Physics A2 - Lecture 5: Wave-Particle Duality - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 5: Wave-Particle Duality - Huynh Quang Linh

1. Lecture 5:Wave-Particle Duality Collector A electrons + Metal Surface V S1 vacuum 3.5 3 S2 2.5 2 (v) 1.5 1 f0 stop V 0.5 0 0 5 10 15 f (x1014 Hz)
2. Wave-particle Duality for Light and Matter In previous lectures we viewed “light” as a wave (i.e. it causes interference and diffraction) Surprise: In the early 1900’s, it was discovered that light has “particle”-like properties in some situations! Furthermore, “matter” (i.e., electrons, protons, etc.) was found to exhibit “wave-like” properties under certain circumstances These two discoveries revolutionized science and technology. What’s the evidence of the wave-particle duality?
3. Photoelectric Effect (2) Incident Light Experiment 1: Measure the (variable frequency f) maximum energy of ejected electrons Collector Bias the “collector” with a negative A charge to repel ejected electrons Increase negative bias voltage until flow of ejected electrons electrons + decreases to zero. Metal Surface V (Current = 0 at V = Vstop) vacuum Measurement of Vstop tells the max kinetic energy, KEmax = eVstop. The Result: The “stopping voltage” is independent of light intensity! Therefore, increasing the intensity I does not increase KE !
4. Photoelectric Effect (4) 3 slope (v) Collector 2 h/e stop A V 1 f0 1 0 0 5 10 15 electrons + f (x1014 Hz) Metal Surface V Summary of Results: vacuum Energy of electrons emitted depends on frequency, not intensity Electrons have a probability to be emitted immediately Electrons are not ejected for frequencies below f0 h is Planck’s constant (measured here) KEmax = eVstop = hf − F F is the “work function” Conclusion: Light comes in “packets” of energy Photons ! -34 with Ephoton = hf h = 6.626 x 10 J • s Increasing I simply increases # photons, not the photon energy.
5. Photoelectric Effect: Example When light of wavelength l = 400 nm shines on a piece of lithium, the stopping voltage of the electrons is Vstop = 0.21 V. What is the work function of lithium? What is the maximum wavelength that can cause the photoelectric effect in lithium? Hint: What is Vstop at the maximum wavelength (minimum frequency)?
6. Lecture 5, exercise 1 Calculating the work function F. 3.5 KEmax = eVstop = hf − F 3 2.5 (v) 2 14 If f0 = 5.5 x 10 Hz, what is F? 1.5 stop 1 (h = 4.14 x 10-15 eV•s) V f0 0.5 0 a) -1.3 V b) -5.5 eV c) +2.3 eV 0 5 10 15 f (x1014 Hz)
7. 2005: World Year of Physics 2005 was officially the international “World Year of Physics” The year was chosen to honor the 100-th anniversary of Einstein’s three world-changing papers in 1905: ➢ Brownian motion (connecting thermal physics with statistical mechanics) ➢ Special Relativity (showing that the notions of time and space depend on relative motions of observers) ➢ Photoelectric effect (introducing the notion that light comes in quantized packets of energy – “photons” [and thus ~starting the field of Quantum Mechanics] {although it took well over a decade until this notion was accepted} Question: For which of these did Einstein get a Nobel Prize (1921)? ➢After Einstein made precise predictions about the photoelectric effect, R. Millikan performed the definitive set of experiments (partly for which he got the Nobel prize in 1923!) ➢It often (mis)stated that the photoelectric effect proves the existence of photons. Actually, it doesn’t, though other results do.
8. exercise 2: Counting photons How do we reconcile this notion that light comes in ‘packets’ with our view of an electromagnetic wave, e.g., from a laser?? Partially transmitting Power input mirror 1. How many photons per second are emitted from a 1-mW laser (l=635nm)? a. 3 x 1010 s-1 hc 1240eV-nm Ephoton = = 2eV b. 3 x 1015 s-1 l 635 nm 20 -1 Power output: P = (# photons/sec) x Ephoton c. 3 x 10 s −3 P 10 J 1eV 1photon 15− 1 (# photons/sec)= = -19 = 3.1 10 s Ephoton s 1.6 10 J 2eV 2. Are more or fewer photons emitted by a cell phone (f = 830 MHz) at the same power? (Cell phones actually emit 0.6 – 3W.) Rate ~ l Ratecelll cell 0.36 m 5  = =−9 =5.7 10 lcell = c/f = 0.36 m Ratelaserl laser 635 10 m
9. Momentum of a Photon (1) Between 1919 and 1923, A.H. Compton showed that x-ray photons collide elastically with electrons in the same way that two particles would elastically collide! “Compton Scattering” Particles of light (i.e., photons) carry momentum ! More recent Glass bead ‘floating’ proof: on a laser beam! (AT&T Bell Labs) Laser
10. Exercise: Optical “Levitation” What laser power is required to suspend a glass bead weighing 0.01 gram? Glass bead ‘floating’ on a laser beam! (AT&T Bell Labs) Assume that the bead absorbs all Laser the incident light. P = DE/Dt = ? Answer: 30 kW
11. Wave-Particle “Duality” We cannot classify EM radiation into distinct categories of “waves” or “particles”. Light exhibits wave-like properties (interference) in certain situations, and particle-like properties (trajectories) in others. We will see next lecture that in fact matter particles (like electrons, protons, etc.) also can display both particle-like and wave-like properties! Important question we need to address: When should we expect to observe wave-like properties, and when should we expect particle-like properties? To help answer these questions, let’s consider 2 examples:
12. Two Slit Interference: Question: what if we reduce the source intensity so that only one particle (photon) goes through the pattern at a time? S1 S2 Photons (wavelength l = h/p)
13. Two Slit Interference: Hold on! This is kind of weird! How do we get an interference pattern from single “particles” going through the slits one at a time? Doesn’t the photon have to go through either slit 1 or slit 2? To answer this, we must deal with statistics, i.e. probabilities. First, define: Probability “amplitude” S1 for going through slit 1: y1 Probability “amplitude” y 2 S for going through slit 2: 2 Pronounced, “sigh” Now, what happens if we cover one of the slits? Do the experiment:
14. Two-Slit Experiment Probability, P1 First, cover slit 2; i.e., only photons that go through slit 1 are transmitted. What do we see on the screen? S1 |y |2 Result: We get a diffraction 1 pattern behind slit 1 S2 Probability Amplitude = y1 2 2 Probability = |Amplitude| = |y1| = P1
15. Two-Slit Experiment Probability, P2 Now instead cover slit 1. What do we see on the screen? S1 Result: We get a diffraction pattern behind slit 2 2 |y2| S2 Probability Amplitude = y2 2 2 Probability = |Amplitude| = |y2| = P2
16. Two-Slit Experiment Now, open both slits. What Probability, P appears on the screen now? S1 Interference ! Why? S2 2 |y1 + y2| Probability Amplitude = y1 + y2 (particle can go through slit 1 or 2) 2 2 2 P = Probability = |y1 + y2 | = |y1| + |y2| + interference term like: [sin(wt) + sin(wt + f)]2 = 2 + 2cos(f) (using phasors) Like EM waves: Add P P1 + P2 amplitudes not intensities
17. Interference – What really counts At the beginning of this course we discussed interference from a classical perspective – there what was interfering was the amplitudes from two or more physical paths. A more modern perspective is that indistinguishable processes interfere. Example: yupper is the amplitude corresponding to the process by which a photon leaves the source, travels through the upper slit, and reaches the point y on the screen. ylower is the amplitude corresponding to the process by which a photon leaves the source, travels through the lower slit, and reaches the point y on the screen. If these processes are in principle indistinguishable, add the amplitudes together and take the absolute square to get the probability: 2 P(y) = |yupper + ylower| If instead these processes are somehow distinguishable (i.e., there’s some measurement we could make in principle that would tell us which one really happened), then we add the probabilities together: P(y) = P(y, by way of upper slit) + P(y, by way of lower slit) 2 2 = |yupper| + |ylower|
18. exercise 3 Suppose we measure with the upper slit covered for half the time and the lower slit covered for the other half of the time. The resulting pattern will be 2 a) |y1 + y2| 2 2 b) |y1| + |y2| Because one slit is covered all the time, at any given time, there is only one contributing amplitude. Therefore, the only pattern we will get is the diffraction pattern - there will be no interference from the two slits. The result will be the sum of the two diffraction patterns.
19. exercise 3’ Now let’s modify the experiment a bit. send in unpolarized photons cover the upper slit with a vertical polarizer and cover the lower slit with a horizontal polarizer V Now the resulting pattern will be ?? Unpolarized 2 A) |y1 + y2| H 2 2 B) |y1| + |y2| In this case, the photon’s polarization serves to label which way it went; because the two processes are in principle distinguishable there is no interference. Note, that we don’t need to actually measure the polarization. It is the mere fact that one could measure it that destroys the interference. Question: How could we recover the interference?
20. More Quantum Weirdness Consider the following interferometer: photons are sent in one at a time the experimenter can choose to leave both paths open, so that there is interference activate switch in the upper path, deflecting that light to a counter What does it mean? Switch OFF → interference → wave-like behavior Switch ON → detector “click” or “no click” and no interference → particle-like behavior (trajectory is identified) What is observed? What kind of behavior you observe depends on what kind of measurement you make. Weird. Principle of Complementarity: You can’t get perfect particle-like and wave-like behavior in the same setup. It gets worse! In the “delayed choice” version of the experiment that was done, the switch could be turned ON and OFF after the photon already passed the first beam splitter! The results depended only on the state of the switch when the photon amplitude passed through it!
21. Supplementary Problem: Microwave photons What’s the energy? A microwave oven generates electromagnetic radiation at a frequency of 2.4 GHz. What is the energy of each microwave photon? -34 9 -1 Ephoton = hf = (6.626 x 10 J • s)(2.4 x 10 s ) = 1.6 x 10-24 J If this is a 625 Watt oven, at what rate does it produce photons? Microwave Energy / second Power Rate( photons / sec) = = Energy / photon Ephoton Rate = (625 J/s)/(1.6 x 10-24 J/photon) = 3.9 x 1026 photons/s