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. Content  Photoelectric Effect light as particles (“Quantization of EM waves”)  Photon momentum/Compton scattering  Wave-particle Duality  Weird Quantumness
3. 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?
4. Photoelectric Effect (1)  Electrons in a metal are “bound” by Binding the energy F, the “work function”. potential If you shine light on a clean metal surface, electrons can emerge the light gives the electrons enough energy (> F) to escape.  perform the experiment in vacuum  measure the flow of emitted electrons with an ammeter  How will the current depend on I and f? We might expect:  Increasing intensity I should increase the current. (By increasing the electric field E, the force on electrons, F = eE, is increased, causing more electrons to be kicked out of the metal.)  Increasing frequency f shouldn’t matter much. Perhaps a decrease in current due to rapid oscillations.  With a weak light, there should be a time delay before current starts to flow (to build up enough energy)
5. 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 !
6. Photoelectric Effect (3) Incident Light  Experiment 2: Measure the maximum energy vs. f (variable frequency f) 3 Collector (v) 2 A stop V 1 f0 electrons 0 + 0 5 10 15 Metal Surface V f (x1014 Hz) vacuum The Results:  Stopping voltage Vstop (and the maximum kinetic energy of electrons) decreases with decreasing f (linear dependence).  Below a certain frequency fo, no electrons are emitted, even for intense light! Makes no sense classically: Increasing E should have an effect.
7. 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.
8. Convenient Units for Ephoton Recall: For EM waves, frequency and wavelength are related by f = c/l. c = 2.9979 x 108 m/s New result: Light comes in “packets” of energy Photons -34 E = hf = hc/l h = 6.626 x 10 J • s photon hc = 1.986 x 10-25 J • m  For light waves it is useful to define wavelength in nanometers (nm)  For electrons it is useful to define energy in electron volts (eV).  1 eV = energy an electron gains moving through a potential energy of one volt = (1.6022 x 10-19 Coulomb)(1 volt) = 1.6022 x 10-19 Joules. -15  Therefore, h = 4.14 x 10 eV-s, and hc = 1240 eV-nm. 1240 eV nm Ephoton in electron volts Ephoton l l in nanometers Example: A red photon with wavelength of 620 nm has an energy of 2 eV.
9. 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)?
10. 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? hc 1240 eV nm Solution: E photon l l F = hf - eVstop 1240 eV 3.1 eV = 3.1eV - 0.21eV 400 For V = 0.21 V, = 2.89 eV stop eVstop = 0.21 eV What is the maximum wavelength that can cause the photoelectric effect in lithium? Hint: What is Vstop at the maximum wavelength (minimum frequency)? Answer: lmax = 429 nm
11. 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)
12. Solution  Calculating the work function F? 3.5 3 KEmax eVstop hf F 2.5 (v) 2 1.5 14 stop 1 If f0 = 5.5 x 10 Hz, what is F? V f0 (h = 4.14 x 10-15 eV•s) 0.5 0 0 5 10 15 a) -1.3 V b) -5.5 eV c) +2.3 eV f (x1014 Hz) -15 14 When Vstop = 0, hf0 = F = 4.1 x 10 eV•s x 5.5 x 10 Hz = 2.3 eV Physical interpretation of the work function:  F is the minimum energy needed to strip an electron from the metal. F  F is defined as positive and is usually given in eV units.  Not all electrons will leave with the maximum kinetic energy (due to losses)
13. 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.
14. 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 b. 3 x 1015 s-1 c. 3 x 1020 s-1 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.)
15. 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
16. Formation of Optical Images  For large light intensities, image formation by an optical system can be described by classical optics.  However, for very low light intensities, one can see the statistical and random nature of image formation.  Use an extremely sensitive CCD camera that can detect single photons. A. Rose, J. Opt. Sci. Am. 43, 715 (1953) Exposure time
17. 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
18. Momentum of a Photon (2) The Photon Wind Radiation ‘pressure’ from sun contributes to comet’s tail v Photons carry momentum! E is the photon energy p = E /c photon c is the speed of light or, using our earlier result that Ephoton = hf = hc/l Sun l is the wavelength of light p = h/l h is Planck’s constant 2 Why not Ephoton = p /2m ? Because photons have no mass. Ephoton = pc comes from Einstein’s special relativity, which more generally says E2 = m2c4 + p2c2. For the photon, m = 0.
19. 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
20. 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 = ? Dp F mg (momentum for each absorbed photon) (# photons/sec) Dt F mg #photons/s ph/ l hc mg Power (energy/photon) (# photons/sec)= mg c llh / 10 5kg) 9.8 m / s 2) 3 10 8 m / s) 30 kW
21. 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:
22. What happens when many photons encounter a slit pattern?  Recall our result for 2-slit interference: S1  Classical view: S2 EM wave (wavelength l) S1  Quantum view: S2 Photons (wavelength l h/p)
23. 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)
24. 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 Exposuretime Photons (wavelength l h/p)  Answer: Just like in the “optical image formation”, given enough time, the classical interference pattern will gradually build up from a huge # of seemingly random “events”!
25. 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:
26. Two-Slit Experiment  First, cover slit 2; i.e., only photons that go through slit 1 are transmitted. What do we see on the screen? S1 S2
27. 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 2 |y1|  Result: We get a diffraction pattern behind slit 1 S2 Probability Amplitude = y1 2 2 Probability = |Amplitude| = |y1| = P1
28. Two-Slit Experiment  Now instead cover slit 1. What do we see on the screen? S1 S2
29. 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
30. Two-Slit Experiment  Now, open both slits. What Probability, P appears on the screen now? S1 S2
31. Two-Slit Experiment Probability, P  Now, open both slits. What 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
32. FYI: Two-Slit Experiment, More Carefully (Supplement for those interested) y1: amplitude to pass through first slit, travel to upper slit, pass through upper slit, and travel to point y y r1 ik r1 y1 ~ e r2 y2: amplitude to pass through first slit, travel to lower slit, pass through lower slit, and travel to point y ik r2 y 2 ~ e Assume that the only difference in y1 Probability Amplitude = y1 + y2 and y2 happens in traveling from the slits to the screen. 2 2 ikr1 ikr 2 ikr 1 ikr 2 ikr 1 ikr 2 P yy12 ~ e e e e) e e ) e ikr1 e ikr 1 e ikr 2 e ikr 2 e ikr 1 e ikr 2 e ikr 1 e ikr 2 11 e iiff e 2 Just like the 2 2cos(ff ),where kr kr ( r r ) classical result! 1 2l 1 2
33. 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|
34. 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. What will be the resulting pattern? 2 a) |y1 y2| 2 2 b) |y1| |y2|
35. 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.
36. 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|
37. 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?
38. Two Slit Interference: Conclusions 1) Particles like photons (or electrons!) can produce interference patterns even one at a time ! In fact, they only produce them one at a time. In the limit of large numbers of particles, we recover the usual classical picture – this is an example of the “correspondence principle”. 2) With one slit closed, image formed is simply a single-slit pattern we “know” (i.e., we have constrained) which way the particle “went”. 3) With both slits open, a particle must interfere “with itself” to produce the observed two-slit interference pattern  This amazing interference effect reflects, in a fundamental way, our inability to know which slit the particle “went through”. We can only state the probability that a particle went through a particular slit.
39. 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!
40. 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? Answer: 1.6 x 10-24 J If this is a 625 Watt oven, at what rate does it produce photons? Answer: 3.9 x 1026 photons/s
41. 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
42. Homework # 3 1) The cutoff wavelength for photoelectric effect of W is of 275 nm. Determine:  Maximum speed of photoelectron emitted from W if the wavelength of the radiated light is of 180 nm?  Work function of W? 2) The energy needed to remove an electron from metallic sodium is 2.28 eV. Does sodium shows a photoelectric effect for red light with the wavelength 678 nm? What is the cutoff wavelength for photoelectric in sodium and to what color does this wavelength correspond? Question: 1) Is energy quantized in classical physics ? 2) How does the photon differ from a material particle ?