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Home , Louis de Broglie, Matter wave, Old quantum theory, Paul Dirac, Point particle, Wave function

LECTURE 6 II

Instructor: Shih-Chieh Hsu Development of Quantum

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¨ 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 ¤ Another experiment raised the concern of the and of the . Theory of of light

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¨ In 1900, resolves the blackbody radiation issues by introducing “quantum” concept of the discrete element ¤ The energy element is discrete and the energy is proportional to the ¤ The invention of h

¨ In 1905, Einstein explained photoelectric effect by using Max Planck’s light quantization concept ¤ is introduced by Gilbert N. Lewis in 1926 More Quantization System

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¨ In 1909, Robert Millikan conducted oildrop experiment and showed that electric is quantized.

¨ In 1911, Ernest Rutherford’s Gold Foil Experiment disproved the of the . Bohr’s classical model of H

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¨ proposed a model of the atom that successfully predicted the observed spectra. ¤ The of the moves in a circular orbit around the positive nucleus according to Coulomb’s law and like the planets orbiting around Sun. Flaw in the classical model

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¨ 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. Spectral Lines

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http://chemistry.tutorvista.com/inorganic-chemistry/spectral-lines.html Hydrogen Spectrum

8 Bohr’s semi-classical model (1913)

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Energy Quantization In Aton

En

|ΔEn| = hf=hc/λ 10

Old

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¨ In 1913, Niels Bohr explains the spectra line of the hydrogen atom – using quantization ideas.

¨ 1918-1923 expansion of research works! e.g. Stern-Gerlach demonstrated property of in 1920

Stern Gerlach Three Failures of

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¨ Black Body Radiation.

¨ The Hydrogen Atom

¨ Photoelectric Effect. 13

New Quantum Theory New Quantum Theory

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¨ In 1924, Louise de Broglie proposes wave theory.

¨ In 1925, Mechanics is invented ¤ Heisenberg Uncertainty was proposed in 1927

Werner Heisenberg Completion of Quantum Mechanics

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¨ In 1925, Erwin Schrödinger invented wave mechanics and non-relativistic Schrödinger as generalization of de Broglie’s theory

¨ 1927, began the process of unifying quantum mechanics with by proposing the for the electron. de Broglie hypothesis

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¨ In 1924 hypothesized: ¤ Since light exhibits particle-like properties and act as a photon, could exhibit wave-like properties and have a definite .

¨ The wavelength and frequency of matter: h E λ = f = p h ¤ For macroscopic objects, de Broglie wavelength is too small to be observed. Example 1

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¨ 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 of about 10-18 kg. What is the de Broglie wavelength for such a particle, if it is moving slowly at 1 mm/s?

−34 h h h 6.626×10 Js 13 λ = = = = = 6.626×10− m p mv 10−18 kg ×10−3 m / s 10−18 kg ×10−3 m / s

h = 6.626 ´ 10-34 J·s = 4.136 ´ 10-15 eV·s of matter

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

X-ray diffraction electron diffraction diffraction Interference and diffraction of matter

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¨ If the 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. Electron microscope

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¨ 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.

Pollen grains Clicker Question 18-1

¨ The electron microscope is a welcome addition to the 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 Classical waves vs. particles

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¨ A classical wave behaves like a sound wave. • It exhibits diffraction and interference. • Its energy is spread out continuously in and time.

¨ A classical particle behaves like a piece of shot.

¤It can be localized and scattered. ¤It exchanges energy suddenly at a in space. ¤It obeys the laws of and in collisions. ¤It does not exhibit interference or diffraction. Wave-particle duality

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¨ 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. Wave-particle duality

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¨ 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 . ¤ If there are a lot of particles, they can be treated as a wave. Clicker Question 19-1

¨ If the wavelength of an electron is equal to the wavelength of a , 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.

h h mv2 p2 h2 λ = = E = = = 2 p mv 2 2m 2mλ

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¨ If we use light with l to measure the of an object, x, its uncertainty, Δx, cannot be less than ~l because of diffraction.

¨ If we use with pg = 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.

 h ΔxΔpx ≥ , where  ≡ 2 2π Quantum Mechanics (1923)

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¨ In quantum mechanics, a particle is described by a wave y that obeys a called the Schrödinger equation.

2  2    ∂  − ∇ Ψ(r ,t) + U (r )Ψ(r ,t) = i Ψ(r ,t) 2m ∂t You absolutely do not need to memorize the formula.

¨ The solution of the equation by itself has no physical meaning. However, the to find a particle in a certain space- time is: Time-Independent Schrodinger Equation

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¨ Solution of the Schrödinger equation. 2  2    ∂  − ∇ Ψ(r ,t) + U (r )Ψ(r ,t) = i Ψ(r ,t) 2m ∂t

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¨ 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 Normalization condition

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¨ 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. Interpretation

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¨ 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 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. Schrödinger’s Cat

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¨ 1935, Austrian 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. Schrödinger’s Poor Cat

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¨ The material doesn’t ¨ The material has decayed. decay. The cat is alive The cat has been killed by the poison. Schrödinger’s Cat: Live or Dead

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¨ According to the Copenhagen interpretation, the cat is both alive and dead. It exists in a state of “superposition” Probability Calculation for a Classical Particle 35

¨ A classical moves back and forth with constant speed between two walls at x = 0 and x = 8.0 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 P0Δx. 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 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

P0Δx = 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. 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

P0Δx. A

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¨ Consider a particle of mass m confined to a one- dimensional 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 y2 describes the probability of finding the particle in some region. A particle in a box: conditions for y

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¨ 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 A particle in a box: allowed wavelengths

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¨ The boundary condition restricts the allowed wavelengths for a particle in a box. ¨ The box length L equals an number of half wavelengths.

λ L = n n ; n =1,2,3, 2 condition A particle in a box: allowed

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2 2 2 mv p h λ E = = = L = n n ; n =1,2,3, 2 2m 2mλ 2 2 ¨ The standing wave condition yields the allowed energies. 2 2 h 2 En = n 2 = n E1 8mL

¨ The lowest allowed energy, E1, is called energy.

¨ Note that E1 is not zero, and depends on the size of the box. Electron bound to an atom

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¨ 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, Ei, to another, Ef, by the emission of a photon (if Ei > Ef). ¨ The frequency and wavelength of the emitted photon are:

E − E c hc f = i f λ = = h f E − E i f

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¨ The number “n” is called a quantum number.

¨ It characterizes yn for a particular state and for the energy of that state, En. ¨ 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 . Standing wave functions: probability densities

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¨ The probability per unit length of finding the particle as a 2 function of position is yn (x). ¨ The particle is most likely to be found near the maxima. The particle cannot be found where y2 = 0. ¨ For very large values of n, the maxima and minima are so closely spaced that y2 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. Large quantum number

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¨ 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 states: In the limit of very large quantum numbers, the classical calculation and the quantum calculation must yield the same results. 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. Example 1

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¨ An electron is in the initial state ni = 3 of an 1-D box of length 100 pm. If it is to make a quantum

jump to the state nf = 6 by absorbing a photon, what must be the energy and wavelength of the photon? h2 E = n2 = n2 E hc hc n 8mL2 1 E = → λ = λ E 2 (hc)2 2 2 h 27 E − E = (6 − 3 ) = 2 2 6 3 8mL2 8(mc )L 1240ev•nm (1240eV •nm)2 = =1.22nm = 27 1015.12eV 8(5.11×105 eV )(0.1nm)2 =1015.12eV Hydrogen Atom

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an electron is bound to a proton by the electrostatic of attraction Timeline

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Backup : conservation of p and E

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¨ From conservation of momentum:       pi = ps + pe → pe = pi − ps       p p p p p p e ⋅ e = ( i − s )⋅( i − s )       = pi ⋅ pi + ps ⋅ ps − 2 pi ⋅ ps = p2 + p2 − 2 p p cosθ i s i s ¨ From conservation of energy:

Ei + Ee, i = Es + Ee, f

2 p c + m c2 = p c + p2c2 + m c2 i e s e ( e ) Compton equation

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¨ Combining the momentum and energy conservation , we get

2 2 2 pe = pi + ps − 2 pi ps cosθ 2 p c m c2 p c p2c2 m c2 i + e = s + e + ( e )

1 1 1 − = (1− cosθ) ps pi mec h λ − λ = (1− cosθ) ≡ λ (1− cosθ) = 2.426 pm(1− cosθ) s i m c C e Expectation Value

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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 Example

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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. Example 3

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¨ 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 of the electron?

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¨ Consider a particle with mass, m, on a spring with force constant, k. ¨ function for a harmonic oscillator is parabolic. U x 1 kx2 1 m 2 x2 ( ) = 2 = 2 ω0 Parabolic well where ω0 = 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.

1 2 2 Eclassic = mω0 A 2 Harmonic oscillator: allowed energies

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¨ Normalizable yn(x) occur only for discrete values of the energy En given by 1 Equally spaced levels: hf E = n + hf 0 n ( 2 ) 0 n = 0, 1, 2, Note that the ground state energy is not 0. Example 3

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¨ 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 + 1 hf n ( 2 ) 0 1240ev•nm E − E = 2hf0 = = 4.96eV 4 2 500nm hc = 1240 eV·nm = 1.988 ´ 10-25 J·m E = 0 + 1 hf = 4.96eV / 4 =1.24eV 0 ( 2 ) 0