Physics A2 - Lecture 6: Introduction to Quantum Physics - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 6: Introduction to Quantum Physics - Huynh Quang Linh

1. “‘Quantum mechanics’ is the description of the behavior of matter and light in all its details and, in particular, of the happenings on an atomic scale. Things on a very small scale behave like nothing that you have any direct experience about. They do not behave like waves, they do not behave like particles, they do not behave like clouds, or billiard balls, or weights on springs, or like anything that you have ever seen.” Richard P. Feynman
2. Lecture 6: Introduction to Quantum Physics: Matter Waves and the Schrửdinger Equation
3. Content  Electron Diffraction particles as waves  Matter-wave Interference  Composite particles  Electron microscopy  Heisenberg Uncertainty Principle  Schrửdinger Equation (SEQ)  Time-independent SEQ gives static solutions for wavefunctions  Physical interpretation of the wavefunction
4. Matter Waves electron gun  DeBroglie (1924) proposed that, like detector photons, particles have a wavelength: Inversely proportional to q l = h/p momentum.  In 1927-8, it was shown (Davisson- Germer) that, like x-rays, ELECTRONS Ni Crystal can also diffract off crystals ! Interference peak ! Electrons can act like waves!! ) q • We will see later that the discrete I( atomic emission lines also arise from o 0 60 the wavelike properties of the q electrons in the field of the nucleus: What does this mean? q Atomic In discussion section: hydrogen
5. “Double-slit” Experiment for Electrons  Electrons are accelerated to 50 keV l = 0.0055 nm  Central wire is positively charged bends electron paths so they overlap  A position-sensitive detector records where they appear.  << 1 electron in system at any time [A. TONOMURA (Hitachi) pioneered electron holography] Exposure time: 1 s 10 s 5 min 20 min
6. Exercise 1: Matter wavelengths  What size wavelengths are we talking about? Consider a photon with energy 3 eV, and therefore momentum p = 3 eV/c. Its wavelength is: h 4.14 10 15 eV  s l c 1.4 10 15 s 3 108 m / s 414 nm p 3 eV  What is the wavelength of an electron with the same momentum? a) le = lp b) le lp
7. Exrcise 1: Matter wavelengths  What size wavelengths are we talking about? Consider a photon with energy 3 eV, and therefore momentum p = 3 eV/c. Its wavelength is: h 4.14 10 15 eV  s l c 1.4 10 15 s 3 108 m / s 414 nm p 3 eV  What is the wavelength of an electron with the same momentum? a) le = lp b) le lp l = h/p Same relation for e e particles and photons.  Note that the kinetic energy of the electron is different from the energy of the photon with the same momentum (and wavelength): 2 p2 h2 6.625 10 34 J  s KE 1.41 10 24 J 2m 2ml2 2( 9.11 10 31kg )( 414 10 9 m )2 1.602 10 19 J / eV 8.8 10 6 eV Compared to the energy of the photon (given above): E pc 3eV
8. Wavelength of an Electron  The DeBroglie wavelength of an electron is inversely related to the electron momentum: l = h/p  Frequently we need to know the relation between the electron’s wavelength l and its kinetic energy E. p and E are related through the classical formula: p2 E m 9.11 10-31 kg 2m e h2 p = h/l E always h 4.14 10-15 eV  s 2ml2 true! 1.505 eV nm2 For m = me: E E in electron volts (electrons) 2 l l in nanometers 1240 eV nm Don’t confuse with E for a photon ! photon l
9. Interference of larger particles  Matter-wave interference has now been demonstrated with electrons, neutrons, atoms, small molecules, BIG molecules, & biological molecules  Recent Example: Interference of C60, a.k.a. “fullerenes”, “buckyballs” Mass = (60 C)(12 g/mole) = 1.2 x 10-24 kg 2 p 3 K. E . kT p 3 kTm 2.1 10 22 kg m / s 22m l = h/p = 2.9 pm (c.f. C60 is ~ 1 nm across!) [A. Zeilinger (U. Vienna), 1999]
10. Interference of larger particles, cont.  Using a velocity selector, they could make the atoms more monochromatic improved interference: Original distribution Narrowed distribution  In 2003 interference was observed with porphyrin, a bio. molecule: Now they’re trying to put a virus into a quantum superposition!!
11. Application of Matter Waves: Electron Microscopy  The ability to “resolve” tiny objects improves as the wavelength decreases. Consider the microscope objective: Objects to d D be resolved diffraction f disks = focal length of lens if image (not interference plane is at a large distance. maxima) Critical angle for l The minimum d for which we f c 1.22 can still resolve two objects dmin f c 1.22l resolution: D D is c times the focal length: A good microscope objective has f/D  2, so with l ~ 500 nm the optical microscope has a resolution of dmin  1 mm. We can do much better with matter waves because, as we shall see, electrons with energies of a few keV have wavelengths less than 1 nm. The instrument is known as an “electron microscope”.
12. Recall from Lecture 4: Transmission of light through slits and circular apertures 1 1I0 diff(x)I0.5 Observation Slit, screen: 0 00 10 l/a 00 l/a 10 width a 12.56 x 12.56q Monochromatic light 1 source at a great 1I0 distance, or a laser diff(x)I0.5 Pinhole, Observation 0 00 screen: 10 0 10 diameter D 1.22l/ 0 1.22l/ 12.56 D x D12.56q 1I1 Object at any 0 distance: diff(x)I0.5 Image Plane: Lens, 0 00 diameter D 101.22l/ 00 1.22l/ 10 12.56 D x D12.56q Laser with pinholes
13. Application of Matter Waves: Electron Microscopy Scientists and engineers - such as those here at the Materials Research Lab and the Beckman Institute - use “electron microscopy” to study nanometer-scale structures in materials and biological systems Cu “islands” Compound eye in a Cu film of a fly (//ntweb.mrl.uiuc.edu/cmm/) (//www.itg.uiuc.edu/)
14. Example: Imaging a Virus*  Electron Microscopy of a Virus: electron gun You wish to observe a virus with a diameter of 20 nm, Electron optics which is much too small to observe with an optical D microscope. Calculate the voltage required to produce an electron DeBroglie wavelength suitable for studying this f virus with a resolution of dmin = 2 nm. The “f-number” for an electron microscope is quite large: f/D 100. (Hint: First find l required to achieve dmin with the given f/D. object Then find E of an electron from l.) Answer: 5.6 kV
15. Solution  Electron Microscopy of a Virus: electron gun You wish to observe a virus with a diameter of 20 nm, Electron optics which is much too small to observe with an optical D microscope. Calculate the voltage required to produce an electron DeBroglie wavelength suitable for studying this f virus with a resolution of dmin = 2 nm. The “f-number” for an electron microscope is quite large: f/D 100. (Hint: First find l required to achieve dmin with the given f/D. object Then find E of an electron from l.) f d 1.22l min D D D l dmin 2nm 0.0164 nm 1.22 f 1.22 f h2 1.505 eV nm2 E 5.6 keV 2ml2 0.0164 nm 2 To accelerate an electron to an energy of 5.6 keV requires 5.6 kilovolts . (The beauty of electron-volt units)
16. Summary: Photons, Matter Waves Light • p = h/l (matter also) 1240 eV nm E • p = E/c photon l • E = hf = hc/l Matter For electrons: • p = h/l (light also) 1.505 eV nm2 E • p 2mE l2 • E = h2/2ml2
17. Heisenberg Uncertainty Principle  So, particles (electrons, photons, etc.) also have wave-like properties – reflect a fundamental uncertainty in our ability to precisely “know” the particle’s location.  It is well known in classical waves that one can produce a localized “wave packet” by superposing waves with a range of wave vectors Dk. We can imagine such a packet in space:  Fourier analysis shows that: DkãDx 1. Dx  Interpretation: To make a short wavepacket requires a broad spread in wavelengths. Conversely, a single-wavelength wave extends forever.  From the quantum relation between momentum and wave vector (p = hk), there is always a trade-off between how well one can measure a particle’s location x (particle-like) and momentum p (wave-like): ħ (DkãDx 1) (ħDk)ãDx ħ DpxãDx ħ This relation is known as the Heisenberg Uncertainty Principle.
18. Uncertainty Principle: Diffraction  Look at a familiar example single-slit diffraction. p q Dpx a  The particle’s transverse location is constrained by the slit width a: Dx a/2  This implies an uncertainty Dpx in the transverse momentum: Dpx ħ/Dx 2ħ/a  Geometry relates Dpx to the total momentum p: Dpx = p sinq = (h/l) sinq  Equating these, we find a sinq l/p .  i.e., slightly inside the location of first diffraction minimum: a sinq = l Confining the particle’s location transversely leads to a bigger spread in the transverse momentum.
19. Heisenberg Uncertainty Principle: Example Consider an electron in the lowest-energy state of a hydrogen atom; its position is known to an accuracy of about 0.05 nm (the radius of the atom). How well is it possible to know the electron’s momentum? Its velocity?
20. Heisenberg Uncertainty Principle: Example Consider an electron in the lowest-energy state of a hydrogen atom; its position is known to an accuracy of about 0.05 nm (the radius of the atom). How well is it possible to know the electron’s momentum? Its velocity? Solution: DxDp  Heisenberg’s uncertainty principle (with  = h/2p). Dp /Dx  = 1.05 10-34 Jãs. = 2.1 10-24 Jãs/m = 2.1 10-24 kg-m/s -30 Dv= Dp/me me = 0.91 10 kg. = 2.3 106 m/s
21. Uncertainty Principle –Implications  The uncertainty principle explains why (negative) electrons in atoms don’t simply fall into the (positive) nucleus: If the electron were “confined” too close to the nucleus (small Dx), it would have a large Dp, and therefore a very large average kinetic energy ( Dp2/2m).  The uncertainty principle does not say “everything is uncertain”. Rather, it tells us very exactly where the limits of uncertainty lie when we make measurements of quantum systems.  Many physicists (and philosophers) believe that every physical concept only has a meaning in terms of the experiments used to measure it, e.g., the path of a particle only has meaning to within the measurement limits.  One interpretation, then, is that such concepts as orbits of electrons do not exist in nature unless and until we observe them. “I believe that the existence of the classical ‘path’ can be pregnantly formulated as follows: The ‘path’ comes into existence only when we observe it.” –Heisenberg
22. Matter Waves Quantitative Having established that matter acts qualitatively like a wave, we want to be able to make precise quantitative predictions, under given conditions.  Usually the “conditions” are specified by giving a potential energy U(x,y,z) in which the particle is located. E.g., electron in the coulomb potential of the nucleus electron in a molecule electron in a solid crystal electron in a semiconductor ‘quantum well’ U(x) Classically, a particle For simplicity, in the lowest energy here is a 1- state would sit right dimensional potential energy at the bottom of the function: well. QM this is not possible. (Why?) x
23. Matter Waves Quantitative  We will see that we can get good predictions (actually, so far they have never been wrong!!) by assuming that the state of a particle is described by a “wave function” (or “probability amplitude”): Y(x,y,z,t) 2  What do we measure? Answer: |Y(x,y,z,t)| the probability density* for detecting a particle near some place (x,y,z), and at some time t.  We need a “wave equation” describing how Y(x,y,z,t) behaves. It should  be as simple as possible  make correct predictions  reduce to the usual classical laws of physics when applied to “classical” objects (e.g., baseballs) *Probability per unit volume = ||2, i.e., ‘psi-squared’ (pronounced “sigh-squared”)
24. Exercise: Classical probability distributions  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? (Hint: Think about the kinetic energy vs position.) Ball in a box: Ball in a valley: U(x) U(x) x x P(x) P(x) ? a. ? a. b. ? b. ? c. c. x x
25. Exercise 2: Classical probability distributions  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? (Hint: Think about the kinetic energy vs position.) Ball in a box: Ball in a valley: U(x) U(x) Total energy Total energy E E E = KE + U(x) = KE + U(x) KE E KE x x P(x) P(x) ? a. ? a. b. ? b. ? c. c. x x
26. Exercise 2: Classical probability distributions  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? Ball in a box: Ball in a valley: U(x) U(x) Total energy E Total energy E = KE + U(x) KE E = KE + U(x) E KE x x P(x) P(x) ? a. b b. ? c. x x Probability is equally distributed
27. Exercise 2: Classical probability distributions  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? Ball in a box: Ball in a valley: U(x) U(x) Total energy E Total energy E E = KE + U(x) KE = KE + U(x) E KE x x P(x) P(x) b c x x Probability is More likely to spend equally distributed time at the edges. To predict a quantum particle’s behavior, we need an equation that tells us exactly how the particle’s wave function, Y(x,y,z,t), changes in space and time
28. The Schrửdinger Equation  In 1926, Erwin Schrửdinger proposed an equation that described the time- and space-dependence of the  wavefunction for matter waves (i.e., electrons, protons, ) The Schrửdinger Equation (SEQ)  There are two important forms for the SEQ First we will focus on a very important special case of the SEQ, the time-independent* SEQ, which is appropriate ONLY when the particle’s wavefunction is associated with a single energy E (we’ll deal with the more general case later). Also simplify (x,y,z) (x). (1-dimension) 2 d 2  (x) h U(x) (x) E (x)  2m dx2 2p *In this important case, which we’ll be primarily concerned with in this course, the probability density | | 2 associated with the particle does not change with time . it is in a “stationary state”.
29. Time-Independent Schrửdinger Eqn.  What does the time-independent SEQ represent? It’s actually not so puzzling it’s just an expression of a familiar result: Kinetic Energy (KE) + Potential Energy (PE) = Total Energy (E) 2 d2(x) U(x)(x) E(x) 2m dx2 KE term PE term Total E term Consider:   cos(kx), p k The kinetic energy of the d2 2 2 k2 cos(kx) KE  d (x) particle is associated with 2 dx term: 2 the curvature of the 2m dx 2 d2 2k2 p2 wavefunction, d2/dx2   2m dx2 2m 2m
30. Particle Wavefunctions: Examples What do the solutions to the SEQ look like for general U(x)? Examples of (x) for a particle in a given potential U(x): (different E) (x)  x  x We call these wavefunctions “states” of the particle. x x x The corresponding probability distributions |(x)|2 of these states are: ||2 ||2 ||2 x x x Key point: Particle cannot be associated with a specific location x. like the uncertainty that a particle went through slit 1 or slit 2.
31. Exercise 3: Particle Wavefunction The three wavefunctions below represent states of a particle in the same potential U(x), and over the same range of x: (a)  x (b)  x (c)  x x x x 1. Which of these wavefunctions represents the particle with the lowest kinetic energy? (Hint: Think “curvature”.) 2. Which corresponds to the highest kinetic energy?
32. Exercise 3: Particle Wavefunction The three wavefunctions below represent states of a particle in the same potential U(x), and over the same range of x: (a)  x (b)  x (c)  x Highest Lowest KE KE x x x 1. Which of these wavefunctions represents the particle with the lowest kinetic energy? (Hint: Think “curvature”.) 2 d 2  (x) p2 The curvature of the wavefunction  represents kinetic energy: 2m dx2 2m Since (b) clearly has the least curvature, that particle has lowest KE. 2. Which corresponds to the highest kinetic energy? (a) has highest curvature highest KE
33. Correspondence Principle Correspondence between classical and quantum regimes for large n: Classical: Quantum: U(x) U(x) (random time sampling) (stationary state) E E x x P(x) P(x) Quantum = 2 particle is also more likely to be found at the edges. x x
34. Supplementary Problem: Wavelengths a) Calculate the wavelength of an electron that has been accelerated from rest across -31 a 3-Volt potential difference (me = 9.11 10 kg). [0.71 nm] -27 b) Do the same for a proton (mp = 1.67 10 kg). [17 pm] c) Calculate the wavelength of a major league fastball -35 (mbaseball = 0.15 kg, v = 50 m/s). [8.8 x 10 m]
35. Supplementary Problem: Wavelengths a) Calculate the wavelength of an electron that has been accelerated from rest across -31 a 3-Volt potential difference (me = 9.11 10 kg). [0.71 nm] -27 b) Do the same for a proton (mp = 1.67 10 kg). [17 pm] c) Calculate the wavelength of a major league fastball -35 (mbaseball = 0.15 kg, v = 50 m/s). [8.8 x 10 m] Solution: a) E = e V = 4.8 10-19 J This is from Physics 212. -25 p = (2meE) = 9.35 10 kg m/s This is Physics 211. l = h/p = 7.1 10-10 m = 0.71 nm This is Physics 214. -23 b) p = (2mpE) = 4.00 10 kg m/s E is the same, because the electric charge is l = h/p = 1.7 10-11 m the same. Mass is bigger l is smaller. c) p = mv = 7.5 kg m/s SI units were designed to be convenient for l = h/p = 8.8 10-35 m macroscopic objects. Quantum mechanical wave effects are negligible in the motion of macroscopic objects. The wavelength is many orders of magnitude smaller than any distance that has ever been measured.