Quantum Nature of Light (Quantum Optics) - Pham Tan Thi

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1. Quantum Nature of Light (Quantum Optics) Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology Courtesy of N. Brunner and J. Simmonds
2. Light behave like a wave or a particle (This is called the duality of the behavior of light)
3. Wave Characteristics
4. Electrically Charged Particles and Electromagnetic Waves Electrons have (-) charge Protons have (+) charge Both have electric fields +- attract ++ and repel • The changing position of a charged particle creates “waves” called electromagnetic waves • The electromagnetic waves travel through empty space eventually interacting with a distant charged particle. • Visible light is an electromagnetic wave.
5. Magnetism (Effect on electric charges) Moving electric charges also produce magnetic fields. Example: electric current passing through a coil. Another interesting example: the Earth’ magnetic field is produced by the spinning of charges in the liquid metal core of the Earth. Conversely, magnetic field force charged particles to move
6. Accelerated Charges (electrons, protons) Produce (Ripples in the Electromagnetic Field) An electromagnetic wave is composed of two oscillating fields, and electric field and a magnetic field perpendicular to each other Antenna receives electromagnetic signal from TV station
7. Wavelength of Electromagnetism means COLOR
8. The Temperature Scale (Conversion from wavelength/energy to temperature) The scale mostly used in sciences, physics and astronomy is Kelvin. The unit is Kelvin (K).
9. Thermal Radiation • The fundamental sources of all electromagnetic radiation (EMR) are electric charges in accelerated motion. • All bodies emit electromagnetic radiation as a result of thermal motion of their molecules. This radiation, called thermal radiationReference, is a mixture #1 of different wavelengths. Reference #2 • Thermal radiation is emitted by all objects An electric heating element emits above absolute zero (-273.15°); but some primarily infrared radiation. But if its temperature is high enough, it also of objects is in visible. emits a discernible amount of visible light.
10. What is a Blackbody A blackbody is an idealized object that absorbs all EMR that falls on it - no EMR passes through it and none is reflected (i.e. perfect emitter and perfect absorber). Because no light (visible EMR) is reflected or transmitted, the object appears black Leave appears green because green color is reflected to human eye
11. Blackbody in Lab Experiment • An object of controlled temperature T contains a cavity, joined to the outside by a small hole. • If the hole is small enough, the radiation in the cavity comes to equilibrium in the walls. • The hole allows a small fraction of the radiation to pass to a spectrometer — the radiation coming out has the same spectrum as what is inside. • The radiancy is the power emitted per unit area per increment of wavelength and so has unit of Wcm-3
12. Blackbody in Lab Experiment • The spectral radiance from the hole is independent of the material used and only depends on the temperature.
13. Rayleigh - Jeans Approximation for Blackbody
14. Kirchhoff’s Law of Thermal Radiation in Thermal Equilibrium
15. Ultraviolet Catastrophe
16. Quantum Theory 1. Stefan-Boltzmann’s Law 2. Wien’s Law 3. Planck’s Theory 4. Planck’s Formula
17. Stefan - Boltzmann’s Law The total emitted radiation (Mλ) from a blackbody is proportional to the fourth power of its absolute temperature. 4 M = T where σ is the Stefan-Boltzmann constant, 5.6697 x 10-8 Wm-2K-4 —> the amount of energy emitted by an object such as the Sun or the Earth is a function of its temperature —> This can be derived by integrating the spectral radiance over entire spectrum 2 4 1 2⇡ k 4 4 L = L d = T OR M = ⇡L = T 5c2h3 Z0
18. Wien’s Displacement Law In addition to computing the total amount of energy existing a theoretical blackbody such as the Sun, we can determine its dominant wavelength (λmax) based on Wien’s displacement law: k = max T where k is the is the Wien’s displacement constant = 2.898 x 10-3 Km, and T is the absolute temperature in K i.e. there is an inverse relationship between the wavelength of the peak of the emission of a blackbody and its temperature. Therefore, as the Sun approximates a 6000 K blackbody, its dominant wavelength (λmax) is 0.48 µm
19. Planck’s Quantum Theory Max Planck found a correct law for the black body radiation by assuming that each oscillator can only exchange energy in discrete portions or quanta The energy exchanges between radiation and matter must be discrete and energy of radiation E = nhν Average energy per standing wave h⌫ " = eh⌫/kT 1 Planck’s modifications 8⇡h ⌫3d⌫ u(⌫)d⌫ = c2 eh⌫/kT 1 h = 6.626 x 10-34 J.s is Planck’s constant
20. Einstein’s Theory of Photoelectric Effect (Provides a direct confirmation for the energy quantization of light) Heinrich Hertz in 1887 later on, it was explained by Albert Einstein in 1905 It was during this study of electromagnetic waves that he saw the if one sends ultraviolet light onto metals, you’ll get sparks coming off. The phenomenon of ejection of electron from the surface of a metal when light of a suitable frequency strikes it is called photoelectric effect. The emitted electrons are called photo-electrons. Experiment: The time between the incidence and emission of a photoelectron is very small, more electrons independent of frequency
21. Einstein’s Interpretation Light comes in packets of energy (photons) E = h⌫ An electron absorbs a single photon to leave the material Work function: W = h⌫0 Larger W needs more energy needed for an electron to leave Classical physics fails: for dependence of the effect on the threshold frequency Photoelectric effect: K = h⌫ W = h⌫ h⌫ 0 The stopping potential: at which all of the electrons will be turned back before reaching the collector
22. Exercise 1 (estimation of the Planck constant When two UV beams of wavelengths λ1 = 280 nm and λ2 = 490 nm fall on a lead surface, they produce photo-electrons with maximum kinetics energies 8.7 eV and 6.67 eV, respectively. (a) Calculate the value of the Planck’s constant (b) Calculate the work function and the cutoff frequency of lead
23. Can Reverse be True?
24. Compton Effect/ScatteringFurther confirmation of (FurtherCompton confirmation effect of photon modelphoton model) 1927 Nobel Compton scattering is another one of those really important events that happenedCompton atscattering the beginningis another of theone 20thof century.those really It indicatedimportant thatevents photonsthat werehappened really. Theyat the reallybeginning do ofbehavethe 20th likecentury a particlethat indicated. that photons were real. They really, like really does behave like a particle. Large wave length small wave length collision E h h 2 h E m0c p h / c c sin Incident photon h c p 0 - cos c E h Target psin p h / c electron p p cos - 2 4 2 2 E m0 c p c p p Scattered electron Scattering of x-rays from electrons in a carbon target and found scattered x-rays with Scatteringa longer wavelength of x-raysthan fromthose electronsincident upon in athe carbontarget. target and found scattered x-rays with a longer wavelength than those incident on target.
25. Further confirmation of Compton effect photon model 1927 Nobel Compton scattering is another one of those really important events that happened at the beginning of the 20th century that indicated that photons were real. They really, like really does behave like a particle. We can to determine what the energy of the photon is and what its momentum is. Large wave length small wave length collision Photon energy: h⌫ Photon momentum: h/ = E/c Collisions: conserve energy and conserve momentum E h h In the original photon 2 direction: h E m0c p h / c c sin Incident photon h c p 0 - cos Initial momentum = final momentum c psin E h h⌫ Target h⌫0 p h / c electron p +0= cos + pcos✓ p cos - 2 4 2 2 c c E m0 c p c p p h⌫0 In the perpendicularScattered direction: 0= sin psin✓ electron c pc(cos✓)=h⌫ h⌫0cos pcsin✓ = h⌫0sin Scattering of x-rays from electrons in a carbon target and found scattered x-rays with a longer wavelength than2 2those incident2 upon the target. 2 p c =(h⌫) 2(h⌫)(h⌫0)cos +(h⌫0) Form the total energy expression, we have: 2 2 2 2 2 p c =(h⌫) 2(h⌫)(h⌫0)+(h⌫0) +2m c (h⌫ h⌫0) 0 2 2m c (h⌫ h⌫0) = 2(h⌫)(h⌫0)(1 cos) 0
26. Compton Effect or Shift h = 0 = (1 cos) m0c In order to conserve energy and momentum, we need to shift the wavelength. That means the outgoing photon has a different wavelength than the incoming one. This is impossible classically. A wave character cannot do that. But in quantum, we can. This shift in wavelength is associated with h/mc and dimensionless. Compton wavelength: h gives the scale of the wavelength change of C = m0c the incident photon If h/m0c = λC > λ, the shift in wavelength is huge (large effect). The Compton scattering is determined by how big the wavelength of the light that we’re sending in comparison to the Compton wavelength.
27. Exercises 2 High energy photons (γ-rays) are scattered from electrons initially at rest. Assume the photons are backscattered and their energies are much larger than the electron’s rest mass energy, E >> mec2 (a) Calculate the wavelength shift, (b) Show that the energy of the scattered photons is half the rest mass energy of the electron regardless of the energy of the incident photons, (c) Calculate kinetic energy of the electrons recoil if the energy of the incident photons is 150 MeV.
28. PairPair production Production p Electron - cos Photon p h / c p cos p Nucleus + Positron ElectromagneticElectromagnetic energyenergy is is converted converted into into matter matter PairEnergy productionand linear requiresmomentum a photoncould not both be conservede- or e+ if pair energyproduction of at leastwere 1.02to occur MeV. in emptymspace,0c2 = so0.51it does MeVnot (restoccur massthere energy) Pair production requires a photon e- or e+ => additional photon energy becomes2 K.E of e- and e+. energy of at least 1.02 MeV. m0c = 0.51 MeV (rest mass energy), - + The pair production: additional direct consequencesphoton energy of thebecomes Einstein Kmass-energy.E of e and relatione . — E = mc2 The pair production : direct consequences of the Einstein mass-energy relation; E = mc2. PairPair annihilation: annihilation: ee- ++ ee++ —> +Υ + Υ
29. Exercise 3 Calculate the minimum energy of a photon so that it converts into an electron-positron pair. Find the photon’s frequency and wavelength.
30. Exercise 3 Calculate the minimum energy of a photon so that it converts into an electron-positron pair. Find the photon’s frequency and wavelength. Answer: The Emin of a photon to produce e- and e+ pair is equal to sum of rest mass energy of both. So Emin = 2mec2 = 1.02 MeV. The photon’s frequency and wavelength can be obtained at once from Emin = hν = 2mec2 and λ = c/ν: => ν = 2.47 x 1020 Hz and λ = 1.2 x 10-12 m