Flashcards
Week 6 Lns
Cards in this deck (29)
- 1Can you recall the following? (Ch11)
- 2Can you recall the following (ch12)
- 3Describe what a Fourier transform spectrometer isBased on the familiar
Michelson interferometer. The measured interferogram I(∆)
is the intensity that exits the interferometer as a function of
the displacement ∆ of the moving mirror #2. In practice, the
slowly-varying average of the transmitted intensity is subtracted
from the measured signal, so that the interferogram is only the
fluctuating part of the signal registered by the photo detector.
It can be shown that the spectrum of the light source in the
usual wavelength or frequency domain is given by the Fourier
transform of the interferogram that was recorded in the spatial
domain. - 4Explain noise that appears in the spectrum of the broadband globular sourcedense series of absorption
lines. These absorption lines, which are not resolved in the
spectrum shown, originate from vibrational modes of water
and C–H bonds in organic molecules in the air, through which
the radiation passed during the measurement
- 5For Fourier transform spectrometers describe the monochromatic light sourceA monochromatic light source such as a laser gives a purely
sinusoidal interferogram. Such a sinusoidal signal has only
one frequency component, and its Fourier transform yields the
wavelength of the laser source. In practice, the measured interferogram will be truncated at low and high displacements,
and the Fourier transform of such a truncated signal yields a
broadened spectrum - 6For Fourier transform spectrometers how is the reflectance spectrum of a solid ultimately foundThe reflectance spectrum of a solid can now be determined by
measuring the spectrum of the light source itself and that of the
same radiation reflected by the sample surface. The reflectance
spectrum is then simply the ratio of these two spectra - 7Can real and imaginary parts of a cmplex dielectric function be found from a single reflectance spectrum?No
In the
measurement process we have lost some information - 8During optical reflectance, describe a major limitation and what becomes impossible to measure from an experimental spectrum. What can be found?The Fresnel equations yield expressions for the amplitude reflection coefficient r(ω) that gives the ratio of the electric field
amplitudes (E) of the reflected and incoming light. In an experiment, we can only measure intensities (I = EE∗
) and thus the
reflectance coefficient R(ω), but not the amplitude reflection
coefficient r(ω). Due to this limitation, we cannot directly determine the phase shift θ, and this in turn makes it impossible
to determine the real and imaginary parts of the complex dielectric function directly from an experimental spectrum. It turns
out, however, that the є1 and є2 spectra are not independent. If
one of them is known over the whole frequency range, then the
other can be calculated via the Kramers-Kronig relations. - 9Describe the derivation of the Kramers Kronig relations to show that the real part of the integral yields the imaginary part of X.In the derivation of the Kramers-Kronig relations, one starts
with Cauchy’s residue theorem, which states that the integral
of an analytic function χ over a closed line path in the complex
plane is zero. This remains true for the integrand χ(ω
′
)/(ω
′−ω)
if one avoids the pole at ω
′ = ω. Therefore, a particular, essentially semi-circular path in the upper half of the complex plane
is used, but with a small semicircle around the pole. Because
the analytic function χ is bounded, the integration over the
top half-circle vanishes as its radius is increased towards infinity. The value of the integral over the small semicircle around
the singularity is given by another residue theorem. P denotes
Cauchy’s principal value, i.e. the value of the improper integral
of χ(ω′)/(ω′−ω) with its pole at ω′ = ω. In the end, one obtains
an expression that relates the integral of χ(ω′)/(ω′ − ω) to χ(ω)
itself. The crucial part is the imaginary prefactor of 1/iπ = −i/π,
so that the real part of the integral yields the imaginary part of
of χ and vice versa.
- 10Answer the following; What does Kramers Kronig formalism apply to? Define the response function What if the Response function is; analytic or bounded Give an example of a response function
- 11Describe obtaining the full complex dielectric function from a reflectance spectrum starting from amplitude reflectance coefficientstart by expressing the
amplitude reflection coefficient r(ω) in terms of the dielectric
function and also in terms of the reflectance R(ω) and phase
shift θ(ω)
As the dielectric function is a response function, the amplitude
reflection coefficient r(ω) is also a response function. We can
therefore apply the Kramers-Kronig relations to obtain the
phase shift θ(ω) from the measured reflectance R(ω). That
gives us the full complex amplitude reflection coefficient r(ω)
and subsequently the dielectric function that we sought
- 12What does the real and imaginary part of dielectric function describeThe imaginary partє2 of the dielectric function describes absorption in a medium, and the є2(ω) spectrum comprises essentially
a series of absorption lines. The real part є1 describes how the
oscillation amplitude of the electrons depends on the frequency
of the driving electric field. It shows not only the magnitude
of the amplitude, but also the phase relation between the electron or ion oscillation and the driving E field. As in the case
of the (mechanical) driven harmonic oscillator, the system follows the driving force below the resonance frequency in-phase
(phase shift θ ≈ 0) but reacts anti-phase above the resonance
frequency (θ ≈ π). Correspondingly, є1 changes from positive
to negative values across every resonance. The whole є1(ω)
spectrum can then be understood as a superposition of several
such “resonance peaks”. - 13Describe optical conductivityThe optical conductivity, σ(ω), is
commonly used in optical studies of solids. It is the extension
of the electrical conductivity to the very high frequencies encountered in optics
(ν = ω/2π = c/λ = 600 THz for green light
of wavelength λ = 500 nm).
OC is the generalised AC conductivity up to optical frequencies - 14Why was the Drude Lorentz model made for the Kramers Kronig approachthe interpretation of reflectance
spectra is not straightforward. The Kramers-Kronig approach
also requires reflectance spectra that have been measured over a
wide frequency range (as a reasonable approximation of 0 → ∞),
which is not always feasible experimentally. The Drude-Lorentz
model has therefore been developed in order to calculate the
dielectric and other optical functions starting from a series
of absorption features or resonances. In addition, it allows
us to model how absorption peaks in є2(ω) translate to the
more complex features in the real part of the dielectric function,
the reflectance spectrum or the frequency-dependent index of
refraction - 15What oscillator represents Type of electric field its driven byWe consider the equation of motion for a classical driven,
damped harmonic oscillator with an undamped frequency ω0,
damping parameter Γ and driving force F(t). This approach
is rather general: The oscillator can represent free electrons
in a metal (ω0 = 0) or bound electrons in a semiconductor
or ions vibrating in a solid, all driven by the electric field of
electromagnetic radiation. Remember that ω0 =
√
K/m for a
spring-mass oscillator with spring constant K and mass m, and
that the damping force is Fd = −mΓ dx/ dt, i.e. proportional to
the particle velocity.
- 16For the Drude Model damped harmonic oscillator describe; Driving force and how the oscillator responds to it What is the response function role here?The driving force is generally not purely sinusoidal, but has a
more complex variation with time. It can then be expressed as
a superposition of sinusoidal signals with a certain frequency
spectrum Fω. To determine how the oscillator responds to a
particular frequency component of the driving force, we write
it as F(ω, t) = Re(Fωe
−iω t) and correspondingly x(ω, t) =
Re(xωe
−iω t) for the displacement, where “Re” denotes the real
part of the complex function. A frequency-dependent response
function α(ω) describes the relation between the two, and it
can be determined by comparison with the equation of motion. - 17For the Drude Model damped harmonic oscillator describe; system with several types of oscillator; total response, oscillator strength, Forces, polarisation,the total response is given by
the sum of the individual contributions. An oscillator strength
fj
is used to characterise how strongly the electric field couples
to every type of oscillator. As for the forces Fω, we consider a
particular Fourier component of the electric field, Eω, so that
Fω = −eEω.
we express
the polarisation in terms of the particle displacement and then
determine the dielectric function. Using the response function
we defined earlier, we obtain an explicit expression for є(ω).
Finally, we include a background dielectric constant є∞ as in
the case of the electron plasma. - 18For the OC derivation; describe the jw lines main resultsUsing again the response function approach, we can relate a
Fourier component jω of the current density j(ω, t) to the corresponding Fourier component of the electric field, Eω. As
always, the proportionality factor is the electrical conductivity,
which becomes now the frequency-dependent optical conductivity, σ(ω).
The main result is that we can express the optical (electrical)
conductivity σ(ω) in terms of the dielectric function є(ω). It is
worth remembering that the real part of one quantity is related
to the imaginary part of the other. The optical conductivity
σ(ω) = −iє0ω[є(ω) − 1] is a generalisation of the electrical
conductivity to optical frequencies
- 19For the DC limit of free electron gas; why is f =1 Tau non zero ResultFor a pure
electron system, the oscillator strength is f = 1, and the damping
constant Γ will be non-zero. We see at the end of the calculation
that the damping parameter is simply the scattering rate for the
electrons, Γ = 1/τ, where τ is the average time between collisions.
In the low-frequency limit (ω → 0), the optical conductivity
σ(ω) becomes ne^2τ/m, which is indeed the expected Drude
dc conductivity
- 20Describe the relations between the various optical functions using the imaginary part of the dielectric functionA good starting point for understanding the relations between
the various optical functions is the imaginary part of the dielectric function. It comprises a series of peaks, possibly overlapping, at excitation energies that correspond to phonon excitations or electronic transitions. The area of each peak is a measure
of the “absorption strength”, or the coupling strength between
that excitation and electromagnetic radiation. - 21Define the real part of the dielectric function and how its determinedThe real part of the dielectric function follows from the imaginary part via the Kramers-Kronig transformation, but can also be
obtained by evaluating the Drude-Lorentz model for the series
of excitations. It is a superposition of “resonance peaks” at the
excitation energies. - 22Define optical conductivity in terms of the dielectric functionThe optical conductivity is essentially the ω-weighted dielectric function with real and imaginary parts swapped — cf. the
definition of optical conductivity. - 23Define the complex refractive index in terms of the complex dielectric functionThe complex refractive index n is the square root of the complex
dielectric function,
n(ω) = √є(ω). The square root mixes the
real and imaginary parts wherever both are non-zero - 24Describe the sharp edges of the experimental reflectance spectra of metals such as aluminium, gold and silverEdges seem to confirm the predications on the basis of the free-electron gas or plasma in that relatively sharp edges are observed. However, these edges should be
located at higher energies or shorter wavelengths if the actual
electron densities of these metals are considered. And the structure of the spectra near the edges is often more complicated
than anticipated - 25Explain the yellow appearance of goldThe shift of the edge of high reflectance causes the coloured
appearance of some metals, such as gold and copper. The reflectance of gold drops from nearly 100% at 700 nm to 50% at
500 nm. The wavelength of 500 nm corresponds to a green colour. If white light is reflected on a gold surface, the blue part of
the spectrum is significantly reduced, which explains its yellow
appearance. - 26Describe the reflectance spectra of silver im part absorption peak coupling of partsThe reflectance spectra and dielectric functions of copper, silver
and gold were first measured by Ehrenreich and Philipp [1].
The imaginary part є2 of the experimental dielectric function
of Ag exhibits a strong sharp peak at 4 eV, which indicates
some absorption process. Inspection of the electronic band
structure of silver reveals this absorption peak to originate from
electronic interband transitions involving states around the L
point in the Brillouin zone. Due to the coupling between the
real and imaginary parts of the dielectric function, the peaks
in є2 translate to features in є1 and consequently the reflectance
spectrum. - 27What contributions make up the experimental spectrum of silverThe experimental є1 spectrum of silver can be decomposed into
contributions from the electron plasma, є
(f )
1
, and from interband transitions, δє(b)
1
, which Ehrenreich & Philipp denoted as
contributions from bound electrons [1]. The modification of the
real part of the dielectric function and the shift of the absorption
edge to lower energies can be seen very clearly. We have here a
situation where the interband transitions shift the zero in є1 to
the energy of the onset of the interband transitions, ωi
. As in
the case of the electron plasma, the edge of high reflectance is
located at the energy where є1 ≈ 0. - 28Explain the red orange appearance of copperThe reflectance of copper drops from over 90% at 2 eV
(∼620 nm) to about 50% at 3 eV (∼400 nm). A photon energy
of 2 eV corresponds to an orange colour. If white light is reflected on a copper surface, the green/blue part of the spectrum is
significantly reduced, which explains its red-orange appearance. - 29For Copper, describe im part coupling of partsThe imaginary part є2 of the experimental dielectric function
of Cu exhibits a strong double-peak feature between 2 and 5 eV,
which indicates some absorption process. Inspection of the electronic band structure of copper reveals these absorption peaks
to originate from electronic interband transitions. As before,
due to the coupling between the real and imaginary parts of the
dielectric function, the peaks in є2 translate to features in є1 and
consequently the reflectance spectrum. The contribution to the
real part of the dielectric function from interband transitions,
δє(b)
1
, yields an overall dielectric function є1 that is strongly modified compared to the free-electron case. In this case, which is
more complicated than that of silver, the zero in the є1
is no
longer shifted to the onset of the interband transitions at frequency ωi
. However, once the real and imaginary part of the
dielectric function are taken into account, the reflectance edge
does get shifted again to approximately ωi.
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- Year 4 Solid State Flashcards on Week 6 Lns, created by R S C on 18/07/2024.
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