement 𝑢𝑠 of any atom must remain
unchanged if we count the label 𝑠 once more around the ring, so that 𝑢𝑠 ≡ 𝑢𝑠+𝑁 . It
follows that the phase of the wave must change by an inconsequential multiple of
2𝜋, which again imposes a restriction on the wave vectors and produces a discrete
set of 𝑁 allowed wave vectors. Note that the two standing-wave solutions with
𝐾 = ±𝑁 𝜋/𝐿 are indistinguishable and count only as one state. This gives the same number of modes (one per mobile atom) as for the standingwave case. However, here we have all wave vectors with positive and negative
sign, which is more convenient in practice. - 31For a 1D linear chain of atoms where atoms at the ends are fixed and it is bent to form a ring; State the values for K for periodic boundary conditions for; 1D
- 32For a 2D chain of atoms where atoms at the ends are fixed and it is bent to form a ring; State the values for K for periodic boundary conditions for; 2D
- 33For a 3D chain of atoms where atoms at the ends are fixed and it is bent to form a ring; State the values for K for periodic boundary conditions for; 3D
- 34For a 1D linear chain of atoms where atoms at the ends are fixed and it is bent to form a ring; What changes in the 2D and 3D cases?
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- Year 4 ICMP Karteikarten am Week 4 LNs, erstellt von R S C am 05/07/2024.
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