MTLE-6120: Advanced Electronic Properties of Materials Basic quantum - PowerPoint PPT Presentation

1 MTLE-6120: Advanced Electronic Properties of Materials Basic quantum mechanics Contents: ◮ Energy quantization: blackbody radiation and photo-electric effect ◮ Momentum quantization: Compton scattering ◮ Wave-particle duality: electron diffraction ◮ Schr¨ odinger equation: motivation, examples, uncertainty principle ◮ Particulate waves: wave packets and group velocity ◮ Spin and quantum statistics Reading: ◮ Kasap: 3.1 - 3.6, 4.6 ◮ Griffiths QM: 1 - 2

2 Blackbody radiation Spectral radiance [kW/(sr m 2 nm)] 14 3000 K 4000 K 12 5000 K 5000 K classical 10 8 6 4 2 0 T 0 1 2 3 4 5 6 Wavelength [ µ m] ◮ Spectrum of light (EM waves) emitted by a perfect absorber (black body) ◮ Experimental realization of blackbody: pinhole in a closed box ◮ Spectrum peaks at a wavelength inversely proportional to T ◮ Solar spectrum ≈ black body radiation at 5800 K ◮ (All this was known before 1900!)

3 1D wave in a box ◮ Find non-trivial solutions of the wave equation ∂ 2 f ∂x 2 = ∂ 2 f v 2 ∂t 2 trapped inside a 1D box i.e. boundary conditions f (0) = f ( L ) = 0 ◮ General solution f ( x, t ) = Ae i ( kx − ωt ) + Be i ( − kx − ωt ) with ω = vk ◮ Apply boundary conditions: f (0) = 0 ⇒ A + B = 0 f ( L ) = 0 ⇒ Ae ikL + Be − ikL = 0 ◮ Solve to get B = − A and sin( kL ) = 0 ◮ Not all wavevectors allowed, k = nπ/L for n = 1 , 2 , . . . ◮ Net solution are standing waves f ( x, t ) = C sin nπx L e − i ( nπv/L ) t

4 EM modes in a box ◮ Standing EM waves in a box: sin nπx in each direction L sin n y πy ◮ Overall modes: sin n x πx sin n z πz L L L � n x π � ◮ Wavevector � L , n y π L , n z π k = ( k x , k y , k z ) = L ◮ Between wavevector magnitude k and k + d k : 4 πk 2 d k ◮ Volume in � k -space: 8 � π � 3 ◮ Volume per � k : L L ◮ Number of modes per � k : 2 polarizations ◮ Number of modes per unit volume: � L � 3 2 · 4 πk 2 d k L 3 = k 2 d k · 1 · π 2 8 π ◮ In terms of wavelength λ = 2 π/k : � 2 2 π d λ � 2 π � 2 � � � 2 π 1 � d2 π � = 1 = 8 π � � λ 4 d λ � � π 2 π 2 λ 2 λ λ λ

5 Classical theory: equipartition theorem ◮ Each mode: classical wave with any amplitude A with energy E = c 0 A 2 ◮ At temperature T , probability of energy E is ∝ e − E/ ( k B T ) ◮ Average energy at temperature T is � ∞ d Ae − E/ ( k B T ) E 0 � E � ≡ � ∞ d Ae − E/ ( k B T ) 0 � ∞ � E/c 0 e − E/ ( k B T ) E d 0 = � ∞ � E/c 0 e − E/ ( k B T ) d 0 � ∞ d Ee − E/ ( k B T ) E 1 / 2 0 = � ∞ d Ee − E/ ( k B T ) E − 1 / 2 0 = ( k B T ) 3 / 2 Γ(3 / 2) ( k B T ) 1 / 2 Γ(1 / 2) = k B T 2 ◮ Oscillators: kinetic and potential energies ⇒ � E � = k B T ◮ EM waves: electric and magnetic fields ⇒ � E � = k B T

6 Rayliegh-Jean’s law 8 π ◮ Number of modes per wavelength: λ 4 ◮ Energy per mode: k B T ◮ Power radiated per surface area: × c 4 ◮ Spectral power per surface area: I λ = 8 π λ 4 · k B T · c 4 = 2 πck B T λ 4 Spectral radiance [kW/(sr m 2 nm)] 14 3000 K 4000 K 12 5000 K 5000 K classical 10 8 6 4 2 0 0 1 2 3 4 5 6 Wavelength [ µ m]

7 Planck hypothesis ◮ Energies for a mode with frequency ν only allowed in increments of hν ◮ In terms of angular frequency ω = 2 πν , in increments of � ω ◮ With an as yet-undetermined constant h (or � = h/ (2 π ) ) ◮ At temperature T , n units of energy hν with probability ∝ e − nhν/ ( k B T ) ◮ Average number of energy units � � n e − nhν/ ( k B T ) n n e − nα n � n � ≡ n e − nhν/ ( k B T ) = ( α ≡ hν/ ( k B T )) � � n e − nα � = − d n e − nα = − d � d α e − nα d α ln � n e − nα n e − α = − d 1 d α ln 1 − e − α = 1 − e − α 1 = e hν/ ( k B T ) − 1 hν � E � = � n � hν = e hν/ ( k B T ) − 1

8 Modification of equipartition theorem ◮ Average energy per mode changes to hν � E � = e hν/ ( k B T ) − 1 ◮ For hν ≪ k B T hν � E � ≈ hν/ ( k B T ) = k B T classical regime with � n � ≫ 1 ◮ For hν ≫ k B T hν e hν/ ( k B T ) = hνe − hν/ ( k B T ) � E � ≈ new regime with � n � ≪ 1

9 Planck’s law ◮ Number of modes per wavelength: 8 π λ 4 (as before) hc/λ ◮ Energy per mode: e hc/ ( λkBT ) − 1 (new, using ν = c/λ ) ◮ Power radiated per surface area: × c 4 (as before) ◮ Spectral power per surface area: 2 πhc 2 I λ = 8 π e hc/ ( λk B T ) − 1 · c hc/λ λ 4 · 4 = λ 5 � � e hc/ ( λk B T ) − 1 Spectral radiance [kW/(sr m 2 nm)] 14 3000 K 4000 K 12 5000 K 5000 K classical 10 8 6 4 2 0 0 1 2 3 4 5 6 Wavelength [ µ m]

10 Planck’s law features ◮ Has a maximum at λ ≈ hc 5 k B T (Wein’s displacement law) ◮ Determine h = 6 . 626 × 10 − 34 Js, � = h/ (2 π ) = 1 . 055 × 10 − 34 Js ◮ Total energy per surface area radiated by black body (Stefan’s law) � ∞ � ∞ 2 πhc 2 P S ≡ d λI λ = d λ λ 5 � � e hc/ ( λk B T ) − 1 0 0 � ∞ � � 2 πhc 2 hc = d ( x ≡ hc/ ( λk B T )) ( hc/ ( xk B T )) 5 ( e x − 1) xk B T 0 � hc � − 4 � ∞ x 5 = 2 πhc 2 x − 2 d x ( e x − 1) k B T 0 = T 4 · 2 πk 4 π 4 B h 3 c 2 15 � �� � σ σ = 2 π 5 k 4 15 h 3 c 2 = 5 . 67 × 10 − 8 W / (m 2 K 4 ) B ◮ Agrees very well with radiated heat measurements. (What is the classical result for σ ?)

11 Photoelectric effect ◮ Light ejects electrons from cathode ⇒ I at V = 0 ◮ V ↑⇒ I ↑ till saturation (all ejected electrons collected) ◮ V ↓⇒ I ↓ till I = 0 : all electrons stopped at V = − V 0 Light ◮ Increase intensity I : higher saturation I but same stopping V I ◮ Increase frequency ω : V higher stopping V ◮ Stopping action: eV 0 = KE max I ◮ Experiment finds eV 0 ∝ ( ω − ω 0 ) ◮ In fact eV 0 = � ( ω − ω 0 ) ◮ Different cathodes ⇒ different ω 0 V but same slope � identical to that from Planck’s law! ◮ Light waves with angular frequency ω behave like particles (photons) with energy � ω (Einstein, 1905) ◮ Why does the saturation I ↓ when ω ↑ at constant I ?

12 Compton scattering ◮ X-ray ejects electron with part of its energy and remainder comes out as secondary X-ray ◮ Energy conservation � ω = � ω ′ + 1 2 mv 2 (assuming ω 0 << ω ) ◮ Output X-ray at angle θ has specific frequency ω ′ . Why? p = � � ◮ Photon also has momentum � k (magnitude � ω/c ) ◮ Momentum conservation c = � ω ′ � ω cos θ + mv cos θ ′ c 0 = � ω ′ spectrometer sin θ − mv sin θ ′ X-ray c ◮ Eliminate electron unknowns ( v, θ ′ ) Sample X-ray source cos θ = ω 2 + ω ′ 2 − 2 mc 2 ( ω − ω ′ ) � 2 ωω ′

13 Wave-particle duality ◮ Light is a wave: electric and magnetic fields oscillating ∼ e − iωt ◮ All of classical wave-optics: diffraction etc. ◮ Light is particulate: photons with energy � ω and momentum � ω/c ◮ Black-body radiation ◮ Photoelectric effect ◮ Compton scattering

14 Electrons ◮ Discovered as ‘cathode rays’ in vacuum tube experiments (1869) ◮ Deflection by magnetic fields to measure charge/mass (1896) ◮ Charge measured in Millikan’s oil drop experiment (1909) ◮ Particle with mass m ≈ 9 × 10 − 31 kg and charge − e ≈ − 1 . 6 × 10 − 19 C known by early days of atomic theory and quantum theory ◮ Is it also a wave?

15 De Broglie wavelength ◮ Yes! With wavelength λ = h/p where p is the momentum ◮ For wavevector � p = � � k (of magnitude 2 π/λ ), this ⇒ � k (same as photon) ◮ In terms of kinetic energy: electrons photons KE [eV] 2 m KE [˚ h 2 m KE+(KE /c ) 2 [˚ h KE [˚ hc √ λ = A] λ = A] λ = A] √ 1 . 24 × 10 4 1 12.3 12.3 1 . 24 × 10 3 10 3.88 3.88 100 1.23 1.23 124 10 3 0.388 0.388 12.4 10 4 0.123 0.122 1.24 10 5 0.0388 0.0370 0.124 10 6 0.0123 0.00872 0.0124 ◮ This rule applies to all particles / matter, not just electrons and photons ◮ What is your typical wavelength when walking? (Why don’t you diffract?)

16 Electron diffraction ◮ Use gold film as grating (GP Thomson, 1927) ◮ Polycrystalline ⇒ rings (like powder X-ray diffraction) ◮ Modern version: transmission electron microscopy (TEM) ◮ ∼ 100 keV energies ⇒ λ ∼ 0 . 05 ˚ A ⇒ atomic resolution Gold Screen fi lm

17 Schrodinger equation ◮ The wave equation for non-relativistic particles with mass m − � 2 ∇ 2 ψ r, t ) ψ = i � ∂ψ + V ( � 2 m ∂t r, t ) : analogous to � r, t ) or � ◮ Wave function ψ ( � E ( � B ( � r, t ) for EM waves ◮ For EM wave, intensity is proportional to | E | 2 ◮ Intepret intensity as the probability of finding light r ) | 2 is the probability density of finding particle ◮ Quantum mechanically, | ψ ( � � r ) | 2 = 1 ) at � r (normalized as d� r | ψ ( � ◮ Note � E ( � r ) actually is the wavefunction of a photon in the quantum theory of EM waves

18 Free particle − � 2 ∇ 2 ψ r, t ) ψ = i � ∂ψ + V ( � 2 m ∂t ◮ Let potential be constant in space and time i.e. V ( � r, t ) = V 0 r, t ) = e i ( � ◮ Solution of the form ψ ( � k · � r − ωt ) � 2 k 2 2 m + V 0 = � ω p 2 + V 0 = E 2 m ���� ���� ���� Total Potential Kinetic ◮ Note how De Broglie and Planck hypothesis connect classical and quantum relations.) ◮ Where is the particle? p = � � Everywhere with a well-defined momentum � k