Spectroscopy PDF - University of Milano-Bicocca

Materials spectroscopy and microscopy - spectroscopy venerdì 7 giugno 2024 12:09 These notes are based on the course Materials spectroscopy and microscopy - spectroscopy, held in A.Y. 2023/2024 at the University of Milano-Bicocca, Master's Degree in Materials Science and Nanotechnology. If you...

Materials spectroscopy and microscopy - spectroscopy venerdì 7 giugno 2024 12:09 These notes are based on the course Materials spectroscopy and microscopy - spectroscopy, held in A.Y. 2023/2024 at the University of Milano-Bicocca, Master's Degree in Materials Science and Nanotechnology. If you find any errors or inaccuracies you can report them by writing to [email protected] The main sources used for these notes are: Lecture notes F. Wooten, “Optical properties of solids”, Academic Press J. G. Solé, L.E. Bausà, D. Jaque, “Optical spectroscopy of Inorganic Solids”, Wiley Pagina 1 Ranges in the EM spectrum lunedì 15 gennaio 2024 15:05 Lower limits for each range are shown Range Wavelength 𝝀 Wavenumber 𝝂 Frequency 𝝂 Energy of the photon 𝑬 𝛾 rays* 10 pm 109 cm-1 30 EHz 124 keV X-rays* 10 nm 106 cm-1 30 PHz 124 eV UV 400 nm 2.5x105 cm-1 750 THz 3.1 eV VIS 700 nm 1.43x105 cm-1 480 THz 1.77 eV NIR** 1 μm 104 cm-1 300 THz 1.24 eV MIR** 10 μm 1000 cm-1 30 THz 124 meV FIR** 100 μm 100 cm-1 3 THz 12.4 meV MW** 1m 0.01 cm-1 300 MHz 1.24 μeV Radio 100 Mm 10-8 cm-1 3 Hz 12.4 feV * Some sources distinguish these two types of radiation based on their source rather than on the spectral range ** NIR: Near InfraRed; MIR: Medium InfraRed; FIR: Far InfraRed; MW: MicroWaves ℎ𝑐 𝐸 = ℎν = ⎯⎯⎯= ℎ𝑐ν λ Pagina 2 (gaussian units) venerdì 8 dicembre 2023 20:00 //superscripts: 𝐺 Gaussian units, 𝐼 SI units Gaussian units use the cgs (centimetre-gram-second) system instead of the MKS (metre-kilogram- second) system: / MKS cgs Conversion mass kilogram gram 1 𝑘𝑔 = 10 𝑔 distance metre centimetre 1 𝑚 = 10 𝑐𝑚 energy Joule erg 1 𝐽 = 10 𝑒𝑟𝑔 force Newton dyne 1 𝑁 = 10 𝑑𝑦𝑛 The dimensional definitions are the same: 1 𝑑𝑦𝑛 = 1 𝑔 𝑐𝑚 𝑠 = 10 𝑘𝑔 ⋅ 10 𝑚 ⋅ 𝑠 = 10 𝑁 1 𝑒𝑟𝑔 = 1 𝑑𝑦𝑛 𝑐𝑚 = 10 𝑁 ⋅ 10 𝑚 = 10 𝐽 Gaussian units are not rationalized, i.e. the factor 4𝜋 appears in Maxwell's equations, as opposed to the SI units (rationalized) in which the factor 4𝜋 appears in force laws (like Coulomb and Biot- Savart) but not in Maxwell's equations. The unit charge in Gaussian units is defined to make Coulomb's law with a force constant equal exactly to 1; the reason is that the quantities ε , μ are not actually fundamental properties of free space but rather artifacts of the SI system, which disappear in Gaussian units. This means that we can also replace μ in Gaussian units using the relationship: 1 ε μ = ⎯⎯ 𝑐 The unit of charge (statcoulomb [𝑠𝑡𝑎𝑡𝐶] or Franklin [𝐹𝑟]) is defined as the charge such that two charges of 1 statC at 1 cm of distance will each feel an electrostatic force of 1 dyne; so dimensionally one obtains that: / 𝑀𝐿 𝑄 = ⎯⎯⎯⎯ 𝑇 That is: / / 1 𝑠𝑡𝑎𝑡𝐶 = 1 𝑐𝑚 𝑔 𝑠 Another thing to keep in mind is that the magnetic induction 𝐵 is defined with an extra factor of 𝑐 even after the statcoulomb is taken into account, so all derived magnetic quantities have different dimensions in SI units than in Gaussian units. This has the benefit though that in the Gaussian system all of the fields 𝐸, 𝑃, 𝐷, 𝐵, 𝑀, 𝐻 have the same dimensions 𝑐𝑚 / 𝑔 / 𝑠 ; also the scalar 𝜙 and vector 𝐴 potentials have the same dimensions 𝑐𝑚 / 𝑔 / 𝑠. A further point is that both the electric and magnetic susceptibilities are dimensionless in both system but will have different numerical values. SOME EQUATIONS Name Gaussian SI Lorentz Force 1 𝐹⃗ = 𝑞 𝐸⃗ + 𝑣⃗ × 𝐵⃗ 𝐹⃗ = 𝑞 𝐸⃗ + ⎯⎯𝑣⃗ × 𝐵⃗ 𝑐 Coulomb's law 𝑞 𝑞 1 𝑞 𝑞 𝐹 = 𝑟 𝐹 = 𝑟 𝑟 4πε 𝑟 Pagina 3 Coulomb's law 𝑞 𝑞 1 𝑞 𝑞 𝐹⃗ = ⎯⎯⎯⎯⎯𝑟̂ 𝐹⃗ = ⎯⎯⎯⎯ ⎯⎯⎯⎯𝑟̂ 𝑟 4πε 𝑟 Electric field of a stationary point-charge 𝑞 1 𝑞 𝐸⃗ = ⎯⎯⎯𝑟̂ 𝐸⃗ = ⎯⎯⎯⎯ ⎯⎯ 𝑟̂ 𝑟 4πε 𝑟 Biot-Savart law 1 𝐼 𝑑ℓ⃗ × 𝑟̂ μ 𝐼 𝑑ℓ⃗ × 𝑟̂ 𝐵⃗ = ⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯ 𝐵⃗ = ⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯ 𝑐 𝑟 4𝜋 𝑟 Poynting vector 𝑐 1 𝑆⃗ = ⎯⎯⎯𝐸⃗ × 𝐵⃗ 𝑆⃗ = ⎯⎯ 𝐸⃗ × 𝐵⃗ 4𝜋 μ Electric displacement 𝐷⃗ = 𝐸⃗ + 4𝜋𝑃⃗ 𝐷⃗ = ε 𝐸⃗ + 𝑃⃗ Polarization 𝑃⃗ = χ 𝐸⃗ 𝑃⃗ = ε χ 𝐸⃗ Permittivity ε = 1 + 4πχ ε = ε (1 + χ ) Magnetic induction 𝐵⃗ = 𝐻⃗ + 4𝜋𝑀⃗ 𝐵⃗ = μ 𝐻⃗ + 𝑀⃗ Magnetization 𝑀⃗ = χ 𝐻⃗ 𝑀⃗ = χ 𝐻⃗ Magnetic permeability μ = 1 + 4πχ μ = μ (1 + χ ) Electric field (with potentials) 1 𝜕𝐴⃗ 𝜕𝐴⃗ 𝐸⃗ = −∇⃗ϕ − ⎯⎯⎯⎯⎯⎯ 𝐸⃗ = −∇⃗ϕ − ⎯⎯⎯ 𝑐 𝜕𝑡 𝜕𝑡 Magnetic field (with potentials) ⃗ ⃗ 𝐵 = ∇×𝐴 ⃗ ⃗ ⃗ 𝐵 =∇×𝐴 ⃗ SOME CONVERSIONS 𝑞 𝑠𝑡𝑎𝑡𝐶 𝑞 = ⎯⎯⎯⎯⎯⎯≈ ⎯⎯⎯⎯ 3 ⋅ 10 ⎯⎯⎯⎯⎯𝑞 4πε 𝐶 𝐼 𝑠𝑡𝑎𝑡𝐶/𝑠 𝐼 = ⎯⎯⎯⎯⎯⎯≈ ⎯⎯⎯⎯ 3 ⋅ 10 ⎯⎯⎯⎯⎯⎯⎯𝑞 4πε 𝐴 ⎯⎯⎯⎯ 1 𝑠𝑡𝑎𝑡𝑉 𝑉 = 4πε 𝑉 ≈ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑉 3 ⋅ 10 𝑉 ⎯⎯⎯⎯ 1 𝑠𝑡𝑎𝑡𝑉/𝑐𝑚 𝐸 = 4πε 𝐸 ≈ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝐸 3 ⋅ 10 𝑉/𝑚 ⎯⎯⎯ 4π 𝑠𝑡𝑎𝑡𝐶/𝑐𝑚 𝐷 = ⎯⎯⎯𝐷 ≈ 4π ⋅ 3 ⋅ 10 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝐷 ε 𝐶/𝑐𝑚 ⎯⎯⎯ 4π 𝐺 𝐵 = ⎯⎯⎯𝐵 = 10 ⎯⎯𝐵 μ 𝑇 ⎯⎯⎯⎯⎯ 𝑂𝑒 𝐻 = 4πμ 𝐻 = 4π ⋅ 10 ⎯⎯⎯⎯𝐻 𝐴/𝑚 1 χ = ⎯⎯⎯χ 4𝜋 1 χ = ⎯⎯⎯χ 4𝜋 [𝑠𝑡𝑎𝑡𝐶]: statcoulomb [𝑠𝑡𝑎𝑡𝑉]: statvolt [𝐺]: gauss [𝑂𝑒]: oersted The "≈ 3" is shorthand for exactly 2.99792458. SOME FUNDAMENTAL CONSTANTS Name Gaussian SI Impedance of free space 4π ⎯⎯ 𝑍 = μ 𝑐 𝑍 =ε 𝑐= ε Pagina 4 Impedance of free space 4π ⎯⎯ 𝑍 = ⎯⎯⎯ μ 𝑐 𝑍 =ε 𝑐= ⎯⎯ ε Electric constant 4π 1 1 = ⎯⎯⎯ ε = ⎯⎯⎯ 𝑍 𝑐 𝑍 𝑐 Magnetic constant 𝑍 𝑐 𝑍 1 = ⎯⎯⎯ μ = ⎯⎯ 4π 𝑐 Fine-structure constant (𝑒 ) 1 (𝑒 ) α = ⎯⎯⎯⎯⎯ α = ⎯⎯⎯⎯ ⎯⎯⎯⎯⎯ ℏ𝑐 4πε ℏ𝑐 Bohr radius ℏ 4πε ℏ 𝑎 = ⎯⎯⎯⎯⎯⎯⎯⎯ 𝑎 = ⎯⎯⎯⎯⎯⎯⎯ 𝑚 (𝑒 ) 𝑚 (𝑒 ) Bohr magneton 𝑒 ℏ 𝑒 ℏ μ = ⎯⎯⎯⎯⎯ μ = ⎯⎯⎯⎯ 2𝑚 𝑐 2𝑚 Pagina 5 Maxwell's equations martedì 28 novembre 2023 09:02 The electromagnetic (EM) wave is a transverse wave, i.e. the direction of propagation of the wave is perpendicular to the oscillations of the electric and magnetic fields. Note that often X-rays and 𝛾-rays are classified not by their wavelenght but by the way they are produced: 𝛾-rays are emitted in radioactive decays, X-rays instead are produced by bremsstrahlung; thus according to this classification they can have the same energy. Both are ionizing radiation, i.e. they are able to ionize a material, to produce free electrons and holes in the material. UV light can be ionizing for some materials but mostly it's not. Scintillators are materials which absorb high-energy photons and, through a conversion process, re-emit them in the UV-vis range, so that they can be detected by common photodetectors. The detection of radiation using scintillators is called "indirect", the radiation is first converted. The main goal of spectroscopy is to create a link between experimental quantities and material parameters. Experimental quantities include transmittance, reflectivity, absorbance, energy loss spectra etc., while material parameters (microscopic and macroscopic) include the complex dielectric function, the complex refractive index, polarizability etc. Microscopic Maxwell's equations (in Gaussian units): ∇⃗ ⋅ 𝑒⃗ = 4𝜋𝜌 , (1) 1 𝜕𝑏⃗ ∇⃗ × 𝑒⃗ = − ⎯⎯⎯⎯⎯, (2) 𝑐 𝜕𝑡 ∇⃗ ⋅ 𝑏⃗ = 0, (3) 1 𝜕𝑒⃗ 4𝜋 ∇⃗ × 𝑏⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯𝚥⃗ , (4) 𝑐 𝜕𝑡 𝑐 The electric field 𝑒⃗(𝑟⃗, 𝑡) and magnetic induction 𝑏⃗(𝑟⃗, 𝑡) are in lowercase to emphasize the microscopic nature, where "microscopic" means we are considering a small volume containing many atoms. In these equations one considers the contribution of each charged particle; for example, in a classical model in which the electrons and atomic nuclei are treated as point charges, the charge density is given by: ρ (𝑟⃗) = 𝑞 δ(𝑟⃗ − 𝑟⃗ ) Where 𝑞 is the charge of the 𝑖th particle. In a quantum mechanical treatment instead the charge density of the electrons would be expressed in terms of their electronic charge −𝑒 and wavefunction ψ(𝑟⃗): ρ , 𝑟 = −𝑒ψ∗ 𝑟 ψ 𝑟 Linear response Pagina 1 ρ , (𝑟⃗) = −𝑒ψ∗ (𝑟⃗)ψ(𝑟⃗) In order to treat a macroscopic system (of volume ΔV much greater than the wavelength of the radiation) we can switch to the macroscopic equations, where the macroscopic quantities are defined in terms of their microscopic counterparts as: Electric field strength: 1 𝐸⃗ (𝑟⃗) = ⟨𝑒⃗(𝑟⃗)⟩ = ⎯⎯⎯ 𝑒⃗(𝑟⃗ + 𝑟⃗ )𝑑𝑟⃗ Δ𝑉 Magnetic induction: 1 𝐵⃗(𝑟⃗) = 𝑏⃗(𝑟⃗) = ⎯⎯⎯ 𝑏⃗(𝑟⃗ + 𝑟⃗ )𝑑𝑟⃗ Δ𝑉 Charge density 1 ρ (𝑟⃗) = ⎯⎯⎯ ρ (𝑟⃗ + 𝑟⃗ )𝑑𝑟⃗ Δ𝑉 Current density 1 𝐽⃗ (𝑟⃗) = ⎯⎯⎯ 𝚥⃗ (𝑟⃗ + 𝑟⃗ )𝑑𝑟⃗ Δ𝑉 Where ∫ 𝑑𝑟⃗ = ∫ 𝑑𝑥 ∫ 𝑑𝑦 ∫ 𝑑𝑧. Thus: ∇⃗ ⋅ 𝐸⃗ = 4πρ , (1) 1 ∂𝐵⃗ ∇⃗ × 𝐸⃗ = − ⎯⎯⎯⎯⎯, (2) 𝑐 𝜕𝑡 ⃗ ⃗ ∇ ⋅ 𝐵 = 0, (3) 1 𝜕𝐸⃗ 4π ∇⃗ × 𝐵⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯𝐽⃗ , (4) 𝑐 𝜕𝑡 𝑐 The current density in a solid may be considered to consist of the contribution of electrons bound to nuclei (thus restricted to localized motion) and the contribution of electrons free to move through the solid: 𝐽⃗ = 𝐽⃗ + 𝐽⃗ In a similar way we can describe the net charge density by the sum of two contributions: one arising from the displacement of charge within the medium and one arising from the introduction of a net extra charge from an external source: ρ =ρ +ρ In the presence of an electric field, the atoms in the solid are polarized; the electronic charge is displaced wrt the nucleus and if the polarization is uniform there is no net charge moved in or out of the region considered. For non-uniform polarization though there is a net change in the charge: ρ = −∇⃗ ⋅ 𝑃⃗ Where the polarization 𝑃⃗ is the electric dipole moment per unit volume. In the presence of a time- dependent electric field the resulting time-dependent polarization gives rise to a current term: 𝑑𝑃⃗ 𝐽⃗ = ⎯⎯⎯ 𝑑𝑡 There may also be a contribution to the current density arising from electron spin, which can be included in terms of the macroscopic magnetization 𝑀⃗, defined as the magnetic dipole moment per unit volume and corresponding to a current density contribution of: 𝐽⃗ = 𝑐∇⃗ × 𝑀⃗ Though magnetic effects are usually small. Thus so far we have that: 𝜕𝑃⃗ 𝐽⃗ = 𝐽⃗ + 𝐽⃗ = ⎯⎯⎯+ 𝑐∇⃗ × 𝑀⃗ 𝜕𝑡 The free current density also consists of two parts: one arising from the motion of conduction electrons in the presence of an electric field 𝐽 and one consisting of a current density Linear response Pagina 2 electrons in the presence of an electric field 𝐽⃗ and one consisting of a current density introduced into the system from an external source 𝐽⃗. We can now rewrite the 1st and 4th of the Maxwell's equations: ∇⃗ ⋅ 𝐸⃗ = −4𝜋∇⃗ ⋅ 𝑃⃗ + 4πρ , (1) ⃗ 1 𝜕𝐸 4π 𝜕𝑃 ⃗ 4π 4π ∇⃗ × 𝐵⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯+ 4π∇⃗ × 𝑀⃗ + ⎯⎯⎯𝐽⃗ + ⎯⎯⎯𝐽⃗ , (4) 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 𝑐 If we define respectively the electric displacement 𝐷⃗ and magnetic field strength 𝐻⃗: 𝐷⃗ = 𝐸⃗ + 4𝜋𝑃⃗ 𝐻⃗ = 𝐵⃗ − 4𝜋𝑀⃗ Then we can rewrite all of the Maxwell's equations in the common macroscopic form: ∇⃗ ⋅ 𝐷⃗ = 4πρ , (1) ⃗ 1 ∂𝐵 ∇⃗ × 𝐸⃗ = − ⎯⎯⎯⎯⎯, (2) 𝑐 𝜕𝑡 ∇⃗ ⋅ 𝐵⃗ = 0, (3) 1 𝜕𝐷⃗ 4π 4π ∇⃗ × 𝐻⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯𝐽⃗ + ⎯⎯⎯𝐽⃗ , (4) 𝑐 𝜕𝑡 𝑐 𝑐 These equations are exact. We now consider an isotropic medium in the linear approximation regime. We can express the polarization 𝑃⃗ and magnetization 𝑀⃗ as: 𝑃⃗ = χ 𝐸⃗ 𝑀⃗ = χ 𝐻⃗ Where χ , χ are respectively the electrical and magnetic susceptibilities. Also we can note that the conduction current density is proportional to the electric field through the conductivity 𝜎: 𝐽⃗ = 𝜎𝐸⃗ In case of pure EM radiation (no electrical charges) 𝜎 is the transverse (or optical) conductivity, which describes the response of a system to a transverse electric field; this can be different from the DC electrical conductivity measured in experiments, which refers to a longitudinal field. The two become equal for cubic isotropic systems in the limit of zero frequency. The three above relationships link some materials properties with the stimulus. Materials don't always have a linear response; when the intensities of the electric and magnetic fields become very strong (e.g. with lasers) these simple laws don't hold anymore, and those relations should contain other terms proportional to higher powers of the fields. We can now rewrite the displacement and magnetic induction as: 𝐷⃗ = 𝐸⃗ + 4𝜋𝑃⃗ = 𝐸⃗ + 4𝜋𝜒 𝐸⃗ = 𝜀𝐸⃗ 𝐵⃗ = 𝐻⃗ + 4𝜋𝑀⃗ = 𝐻⃗ + 4𝜋𝜒 𝐻⃗ = 𝜇𝐻⃗ Where 𝜀 is the dielectric function and 𝜇 is the magnetic permeability, defined respectively as: ε ≔ 1 + 4𝜋χ μ ≔ 1 + 4𝜋χ The dielectric function is crucial, it expresses the capability of the materials to react by polarizing itself. It is a function of the frequency ω; the common dielectric constant is ε(ω = 0). Finally we can write another different version of the Maxwell's equations: ∇ ⋅ ε𝐸 = 4πρ , 1 μ 𝜕𝐻 Linear response Pagina 3 ∇⃗ ⋅ ε𝐸⃗ = 4πρ , (1) μ 𝜕𝐻⃗ ∇⃗ × 𝐸⃗ = − ⎯⎯⎯⎯⎯, (2) 𝑐 𝜕𝑡 ∇⃗ ⋅ 𝜇𝐻⃗ = 0, (3) 1 𝜕 𝜀𝐸⃗ 4π𝜎 4π ∇⃗ × 𝐻⃗ = ⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯𝐸⃗ + ⎯⎯⎯𝐽⃗ , (4) 𝑐 𝜕𝑡 𝑐 𝑐 These hold for any stimulus, EM radiation but also external charges or currents. We will now focus only on EM radiation, thus excluding external charges or currents. This means that the 1st and 4th equations become: ∇⃗ ⋅ ε𝐸⃗ = 0, (1) 1 𝜕 𝜀𝐸⃗ 4π𝜎 ∇⃗ × 𝐻⃗ = ⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯𝐸⃗ , (4) 𝑐 𝜕𝑡 𝑐 Finally we remark that we can split the total electric field into an external field and an induced response: 𝐸⃗ = 𝐸⃗ + 𝐸⃗ In this view 𝐸⃗ = 𝐸⃗ is the field that appears in the usual version of the Maxwell's equations, while 𝐷⃗ = 𝐸⃗ is the external field, it is the source. 𝑃⃗ is the induced response. In a similar way, as we have already seen, we can split the current density: 𝐽⃗ = 𝐽⃗ = 𝐽⃗ + 𝐽⃗ = 𝐽⃗ + 𝐽⃗ + 𝐽⃗ + 𝐽⃗ The first three are induced currents, while the third is the external one. The same is true for the charge density: ρ =ρ +ρ Where ρ is just the induced charge density. Linear response Pagina 4 (Formal solution of Maxwell's equations) mercoledì 7 febbraio 2024 20:58 //This part is not mandatory We recall the common form of the Maxwell's equations: ∇⃗ ⋅ 𝐸⃗ = 4πρ , (1) ⃗ 1 ∂𝐵 ∇⃗ × 𝐸⃗ = − ⎯⎯⎯⎯⎯, (2) 𝑐 𝜕𝑡 ∇⃗ ⋅ 𝐵⃗ = 0, (3) 1 𝜕𝐸⃗ 4π ∇⃗ × 𝐵⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯𝐽⃗ , (4) 𝑐 𝜕𝑡 𝑐 Equation (3) implies the existence of a vector potential 𝐴⃗ such that: 𝐵⃗ = ∇⃗ × 𝐴⃗ The equation (2) becomes: 1∂ ∇⃗ × 𝐸⃗ = − ⎯⎯⎯⎯ ∇⃗ × 𝐴⃗ 𝑐 𝜕𝑡 1 𝜕𝐴⃗ ⇒ ∇⃗ × 𝐸⃗ + ⎯⎯⎯⎯⎯ = 0 𝑐 𝜕𝑡 A vector whose curl is zero must be the gradient of some scalar potential function 𝜙: 1 𝜕𝐴⃗ 𝐸⃗ + ⎯⎯⎯⎯⎯= −∇⃗𝜙, (5) 𝑐 𝜕𝑡 Note that it's possible to define a new vector potential: 𝐴⃗ = 𝐴⃗ + ∇𝜓, (6) Such that it leaves the magnetic induction unchanged: 𝐵⃗ = ∇⃗ × 𝐴⃗ ≡ ∇⃗ × 𝐴⃗ Since ∇⃗ × ∇⃗ψ ≡ 0. There is no unique choice of vector potential; an arbitrary shift by a gradient of a scalar field (equation 6) is called "gauge transformation". Note that neither 𝐵⃗ nor 𝐸⃗ are changed by a gauge transformation as soon as we redefine ϕ = ϕ − ⎯ ⎯⎯⎯. One way to restrict the choice of vector potential is to specify its divergence; a common choice is the Coulomb gauge, for which we have: ∇⃗ ⋅ 𝐴⃗ = 0 Still it's possible to make transformations, for example: ∇⃗ ⋅ 𝐴⃗ = ∇⃗ ⋅ 𝐴⃗ + ∇𝜓 = ∇ 𝜓 So it is only necessary to require ∇ ψ = 0 everywhere. The most convenient choice of 𝐴⃗ depends on the problem to be solved; for optical properties of solids the Colulomb gauge is usually preferrable. If we substitute equation (5) into (1) we get: 1 𝜕𝐴⃗ 1∂ ∇⃗ ⋅ 𝐸⃗ = −∇⃗ ⋅ ⎯⎯⎯⎯⎯+ ∇⃗𝜙 = −∇ 𝜙 − ⎯⎯⎯⎯ ∇⃗ ⋅ 𝐴⃗ = 4πρ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 Since ∇⃗ ⋅ 𝐴⃗ = 0 in the Coulomb gauge then the scalar potential satisfies the Poisson's equation: ∇ ϕ = 4πρ So we have found that 𝜙 is actually the familiar potential from electrostatics; that is why this gauge is referred to as "Coulomb gauge". Linear response Pagina 5 is referred to as "Coulomb gauge". If we substitute equation (5) into (4) we get: 1 𝜕𝐸⃗ 4π 1 𝜕 1 𝜕𝐴⃗ 4π 1 ∂ 𝐴⃗ 1 ∂ 4π ∇⃗ × 𝐵⃗ = ⎯⎯⎯⎯⎯+ ⎯⎯⎯𝐽⃗ = − ⎯⎯⎯⎯ ⎯⎯⎯⎯⎯+ ∇⃗𝜙 + ⎯⎯⎯𝐽⃗ = − ⎯⎯ ⎯⎯⎯ − ⎯⎯⎯⎯ ∇⃗𝜙 + ⎯⎯⎯𝐽⃗ 𝑐 𝜕𝑡 𝑐 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 Substituting 𝐵⃗ = ∇⃗ × A⃗ and using ∇⃗ × ∇⃗ × 𝐴⃗ = ∇⃗ ∇⃗ ⋅ 𝐴⃗ − ∇ 𝐴⃗ we get: 1 ∂ 𝐴⃗ 1 ∂ 4π ∇⃗ ∇⃗ ⋅ 𝐴⃗ − ∇ 𝐴⃗ = − ⎯⎯ ⎯⎯⎯ − ⎯⎯⎯⎯ ∇⃗𝜙 + ⎯⎯⎯𝐽⃗ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 1∂ 𝐴 ⃗ 1 ∂𝜙 4π ∇ 𝐴⃗ − ⎯⎯ ⎯⎯⎯ − ∇⃗ ⎯⎯⎯⎯⎯+ ∇⃗ ⋅ 𝐴⃗ = − ⎯⎯⎯𝐽⃗ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 Choose once again the Coulomb gauge and we get: 1 ∂ 𝐴⃗ 1 ∂𝜙 4π ∇ 𝐴⃗ − ⎯⎯ ⎯⎯⎯ = ⎯⎯∇⃗ ⎯⎯⎯− ⎯⎯⎯𝐽⃗ , (7) 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 Now we digress briefly on the resolution of fields and currents into longitudinal and transverse parts. Recall that for electromagnetic waves in homogeneous media 𝐸⃗ , 𝐵⃗ are transverse, while static fields are longitudinal. The decomposition of a vector field 𝜉⃗ into longitudinal and transverse parts can be done in the following way: consider a vector field that satisfies the equation: ∇ 𝑊⃗ = −𝜉⃗ The solution is (see //Green's function): 1 𝜉⃗(𝑟⃗ ) 𝑊⃗ (𝑟⃗) = ⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗ 4π |𝑟⃗ − 𝑟⃗ | Using ∇⃗ × ∇⃗ × 𝑊⃗ = ∇⃗ ∇⃗ ⋅ 𝑊⃗ − ∇ 𝑊⃗ and defining: ∇⃗ ⋅ 𝑊⃗ = −𝑉, ∇⃗ × 𝑊⃗ = 𝐴⃗ We get: 𝜉⃗ = ∇⃗𝑉 + ∇⃗ × 𝐴⃗ This is known as Helmholtz's theorem. We can now define the longitudinal and transverse parts of 𝜉⃗ respectively as: 𝜉⃗ = ∇⃗𝑉, 𝜉⃗ = ∇⃗ × 𝐴⃗ So that 𝜉⃗ = 𝜉⃗ + 𝜉⃗ and: ∇⃗ × 𝜉⃗ = 0, ∇⃗ ⋅ 𝜉⃗ = ∇⃗ ⋅ 𝜉⃗ ∇⃗ ⋅ 𝜉⃗ = 0, ∇⃗ × 𝜉⃗ = ∇⃗ × 𝜉⃗ These in particular hold also for 𝐸⃗ , for which we have: 𝑖𝑞⃗ × 𝐸⃗ = 0 𝑖𝑞⃗ ⋅ 𝐸⃗ = 0 As 𝐸⃗ , 𝐸⃗ are respectively parallel and perpendicular to the direction of propagation, given by the wavecector 𝑞⃗. If we now get back to equation (7) we can note that ⎯∇⃗ ⎯⎯ must correspond to a current density related only to the longitudinal part of 𝐽⃗ ; this is because only the longitudinal part of the electric field is derivable from a potential function 𝜙, since ∇⃗ × E⃗ = 0. We can show this explicitly; the solution to the Poisson's equation is obtained from electrostatics (see //Green's function), thus: ∂ ρ (𝑟⃗ , 𝑡) ∂ϕ(𝑟⃗, 𝑡) ⎯⎯ ρ (𝑟⃗ , 𝑡) ϕ(𝑟⃗, 𝑡) = ⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗ ⇒ ⎯⎯⎯⎯⎯⎯⎯= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗𝜕𝑡 |𝑟⃗ − 𝑟⃗ | 𝜕𝑡 |𝑟⃗ − 𝑟⃗ | By using the continuity equation: ∂ρ ∇ ⋅𝐽 =− 𝜕𝑡 Linear response Pagina 6  ∂ρ ∇⃗ ⃗ ⋅ 𝐽⃗ = − ⎯⎯⎯⎯⎯ 𝜕𝑡 Where ∇⃗ ⃗ indicates the operation is wrt 𝑟⃗ , then: ∂ϕ(𝑟⃗, 𝑡) ∇⃗ ⃗ ⋅ 𝐽⃗ ⎯⎯⎯⎯⎯⎯⎯= − ⎯⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗ 𝜕𝑡 |𝑟⃗ − 𝑟⃗ | We can decompose 𝐽⃗ = 𝐽⃗ + 𝐽⃗ , then as usual we have that the divergence of a transverse vector field is zero ∇⃗ ⋅ 𝐽⃗ = 0 , so that by applying on both sides ∇ (𝑟⃗): ∂ϕ(𝑟⃗, 𝑡) ∇⃗ ⃗ ⋅ 𝐽⃗ ∇ ⎯⎯⎯⎯⎯⎯⎯= −∇ ⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗ 𝜕𝑡 |𝑟⃗ − 𝑟⃗ | Finally we exploit the following property (see //here): 1 ∇ − ⎯⎯⎯⎯⎯⎯⎯ = 4πδ(𝑟⃗ − 𝑟⃗ ) |𝑟⃗ − 𝑟⃗ | So that: ∂ϕ(𝑟⃗, 𝑡) ∇⃗ ⋅ ∇⃗ ⎯⎯⎯⎯⎯⎯⎯= 4𝜋∇⃗ ⋅ 𝐽⃗ (𝑟⃗, 𝑡) 𝜕𝑡 ∂ϕ ⇒ ∇⃗ ⎯⎯⎯= 4𝜋𝐽⃗ 𝜕𝑡 By using this result we can rewrite equation (7) as a wave equation for 𝐴⃗ expressed solely in terms of the transverse current density: 1 ∂ 𝐴⃗ 1 4π ∇ 𝐴⃗ − ⎯⎯ ⎯⎯⎯ = ⎯⎯4𝜋𝐽⃗ − ⎯⎯⎯𝐽⃗ 𝑐 𝜕𝑡 𝑐 𝑐 1 ∂ 𝐴⃗ 4π ⇒ ∇ 𝐴⃗ − ⎯⎯ ⎯⎯⎯ = − ⎯⎯⎯𝐽⃗ , (8) 𝑐 𝜕𝑡 𝑐 The Coulomb gauge is sometimes referred to as the "transverse gauge", and the reason is that in this gauge the vector potential is also transverse, as is clear from equation (8). Linear response Pagina 7 Wave equation martedì 12 dicembre 2023 09:46 Recall the following vector identity: ∇⃗ × ∇⃗ × 𝐸⃗ = ∇⃗ ∇⃗ ⋅ 𝐸⃗ − ∇ 𝐸⃗ By using this and the 2nd and 4th of the Maxwell's equations (without external charges or currents) we can write (∗): εμ ∂ 𝐸⃗ 4𝜋σμ 𝜕𝐸⃗ ∇⃗ × ∇⃗ × 𝐸⃗ = − ⎯⎯ ⎯⎯⎯ − ⎯⎯⎯⎯⎯⎯⎯⎯ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 Finally we also recall that ∇⃗ ⋅ 𝐸⃗ = 0 since there are no external sources (ρ = 0), so we get: εμ ∂ 𝐸⃗ 4𝜋σμ 𝜕𝐸⃗ ∇ 𝐸⃗ = ⎯⎯ ⎯⎯⎯ + ⎯⎯⎯⎯⎯⎯⎯⎯ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 This is the fundamental wave equation obtained in case of no external charges or currents. The solutions are restricted to transverse plane waves since ∇⃗ ⋅ E⃗ = 0 in the absence of a net charge density. The conductivity 𝜎 is the "optical conductivity", since the energy absorption related to electronic transitions corresponds to a transverse current that does not include the conventional longitudinal current obtained with a battery connected across the sample. At sufficiently long wavelenghts the transverse optical conductivity approaches the ordinary electrical conductivity for isotropic materials. For anisotropic materials though the optical conductivity and dielectric functions must be treated as tensors. We consider the propagation of a plane monochromatic wave within an isotropic medium: 𝐸⃗ = 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 𝑞⃗ ⊥ 𝐸⃗ is the wavevector and we take it as a complex quantity. We want to describe the real propagation of waves in matter, and defining 𝑞 as complex is useful to describe the different things that can happen to a wave (propagation, reflection, absorption). If we substitute this solution in the wave equation we find the dispersion relation (∗∗): ω 4πσ 𝑞 = μ ⎯⎯⎯ ε + 𝑖 ⎯⎯⎯⎯ 𝑐 ω 𝑞 is frequency-dependent, it contains the meaning of spectroscopy. It's also possible to define a complex refractive index 𝑛 ≔ (𝑛 + 𝑖𝑘) in such a way that: ω ω 𝑞 = ⎯⎯𝑛 = ⎯⎯(𝑛 + 𝑖𝑘) 𝑐 𝑐 𝑛 is the "usual" refractive index found in geometrical optics, 𝑘 instead is called "extinction coefficient" and is related to absorption. Now we compare this relationship with the one of 𝑞 : 4πσ 𝑛 = (𝑛 + 𝑖𝑘) = μ ε + 𝑖 ⎯⎯⎯⎯ ω 4πσ ⇒ 𝑛 − 𝑘 + 𝑖2𝑛𝑘 = με + 𝑖μ ⎯⎯⎯⎯ ω Thus we can put equal the real and imaginary parts: 𝑛 −𝑘 ε = ⎯⎯⎯⎯⎯⎯⎯ 𝜇 4πσ 2𝑛𝑘 ⎯⎯⎯⎯= ⎯⎯⎯⎯ ω 𝜇 We now define a complex dielectric function: 𝑛 ε ≔ ε + 𝑖ε = ⎯⎯⎯ 𝜇 And so: Linear response Pagina 8 And so: 𝑛 −𝑘 ε = ⎯⎯⎯⎯⎯⎯⎯ 𝜇 2𝑛𝑘 4πσ ε = ⎯⎯⎯⎯= ⎯⎯⎯⎯ μ ω Where ε is the old ε used in the equations above. Thus: ω ω 𝑞 = 𝜇 ⎯⎯⎯ε = 𝜇 ⎯⎯⎯(ε + 𝑖ε ) 𝑐 𝑐 We will see that also 𝑛, 𝑘 depend on frequency. We now substitute in 𝐸⃗ the form of 𝑞⃗ we found: ω ω 𝐸⃗ = 𝐸⃗ exp − ⎯⎯𝑘⃗ ⋅ 𝑟⃗ exp 𝑖 ⎯⎯𝑛⃗ ⋅ 𝑟⃗ − 𝜔𝑡 𝑐 𝑐 The first term describes the attenuation of the wave in the material, while the second describes propagation of a wave with phase velocity 𝑐/𝑛. We define the absorption coefficient as the relative reduction of the intensity of the beam as a function of the distance 𝑟 travelled by the beam: 1 𝑑𝐼 α = − ⎯⎯⎯⎯⎯ 𝐼 𝑑𝑟 From this we can obtain the Lambert-Beer law: 𝐼=𝐼 𝑒 Recall that 𝐼 ∝ 𝐸⃗ : 2ω 𝐼=𝐼 𝑒 ∝ 𝐸⃗ exp − ⎯⎯⎯𝑘⃗ ⋅ 𝑟⃗ 𝑐 Therefore we found that: 2𝜔𝑘 4𝜋𝑘 2𝜋𝑐 α = ⎯⎯⎯⎯= ⎯⎯⎯⎯, λ = ⎯⎯⎯ 𝑐 𝜆 𝜔 If the material is isotropic all of these quantities don't have any dependence on the direction of the field. If instead the material is anisotropic these quantities change in accordance with the direction of propagation, they'll become vectors or tensors. (∗): μ 𝜕𝐻⃗ μ∂ μ ∂ 𝜀 𝜕𝐸⃗ 4π𝜎 εμ ∂ 𝐸⃗ 4πσμ 𝜕𝐸⃗ ∇⃗ × ∇⃗ × 𝐸⃗ = ∇⃗ × − ⎯⎯⎯⎯⎯ = − ⎯⎯⎯⎯ ∇⃗ × 𝐻⃗ = − ⎯⎯⎯⎯ ⎯ ⎯⎯⎯+ ⎯⎯⎯⎯𝐸⃗ = − ⎯⎯ ⎯⎯⎯ − ⎯⎯⎯⎯⎯⎯⎯⎯ 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 𝜕𝑡 𝑐 𝑐 𝜕𝑡 𝑐 𝜕𝑡 (∗∗): εμ ∂ 4𝜋σμ 𝜕 ∇ 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 = ⎯⎯ ⎯⎯⎯ 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 + ⎯⎯⎯⎯⎯⎯⎯ 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 𝑐 𝜕𝑡 𝑐 𝜕𝑡 εμ 4𝜋σμ −𝑞 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 = ⎯⎯ (−ω ) 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 + ⎯⎯⎯⎯⎯(−𝑖ω) 𝐸⃗ exp 𝑖 𝑞⃗ ⋅ 𝑟⃗ − 𝜔𝑡 𝑐 𝑐 εμ 4πσμ 𝑞 = ⎯⎯ ω + 𝑖 ⎯⎯⎯⎯⎯ω 𝑐 𝑐 ω 4πσ 𝑞 = μ ⎯⎯⎯ ε + 𝑖 ⎯⎯⎯⎯ 𝑐 ω Linear response Pagina 9 Lorentz model martedì 12 dicembre 2023 10:10 The Lorentz model is for insulators and semiconductors, so materials which have a band gap and there are no free charges (for non-doped semiconductors). The core of this model is the harmonic oscillator; we depict a material as an ensemble of harmonic oscillators, where (generally) one electron is bound to the nucleus, called "optical electron". This system can oscillate when subjected to some stimulus, like an optical wave. This is a classical harmonic oscillator. Here we have two strong approximations: the first is that we're not considering any quantum mechanical effects, the second is that we're not considering any anharmonicity of the oscillator. This is valid when the intensity of the stimulus is moderately low. The equation of motion is: 𝑚𝑟̈⃗ + 𝑚Γ𝑟̇⃗ + 𝑚ω 𝑟⃗ = −𝑒𝐸⃗ (ω) Where 𝑟⃗ is the displacement of the electron from its equilibrium position and Γ is the damping term; it represents any kind of phenomena that hampers the motion of the electrons (scattering etc.). ω is the resonance frequency of the oscillator, while 𝐸⃗ (ω) is the stimulus and we write "local" because we consider the field in the point where the electron is. This is the equation of a harmonic oscillator damped and forced. We remark that: 𝐸⃗ = 𝐸⃗ 𝑒 In principle we would have to consider two masses, the one of the electron and the one of the nucleus, thus using a two-body oscillator approach, but the nucleus is much heavier than the single electron, and so we can consider that it does not move (it has an infinite mass). We also neglect the Lorentz force, because the speed of the electrons is much smaller than the speed of light: 𝑒 𝐹⃗ = − ⎯ 𝑣⃗ × 𝐵⃗ ≃ 0 𝑐 It can be demonstrated that a solution of that differential equation is: 1 𝑒𝐸⃗ 𝑟̂⃗ = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) − 𝑖Γ𝜔 The dipole moment due to the polarization is just: 1 𝑒 𝐸⃗ 𝑝̂⃗ = −𝑒𝑟̂⃗ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) − 𝑖Γ𝜔 Now we want to introduce a response function that relates the electric field to the dipole moment, and this is the atomic polarizability (α): 𝑝̂⃗ = 𝛼 (ω)𝐸⃗ This is a response function because it relates the stimulus 𝐸⃗ to a response of the material (𝑝⃗). This linearity is valid when the stimulus is not too intense; if the stimulus becomes too strong the linearity will break and we'll need additional terms. By the equation above we can write the polarizability as: 1 𝑒 α ω = 𝑚 ω − ω − 𝑖Γ𝜔 Linear response Pagina 10  1 𝑒 α(ω) = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) − 𝑖Γ𝜔 This quantity depends on frequency. If we consider to have 𝑁 atoms per unit volume we can define the macroscopic polarization 𝑃⃗ as: 𝑃⃗ = 𝑁⟨𝑝⃗⟩ = 𝑁α 𝐸⃗ = χ 𝐸⃗ Where 𝐸⃗ is not the local field anymore but the total field and χ = 𝑁α is the electric susceptibility. Now we can recover the dielectric function: 4𝜋𝑁𝑒 1 ε(ω) = 1 + 4πχ = 1 + 4𝜋𝑁α = 1 + ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) − 𝑖Γ𝜔 This is also dependent on frequency. In these last steps we have replaced the local field with the total field, but this assumption is not trivial. This is reasonable in case of metals, since the conduction electrons are free to move over all regions of the material, so they experience an average total electric field. This model though is meant for materials in which the electrons are bound, so it actually not so good for semiconductors and intrinsically incorrect for insulator. Nevertheless this model can be employed for first approximations. We can also say that the difference between the local and total field is not very big, still this is a simplified description of reality only useful in some applications. As a further assumption we consider non-magnetic materials (μ = 1). We can split the real and imaginary part of the dielectric function, and we get: 4𝜋𝑁𝑒 ω −ω ε = 𝑛 − 𝑘 = 1 + ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) + Γ ω 4𝜋𝑁𝑒 Γω ε = 2𝑛𝑘 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) + Γ ω Dispersion (𝑛) and absorption (𝑘) are mixed together in the dielectric function, so they are not independend of each other. Until now we have only considered one kind of electron, but in the general case we can write: 4𝜋𝑒 𝑁 ε = 1 + ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 ω − ω − 𝑖Γω 𝑁 is the number of electrons per unit volume that have natural frequency ω. This model is completely classical and so we assume any natural frequency is ok, but a more complete description would consider the quantum mechanics of the system and that natural frequencies are related to the jumps in energy levels. The quantum mechanical analogue of this formula is: 4𝜋𝑒 𝑁𝑓 ε = 1 + ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 ω − ω − 𝑖Γω Where 𝑓 is the "oscillator strength" and represents the probability of an electronic transition between two levels separated by ℏω. It is possible to show that the reflectance wave amplitude 𝑟̂ (ratio of the strength of the reflected field 𝐸 to the strength of the incident field 𝐸 ) at normal incidence and for a non-magnetic material is: 𝐸 1−𝑛 𝑟̂ = ⎯⎯⎯= ⎯⎯⎯⎯⎯ 𝐸 1+𝑛 The reflectivity 𝑅 then is just given by: 1−𝑛 (1 − 𝑛) + 𝑘 𝑅 = 𝑟̂ ∗ 𝑟̂ = ⎯⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 1+𝑛 (1 + 𝑛) + 𝑘 Linear response Pagina 11 We can also recover the expressions of 𝑛, 𝑘 in terms of the components of the dielectric contstant: ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 1 ⎯⎯⎯⎯⎯⎯ 𝑛 = ⎯⎯ ε + ε + ε 2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 1 ⎯⎯⎯⎯⎯⎯ 𝑘 = ⎯⎯ ε + ε − ε 2 Now we can examine to these quantities when we vary the frequency of the incident field. In a material with only one resonant frequency ω we would get: If we look at the trend of ε , ε , 𝑛, 𝑘, 𝑅 all together: Curves calculated for ℏω = 4 𝑒𝑉, ℏΓ = 1 𝑒𝑉 and ⎯⎯⎯⎯⎯= 60. Region I (T): ε > 0, ε ≃0 𝑛 dominates whike 𝑘 is very small; the material is highly dispersive but not absorptive, thus the material is transparent. Region II (A): ε variable, ε >0 Both 𝑛, 𝑘 are strong, so we have both strong abrosption and some reflection. Transmission is very low. Region III (R): ε < 0, ε ≃0 We have a small 𝑛 but a large 𝑘; we still have absorption but the predominant phenomenon is relfection. The point where 𝑛 = 𝑘 is called "plasma frequency"; the real part of ε is zero. Linear response Pagina 12  relfection. The point where 𝑛 = 𝑘 is called "plasma frequency"; the real part of ε is zero. Region IV (T): ε → 1, ε ≃0 𝑘 tends to zero and 𝑛 tends to one, so again we have transparency. Note that this distinction between regions is not too strong, we always have some of all phenomena, but some are stronger than others. Reflectivity is never really zero, but in the first and last regions it's quite small, while it increases in the second region and it becomes the dominating phenomenon in the third region. Some reflection is always there, even if the material is transparent, so transmission is never 100%. At the plasma frequency as we said we have: ε =𝑛 −𝑘 = 0 So this means that: 4𝜋𝑁𝑒 ω −ω ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −1 𝑚 (ω − ω ) + Γ ω We can find the frequency ω at which this happens; this frequency is such that ω ≫ ω ≫ Γ and by making the full calculations we get: ⎯⎯⎯⎯⎯⎯ 4𝜋𝑁𝑒 ω = ⎯⎯⎯⎯⎯⎯ 𝑚 Linear response Pagina 13 Drude model martedì 19 dicembre 2023 09:52 This model is for metals, materials in which at least a fraction of electrons are able to respond to EM radiation by moving through the material. Inner, bound electrons are not considered in this model. The starting point is again the Lorentz model; to neglect the existence of a resonance frequency we can just put it to zero, moreover here the mean value of the local field is really the total field. We also introduce the mean time between electronic collision 𝜏 defined as Γ = τ. The results are easily obtained from the Lorentz model for ω = 0: 4𝜋𝑁𝑒 1 ε = 1 − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 ω +Γ 4𝜋𝑁𝑒 Γ ε = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 ω(ω + Γ ) Recovering the definition of plasma frequency we can simplify these: ω τ ε = 1 − ⎯⎯⎯⎯⎯⎯⎯⎯ 1+ω τ ω τ ε = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ω(1 + ω τ ) Calculations for ⎯⎯⎯⎯⎯= 36 𝑒𝑉 and ℏΓ = 0.02 𝑒𝑉 In principle an ideal metal without strong damping terms would be strongly reflective before the plasma frequency and then become transparent. This means we can use metals as filters. In a metal the plasma frequency is on the order of ω ∼ 10 𝑠 , while the mean time between collision is τ ∼ 10 𝑠, so: ω τ≫1 Thus for ω > ω we can write: ω ε = 𝑛 − 𝑘 ≃ 1 − ⎯⎯⎯ ω In this region the metal is transparent; moreover we have 𝑛 ≫ 𝑘, and so: Linear response Pagina 14 In this region the metal is transparent; moreover we have 𝑛 ≫ 𝑘, and so: ω 𝑛 ≃ 1 − ⎯⎯⎯ ω This means that for ω = ω we have 𝑛 = 0. Since the phase velocity of a wave in a medium is 𝑣 = ⎯ then at the plasma frequency we have 𝑣 → ∞ and λ → ∞. That the wavelength becomes infinite means the electrons are all oscillating in phase; however, there is no polarization charge density as with a true plasma oscillation. Let's consider a slab of metal and an initially applied field which generates a charge separation: Because of the force between the two regions of unbalanced charge, they will be attracted towards each other. In the absence of a damping mechanism, there will be an overshoot, leading to the situation depicted in the figure, but with the charges reversed. The system will continue to oscillate in this manner at a characteristic frequency known as the plasma frequency. The oscillations are called plasma oscillations. At the plasma frequency all the electrons move together, they oscillate in phase. The polarization is: 𝑁𝑒𝐴(𝛿𝑥)𝐿 𝑃 = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯= −𝑁𝑒(𝛿𝑥) 𝐴𝐿 Where 𝑁 is the number of charges created and 𝐴 is the area of the slab perpendicular to 𝐿. The electric field generated by this polarization is: 𝐸 = −4𝜋𝑃 = 4𝜋𝑁𝑒(𝛿𝑥) Each electron feels a force equal to: 𝑚𝑥̈ = −𝑒𝐸 = −4𝜋𝑁𝑒 (𝛿𝑥) This is just an harmonic oscillator, with: 4𝜋𝑁𝑒 ω = ⎯⎯⎯⎯⎯⎯ 𝑚 We consider the case of sodium and we try to calculate ω , the region of transparency and the absorption coefficient at low frequencies. For sodium we have that the density of conduction electrons is 𝑁 = 2.65 ⋅ 10 𝑐𝑚 , and thus: ⎯⎯⎯⎯⎯⎯ 4𝜋𝑁𝑒 ω = ⎯⎯⎯⎯⎯⎯ = 9.18 ⋅ 10 𝑠 𝑚 The wavelength at which transparency starts is: 2𝜋𝑐 λ = ⎯⎯⎯ = 2.05 ⋅ 10 𝑐𝑚 = 205 𝑛𝑚 ω This is the UV, so sodium metal becomes transparent below 205 nm. For ω < ω we have 𝑛 ≪ 𝑘 and so we can write: ω 𝑘 = −1 ω Linear response Pagina 15  ω 𝑘 = ⎯⎯⎯− 1 ω Then: ⎯⎯⎯⎯⎯⎯ 2𝜔𝑘 2ω ω 2ω α = ⎯⎯⎯⎯= ⎯⎯⎯ ⎯⎯⎯− 1 ≈ ⎯⎯⎯⎯= 6.12 ⋅ 10 𝑐𝑚 𝑐 𝑐 ω 𝑐 Below the plasma frequency the metal is also able to absorb, even if reflection is the main phenomenon occurring. We can list the 𝜆 of onset of the region of transparency for various metals: Metal Calculated (nm) Experimental (nm) 𝐿𝑖 154 205 𝑁𝑎 205 210 𝐾 282 315 𝑅𝑏 312 360 𝐶𝑠 350 440 The discrepancy arises from the fact that the Drude model is a rather crude description of reality, for example we're not taking into consideration the contribution of bound electrons; nevertheless the trend is correct. Linear response Pagina 16 Real materials martedì 9 gennaio 2024 09:22 Reflectivity of silicon (semiconductor) Reflectivity of 𝐾𝐶𝑙 (insulator) In silicon it looks a bit like the ideal reflectivity (TART model); the peaks are due possibly to the response of deeper electrons. The case of 𝐾𝐶𝑙 is much more complicated, we cannot appreciate all the 4 regions. We have a first transparency region and then we have a set of peaks, that are actually several resonance peaks. In the Lorentz model we have considered only the presence of a single resonance frequency, whereas real materials have several resonant frequencies, so the region of absorption is very complicated. Aluminium The reason for the peak at low frequencies in aluminium resides in the presence of inter-band transitions; metals also have deeper bands, that give rise to absorption processes. Linear response Pagina 17 transitions; metals also have deeper bands, that give rise to absorption processes. This is called "parallel band effect", and the transitions which give rise to that peak are marked by arrows in the image on the right. These transition are not very probable but occur all in the same region, so they sum up. Copper Silver In these two cases the reflectivity drops well before ω ; both metal have only one 𝑛𝑠 electron and so the reason for this is the occurrence of transitions between (𝑛 − 1)𝑑 and 𝑛𝑠 orbitals, for example for copper we have: Silver has a completely reflecting appearance, this discrepancy from ideal reflectivity starts in the UV, whereas in the case of copper it begins already in the visible region, in fact copper appears not as reflective as other metals but has a reddish colour. Linear response Pagina 18 Kramers-Kronig relations martedì 9 gennaio 2024 10:13 Here we are going to show that the quantities ε , ε , 𝑛, 𝑘 are not independent of each other. Let's start from a general mathematical relation that links a stimulus 𝑓(𝑟 , 𝑡 ) to a response 𝑋(𝑟, 𝑡) through a response function 𝐺(𝑟, 𝑟 , 𝑡, 𝑡 ): 𝑋(𝑟⃗, 𝑡) = 𝐺(𝑟⃗, 𝑟⃗ , 𝑡, 𝑡 )𝑓(𝑟⃗ , 𝑡 )𝑑𝑟⃗ 𝑑𝑡 ℝ What this formula is saying is that the response 𝑋 at point 𝑟 and time 𝑡 can depend on a stimulus 𝑓 present in any position 𝑟 and time 𝑡 through the response function 𝐺. We make two important hypotheses: Local approximation: the response depends on the stimulus only in the point where the response is considered. If we also realize that the flow of time is uniform then: 𝐺(𝑟⃗, 𝑟⃗ , 𝑡, 𝑡 ) = δ(𝑟⃗ − 𝑟⃗ )𝐺(𝑡 − 𝑡 ) This is the same as considering that the wavelengths of the variable fields are sufficiently long that spatial dispersion can be neglected. The system is causal, that is there can be no response before there is a stimulus: 𝐺(𝑡 − 𝑡 ) ≡ 0 for 𝑡 < 𝑡 Thus we can rewrite: 𝑋(𝑡) = 𝐺(𝑡 − 𝑡 )𝑓(𝑡 )𝑑𝑡 ℝ Taking 𝑓(𝑡 ) = δ(𝑡 − 𝑡 ) we obtain: 𝑋(𝑡) = 𝐺(𝑡 − 𝑡 ) Thus we see that the response function is a Green's function (see //Green's function) which describes the response of a system at time 𝑡 to a stimulus at some earlier time 𝑡. To consider the response in function of time is not practical from an experimental POV, apparatuses usually work as a function of frequency. We can convert all these quantities by Fourier transform: 𝑓(ω) = 𝑓(𝑡)𝑒 𝑑𝑡 ℝ 𝑋(ω) = 𝑋(𝑡)𝑒 𝑑𝑡 ℝ 𝐺(ω) = 𝐺(𝑡 − 𝑡 )𝑒 𝑑𝑡 ℝ By replacing these in the formula above and performing the full calculations we get: 𝑋(ω) = 𝐺(ω)𝑓(ω) In terms of the FTs, a monochromatic stimulus 𝑓(ω) is just multiplied by some number 𝐺(ω) to give the response 𝑋(ω). If 𝐺(ω) has a pole at ω = ω (i.e. the denominator of 𝐺(ω) vanishes) then there can be a finite response in the absence of a stimulus. This means that ω corresponds to the frequency of a normal mode of the system. Let ω = ω + 𝑖ω , then: 𝐺(ω) = 𝐺(𝑡 − 𝑡 )𝑒 𝑑𝑡 = 𝐺(𝑡 − 𝑡 )𝑒 𝑒 𝑑𝑡 ℝ ℝ For 𝑡 − 𝑡 > 0, the factor 𝑒 is bounded only in the upper-half plane, thus the assumption of causality requires that the integral be evaluated in the upper-half plane. Now let 𝜔 be on the real axis, then according to the Cauchy theorem we can write: 1 𝐺(ω ) 𝐺(ω) = ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯𝑑ω 𝑖𝜋 ℝω − ω Linear response Pagina 19 Where 𝒫 denotes the principal value. This result is obtained by integrating over the following contour: Moreover we assume that 𝐺(ω) decreases such that as the radius of the semicircle approaches infinity the contribution to the integral over the semicircle approaches zero. If we split 𝐺(𝜔) into its real and imaginary parts we get: 1 ℑ𝔪 𝐺(ω ) ℜ𝔢 𝐺(ω) = ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑ω π ℝ ω −ω 1 ℜ𝔢 𝐺(ω ) ℑ𝔪 𝐺(ω) = − ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑑ω 𝜋 ℝ ω −ω These two expressions connect the real and imaginary parts of the response function. Let's take for example the dielectric function ε = ε + iε , which links the external and internal fields 𝐷 = ε𝐸 (to be precise, since 𝐷 is the external field the response function would be ε ). The refractive index 𝑛 = 𝑛 + 𝑖𝑘 could also be considered a response function. Recall the expression of the polarization we have chosen in the Lorentz model: 𝑁𝑒 1 𝑃⃗(ω) = 𝑁α𝐸⃗ (ω) = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝐸⃗ (ω) 𝑚 (ω − ω ) − 𝑖Γω Then the response function is: 𝑁𝑒 1 𝐺(ω) = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑚 (ω − ω ) − 𝑖Γω It can be demonstrated that this 𝐺(ω) is causal (//see Wooten book, section 6.1), i.e. no response of the system (polarization) can occur before an input. Since this model for polarizability is a causal one then 𝜀̂ − 1 = 𝜀 − 1 + 𝑖𝜀 = 4𝜋𝑁𝛼 must also be causal. By comparison with the two above equations for the real and imaginary parts of 𝐺(ω) we get: 1 ε (ω ) ε (ω) − 1 = ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯𝑑ω π ω −ω 1 ε (ω ) − 1 ε (ω) = − ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑ω π ω −ω We are interested in physical systems with real inputs and outputs, so we have: 𝐺(−ω) = 𝐺 ∗ (ω) This means that: ε(−ω) = ε∗ (ω) ⇒ ε (−ω) = ε (ω), ε (−ω) = −ε (ω) We can use these to rewrite the dispersion relations in the usual form in terms of integrals over positive frequencies: 2 ω ε (ω ) ε (ω) − 1 = ⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯⎯⎯𝑑ω π ω −ω 2ω ε (ω ) − 1 ε (ω) = − ⎯⎯⎯𝒫 ⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑ω π ω −ω The link between real and imaginary parts has two important consequences. The first one is of fundamental nature: a modification of the real part implies necessarily a modification of the imaginary part and vice versa. Given for example a change in ε around frequency ω, the Linear response Pagina 20 imaginary part and vice versa. Given for example a change in ε around frequency ω, the denominator of the above relations implies that the modification in ε should not happen very far from ω too, otherwise the integral goes to zero. ε is more related to dispersion, ε is more related to absorption, but both are linked. The second important consequence is practical: because of this link we can imagine that if we measure well the real quantity we can calculate the imaginary part through that integral, which is in principle very difficult but we can reasonably link the integrals in a region not too far from ω, thanks to the denominator. Imagine we have an input formed as the superposition (a) of many Fourier components (b) and a system able to absorb only one of them. If none of the other non-absorbed components is affected by this absorption then the output (c) would contain the complement of the absorbed component B during times before the onset of the input wave, in contradiction with the causality principle: The causality principle implies that absorption of one frequency must be accompanied by a compensating shift of phase of all the other frequencies as prescribed by dispersion relations. Thus the response function must have real and imaginary parts to describe both absorption and phase shifts, and the real part at a single frequency must be related to the imaginary part at all the other frequencies and vice versa. Linear response Pagina 21 //Green's function sabato 17 febbraio 2024 15:06 //This is not part of the program A differential operator is an operator defined as a function of the differentiation operator. In particular, a linear differential operator ℒ is a linear combination of differentiation operators. For example: 𝑑 ℒ = ⎯⎯⎯ + 𝑥 𝑑𝑥 An ordinary differential equation (ODE) is said to be linear if we can write: 𝑦( ) = 𝑎 (𝑥)𝑦 ( ) + 𝑟(𝑥) Where 𝑦, 𝑎 (𝑥), 𝑟(𝑥) are functions of the independent variable 𝑥, in particular 𝑟(𝑥) is called "source term"; a linear ODE is said to be "inhomogeneous" if 𝑟(𝑥) ≠ 0. A Green's function 𝐺 is the impulse response of an inhomogeneous linear differential operator ℒ defined on a domain with specified initial or boundary conditions. If ℒ is a linear differential operator then: The Green's function 𝐺 is the solution of the equation ℒ𝐺 = 𝛿, where 𝛿 is Dirac's delta The solution of the initial-value problem ℒ𝑦 = 𝑓 is the convolution (𝐺 ∗ 𝑓) Formally, a Green's function 𝐺(𝑥, 𝑠) of a linear differential operator ℒ(𝑥) acting on distributions over a subset of the Euclidean space ℝ , at a point 𝑠, is any solution of: ℒ𝐺(𝑥, 𝑠) = δ(𝑥 − 𝑠), (1) This says that the Green's function is the solution to the differential equation with a forcing term given by a point source. The properties of a Green's function can be exploited to solve arbitrary differential equations of the form: ℒ𝑢(𝑥) = 𝑓(𝑥), (2) If we multiply equation (1) by 𝑓(𝑠) and the integrate wrt 𝑠 we obtain: ℒ𝐺(𝑥, 𝑠)𝑓(𝑠)𝑑𝑠 = δ(𝑥 − 𝑠)𝑓(𝑠)𝑑𝑠 = 𝑓(𝑥) Because ℒ(𝑥) is linear and only acts on 𝑥 we can take it out of the integral: ℒ 𝐺(𝑥, 𝑠)𝑓(𝑠)𝑑𝑠 = 𝑓(𝑥) By comparing this with equation (2) one obtains that: 𝑢(𝑥) = 𝐺(𝑥, 𝑠)𝑓(𝑠)𝑑𝑠 , (3) This is a solution of ℒ𝑢(𝑥) = 𝑓(𝑥), where 𝑓(𝑥) is the source term. In general Green's functions are distributions. GREEN'S FUNCTION FOR THE LAPLACIAN Suppose that the linear differential operator ℒ is the Laplacian ∇ and that there is a Green's function 𝐺 for the Laplacian. The defining property of the Green's function still holds: ℒ𝐺(𝑟⃗, 𝑟⃗ ) = ∇ 𝐺(𝑟⃗, 𝑟⃗ ) = δ(𝑟⃗ − 𝑟⃗ ) It can be demonstrated that, with no boundary conditions, the Green's function for the Laplacian is: 1 𝐺 𝑟, 𝑟 = − 4π 𝑟 − 𝑟 Linear response Pagina 22  1 𝐺(𝑟⃗, 𝑟⃗ ) = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯ 4π|𝑟⃗ − 𝑟⃗ | Using this one may solve the Poisson equation (∇ ϕ = 4πρ) to yield: ρ(𝑟⃗ ) ϕ(𝑟⃗) = − ⎯⎯⎯⎯⎯⎯⎯𝑑𝑟⃗ |𝑟⃗ − 𝑟⃗ | Linear response Pagina 23 Optical absorption giovedì 11 gennaio 2024 09:09 //Note: here the frequency is denoted by ν = ⎯⎯ It is possible to write the absorption coefficient 𝛼 as: α(ν) = σ(ν)(𝑁 − 𝑁 ) Where σ(ν) is the cross section of the transition and it represents the ability of the system to absorb radiation of a given frequency. 𝑁 is the population of the ground state while 𝑁 is the population of the excited state. For weak incident beams 𝑁 ≫ 𝑁 , thus: α(ν) = σ(ν)𝑁 This definition is general but it may not hold anymore for very intense beams such as lasers. To measure absorption we must compare the intensities of two beams: the incident one and the transmitted one. The apparatus can be of two kinds: single-beam spectrophotometer or double-beam spectrophotometer. The difference between the two is that in the first type the measurement of the baseline (no sample) and of the sample must be done separately, while the second spectrophotometer allows for the simultaneous measurement. The spectra obtained with the single- beam spectrophotometer are affected by spectral and temporal variations in the illumination intensity. The spectral variations are due to the combined effects of the lamp spectrum and the monochromator response, while the temporal variations occur because of lamp stability. Plotting the first and second derivatives of the spectra usually allows for a better interpretation of the absorption peaks. Optical spectroscopy deals with interaction of matter with UV-vis and NIR (near infrared) electromagnetic radiation. There are a number of quantities that can be directly measured by the spectrophotometer, such as (neglecting reflection): Transmittance: 𝐼 𝑇 = ⎯⎯= 𝑒 𝐼 Absorbance: 𝐴=1−𝑇 Optical density (or absorbance): 𝐼 𝑂𝐷 = 𝐴𝑏𝑠 = log ⎯⎯= − log 𝑇 𝐼 𝐼 is the incident intensity while 𝐼 is the transmitted intensity. All of the above quantities depend on the sample thickness 𝑥. We can rewrite a number of quantities in terms of the optical density: 𝑇 = 10 𝐴 = 1 − 10 𝑂𝐷 2.303 𝑂𝐷 α = ⎯⎯⎯⎯⎯⎯≈ ⎯⎯⎯⎯⎯⎯⎯⎯ 𝑥 log 𝑒 𝑥 Note that "log" is the base-10 logarithm. 𝛼 is usually given in 𝑐𝑚 and it is independent of sample Absorption and reflection Pagina 1 Note that "log" is the base-10 logarithm. 𝛼 is usually given in 𝑐𝑚 and it is independent of sample thickness, therefore it's the suitable quantity to be used especially for comparisons. Usually the range of measurable OD is 5 ⋅ 10 ÷ 3. This range is given by instrumental limitations: at very low OD it's difficult to compare the two beams, while at high OD the transmitted beam is very weak. Absorption and reflection Pagina 2 Reflectivity giovedì 11 gennaio 2024 09:54 Reflectivity is defined as: 𝐼 𝑅 = ⎯⎯ 𝐼 Where 𝐼 is the intensity of the reflected beam. We have two ways of measuring reflectivity: by direct reflectivity measurements (a) and by diffuse reflectivity (b): In direct reflectivity we measure the beam reflected back. For this you must have a sample with a very high quality of the surface. It's very difficult to make these type of measurement. Monochromatic light is passed through a beam splitter which deviates the light reflected in the sample toward a detector. By diffuse reflectivity we can analyse non-planar samples and powders or highly-absorbing systems. In this method we use the "integrating sphere" (or Ulbricht sphere), a spherical chamber with a hole for the incident beam. The inner surface is covered by a very highly-diffusing coating (e.g. 𝑀𝑔𝑂), very white. We can put the sample in the integrating sphere so that when we shine light on it the reflected beams are reflected on the inner walls of the chamber, which in turn can reflect them back on a detector. Even though the detector is not covering all the walls of the sphere we can consider the solid angle it monitors and we suppose that the light is homogeneously diffused, so that even only a small portion can be representative of the whole reflected light; if you're monitoring a solid angle Ω and you detect 𝑁 photons then the total number of photons is: 4π 𝑁 = 𝑁 ⎯⎯⎯ Ω The integrating sphere can be used also to monitor absorption and luminescence; for absorption the sample is put in the centre of the sphere and by comparing the data obtained in this way and those obtained with reflectivity measurements we can retrieve absorption. Reflectivity depends on both 𝑛 and 𝑘 as we have already seen: (1 − 𝑛) + 𝑘 𝑅 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ (1 + 𝑛) + 𝑘 Remembering that: ω ω 𝐸⃗ = 𝐸⃗ exp − ⎯⎯𝑘⃗ ⋅ 𝑟⃗ exp 𝑖 ⎯⎯𝑛⃗ ⋅ 𝑟⃗ − 𝜔𝑡 𝑐 𝑐 The combination of absorption and reflectivity measurements allows to evaluate the parameters that govern light propagation in a solid, which depends on its electronic structure as well as the presence of impurities. Absorption and reflection Pagina 3 FT spectrophotometer giovedì 11 gennaio 2024 10:04 Up until now we have seen "scanning" instruments: a property is measured at a certain wavelength, which is then changed and the measurement is repeated for each wavelength. There exists also another method of measurement, especially used for IR spectra, that is the Fourier Transform (FT) spectrophotometer: There is no monochromator. The configuration is similar to the Michelson interferometer. This measures the intensity of the beam in function of the displacement of the moving mirror; the result is an interferogram. Maxima and minima are given by the constructive/destructive interference of the beam reflected by the moving mirror and the beam reflected by the fixed mirror. Without any sample, a monochromatic source would give a sine wave in the interferogram (a), while a lamp gives a more complicated line (b). By FT of these interferograms (figures on the right) one obtains the spectrum of the source (figures on the left). For a monochromatic source we have maxima for 𝑥 = 𝑚𝜆 and minima for 𝑥 = ⎯ + 𝑚 𝜆, where 𝑥 is the displacement of the moving mirror and 𝑚 ∈ ℕ. A measurement of the sample transmittance 𝑇(ν) is obtained by dividing the sample spectra 𝑆(ν) by the reference spectra 𝑅(𝜈): 𝑆(ν) 𝑇(ν) = ⎯⎯⎯⎯ 𝑅(ν) This method does not involve scanning, so the collection of data is much faster and the S/N ratio is higher. The most important thing though is that the resolution of the peaks is normally much higher wrt other methods. Among the disadvantages of these spectrometers we can find their usually high prices and the requirement of computational work to determine the spectra from the interferograms, which can introduce some artifacts. The problem of diffusion from surfaces in the IR region is not so dramatic since the wavelengths become very long wrt the average dimension of the surface roughness. Absorption and reflection Pagina 4 Line broadening giovedì 11 gennaio 2024 10:19 Normally we have a transition like this: In a case like this the spectra would ideally be a delta function, while experimentally we observe line broadening. Mostly there are three causes: Natural broadening: given by Heisenberg's uncertainty principle, since Δ𝜈Δ𝑡 ≥ ⎯⎯. This introduces an uncertainty in frequency due to the fact that there is an uncertainty in the time in which the excited states decay. Broadening due to lattice vibration: the crystal lattice is vibrating and this causes some oscillations also of the electronic centres, which in turn causes some broadening of the electronic bands. This phenomenon can be reduced by cooling down the system to low temperatures. Broadening due to lattice disorder: different absorbing centres have slightly different resonance frequencies due to slightly different surroundings. This is typical of disordered systems, such as glasses. The first two are called "homogeneous broadening", since they involve every atom in the material, while the third one is called "inhomogeneous broadening". Ideal delta-like spectra Homogeneous broadening Lorentzian shape Inhomogeneous broadening Gaussian shape Absorption and reflection Pagina 5 Exercises lunedì 15 gennaio 2024 14:57 1 An allowed transition of an optical ion in a material has a lifetime of Δ𝑡 ≈ 10 𝑛𝑠. Estimate the natural broadening. 1 Δν = ⎯⎯⎯⎯⎯≈ 16 𝑀𝐻𝑧 2𝜋Δ𝑡 This number is actually very small; optical transition means that we are in the visible range, which frequencies are in the order of 10 − 10 𝐻𝑧, much bigger than 10 𝑀𝐻𝑧 = 10 𝐻𝑧. 2 A neodymium 𝑁𝑑 ion in different materials has several absorption lines, with different widths. For example in neodymium-doped yttrium vanadate 𝑌𝑉𝑂 , the 𝑁𝑑 ions replace some 𝑌 ions (𝑌𝑉𝑂 : 𝑁𝑑 ); this has an absorption line at 809 nm with a natural broadening of 18 GHz. Estimate the natural broadening in wavelength and the lifetime of the excited state. Remember: ℎ ≈ 4.136 ⋅ 10 eV 𝑠 𝑐 = 3 ⋅ 10 𝑚 𝑠 ℎ𝑐 𝐸 = ⎯⎯⎯= 1.53 𝑒𝑉 λ 𝐸 ⇒ ν = ⎯⎯= 3.7 ⋅ 10 𝐻𝑧 ℎ We know that: ℎ𝑐 ⎯⎯⎯= ℎ𝜈 λ We take the differential: 𝑐 ⎯⎯𝑑𝜆 = 𝑑𝜈 λ λ ⇒ Δλ = Δν ⎯⎯ = 3.9 ⋅ 10 𝑛𝑚 𝑐 The lifetime of the excited state is: 1 Δ𝑡 = ⎯⎯⎯⎯⎯= 4.4 ⋅ 10 𝑠 = 4.4 𝑝𝑠 2πΔν 3 An optical material has absorption coefficient 𝛼 = 10 𝑐𝑚 at 400 nm. Determine the thickness 𝑥 of the material necessary to make a filter with 𝑂𝐷 = 3 at 400 nm. Suppose we have a laser of power 1 W at 400 nm and we want to illuminate the sample with a power 1 mW; determine the number of necessary filters. 𝑂𝐷 = α𝑥 log 𝑒 ⇒ 𝑥 = 0.69 𝑐𝑚 𝑂𝐷 = 3 means that the intensity of the incident beam is reduced by 10 times. To reduce the power to 1 mW we just need one filter. 4 We have a filter of 𝑂𝐷 = 4 at 633 nm. Determine the transmittance at the same wavelength. If a laser of 1 mW power is able to pass through the filter, determine the power of the transmitted Absorption and reflection Pagina 6 If a laser of 1 mW power is able to pass through the filter, determine the power of the transmitted beam. 𝑇 = 10 = 10 𝑃 = 1𝑚𝑊 ⋅ 10 = 0.1 𝜇𝑊 Absorption and reflection Pagina 7 Examples of absorption spectra lunedì 15 gennaio 2024 15:47 Dopants and impurities introduce localized energy levels in the band gap and give rise to optical absorption features in an otherwise transparent material: The levels below the Fermi energy (𝐸 ) are normally occupied by electrons The dopants are normally active ions; the response of the material depends on the concentration of the active ions, on crystal field effects and on temperature. Rare earths (RE) display a great variety of optical transitions. This is a Dieke diagram of 4𝑓 levels for trivalent rare earth ions in 𝐿𝑎𝐶𝑙 The 4𝑓 levels are weakly dependent on crystal field, it means that we can replace 𝐿𝑎 with any of the other RE and the scheme is still approximately valid. 4𝑓 − 4𝑓 transitions give rise to lines, while 4𝑓 − 5𝑑 transitions give rise to bands. The states with dots are those from which radiative emission can occur. 4𝑓 − 4𝑓 transitions would be usually prohibited on a single atom, but in a material they become partially allowed, though somewhat rare. Absorption and reflection Pagina 8  Optical absorption (OA) spectra of 𝑁𝑑 in 𝐿𝑖𝑁𝑏𝑂. The 4𝑓 levels are shielded from the sorroundings by 5𝑑 levels, so their lines are very narrow, they look more like isolated atoms. There is a weak crystal field coupling due to the shielding of 4𝑓 levels by outer shells (e.g.5𝑑 levels) OA spectrum of 𝐸𝑢 in 𝑁𝑎𝐶𝑙. Europium displays transitions involving 5𝑑 levels, which are much more interacting with the lattice so broadening due to interactions and lattice vibrations is stronger, there is a strong lattice coupling. Here europium ions replace 𝑁𝑎 ; it's very difficult for it to become monovalent, usually it becomes divalent, very rarely trivalent. Cerium in sol-gel glasses; cerium wants to become 𝐶𝑒 since silicon is has valence IV, but in a reducing atmosphere it can become 𝐶𝑒 due to the fact that some oxygen atoms leave the material. If we anneal the material in an oxygen atmosphere we're only left with a broad line that is a charge- transfer transition, an electron from oxygen is transferred to a cerium ion. Absorption and reflection Pagina 9  Same host as before but we put ytterbium 𝑌𝑏 , which has only one sharp transition in the IR region. We can also have defects that introduce some energy levels, for example in 𝑁𝑎𝐶𝑙: The 𝐹 centre is an anion vacancy. All of these can give rise to colour. All of these suffer from heavy broadening due to lattice vibrations. Radiation-induced absorption bands can occur in many materials. Many optical materials work in harsh conditions, they must be "radiation-hard", which means they must preserve their characteristics after the exposure to high-energy radiation. This material in particular exhibits the presence of a new absorption band after exposure to radiation. Numerical fit of the bands can be done with Gaussian components. Bands can also be temperature-dependent and exhibit homogeneous broadening: Absorption and reflection Pagina 10 Absorption and reflection Pagina 11 Luminescence martedì 16 gennaio 2024 08:47 Luminescence studies what happens when an electron, already raised to an excited state during absorption, decays to a lower state. The selection rules are the same for absorption and emission, allowed transitions decay faster. Luminescence spectroscopy occurs when the return path is radiative, i.e. a photon is emitted. Other non- radiative recombination mechanisms can occur, for example the electron can go back by just emitting phonons, or it can be transferred to another centre. We can choose the Morse potential to describe the interatomic potential. The ground and excited state are shifted in the configurational coordinate, so that the distance between the central ion and the ligand increases when we are in an electronic excited state. When vertical absorption occurs the electron is promoted to some vibrational level of the excited electronic state. Energy dissipation occurs due to phonon emission, then the electron can decay radiatively to some vibrational level of the ground state. There the electron can still emit some phonons to reach the vibrational ground state. In this way the energy of absorption is usually higher than that of emission. Absorption is called "excitation" when we deal with luminescence, "emission" is the return path. Luminescence is very important in several fields. There are several kinds of luminescence processes according to the excitation mechanism: Light: photoluminescence Electrons: cathodoluminescence X- and gamma rays: radioluminescence Heat, following irradiation with ionizing radiation: Thermoluminescence Strong electric fields: electroluminescence Mechanical energy, e.g. friction: mechanoluminescence or triboluminescence Sound: sonoluminescence Chemical reactions: chemiluminescence Etc. In a typical spectrofluorimeter we have a lamp and an excitation monochromator, which selects a small range of wavelengths. Then through some lenses the beam is shone on the sample, which emits light (normally in an isotropic way); we collect a specific portion of it in a specific solid angle. We have also an emission monochromator which selects a specific emitted wavelength. We can collect two kinds of spectra with this apparatus: if select an excitation wavelength and vary the emission wavelength we collect an emission spectrum; if instead we fix the emission wavelength and we change the excitation one we obtain an excitation spectrum. To obtain the whole spectrum simultaneously we can use an integrator. Each detector has its sensitivity vs. wavelength; many devices have a lower limit of detection of 1 eV because they're made of silicon, which has a gap around 1 eV, below this value we cannot generate free carriers. There are specific detectors for IR radiation based for example on germanium. Each time we collect a spectrum we must correct it wrt the spectral sensitivity of the detector. Luminescence Pagina 1 collect a spectrum we must correct it wrt the spectral sensitivity of the detector. Suppose we have a 3-level materials and we suppose all emission transitions are allowed and radiative. a) Optical absorption, the system can absorb at two different energies. First we fix the excitation wavelength (emission spectra): b) If we fix ν we just have an emission peak, in the same position as absorption. c) If we fix ν instead we can have many peaks, one stems from a composite process. Now we fix the emission wavelength and we must move the excitation wavelength (excitation spectra): d) The only transition able to excite the 1 ← 2 emission is the 0 → 2 e) For the 0 ← 1 emission there are two possible excitation transitions (see point c) Very strong absorption is not always efficient in producing luminescence, if absorption occurs very rapidly the light is not able to reach all the luminescence centres in the material. Luminescence Pagina 2 Examples of spectra martedì 13 febbraio 2024 10:41 RL intensity of 𝐻𝑓𝑂 : 𝑇𝑖 vs sintering temperature. 𝐻𝑓𝑂 is an insulator, in itself it does not have emission, but it can if doped with 𝑇𝑖. It's able to stop ionizing radiation, it can be used as a scintillator. These are nanopowders, if the powders are not sintered the emission spectrum is not as good and the intensity is lower (these spectrums are normalized). In the non-sintered material 𝑇𝑖 is not as evident. 𝑆𝑖𝑂 is an insulator, poor optical activity. The material can be doped with several rare earths, which place different levels in accordance with the kind of dopants and thus we obtain a different emission spectrum for each one. If the concentration of dopants is high the rare earths can form clusters which are detrimental to optical properties, they lower the luminescence of the material Absorption and emission of 𝐶𝑎𝐹 and 𝐿𝑎𝐹 doped with 𝑃𝑟. Praseodymium displays several emission bands. The two spectra are similar but not identical because of the different crystal field. YAP is an yttrium-aluminium perovskite. The second crystal has a slightly lower concentration of cerium. When we make an absorption measurement that has a very strong luminescence this can happen; there is no monochromator between the sample and the detector, so that in a transmission experiment intense luminescence from the sample can give a false signal, thus the absorption band is seemingly lowered/cut. The solution is to put the sample as far as possible from the detector, use a screening hole or optical filter between the sample and the detector. Luminescence Pagina 3 Quantum efficiency martedì 16 gennaio 2024 09:44 The quantum (or luminescence) efficiency 𝜂 (or Q.E.) represents the efficacy of having luminescence from a specific transition. It is defined as: 𝐼 𝐼 η = ⎯⎯⎯ = ⎯⎯⎯⎯⎯< 1 𝐼 𝐼 −𝐼 Where 𝐼 has units of energy per unit time per unit area. This parameter is always strictly lower than 1 due to the fact that some energy is lost by heat during the emission process. It can be close to 1 when transitions are very sharp because the ground and excited states are not much separated in the configurational coordinate. We can also define this quantity as a ratio of the number of photons: 𝑁.. η = ⎯⎯⎯⎯⎯⎯⎯≤ 1 𝑁.. In this case it's possible for η to be 1. The intensity of emission can be written as: 𝐼 / 𝐼 = η𝐼 = η(𝐼 − 𝐼) = η𝐼 1 − ⎯⎯ = η𝐼 (1 − 10 ) 𝐼 Experimentally we should add a geometric factor 𝑘 , since we are only monitoring a part of the full solid angle: 𝐼 = 𝑘 η𝐼 (1 − 10 ) For low 𝑂𝐷 we can replace 10 by a series expansion at the first term: 𝐼 ≃ 𝑘 η𝐼 [1 − (1 − 𝑂𝐷 ln 10)] ≈ 𝑘 η𝐼 𝑂𝐷 𝑂𝐷 is the optical density of the transition in absorption. 𝐼 is proportional to 𝑂𝐷, so a transition can be monitored either using absorption or emission, but the two measurement don't have the same sensitivity; luminescence is far more sensitive. For example: consider a source of 100 𝜇𝑊 and a working wavelength of λ = 400 𝑛𝑚. The quantum efficiency is η = 0.1, the geometrical factor is 𝑘 = 10 (i.e. only a thousandth of the light reaches the detector) and the minimum detector sensitivity is 10 𝑝ℎ 𝑠 ; these are quite bad conditions. Still we can determine the minimum optical density to monitor luminescence: ℎ𝑐 𝐸 = ⎯⎯⎯= 4.96 ⋅ 10 𝐽 λ 10 𝑊 ⇒ 𝐼 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 2 ⋅ 10 𝑝ℎ 𝑠 4.96 ⋅ 10 𝐽 (𝐼 ) 10 𝑝ℎ 𝑠 𝐼 ≈ 𝑘 η𝐼 𝑂𝐷 ⇔ (𝑂𝐷) = ⎯⎯⎯⎯⎯⎯⎯⎯= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 5 ⋅ 10 𝑘 η𝐼 10 ⋅ 0.1 ⋅ 2 ⋅ 10 𝑝ℎ 𝑠 In this not favourable conditions the minimum OD that can be monitored is on the order of 10 , while usually the minimum OD that can be monitored in absorption is not lower than 10. The reason is that absorption monitors the difference between two beams, and if the difference is very small the detector cannot appreciate it, while luminescence just monitors a signal with no background. There are several types of ways to evaluate quantum efficiency experimentally. Luminescence Pagina 4  In the second case the crossing point is below the excited vibrational level, so the quantum efficiency is much lower as intersystem crossing has a high probability. An electron can still decay radiatively Luminescence Pagina 5 Photoacoustic spectroscopy martedì 16 gennaio 2024 10:12 This is a method for the determination of 𝜂, particularly suited for cases in which there are no energy transfer processes but only losses by multiphonon interactions: the concentration of luminescence centres must be low enough for energy transfer and migration processes to be neglected. The laser emits pulses of light, which shine on a sample tightly attached to a piezoelectric transducer (PZT), which converts the heat (acoustic waves) produced by the sample in an electrical signal. We see an instantaneous signal due to the laser, then after some time there is a photoacoustic signal related to the heat produced during the recombination transition. After suitable calibrations it is possible to evaluate the heat proportional to 𝜂. Consider a 2-level system (above image, left); the heat produced per unit volume 𝐻 is: 𝐻 = 𝑁 𝐸 (1 − η) Where 𝑁 is the population of level 1. The signal 𝑆 is proportional to 𝐻 through an experimental instrumental proportionality constant 𝐾, which depends on the volume of the sample, on the coupling of the sample with the PZT and on the photoacoustic response of the system: 𝑆 = 𝐾𝑁 𝐸 (1 − η) We can show that: 𝑁 ≃𝐼 τ α τ is the timescale of the laser pulses and 𝐼 is the excitation intensity (𝑝ℎ 𝑠 𝑚 ). Thus: 𝑆 = 𝐾𝐼 τ α 𝐸 (1 − η) η, 𝐾 are two unknown but we only have one equation. This problem can be solved if we consider a 3-level system (above image, right); this is not an uncommon situation, especially in rare earths. We perform a 0 → 2 excitation and we suppose that the 1 ← 2 transition is non-radiative. 𝐾 is going to be the same in the two cases, so: 𝑆 = 𝐾𝐼 τ α (𝐸 − η𝐸 ) Thus if we make the ratio: 𝑆 𝐾𝐼 τ α 𝐸 1−η α 𝐸 1−η = = 𝑆 𝐾𝐼 τ α 𝐸 − η𝐸 α 𝐸 − η𝐸 Luminescence Pagina 6 𝑆 𝐾𝐼 τ α 𝐸 (1 − η) α 𝐸 (1 − η) ⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑆 𝐾𝐼 τ α (𝐸 − η𝐸 ) α (𝐸 − η𝐸 ) Now the only unknown left is 𝜂. Luminescence Pagina 7 Integrating sphere method martedì 16 gennaio 2024 10:24 For a more general evaluation of 𝜂 there is another method, which involves the use of the integrating sphere. We combine three different conditions: a. A laser is shone inside the sphere with no sample; the collected light is just the laser light, which gives rise to a very sharp line. b. The sample is put not in the path of the laser beam but a bit off-centre; we still have the laser signal (a bit weaker) but also luminescent emission, caused by the indirect excitation of the sample by the diffused light. c. The sample is put directly on the path of the laser; we still measure some laser light but then