Physics 2 - Lecture 9: Quantum Nature of Light (Quantum Optics) - Pham Tan Thi

Nội dung text: Physics 2 - Lecture 9: Quantum Nature of Light (Quantum Optics) - Pham Tan Thi

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. Wave Characteristics
3. 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
4. Wavelength of Electromagnetism means COLOR
5. 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.
6. 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
7. Rayleigh - Jeans Approximation for Blackbody
8. Ultraviolet Catastrophe
9. 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
10. 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
11. 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
12. Can Reverse be True?
13. 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
14. 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.
15. 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.