LECTURE 6 QUANTUM PHYSICS II. Instructor: Shih-Chieh Hsu

Transcription

1 LECTURE 6 QUANTUM PHYSICS II Instructor: Shih-Chieh Hsu

2 Development of Quantum Mechanics 2 In 1862, Kirchhoff coined black body radiation or known as cavity radiation The experiments raised the question of the failure of classical EM theories In 1887, Heinrich Hertz discovers photoelectric effect Another experiment raised the concern of the wave and particle nature of the light.

3 Theory of quantization of light 3 In 1900, Max Planck resolves the blackbody radiation issues by introducing quantum concept of the discrete energy element The energy element is discrete and the energy is proportional to the frequency The invention of Planck constant h In 1905, Einstein explained photoelectric effect by using Max Planck s light quantization concept Photon is introduced by Gilbert N. Lewis in 1926

4 More Quantization System 4 In 1909, Robert Millikan conducted oildrop experiment and showed that electric charge is quantized. In 1911, Ernest Rutherford s Gold Foil Experiment disproved the plum pudding model of the atom.

5 Bohr s classical model of H 5 Niels Bohr proposed a model of the hydrogen atom that successfully predicted the observed spectra. The electron of the hydrogen atom moves in a circular orbit around the positive nucleus according to Coulomb s law and classical mechanics like the planets orbiting around Sun.

6 Flaw in the classical model 6 Classical EM theory says that an electron in a circular orbit is accelerating, so it would radiate an EM wave and loses its energy. This atom would quickly collapse as the electron spirals into the nucleus and radiates away the energy.

7 Spectral Lines 7

8 8 Hydrogen Spectrum

9 Bohr s semi-classical model (1913) 9 Energy Quantization In Aton E n energy level ΔE n = hf=hc/λ

10 10 Old Quantum Theory

11 Old Quantum Theory 11 In 1913, Niels Bohr explains the spectra line of the hydrogen atom using quantization ideas expansion of quantum mechanics research works! e.g. Stern-Gerlach demonstrated spin property of electrons in 1920 Stern Gerlach

12 Three Failures of Classical Physics 12 Black Body Radiation. The Hydrogen Atom Photoelectric Effect.

13 13 New Quantum Theory

14 New Quantum Theory 14 In 1924, Louise de Broglie proposes matter wave theory. In 1925, Matrix Mechanics is invented Heisenberg Uncertainty was proposed in 1927 Werner Heisenberg Max Born Pascual Jordan

15 Completion of Quantum Mechanics 15 In 1925, Erwin Schrödinger invented wave mechanics and non-relativistic Schrödinger equation as generalization of de Broglie s theory 1927, Paul Dirac began the process of unifying quantum mechanics with special relativity by proposing the Dirac equation for the electron.

16 de Broglie hypothesis 16 In 1924 Louis de Broglie hypothesized: Since light exhibits particle-like properties and act as a photon, particles could exhibit wave-like properties and have a definite wavelength. The wavelength and frequency of matter: λ = h p f = E h For macroscopic objects, de Broglie wavelength is too small to be observed.

17 Example 1 17 One of the smallest composite microscopic particles we could imagine using in an experiment would be a particle of smoke or soot. These are about 1 µm in diameter, barely at the resolution limit of most microscopes. A particle of this size with the density of carbon has a mass of about kg. What is the de Broglie wavelength for such a particle, if it is moving slowly at 1 mm/s? λ = h p = h mv = h kg 10 3 m / s = Js kg 10 3 m / s = m h = J s = ev s

18 Diffraction of matter 18 In 1927, C. J. Davisson and L. H. Germer first observed the diffraction of electron waves using electrons scattered from a particular nickel crystal. G. P. Thomson (son of J. J. Thomson) showed electron diffraction when the electrons pass through a thin metal foils. Diffraction has been seen for neutrons, hydrogen atoms, and alpha particles. In all cases, the measured l matched de Broglie s prediction. X-ray diffraction electron diffraction neutron diffraction

19 Interference and diffraction of matter 19 If the wavelengths are made long enough (by using very slow moving particles), interference patters of particles can be observed. These figures show the build up of the electron two-slit interference pattern as the electrons arrive at the detector.

20 Pollen grains Electron microscope 20 When viewing details of objects with visible light, the details can be resolved only if they are larger than the wavelength of the light. In electron microscopes, beams of electrons, with small wavelength is used to see small objects.

21 Clicker Question 18-1 The electron microscope is a welcome addition to the field of microscopy because electrons have a wavelength than light, thereby increasing the of the microscope. longer;; resolving power longer;; breadth of field shorter;; resolving power longer;; intensity

22 Classical waves vs. particles 22 A classical wave behaves like a sound wave. It exhibits diffraction and interference. Its energy is spread out continuously in space and time. A classical particle behaves like a piece of shot. It can be localized and scattered. It exchanges energy suddenly at a point in space. It obeys the laws of conservation of energy and momentum in collisions. It does not exhibit interference or diffraction.

23 Wave-particle duality 23 Light, normally thought of as a wave, exhibits particle properties when it interacts with matter. photoelectric effect Electrons, normally thought of as particles, exhibit the wave properties when they pass near the edges of obstacles. interference and diffraction All carriers of p and E exhibit both wave and particle characteristics. In classical physics, the concepts of waves and particles are mutually exclusive.

24 Wave-particle duality 24 The classical concepts of waves and particles do not adequately describe the complete behavior of any phenomenon. Everything propagates like a wave and exchanges energy like a particle. Often the concepts of the classical particle and the classical wave give the same results. If l is very small, diffraction and interference are not observable. If there are a lot of particles, they can be treated as a wave.

25 Clicker Question 19-1 If the wavelength of an electron is equal to the wavelength of a proton, then. 1. the speed of the proton is greater than the speed of the electron 2. the speeds of the proton and the electron are equal 3. the speed of the proton is less than the speed of the electron 4. the energy of the proton is greater than the energy of the electron, 5. both (1) and (4) are correct. λ = h p = h mv E = mv2 2 = p2 2m = h2 2mλ 2

26 Uncertainty principle 26 If we use light with l to measure the position of an object, x, its uncertainty, Δx, cannot be less than ~l because of diffraction. If we use photons with p g = h/λ to measure the momentum of an object, p, Δp of the object cannot be less than ~h/λ since the photon changes the momentum of the object upon scattering. The Heisenberg uncertainty principle states that: It is impossible to simultaneously measure both the position and the momentum of a particle with unlimited precision. ΔxΔp x 2, where h 2π

27 Quantum Mechanics (1923) 27 In quantum mechanics, a particle is described by a wave function y that obeys a wave equation called the Schrödinger equation. 2 2m 2 Ψ( r,t) + U ( r )Ψ( r,t) = i t Ψ ( r,t) You absolutely do not need to memorize the formula. The solution of the equation by itself has no physical meaning. However, the probability to find a particle in a certain spacetime is:

28 Time-Independent Schrodinger Equation 28 Solution of the Schrödinger equation. 2 2m 2 Ψ( r,t) + U ( r )Ψ( r,t) = i t Ψ ( r,t)

29 Wave function 29 The Schrödinger equation describes a single particle. The probability density P(x), the probability per unit volume (or length in 1-D), of finding the particle as a function of position is given by P( x) = ψ 2 ( x) The probability is probability times unit volume, i.e. P(x) Δx

30 Normalization condition 30 If we have a particle, the probability of finding the particle somewhere must be 1. Therefore the wave function must satisfy the normalization condition. P( x)dx = ψ 2 ( x)dx = 1 For y to satisfy the normalization condition, it must approach zero as x approaches infinity.

31 31 Copenhagen Interpretation 1927, Bohr, Heisenberg, Pauli had converged to a consensus based on Bohr's concept of complementarity, which states that a physical phenomenon may manifest itself in two different complementary' ways depending on the experiment set up to investigate it. Thus light, for example, could appear sometimes as a wave and sometimes as a particle. Although mutually exclusive, both pictures were necessary to obtain a full description of the phenomenon.

32 32 Schrödinger s Cat 1935, Austrian physicist Erwin Schrödinger proposed this thought experiment, often described as a paradox, to illustrate what he saw as the problem of the Copenhagen interpretation of quantum mechanics applied to everyday objects.

33 Schrödinger s Poor Cat 33 The material doesn t decay. The cat is alive The material has decayed. The cat has been killed by the poison.

34 Schrödinger s Cat: Live or Dead 34 According to the Copenhagen interpretation, the cat is both alive and dead. It exists in a state of superposition

35 Because the probability density is uniform, the probability of a particle being in some range Δx in the region 0 < x < 8.0 cm is P 0 Δx. Probability Calculation for a Classical Particle 35 A classical point particle moves back and forth with constant speed between two walls at x = 0 and x = 8.0 cm.

36 Probability Calculation for a Classical Particle 36 What is the probability density P(x)?. The probability density P(x) is uniform between the walls and zero elsewhere: probability density x Total length = 1 = P0 x 8cm

37 Probability Calculation for a Classical Particle 37 What is the probability of finding the particle at the point where x equals exactly 2 cm? On the interval 0 < x < 8.0 cm, the probability of finding the particle in some range Δx is proportional to P 0 Δx = Δx/(8 cm). The probability of finding the particle at the point x = 2 cm is zero because Δx is zero (no range exists). Alternatively, because an infinite number of points exists between x = 0 and x = 8 cm, and the particle is equally likely to be at any point, the chance that the particle will be at any one particular point must be zero.

38 Probability Calculation for a Classical Particle 38 What is the probability of finding the particle between x = 3.0 cm and x = 3.4 cm? Because the probability density is uniform, the probability of a particle being in some range Δx in the region 0 < x < 8.0 cm is P 0 Δx.

39 A particle in a box 39 Consider a particle of mass m confined to a onedimensional box of length L. Classical Mechanics: The particle with any values of energy and momentum bounces back and forth between the walls of the box. Quantum Mechanics: The particle is described by a wave function y, and y 2 describes the probability of finding the particle in some region.

40 A particle in a box: conditions for y 40 The particle is somewhere in the box. L ψ 2 ( x)dx = 1 0 The particle is not outside the box. ψ ( x) = 0 for x 0 and x L y is continuous everywhere. ψ ( 0) = 0 and ψ ( L) = 0 This is the same boundary condition as the condition for standing waves on a string fixed at x = 0 and x = L and satisfies following equation

41 A particle in a box: allowed wavelengths 41 The boundary condition restricts the allowed wavelengths for a particle in a box. The box length L equals an integral number of half wavelengths. L = n λ n 2 ; n =1,2,3, Standing wave condition

42 A particle in a box: allowed energies 42 E = mv2 2 The standing wave condition yields the allowed energies. E n = n 2 h 2 = p2 2m = h2 2mλ 2 8mL 2 = n2 E 1 L = n λ n 2 ; n =1,2,3, The lowest allowed energy, E 1, is called ground state energy. Note that E 1 is not zero, and depends on the size of the box.

43 Electron bound to an atom 43 If an electron is constrained to be within an atom, the electron is confined in one of the allowed energy states. The electron can make a transition to and from one energy state, E i, to another, E f, by the emission of a photon (if E i > E f ). The frequency and wavelength of the emitted photon are: f = E i E f h λ = c f = hc E i E f

44 Quantum number 44 The number n is called a quantum number. It characterizes y n for a particular state and for the energy of that state, E n. For a particle in a 1-D box, a quantum number arises from the boundary condition on y : y (0) = 0 and y (L) = 0. For a particle in a 3-D box, three quantum numbers arise, one associated with a boundary condition in each dimension.

45 Standing wave functions: probability densities 45 The probability per unit length of finding the particle as a function of position is y n2 (x). The particle is most likely to be found near the maxima. The particle cannot be found where y 2 = 0. For very large values of n, the maxima and minima are so closely spaced that y 2 cannot be distinguished from its average value. The particle is equally likely to be found anywhere in the box, the same as in the classical result.

46 Large quantum number 46 The fractional energy difference of adjacent states becomes very small as the quantum number increases. For a very large n, energy quantization is not important. Bohr s correspondence principle states: In the limit of very large quantum numbers, the classical calculation and the quantum calculation must yield the same results.

47 Clicker Question 19-2 There are three 1-D boxes, B1, B2, and B3, with length L, 2L, and 3L, respectively. Each box contains an electron in the state for n = 10. Rank the boxes according the number of maxima for the probability density of the electron, greatest first. B1, B2, B3 B3, B2, B1 B2, B3, B1 They are all tie.

48 Example 1 48 An electron is in the initial state n i = 3 of an 1-D box of length 100 pm. If it is to make a quantum jump to the state n f = 6 by absorbing a photon, what must be the energy and wavelength of the photon? E n = n 2 h 2 8mL 2 = n2 E 1 E 6 E 3 = ( ) h 2 = 27 (hc) 2 8mL 2 8(mc 2 )L 2 (1240eV nm) 2 = 27 8( ev )(0.1nm) 2 = eV E = hc λ λ = hc E = 1240ev nm eV =1.22nm

49 Hydrogen Atom 49 an electron is bound to a proton by the electrostatic force of attraction

50 50 Timeline

51 51 Backup

52 Compton scattering: conservation of p and E 52 From conservation of momentum: p i = p s + p e p e = p i p s p e p e = ( p i p ) s ( p i p ) s = p i p i + p s p s 2 p i p s = p i 2 + p s 2 2 p i p s cosθ From conservation of energy: E i + E e, i = E s + E e, f p i c + m e c 2 = p s c + p 2 e c 2 + ( m e c 2 ) 2

53 Compton equation 53 Combining the momentum and energy conservation equations, we get p e 2 = p i 2 + p s 2 2 p i p s cosθ p i c + m e c 2 = p s c + p 2 e c 2 + ( m e c 2 ) = 1 (1 cosθ) p s p i m e c λ s λ i = h m e c (1 cosθ) λ (1 cosθ) = pm(1 cosθ) C

54 54 Expectation Value

55 Example 55 A particle in a one-dimensional box of length L is in the ground state. Find the probability of finding the particle (a) in the region that has a length Δx = 0.01L and is centered at x = L and (b) in the region 0 < x < L. (a) 0 (b) 1

56 Example 56 A particle in a one-dimensional box of length L is in the ground state. Find the probability of finding the particle (a) in the region that has a length Δx = 0.01L and is centered at x = L/2 and (b) in the region 0 < x < L/4.

57 Example 3 57 The photons in a monochromatic beam are scattered by electrons. The wavelength of the photons that are scattered at an angle of 135 with the direction of the incident photon beam is 2.3 percent more than the wavelength of the incident photons. a) What is the wavelength of the incident photons? b) What is the kinetic energy of the electron?

58 Harmonic oscillator potential well 58 Consider a particle with mass, m, on a spring with force constant, k. Potential energy function for a harmonic oscillator is parabolic. U x ( ) = 1 2 kx2 = 1 2 mω 0 where ω 0 = 2 x 2 k m is the natural frequency of the oscillator. Classically, the object oscillates between ±A, and its total energy, E, can have any nonnegative value, including zero. E classic = 1 2 mω 2 A 2 0 Parabolic well

59 Harmonic oscillator: allowed energies 59 Normalizable y n (x) occur only for discrete values of the energy E n given by E n = ( n )hf 0 n = 0, 1, 2, Equally spaced levels: hf 0 Note that the ground state energy is not 0.

60 Example 3 60 An electron in a harmonic oscillator is initially in the n = 4 state. It drops to n = 2 state and emits a photon with wavelength 500 nm. What is the ground state energy of this harmonic oscillator? E n = ( n + 1 2)hf ev nm = 500nm = 4.96eV hc = 1240 ev nm = J m E 4 E 2 = 2hf 0 E 0 = ( )hf 0 = 4.96eV / 4 =1.24eV