Physics A2 - Lecture 6 - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 6 - 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 Schrödinger Equation
3. 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
4. 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
5. 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
6. 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!!
7. 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
8. 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
9. 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
10. 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.
11. 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 J·s. = 2.1 10-24 J·s/m = 2.1 10-24 kg-m/s -30 Dv= Dp/me me = 0.91 10 kg. = 2.3 106 m/s
12. 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
13. 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
14. 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
15. The Schrödinger Equation In 1926, Erwin Schrödinger proposed an equation that described the time- and space-dependence of the  wavefunction for matter waves (i.e., electrons, protons, ) The Schrödinger 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”.
16. 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.
17. 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
18. 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]