# ch39 young freedman - Waves As we have seen, light has a duality of being a wave and a particle. By a symmetry argument, de Broglie in 1924 proposed that all form of matter should ...

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<ul><li><p>Heisenbergs Uncertainty Principle</p><p>Probability in finding where they land requires a wave description</p><p>Photons going through the slit as particles</p><p>THEN the spreading-out of the photons implies that the photons reaching the screen must have a small uncertainty in the vertical ( ) direction of its momentum as it exits the slit.y</p><p>yp</p></li><li><p>We will try to estimate the uncertainty for these photons by</p><p>Heisenbergs Uncertainty Principle</p><p>yp</p><p>yp</p><p>The central max dominates the diffraction pattern and a majority of the photons will land in the central max.</p></li><li><p>Considering the diffraction pattern, the width of the central max is given by the location of its 1st order minimum ,</p><p>Heisenbergs Uncertainty Principle</p><p>1sin a </p><p>1</p><p>Assuming that so that is small so that we have 1a 1 1sin a </p><p>yp</p></li><li><p>yp</p><p>- And, from the diagram, </p><p>Heisenbergs Uncertainty Principle</p><p>- For the photons reaching , they must have a range of y-dir momentum: </p><p>,y yp p </p><p>1y xp p </p><p> So, we can estimate the uncertainty needs to be as large as 1xp yp</p><p>1</p><p>1</p><p>1 1tany</p><p>x</p><p>pp</p><p>for small</p></li><li><p>Heisenbergs Uncertainty Principle</p><p>With Then, putting in , we have 1y x xp p p a 1 a</p><p>The incoming photons moving along the x-dir with momentum in the x-dir</p><p>will have a wavelength given by . Substituting this into the </p><p>inequality, we have</p><p>xp</p><p>y xp p x</p><p>hp</p><p>haa</p><p>xh p </p><p>1y xp p </p><p>yp</p></li><li><p>Heisenbergs Uncertainty Principle</p><p>For photons going through the slit with width a, one can also say that the y-</p><p>position of the photons has an uncertain of </p><p>yy p h </p><p>y a </p><p>Putting everything together, we arrive at a constraint relating the conjugate pair </p><p>of dynamic variables such that, , yy p (where h is the Plancks constant)</p><p>yp</p></li><li><p>Heisenbergs Uncertainty Principle</p><p>Our discussion is not specific to the direction. In general, we will have </p><p>similar uncertainty relations for all three spatial directions, i.e.,</p><p>2yy p </p><p>We have use heuristic argument in getting to previous inequality. A more </p><p>precise inequality can be derivate from full QM calculations and it should read</p><p>where / 2h </p><p>, ,2 2 2x y z</p><p>x p y p z p </p><p>y</p><p>This is called the Heisenbergs Uncertainty Principle.</p><p>But, there are no restrictions for unconjugated variables: , .yx p or x y etc </p></li><li><p>Uncertainty Relation for E &amp; t</p><p>Intuitively, the Heisenbergs Uncertainty Principle enforces an inverseproportional relation on the two conjugate pairs of dynamics variables:</p><p> ,xp x ,E t and</p><p>By decreasing the uncertainty in one of the variables (x or t), its corresponding conjugate variable must increases accordingly ! orxp E</p><p>2xp x </p><p>2E t </p><p>And, additionally, there is an uncertainty relation for and :E t</p><p>similar to</p></li><li><p>Heisenberg Uncertainty Principle againThe Heisenberg Uncertainty Principle is not a statement on the accuracy of the experimental instruments. It is a statement on the fundamental limitations in making measurements ! </p><p>To understand this, there are two important factors:</p><p>1. The wave-particle duality2. The unavoidable interactions between the observer and the object </p><p>being observed</p></li><li><p>Uncertainty Principle: Conceptual ExampleSuppose we want to measure the position of a particle (originally at rest) by a laser light.</p><p>The measurement is accomplished by the scattering of photons off this particle (shining a light) :</p><p>photon</p><p>particle</p><p>scattered phone</p></li><li><p>Uncertainty Principle: Conceptual ExampleBut, photons carry momentum and the particle after scattering will recoil !</p><p>particlescattered phone</p><p>recoil</p><p>While the scattered photon gives us a precise position of the particle, the photon will also inevitably impart momentum onto the particle.</p></li><li><p>Uncertainty Principle: Conceptual ExampleIf the measurement is made by a photon with wavelength , then at best,</p><p>we expect our position measurement to be accurate up to an uncertainty, </p><p>x </p><p>And, the photon will transfer parts of its momentum to the particle along the measurement direction x. The amount transfer is given by,</p><p>xhp</p><p>Combining these two expressions gives: xx p h </p><p>Note: By improving x using a smaller does not help because px ~ h/gets larger in inverse proportion to !.</p></li><li><p>Waves and UncertaintyConsider a plane wave with polarization in the y-dir and propagating in the x-dir:</p><p> , sinyE x t A kx t </p><p>at a fixed time: 0 0t t ,0 siny kxE x A </p><p>at a fixed location: 0 0x x 0, sinyE t A t </p><p>x</p><p>yE</p><p>t</p><p>yE</p><p>wave with a definite wavelength :2 k </p><p>wave with a definite frequency :2f </p><p> f</p></li><li><p>Waves and UncertaintyRecall that is the wave number and is the angular frequency of the wave and we can express the x-momentum and energy of a photon in terms of </p><p>22x</p><p>hp kh </p><p>k </p><p>andk </p><p>22</p><p>hf fE h </p><p> and</p><p>Substituting these relations into our equation for , ,yE x t</p><p> , sinyE x t A kx t </p><p>This wave function for a (plane wave) photon expressed in this way will have definite values of x-momentum and energy , i.e.,xp E</p><p>0xp 0E and</p><p>There is no uncertainty in the values for and for a plane wave.xp E</p><p> , siny xp EE x t A x t we have, </p></li><li><p>x</p><p>yE</p><p>Waves and UncertaintyOne might then ask if , can be still be satisfied?</p><p>YES! In fact, the Heisenbergs Uncertainty Principle implies that:</p><p>so that the uncertainty in the position for a plane wave must be infinite !</p><p>0xp </p><p>x </p><p>Looking back at our plane wave with a definite wavelength (x-momentum ), it indeed extends over the entire range of x and one cannot localize where the photon might be.</p><p>2xp x </p><p>xp</p></li><li><p>t</p><p>yE</p><p>Waves and UncertaintyCorrespondingly, one can argue that a similar uncertainty relation applies to the uncertainties in energy and time, i.e.,</p><p>And, in particular, for a plane wave, we have,</p><p>For a plane wave with a definite frequency (energy E), it will extend over all time t and one cannot localize the photon in time.</p><p>2E t </p><p>0E t and</p><p>(Heisenbergs Uncertainty Principle for energy and time)</p></li><li><p>Waves and UncertaintyObviously, a plane wave with infinite extents in space &amp; time is an idealization. </p><p>In practice, wave function of a photon will have limited extents in space &amp; time (i.e., we know roughly where it is located in space and time),</p><p>x t and</p><p>Then, correspondingly, according to the Heisenbergs Uncertainty Principle, the uncertainties for the x-momentum and energy for a realistic photon cannot be zero,</p><p>0xp 0E and</p><p>such that and are satisfied !2x</p><p>p x 2</p><p>E t </p></li><li><p>Waves and Uncertainty</p><p>Intuitively, the Heisenbergs Uncertainty Principle enforces an inverseproportional relation on the two conjugate pairs of dynamics variables:</p><p> ,xp x ,E tand</p><p>By decreasing the uncertainty in one of the variables (x or t), its corresponding conjugate variable must increases accordingly ! orxp E</p><p>2xp x </p><p>2E t </p></li><li><p>Waves and UncertaintyTo visualize a localized wave function, one can consider a superposition of multiple plane waves</p><p> 1 1 1 2 2 2, sin siny x xE x t A p x E t A p x E t </p><p>For simplicity, let consider the sum of two of the same plane waves that we have been considering BUT with slightly different x-momentum and E.</p><p>Looking at the wave at a fixed time and0 0t t </p><p>2 1A A we have</p><p>(black) (red)(green)</p></li><li><p>Waves and Uncertainty</p><p>The resultant electric field are localized (beat pattern). With superposition of waves with many wavelengths/frequencies (x-momentum/energy), the localization can be stronger. The result is called a wave packet.</p><p>NOTE: This wave packet is localized in space BUT it contains many and as required by the HUP.</p><p>x xp 0xp </p></li><li><p>PHYS 262George Mason University</p><p>Prof. Paul So</p></li><li><p>Chapter 39: Particles Behaving as Waves Matter Waves Atomic Line-Spectra </p><p>and Energy Levels Bohrs Model of H-</p><p>atom The Laser Continuous Spectra &amp; </p><p>Blackbody Radiation</p></li><li><p>Matter WavesAs we have seen, light has a duality of being a wave and a particle. </p><p>By a symmetry argument, de Broglie in 1924 proposed that all form of matter should process this duality also.</p><p>Recall for photons, we have:</p><p>For a particle with momentum p = mv (or mv) and total energy E, de Broglie proposed:</p><p>h h hp mv mv</p><p>Efh</p><p>(de Broglie wavelength)</p><p>hp</p></li><li><p>Electron DiffractionClinton Davisson and Lester Germer (1927)</p><p>sin 50od </p><p>Nickel crystal acts as a diffraction grading.</p><p>Constructive interference is expected at:</p><p>sin , 1, 2,d m m </p><p>2 / 2</p><p>/ / 2a</p><p>a</p><p>eV p m</p><p>h p h meV</p><p>54aV V peak</p><p>Va</p></li><li><p>Diffraction of e off Aluminum FoilIn 1928, G.P. Thomson (son of J.J. Thomson) performed another demonstrative experiment in showing electrons can act as a wave and diffract from a polycrystalline aluminum foil.</p><p>Debye and Sherrer: X-Ray diffraction experiment done a few years earlier using a similar set up.</p><p>G.P. Thomsons electron diffraction experiment produced a qualitatively similar result.</p></li><li><p>Electron Two-Slit Interference</p><p>Similar to the two-slit experiment with photons, electrons demonstrate both particle (detection of single es on film) and wave (interference pattern) characteristics.</p><p>The probability of arrival vs. y can only be explained by two interfering matter waves !</p><p>Jonsson, Claus (Germany) Journal of Physics 161 (1961), p. 454-474 (first experiment)</p></li><li><p>de Broglie wavelength: an electron vs. a baseball</p><p>e-</p><p>an electronmoving at</p><p>a baseballmoving at</p><p>71.00 10 /v m s </p><p>40.0 /v m s</p><p> 34</p><p>31 7</p><p>11</p><p>6.63 109.11 10 1.00 10 /</p><p>7.28 10</p><p>h J smv kg m s</p><p>m</p><p>(experimentally assessable)</p><p> 34</p><p>3</p><p>34</p><p>6.63 10145 10 40.0 /</p><p>1.14 10</p><p>h J smv kg m s</p><p>m</p><p>(immeasurably small !)</p><p>Since h is so small, typical daily objects wave characteristic is insignificant !</p></li><li><p>- Similar to a light beam, a beam of electrons can be bent by reflection and refraction using electric and/or magnetic fields.</p><p>The Electron Microscope</p><p> Electron microscope is better than regular microscope since </p><p>- Using electric/magnetic fields as lens, an electron beam can be used as a microscope to form magnified images of an object.</p><p>- Resolution of a microscope is limited by diffraction effects !</p><p>electron visible light </p></li><li><p>From the electron diffraction experiment, we know how to relate the wavelength for a fast moving electron to the accelerating voltage</p><p>The Electron Microscope</p><p>aV</p><p>234</p><p>231 19 12</p><p>6.626 10</p><p>2 9.109 10 1.602 10 10 10a</p><p>J sV</p><p>kg C m</p><p>(Example 39.3)</p><p>aV</p><p>2 / 2/</p><p>aeV p mp h </p><p>2</p><p>22ahV</p><p>me</p><p>41.5 10 15,000V V </p><p>What accelerating voltage is needed to produce e with ? 0.010nm </p></li><li><p>Electric Discharge Tube with Diluted Gas</p><p>Diluted gas containing trace elements such as H, He, Na, Hg, </p><p>Observation: Energetic electrons from cathode excite gaseous atoms in the tube, light can be emitted.</p><p>Ddischarge tube</p><p>Gas is diluted so that the emission process is by individual atoms. The spectrum of light emitted are sets of unique lines characteristic of the specific type of atoms in the gas.</p></li><li><p>Emission Line Spectra</p></li><li><p>Continuous vs. Line SpectraWhen light emitted by a hot solid (e.g. filament in a light bulb) or liquid, the spectrum (from a diffraction grating or prism) is continuous</p><p>But light from the spectrum of a gas discharge tube is composed of sharp lines. Atoms/molecules in their gaseous state will give a set of unique wavelengths.</p><p>Understanding the reason why they are different</p><p>Will require our understanding of:</p><p>- light behaving like particles- electron behaving like waves</p></li><li><p>The Nuclear Atom (a bit of history)What were known by 1910:</p><p> 1897: J.J. Thomson discovered e and measured e/m ratio. 1909: Millikan completed the measurement of the electron charge e. Size of atom is on the order of 10-10 m. Almost all mass of an atom is associated with the + charge.</p><p>But, the mass and charge distributions inside the atom were not known. The leading assumption was by J.J. Thomson and an atom was modeled as a sphere with positive charge and the electrons were thought to be embedded within it like raisins in a muffin.</p><p>Ernest Rutherford designed the first experiment to probe the interior structure of an atom in 1910-1911. (together with Hans Geiger and Ernest Marsden)</p></li><li><p>Rutherford Scattering Experiments</p><p>Alpha particles: high energy charged helium nuclei. It can travels several centimeter through air and ~0.1mm through solid matter.</p><p>Target: Thin gold, sliver, or copper foils.</p><p>The alpha particle is about 7300 times heavier than an electron. So, by momentum considerations alone, it will have only minimum interactions with the much lighterelectron. Only the positive charge within the atom associated with the majority of its mass can produce significant deflections in scattering the alpha particles.</p></li><li><p>Rutherford Scattering Experiments</p><p>Positive charge inside atom is distributed throughout the volume so that electric field inside the atom is expected to be diffused (small) and the force that it exerts on the will also be small deflection angles are expected to be small.</p><p>BUT, there were large scattering angles (~ 180o, almost backward) observed !</p><p>Indicating that Thomsons model was not correct.</p><p>Nucleus</p></li><li><p>Rutherford Scattering ExperimentsThese large angle deflections can only be possible if the positive charge within an atom is concentrated in a small and dense space called the nucleus.</p><p>The following is a computer simulation of the deflection angles with two different nucleus size (done much later). The experimental results agreed with simulation data using small ( ~10-15 m) nucleus sizes.</p><p>The nucleus occupies only about 10-12 of the total volume but it contains 99.95% of the total mass.</p></li><li><p>Failure of Classical PhysicsFollowing his discovery of the atomic nucleus, Rutherford also suggested that the negatively charged electron might revolve around the positively charged nucleus similar to a planet orbiting the sun.</p><p>nucleus</p><p>e-</p></li><li><p>Failure of Classical PhysicsFollowing his discovery of the atomic nucleus, Rutherford also suggested that the negatively charged electron might revolve around the positively charged nucleus similar to a planet orbiting the sun.</p><p>However, there are conceptual problems with the classical circular orbits idea:</p><p>According to classical EM, an accelerating electron in circular orbit around the nucleus will radiate EM radiations. Thus, its energy will decrease continuously and spiral rapidly toward the nucleus.The radiated frequency will also depend on the frequency of revolution and since its orbit is expected to decay continuously, the radiated spectrum will also be continuous with a mixture of frequencies.</p></li><li><p>Photon Emission by AtomsIn order to explain the observed discrete spectra lines from atomic emissions, Niels Bohr in 1913 combined the following two central ideas in his model:</p><p>electrons as waves &amp; light as packets of energy</p><p>E</p><p>E1</p><p>E2</p><p>E3</p><p>Energy Levels in a typical atom</p><p>i fhchf E E</p><p> Each atom has a specific set of possible internal energy states. An atom can possess any one of these levels but cannot take on any intermediate values. </p><p>a photon</p><p>(photons) (energy levels in atoms) </p></li><li><p>The Hydrogen Spectrum</p><p>Hydrogen is the simplest atom and it also has the easiest spectrum to analyze.In 1885, Johann Balmer derives an empirical relation to accurately describe the wavelengths for these observed spectral lines.</p><p>2 2</p><p>1...</p></li></ul>