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Scientific Methods • • • • • Observations (experiments) Hypothesis to explain (modified and refined) Models and Theory Laws (general behaviors) Assumptions/conditions, time-space scopes, definitions - limitations Atkins / Paula Physical Chemistry, 9th Edition Chapter 7 (Ch.8 of 8th Ed.) Quantum theory : introduction and principles http://ebooks.bfwpub.com/pchemoup.php http://ebooks.bfwpub.com/pchemoup Atkins & Paula ,“Physical Chemistry,” 9th Ed., Oxford, 2010 The origins of quantum mechanics 7.1 The failures of classical physics and Energy quantization 7.2 Wave-particle duality The dynamics of microscopic systems 7.3 The Schrödinger equation 7.4 The Born interpretation of the wavefunction Quantum mechanical principles 7.5 The information in a wavefunction 7.6 The uncertainty principle 7.7 The postulates of quantum mechanics P.249 Objectives - From Classic to Quantum Mechanics • Introduction of Quantum Mechanics • Understand the difference of classical theory and experimental observations of quantum mechanics (Physical Chemistry, 2nd Ed., Thomas Engel, Philip Reid) Outline 1. Why Study Quantum Mechanics? 2. Quantum Mechanics Arose Out of the Interplay of Experiments and Theory 3. Blackbody Radiation 4. The Photoelectric Effect 5. Particles Exhibit Wave-Like Behavior 6. Diffraction by a Double Slit 7. Atomic Spectra and the Bohr Model of the Hydrogen Atom Why Study Quantum Mechanics? • Quantum mechanics predicts that atoms and molecules can only have discrete (分離的, 離散 的)energies. • Quantum mechanical calculations of chemical properties of molecules are reasonably accurate. Quantum Mechanics Arose Out of the Interplay of Experiments and Theory • Two key properties are used to distinguish classical and quantum physics. 1. Quantization (量子化)- Energy at the atomic level is not a continuous variable, but in discrete packets called quanta (量子) . 2. Wave-particle duality (波粒二象性)- At the atomic level, light waves have particle-like properties, while atoms and subatomic particles have wave-like properties. The origins of quantum mechanics λν=c (7.1) ν – frequency c - speed of light Wavenumber (unit: cm-1): ν 1 ~ ν = = (7.2) c λ Fig.7.1 The wavelength, λ, of a wave is the peak-to-peak distance. (b) The wave is shown travelling to the right at a speed c. At a given location, the instantaneous amplitude of the wave changes through a complete cycle (the four dots show half a cycle). The frequency, ν, is the number of cycles per second that occur at a given Chapter 7. Quantum theory: point. introduction and principles P.250 Fig.7.2 The electromagnetic spectrum and the classification of the spectral regions Chapter 7. Quantum theory: introduction and principles P.250 7.1 Energy quantization - Key points • • • • (a) The classical approach to the description of black-body radiation results in the ultraviolet catastrophe. (b) To avoid this catastrophe, Planck proposed that the electromagnetic field could take up energy only in discrete amounts. (c) The thermal properties of solids, specifically their heat capacities, also provide evidence that the vibrations of atoms can take up energy only in discrete amounts. (d) Atomic and molecular spectra show that atoms and molecules can take up energy only in discrete amounts. Blackbody Radiation (黑體輻射) • • An ideal blackbody is a cubical solid at a high temperature emits photons from an interior spherical surface. The reflected photons ensure that the radiation is in thermal equilibrium (熱平衡) with the solid. Black-body Radiation (黑體輻射, cont’d) A black body is an object of emitting and absorbing all frequencies of radiation uniformly. Fig.7.4 An experimental representation of a black-body is a pinhole in an otherwise closed container. The radiation is reflected many times within the container and comes to thermal equilibrium with the walls at a temperature T. Radiation leaking out through the pinhole is characteristic of the radiation within the container. Chapter 7. Quantum theory: introduction and principles P.251 Black-body Radiation (黑體輻射, cont’d) Energy density (Jm-3): dε(λ,T) = ρ(λ,T) dλ (7.3) Total energy density in a region: ∞ (7.4) ε (T ) = ρ (λ, T )dλ ∫ 0 Total energy: E(T) = V ε(T) (7.5) Fig.7.3 The energy distribution in a black-body cavity at several temperatures. Note how the energy density increases in the region of shorter wavelengths as the temperature is raised, and how the peak shifts to shorter wavelengths. The total energy density (the area under the curve) increases as the temperature is increased (as T4). Chapter 7. Quantum theory: introduction and principles P.251 Black-body Radiation (黑體輻射, cont’d) Fig.7.5 The electromagnetic vacuum can be regarded as able to support oscillations of the electromagnetic field. When a high frequency, short wavelength oscillator (a) is excited, that frequency of radiation is present. The presence of low frequency, long wavelength radiation (b) signifies that an oscillator of the corresponding frequency has been excited. Chapter 7. Quantum theory: introduction and principles P.252 Black-body Radiation (黑體輻射, cont’d) Energy density (Jm-3): dε(λ,T) = ρ(λ,T) dλ (7.3) Rayleigh-Jeans law - density of states (Jm-4) : ρ (λ,T) =8πkT/ λ4 (7.6) k – Boltzmann’s constant: 1.381 × 10-23 JK-1 Fig.7.6 The Rayleigh–Jeans law (eqn 7.6) predicts an infinite energy density at short wavelengths. This approach to infinity is called the ultraviolet catastrophe. Chapter 7. Quantum theory: introduction and principles P.252 Blackbody Radiation (黑體輻射, cont’d) • The limitation of energy to discrete values is called the quantization of energy. • E=nhν (n=0,1,2,…) (7.7) h- Planck’s constant 6.626x10-34Js Planck distribution: 8πhc dE= ρ d λ (7.8) ρ (λ , T ) = λ5 (e hc / λkT − 1) hc / λ kT → ∞ faster than λ5->0 For short wavelengthλ, hc/λkT» 1, e ρ →0 as λ →0 For long wavelengthλ, hc/λkT«1, or ν →∞ hc hc e hc / λkT − 1 = 1 + + ... − 1 ≈ λkT λkT Taylor series: f(x) = f(x0) + f’(x0)(x-x0)/1! + f”(x0)(x-x0)2/2! +… Blackbody Radiation (cont’d) • Max Planck derived the agreement between theory and experiment on radiation energy. E = nhv where h = Planck’s constant n = a positive integer (n 0, 1, 2, . . . ) • The theory states that the energies radiated by a blackbody are not continuous, but can take discrete values for each frequency. Blackbody Radiation (黑體輻射, cont’d) energy density ∞ 8πhc d λ = aT 4 ε (T) = ∫ 5 hc / λkT 0 λ e −1 ( with ) 8π 5 k 4 a= 15(hc) 3 (7.9 ) Fig.7.7 The Planck distribution (eqn 7.8) accounts very well for the experimentally determined distribution of black-body radiation. Planck ’s quantization hypothesis essentially quenches the contributions of high frequency, short wavelength oscillators. The distribution coincides with the Rayleigh–Jeans distribution at long wavelengths. Chapter 7. Quantum theory: introduction and principles P.246 Blackbody Radiation (cont’d) • Spectral density (譜密度))is the energy stored in the electromagnetic field of the blackbody radiator. Heat Capacity • • • • Molar heat capacities Cv=(∂U/∂T)V of all monatomic solids ~25JK-1mol-1 (Dulong & Petit, early 19th century) Equipartition principle – mean energy of an atom as it oscillates about its mean position in a solid is kT for each direction of displacement Cv,m=3NA K=3R= 24.9 JK-1mol-1 (7.10) It was found that the molar heat capacities of all monatomic solids are lower than 3R at low temperatures, and that the values approaches zero as T→0 Einstein assumed that each atom oscillated about its equilibrium position with a single frequency νin 1905 Heat capacities (cont’d) molar internal energy : U m = 3N AkT = 3RT (7 .10 a ) molar constant v olume heat capacity : ∂U m −1 −1 CV , m = (7.10b) = 3 R = 24.9 JK mol ∂T V Einstein formula : CV ,m (T ) = 3 Rf E (T ) Fig.7.8 Experimental low-temperature molar heat capacities and the temperature dependence predicted on the basis of Einstein ’s theory. His equation (eqn 7.11) accounts for the dependence fairly well, but is everywhere too low. 2 θ e f E (T ) = E θ E / T T e − 1 Einstein temperatur e θ E = hν / k 1) At high temperatur es (T >> θ E ) : θ E / 2T 2 (7.11) 2 θ 1 +θ E/ 2T + L f E (T ) = E ≈ 1 (7.12a); T (1 +θ E/ T + L) − 1 2) At low temperatur es (T << θ E ) : 2 θ θE e 2 θ f E (T ) ≈ θE / T = E e -θE / T (7.12b) Chapter 7. Quantum theory: introduction P.248 T and e principles T 2 E / 2T 2 Heat capacities (cont’d) By averaging over all the frequencies present from zero up to a maximum value νD: Debye formula : CV ,m (T ) = 3Rf D (T ) 3 T θ D / T x 4e x f D (T ) = 3 ∫ dx ( 7.13) 2 0 x θ e −1 D Debye temperature : θ D = h ν D / k ( ) Fig.7.9 Debye’s modification of Einstein’s calculation (eqn 7.13) gives very good agreement with experiment. For copper, T/θD=2 corresponds to about 170 K, so the detection of deviations from Dulong and Petit’s law had to await advances in lowtemperature physics. Chapter 7. Quantum theory: introduction and principles P.249 The Photoelectric Effect (光電效應) • • The electrons emitted by the surface upon illumination are incident on the collector, which is at an appropriate electrical potential to attract them. This is called the photoelectric effect. The Photoelectric Effect (cont’d) • Albert Einstein states that the energy of light, E = βv • where β = constant v = frequency From energy conservation the energy of the electron, Ee, is Ee = βv − φ (7.15) where Ф = work function, characteristic of the metal, the energy required to remove an electron from the metal to infinity, the analogue of the Ionization Energy of an atom or molecule. The Photoelectric Effect (cont’d) • The results of β is identical to Planck’s constant, h, thus E = hv The Photoelectric Effect (cont’d) 1) No electrons are ejected, regardless of the intensity of the radiation, unless its frequency exceeds a threshold value characteristic of the metal. 2) The kinetic energy of the ejected electrons increases linearly with the frequency of the incident radiation but is independent of the intensity of the radiation. 3) Even at low light intensities, electrons are ejected immediately if the frequency is above the threshold. Fig.7.13 In the photoelectric effect, it is found that no electrons are ejected when the incident radiation has a frequency below a value characteristic of the metal and, above that value, the kinetic energy of the photoelectrons varies linearly with the frequency of the incident radiation. Chapter 7. Quantum theory: introduction and principles P.251 The Photoelectric Effect (cont’d) 1 mev 2 = hν − Φ (eqn 7.15) provides a technique for 2 the determination of Planck’s constant, for the slopes of the lines in Fig.7.13 are all equal to h. 1) 2) 3) Photoejection cannot occur if hν < Φ because the photon brings insufficient energy; 1 me v2 = hν − Φ 2 (Eq.7.15) predicts that the kinetic energy of an ejected electron should increase linearly with frequency; When a photon collides with an electron, it gives up all its energy, so we should expect electrons to appear as soon as the collisions begin, provided the photons have sufficient energy. Fig.7.14 The photoelectric effect can be explained if it is supposed that the incident radiation is composed of photons that have energy proportional to the frequency of the radiation. (a) The energy of the photon is insufficient to drive an electron out of the metal. (b) The energy of the photon is more than enough to eject an electron, and the excess energy is carried away as the kinetic energy of the photoelectron (the ejected electron). Chapter 7. Quantum theory: introduction and principles P.251 Example 1 Light with a wavelength of 300 nm is incident on a potassium surface for which the work function, φ , is 2.26 eV. Calculate the kinetic energy and speed of the ejected electrons. Solution We write Ee = hv − φ = (hc / λ ) − φ and convert the units of φ from electron-volts to joules: φ = (2.26eV )(1.602 ×10 −19 J / eV ) = 3.62 ×10−19 J Electrons will only be ejected if the photon energy, hv, is greater than φ . The photon energy is calculated to be hc λ ( 6.626 ×10 )(2.998 ×10 ) = = 6.62 ×10 34 8 −9 300 ×10 which is sufficient to eject electrons. −19 J Solution (cont’d) We can obtain Ee = (hc / λ ) − φ = 2.99 ×10−19 J . Using Ee = 1 / 2mv 2 , we calculate that ( ) 2 Ee 2 2.99 ×10 −19 J 5 v= = = 8 . 10 × 10 m/ s −31 m 9.109 ×10 Example 2 Calculating the number of photons - The particle character of electromagnetic radiation • Calculate # of photons emitted by a 100 W yellow lamp in 1.0s. (wavelength of yellow light ~ 560 nm, assume 100% efficiency). • Method: Each photon has an energy hν, so the total # of photons needed to produce energy E is E/hν. The frequency of the radiation (ν = c/λ=?) and the total energy (?) emitted by the lamp E =?. E = P∆t, P - the power (in watts), ∆t - the time interval for which the lamp is turned on. • Answer: The number of photons is E P∆t λP∆t N= = = h ν h (c λ ) hc ( 5.60×10 = (6.626×10 −7 )( ) Js )×(2.998×10 m × 100 Js −1 ×(1.0 s ) −34 20 = 2 . 8 × 10 8 ms−1 ) • Note: it would take ~40 min to produce 1 mol of these photons. Self Test - The particle character of electromagnetic radiation • How many photons does a monochromatic (single frequency) infrared rangefinder of power 1 mW and wavelength 1000 nm emit in 0.1 s? • Answer: [5×1014] Particles Exhibit Wave-Like Behavior • • In 1924, Louis de Broglie suggested a relationship between momentum and wavelength for light applying to particles - any particle, not only photons, travelling with a linear momentum p should have (in some sense) a wavelength given by the de Broglie relation: h λ= p where p = mv (particle momentum) (7.16) Not only has electromagnetic radiation the character classically ascribed to particles, but electrons (and all other particles) have the characteristics classically ascribed to waves. This joint particle and wave character of matter and radiation is called wave–particle duality. 7.2 Wave-particle duality • Key points: (a) The photoelectric effect establishes the view that electromagnetic radiation, regarded in classical physics as wave-like, consists of particles (photons). (b) The diffraction of electrons establishes the view that electrons, regarded in classical physics as particles, are wave-like with a wavelength given by the de Broglie relation. Fig.7.16 An illustration of the de Broglie relation between momentum and wavelength. The wave is associated with a particle (shortly this wave will be seen to be the wavefunction of the particle). A particle with high momentum has a wavefunction with a short wavelength, and vice versa. Chapter 7. Quantum theory: introduction and principles P.253 Example 7.3 Estimating the de Broglie wavelength • Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 40 kV. • Method: λ=h/p (7.16) The linear momentum of the electrons, p=? E=? The energy acquired by an electron accelerated through a potential difference ∆Φ is e ∆Φ, where e is the magnitude of its charge. At the end of the period of acceleration, all the acquired energy is in the form of kinetic energy, EK = 1/2mev2=p2/2me Let p2/2me = e∆Φ → p Carry through the calculation algebraically before substituting the data. Example 7.3 Estimating the de Broglie wavelength • Answer: p2/2me = e ∆Φ p = (2mee ∆Φ)1/2 de Broglie relation λ = h/p h λ= 1 (2me e∆φ ) 2 λ= {2 × (9.109 ×10 6.626 ×10 − 34 J ⋅ s − 31 ) ( ) ( kg × 1.602 × 10 −19 C × 4.0 × 10 4V )} 1/ 2 = 6.1×10 −12 m • 1 V C = 1 J and 1 J = 1 kg m2 s−2. • The wavelength of 6.1 pm is shorter than typical bond lengths in molecules (about 100 pm). Electrons accelerated in this way are used in the technique of electron diffraction for the determination of molecular structure • (see Section 23.3 structure of solid surface p.885). Self Test 7.2 • Calculate: (neutron mass: mn=1.675×10-27kg Planck’s const.: h=6.626×10-34Js Boltzmann’s const.: k=1.381×10-23J/K) • (a) the wavelength of a neutron with a translational kinetic energy equal to kT at 300 K, • (b) a tennis ball of mass 57 g travelling at 80 km/h. • Answer: • (a) 178 pm, • (b) 5.2 × 10−34 m Example 4 Electrons are used to determine the structure of crystal surfaces. To have diffraction, the wavelength of the electrons should be on the order of the lattice constant, which is typically 0.30 nm. What energy do such electrons have, expressed in electron-volts and joules? Solution: Using E=p2/2m for the kinetic energy, we obtain E= ( ) − 34 2 p h 6.626 ×10 −18 = = = 2 . 7 × 10 or 17eV 2 −31 −10 2 m 2 mλ 2 9.109 × 10 3.0 ×10 2 2 ( )( ) Diffraction by a Double Slit (雙縫衍射) Diffraction of Light • Diffraction is a phenomenon that can occur with any waves, including sound waves, water waves, and electromagnetic (light) waves. Chapter 7. Quantum theory: introduction and principles P.252 Diffraction by a Double Slit (cont’d) • For diffraction of light from a thin slit, b >> a. Diffraction by a Double Slit (cont’d) • Maxima and minima arise as a result of a path difference between the sources of the cylindrical waves and the screen. Diffraction by a Double Slit (cont’d) • The condition that the minima satisfy is nλ sin θ = , n = ±1,±2,±3,±..... a where λ = wavelength Diffraction by a Double Slit (cont’d) • For double-slit diffraction experiment, One slit or the other blocked both slits open Physical Chemistry Fundamentals: Figure 7.15 Fig. 7.15 The Davisson–Germer experiment. The scattering of an electron beam from a nickel crystal shows a variation of intensity characteristic of a diffraction experiment in which waves interfere constructively and destructively in different directions. Atomic Spectra and the Bohr Model of the Hydrogen Atom • Light is only observed at certain discrete wavelengths, which is quantized. For the emission spectra, the inverse of the wavelength, 1 / λ = v~ of all lines in an atomic hydrogen spectrum is given by • 1 −1 −1 1 ~ v cm = RH cm 2 − 2 , n > n1 n1 n ( ) ( ) Physical Chemistry Fundamentals: Figure 7.10 Atomic spectra Bohr frequency condition: ∆E = hν (7.14) Fig. 7.10 A region of the spectrum of radiation emitted by excited iron atoms consists of radiation at a series of discrete wavelengths (or frequencies). Physical Chemistry Fundamentals: Figure 7.11 Molecular spectra Fig. 7.11 When a molecule changes its state, it does so by absorbing radiation at definite frequencies. This spectrum is part of that due to the electronic, vibrational, and rotational excitation of sulfur dioxide (SO2) molecules. This observation suggests that molecules can possess only discrete energies, not an arbitrary energy. Physical Chemistry Fundamentals: Figure 7.12 The failures of classical physics - Atomic and molecular spectra Fig. 7.12 Spectroscopic transitions, such as those shown above, can be accounted for if we assume that a molecule emits a photon as it changes between discrete energy levels. Note that high-frequency radiation is emitted when the energy change is large. SUMMARY: Energy quantization - Key points • • • • (a) The classical approach to the description of black-body radiation results in the ultraviolet catastrophe. (b) To avoid this catastrophe, Planck proposed that the electromagnetic field could take up energy only in discrete amounts. (c) The thermal properties of solids, specifically their heat capacities, also provide evidence that the vibrations of atoms can take up energy only in discrete amounts. (d) Atomic and molecular spectra show that atoms and molecules can take up energy only in discrete amounts. Electron microscopy (電子顯微鏡) • Traditional light microscopy - illuminating a small area of a sample and collecting light with a microscope. • The resolution of a microscope, the minimum distance between two objects that leads to two distinct images, is on the order of the wavelength of light used as a probe. • Conventional microscopes employing visible light have resolutions in the micrometre range and are blind to features on a scale of nanometres. • Electron microscopy, in which a beam of electrons with a well defined de Broglie wavelength replaces the lamp found in traditional light microscopes. Instead of glass or quartz lenses, magnetic fields are used to focus the beam. • In transmission electron microscopy (透射電子顯微鏡 TEM), the electron beam passes through the specimen and the image is collected on a screen. • In scanning electron microscopy (掃描電子顯微鏡 SEM), electrons scattered back from a small irradiated area of the sample are detected and the electrical signal is sent to a video screen. An image of the surface is then obtained by scanning the electron beam across the sample. Electron microscopy (電子顯微鏡cont’d) • The wavelength of and the ability to focus the incident beam— electrons—govern the resolution. • Electron wavelengths in typical electron microscopes can be as short as 10 pm, but it is not possible to focus electrons well with magnetic lenses so. Typical resolutions of TEM and SEM instruments are about 2 nm and 50 nm, respectively. • Electron microscopes cannot resolve individual atoms, whose diameters ~ 0.2 nm=2Å. • Only certain samples can be observed under certain conditions. The measurements must be conducted under high vacuum. • For TEM observations, the samples must be very thin crosssections of a specimen and SEM observations must be made on dry samples. Neither technique can be used to study living cells. Physical Chemistry Fundamentals: Figure 7.17 Electron microscopy is very useful in studies of the internal structure of cells Fig. 7.17 A TEM image of a cross-section of a plant cell showing chloroplasts, organelles responsible for the reactions of photosynthesis. Chloroplasts are typically 5 µm long. (Brian Bowes) Objectives - The Schrödinger Equation • Key concepts of operators, eigenfunctions, wave functions, and eigenvalues. Outline 1. What Determines If a System Needs to Be Described Using Quantum Mechanics? 2. Classical Waves and the Nondispersive(非色 散) Wave Equation 3. Waves Are Conveniently Represented as Complex Functions 4. Quantum Mechanical Waves and the Schrödinger Equation Outline 1. Solving the Schrödinger Equation: Operators, Observables, Eigenfunctions, and Eigenvalues 2. The Eigenfunctions of a Quantum Mechanical Operator Are Orthogonal 3. The Eigenfunctions of a Quantum Mechanical Operator Form a Complete Set 4. Summing Up the New Concepts What Determines If a System Needs to Be Described Using Quantum Mechanics? • • Particles and waves in quantum mechanics are not separate and distinct entities. Waves can show particle-like properties and particles can also show wave-like properties. What Determines If a System Needs to Be Described Using Quantum Mechanics? (cont’d) • • In a quantum mechanical system, only certain values of the energy are allowed, and such system has a discrete energy spectrum. Thus, Boltzmann distribution is used. ni g i − [ei − e j ]/ kT = e nj g j where ni = number of atoms or molecules εi = energy of atoms or molecules gi = degeneracy (簡並度) at energy level i (energy level vs. energy state) Example 5 Consider a system of 1000 particles that can only have two energies, ∆ε = ε 2 − ε 1, with ε1 and ε 2 . The difference in the energy between these two values is ε 2 > ε1 . Assume that g1=g2=1. a. Graph the number of particles, n1 and n2, in states ε1 and ε 2 as a function of kT / ∆ε . Explain your result. b. At what value of kT / ∆ε do 750 of the particles have the energy ε 1 ? Solution We can write down the following two equations: n2 / n1 = e − ∆ε / kT and n1 + n2 = 1000 Solve these two equations for n2 and n1 to obtain 1000 e − ∆ε / kT n2 = 1 + e − ∆ε / kT 1000 n1 = 1 + e − ∆ε / kT Solution Part (b) is solved graphically. The parameter n1 is shown as a function of kT / ∆ε on an expanded scale on the right side of the preceding graphs, which shows that n1=750 for kT / ∆ε = 0.91 . Classical Waves and the Nondispersive Wave Equation • Transverse(橫向 ), Longitudinal(縱向 ), and Surface Waves • A wave can be represented pictorially by a succession of wave fronts, where the amplitude has a maximum or minimum value. transverse wave (橫波 橫波) 橫波 • A transverse wave is a moving wave that consists of oscillations occurring perpendicular to the direction of energy transfer. Transverse plane wave Longitudinal wave (縱波 縱波) 縱波 • Longitudinal waves are waves that have the same direction of oscillation or vibration along their direction of travel, which means that the oscillation of the medium (particle) is in the same direction or opposite direction as the motion of the wave. Plane pressure wave • Longitudinal harmonic Sound waves: x y(x, t) = y0 sin ω t − c Where y is the displacement of the point on the traveling sound wave; x is the distance the point has traveled from the wave's source; t is the time elapsed; y0 is the amplitude of the oscillations, c is the speed and ω is the angular frequency of the wave. The quantity x/c is the time that the wave takes to travel the distance x. Classical Waves and the Nondispersive Wave Equation • The wave amplitude ψ is: x t Ψ ( x, t ) = A sin 2π − λ T • It is convenient to combine constants and variables to write the wave amplitude as Ψ ( x, t ) = A sin (kx − wt ) where k = 2π/λ (wave vector) ω = 2πv (angular frequency) p = hk , h≡ h 2π Classical Waves and the Wave Equation • Interference of Two Traveling Waves • For wave propagation in a medium where frequencies have the same velocity (a nondispersive medium), we can write ∂Ψ ( x , t ) 1 ∂ 2 Ψ ( x , t ) = 2 2 ∂x v ∂t 2 where v = velocity at which the wave propagates Example 6 The nondispersive wave equation in one dimension is given by ∂ 2ψ ( x, t ) 1 ∂ 2ψ (x, t ) = 2 2 ∂x v ∂t 2 Show that the traveling wave ψ (x, t ) = A sin (kx − ωt + φ ) is a solution of the nondispersive wave equation. How is the velocity of the wave related to k and ω? Solution: We have ∂ 2ψ (x, t ) 1 ∂ 2ψ (x, t ) = 2 2 ∂x v ∂t 2 ∂ 2 A sin (kx − ω t + φ ) 2 = − k A sin (kx − ωt + φ ) 2 ∂x ∂ 2 A sin (kx − ω t + φ ) − ω 2 = 2 A sin (kx − ωt + φ ) 2 2 v ∂t v Equating these two results gives v = ω / k Waves Are Conveniently Represented as Complex Functions • It is easier to work with the whole complex function knowing as we can extract the real part of wave function. Example 7 a. Express the complex number (4+4i) in the form reiθ = r (cosθ + i sin θ ) π 3 e b. Express the complex number in the form (a+ib) i3 / 2 Solution a. The magnitude of 4+4i is [(4 + 4i )(4 − 4i)]1/ 2 = 4 The phase is given by cos θ = 4 4 2 = 1 1 π or θ = cos −1 = 2 2 4 Therefore, 4+4i can be written 4 2ei (π / 4) 2 Solution b. Using the relation e iα can be written = exp(iα ) = cos α + i sin α , 3ei 3π / 2 3π 3π 3 cos + i sin = 3(0 − i ) = −3i 2 2 The dynamics of microscopic systems • Quantum mechanics acknowledges the wave– particle duality of matter by supposing that, rather than travelling along a definite path, a particle is distributed through space like a wave. • The mathematical representation of the wave that in quantum mechanics replaces the classical concept of trajectory is called a wavefunction, ψ (psi). 7.3 Quantum Mechanical Waves and The Schrödinger equation • Key point: Schrödinger equation is a second-order differential(微分) eqn used to calculate the wavefunction of a system • (1926) The time-independent Schrödinger equation for a particle of mass m moving in one dimension with energy E is h 2 d 2ψ ( x) − + V ( x)ψ ( x) = Eψ ( x ) 2 2m dx (7.17) • Study the stationary states of quantum mechanical systems. Quantum Mechanical Waves and the Schrödinger Equation • An analogous quantum mechanical form of time-dependent classical nondispersive wave equation is the time-dependent Schrödinger equation, given as ∂Ψ ( x, t ) h 2 ∂ 2 Ψ( x, t ) ih =− + V ( x, t )Ψ ( x, t ) 2 ∂t 2m ∂ x • where V(x,t) = potential energy function This equation relates the temporal and spatial derivatives of ψ(x,t) and applied in systems where energy changes with time. Quantum Mechanical Waves and the Schrödinger Equation • • For stationary states of a quantum mechanical system, we have ∂Ψ( x, t ) ih = E Ψ( x, t ) ∂t Since , we can show that that wave functions whose energy is independent of time have the form of Ψ ( x, t ) = ψ ( x)e Et / ih = ψ ( x)e − i( E / h ) t Using the Schrödinger equation to develop the de Broglie relation d 2ψ 2m = − 2 (E − V )ψ 2 dx h ψ = cos kx = cos( 2πx / λ ) (harmonic 2m(E − V ) 2mEk k = = 2 2 h h Ek = k 2h 2 / 2 m = p 2 / 2 m 1/ 2 2π h h p = kh = × = λ 2π λ 1/ 2 wave) Born interpretation of the wavefunction • Key points: According to Born interpretation, the probability density is proportional to the square of the wavefunction. • (a) A wavefunction is normalized if the integral of its square is equal to 1; • (b) The quantization of energy stems from the constraints that an acceptable wavefunction must satisfy. Born interpretation of the wavefunction (cont’d) • If the wavefunction of a particle has the value ψ at some point x, then the probability of finding the particle between x and x + dx is proportional to |ψ|2dx. • |ψ|2 is the probability density. The wavefunction ψ itself is called the probability amplitude. • If the wavefunction of a particle has the value ψ at some point r, then the probability of finding the particle in an infinitesimal volume dτ = dxdydz at that point is proportional to |ψ|2dτ. Born interpretation of the wavefunction (cont’d) Fig. 7.18 The wavefunction ψ is a probability amplitude in the sense that its square modulus (ψ*ψ or |ψ|2) is a probability density. The probability of finding a particle in the region dx located at x is proportional to |ψ|2dx. We represent the probability density by the density of shading in the superimposed band. Born interpretation of the wavefunction (cont’d) For a particle free to move in 3-dimensions (e.g., an electron near a nucleus in an atom), the wave-function depends on the point dr with coordinates (x, y, z) Fig. 7.19 The Born interpretation of the wavefunction in three-dimensional space implies that the probability of finding the particle in the volume element dτ = dxdydz at some location r is proportional to the product of dτ and the value of |ψ|2 at that location. Physical Chemistry Fundamentals: Figure 7.20 Fig. 7.20 The sign of a wavefunction has no direct physical significance: the positive and negative regions of this wavefunction both correspond to the same probability distribution (as given by the square modulus of ψ and depicted by the density of shading). Example 7.3 Interpreting a wavefunction • The wavefunction of an electron in the lowest energy state of a hydrogen atom is proportional to exp(−r/a0), with a0 a constant (Bohr radius) and r the distance from the nucleus. Calculate the relative probabilities of finding the electron inside a region of volume 1.0 pm3, located at (a) the nucleus, (b) a distance a0 from the nucleus. • Method The region of interest is small on the scale of the atom. Ignore the variation of ψ within it and write the probability, P, as proportional to the probability density (ψ2) evaluated at the point of interest multiplied by the volume of interest, δV. i.e., P ∝ ψ2δV, with ψ2 ∝ exp(−2r/a0). Example 7.3 Interpreting a wavefunction (cont’d) • Answer In each case δV = 1.0 pm3. • (a) At the nucleus, r = 0, ( ) ( P ∝ e 0 × 1.0 pm3 = (1.0 )× 1.0 pm3 ) • (b) At a distance r = a0 in an arbitrary direction, P ∝ e−2 × (1.0 pm3 ) = (0.14 ) × (1.0 pm3 ) The ratio of probabilities is 1.0/0.14 = 7.1. Note: it is more probable (by a factor of 7) that the electron will be found at the nucleus than in a volume element of the same size located at a distance a0 from the nucleus. The negatively charged electron is attracted to the positively charged nucleus, and is likely to be found close to it. Test 7.3 • The wavefunction for the electron in its lowest energy state in the ion He+ is proportional to exp(−2r/a0). Repeat the calculation for this ion. Any comment? • Answer: 55; more compact wavefucntion A note on good practice: The square of a wavefunction is not a probability: it is a probability density, and (in 3-dimensions) has the dimensions of 1/length3. It becomes a (unitless) probability when multiplied by a volume. In general, we have to take into account the variation of the amplitude of the wavefunction over the volume of interest, but here we are supposing that the volume is so small that the variation of ψ in the region can be ignored. Normalization of wave-function N - normalization constant ∞ N 2 ∫ ψ ∗ψdx = 1 (7 .18) −∞ 1 N= 1/ 2 ∞ ψ ∗ψdx ∫− ∞ (7 .19 ) A normalized wavefunction satisfies: In 1 − Dimension: ∫ ∞ ψ ∗ψdx = 1 −∞ (7. 20a) In 3 − Dimensions : ∞ ∞ ∞ ψ ∗ψdxdydz = 1 ∫ ∫ ∫ ∫ψ ψdτ = 1 −∞ −∞ −∞ ∗ where dτ = dxdydz (7. 20b) (7.20c) Spherical polar coordinates for system with spherical symmetry x=rsinθcosφ, y=rsinθsinφ, z=rcosθ r – radius [0,∞) θ - colatitude(餘緯度), [0, π] φ - azimuth(方位), [0,2π] dτ=r2sinθdrdθdφ ∞ π 0 0 ∫ ∫ ∫ 2π 0 ψ ∗ψ r 2dr sin θdθdφ = 1 (7.20d ) Fig. 7.21 The spherical polar coordinates used for discussing systems with spherical symmetry. Physical Chemistry Fundamentals: Figure 7.22 Fig. 7.22 The surface of a sphere is covered by allowing θ to range from 0 to π, and then sweeping that arc around a complete circle by allowing φ to range from 0 to 2π. The Eigenfunctions of a Quantum Mechanical Operator Are Orthogonal • • 3-D system is importance to us is the atom. Atomic wave functions are best described by spherical coordinates. Example 7.4 Normalizing a wavefunction • Normalize the wavefunction for the hydrogen atom in Ex.7.3. • Method: Find the normalization factor N in eqn 7.20c. For a spherical system is, use spherical coordinates and carry out the integrations in eqn 7.20d. Note: the limits on the 1st integral sign - r, 2nd - θ, the 3rd φ. A useful integral: ∫ ∞ 0 n! x e dx = n+1 a n − ax where n! denotes a factorial: n! = n(n − 1)(n − 2)...1. Answer to Ex.7.4 Normalizing a wavefunction: The integration required is the product of three factors: ∫ψ ψdτ = N ∫ ∗ 2 ∞ 0 2 −2 r / a 0 r e π 2π 0 0 dr ∫ sin θdθ ∫ dπ 1 = N 2 • a03 • 2 • 2π = πa03 N 2 = 1 4 • and the normalized wavefunction is 1 N = 3 πa0 1/ 2 1/ 2 1 − r / a0 ψ = 3 e πa0 • Note: a0 is a length, the dimensions of ψ are 1/length3/2 and those of ψ2 are 1/length3 - probability density. • Repeating Ex.7.3, actual probabilities of finding the electron in the volume element at each location (a0=52.9pm): (a) 2.2×10−6, ~ 1 chance in about 500,000 inspections of finding the electron in the test volume; (b) 2.9 × 10−7, ~ 1 chance in 3.4 million. Example −r Normalize the function e over the interval 0 ≤ r ≤ ∞; 0 ≤ θ ≤ π ; 0 ≤ φ ≤ 2π Solution: Volume element in spherical coordinates is r 2 sin θdrdθdφ , thus 2π π ∞ 0 0 0 N 2 ∫ dφ ∫ sin θdθ ∫ r 2e −2 r dr = 1 ∞ 4πN 2 ∫ r 2 e−2 r dr = 1 0 Solution e− r Using the standard integral , ∫ ∞ 0 x n e − ax dx = n!/ a n +1 (a > 0, n is a positive integer) 2 π 4 N we obtain 2! 1 = 1 so that N = 23 π The normalized wave function is 1 π e −r Note that the integration of any function involving r, even if it does not explicitly involve φ or θ , requires integration over all three variables. Test 7.4 Normalize the wavefunction given in test 7.3. He+ g.s.~ exp(−2r/a0). • Correct Answer: N = (8/πa03)1/2 • 7.4 Born interpretation of wavefunction (b) Quantization • ψ must be continuous, have a continuous slope (the 2nd derivative of ψ must be well-defined), be single-valued, and be square-integrable - The wavefunction ψ must not be infinite anywhere (the wavefunction must not be infinite over any finite region). • An acceptable wavefunction cannot be zero everywhere, because the particle it describes must be somewhere. • These are such severe restrictions that acceptable solutions of the Schrödinger eqn do not in general exist for arbitrary values of the energy E. In other words, a particle may possess only certain energies, for otherwise its wavefunction would be physically unacceptable. i.e., the energy of a particle is quantized. Physical Chemistry Fundamentals: Figure 7.23 Fig. 7.23 The wavefunction must satisfy stringent conditions for it to be acceptable. (a) Unacceptable because it is not continuous; (b) unacceptable because its slope is discontinuous; (c) unacceptable because it is not single-valued; (d) unacceptable because it is infinite over a finite region. Quantum Mechanical principles - 7.5 The information in a wavefunction • Key points: (a) The wavefunction (波函數) of a free particle with a specific linear momentum corresponds to a uniform probability density (均勻概率密度). • (b) The Schrodinger eqn is an eigen value eqn (本徵方程) in which the wavefunction is an eigenfunction (本徵函數) of the Hamiltonian operator(算符). • (c) Observables (可觀測量) are represented by operators (算符); the value of an observable is an eigenvalue (本徵值) of the corresponding operator constructed from the operators for position and linear momentum (線性動量). • (d) All operators that correspond to observables are hermitian; their eigenvalues are real and their eigenfunctions are mutually orthogonal (正 交). Sets of functions that are normalized and mutually orthogonal are called orthonormal (正交歸一). • (e) When the system is not described by an eigenfunction of an operator, it may be expressed as a superposition (線性疊加) of such eigenfunctions. The mean value (平均值) of a series of observations is given by the expectation value (期望值) of the corresponding operator. QM principles - 7.5 The information in a wavefunction (cont’d) h 2 d 2ψ − = Eψ 2 2m dx ( 7.21) Solutions :ψ = Ae + Be B =0: ψ = Ae ikx ik x ( −ik x , k 2h 2 E= 2m ) (Ae ) = (A e )(Ae ) = A ψ = Ae ikx ∗ A = B: ψ = A(e ikx + e −ikx ) = 2Acoskx 2 ikx ∗ −ikx ikx (7.22) (7.23) 2 (7.24) (7.25) probabilit y density : ∗ ψ = (2Acoskx) (2Acoskx) = 4 A cos 2 kx 2 node 2 (7.26) Fig.7.24 (a) The square modulus of a wavefunction corresponding to a definite state of linear momentum is a constant; so it corresponds to a uniform probability of finding the particle anywhere. (b) The probability distribution corresponding to the superposition of states of equal magnitude of linear momentum but opposite direction of travel. Chapter 7. Quantum theory: introduction and principles P.267 7.5 The information in a wavefunction (b) Operators, eigenvalues, and eigenfunctions Schr&o&dinger eqn (in operator form) : Hamiltonia n operator (in one - dimension) : Hˆ ψ = E ψ h2 d 2 ˆ H =− + V ( x) 2 m dx 2 (Operator)(function) = (constant factor) × (same function) (7.28a) ˆ ψ = ωψ Eigenvalue equation : Ω (7.28b) (Operator)(eigenfunction) = (eigenvalu e) × (eigenfunc tion) (7.28c) ( 7.27 a) ( 7.27 b) Solving the Schrödinger Equation: Operators, Observables, Eigenfunctions, and Eigenvalues • • We would need to use operators, observables, eigenfunctions, and eigenvalues for quantum mechanical wave equation. The time-independent Schrödinger equation is an eigenvalue equation for the total energy, E h 2 ∂2 + V ( x )ψ n ( x ) = E n ψ n ( x ) − 2 2 m ∂x • ) where {} = total energy operator or H It can be simplified as Hˆ ψ n = Enψ n Example 2 2 d / dx and d / dx Consider the operators . Is the function ψ (x ) = Aeikx + Be −ikx an eigenfunction of these operators? If so, what are the eigenvalues? Note that A, B, and k are real numbers. Solution To test if a function is an eigenfunction of an operator, we carry out the operation and see if the result is the same function multiplied by a constant: ( ) d Ae ikx + Be − ikx = ikAeikx − ikB −ikx = ik Ae ikx − Be −ikx dx ( ) In this case, the result is not ψ (x) multiplied by a constant, so ψ (x) is not an eigenfunction of the operator d/dx unless either A or B is zero. Solution This equation shows that ψ (x ) is an eigenfunction of the operator d 2 / dx 2 with the eigenvalue k2. ( ) d 2 Aeikx + Be − ikx 2 2 −ikx ikx = ( ik ) Ae + ( − ik ) B 2 dx = −k 2 Ae ikx + Be −ikx ( = −k 2ψ (x ) ) Example 7.5 Identifying an eigenfunction (p.268) • Show that eax is an eigenfunction of the operator d/dx ; find the corresponding eigenvalue. Show that exp(ax2) is not an eigenfunction of d/dx. • Method: Operate on the function with the operator and check whether the result is a constant factor times the original function. • Answer: d ax ˆ Ωψ = e = ae ax = aψ dx d ax2 ax 2 ˆ Ωψ = e = 2axe = 2ax ×ψ dx • Test 7.5: Is the function cos ax an eigenfunction of (a) d/dx, (b) d2/dx2? • Correct Answer: (a) No, (b) yes (c) The construction of operators (p.268) ˆ, Observables, Ω, are represente d by operators, Ω built from the following position and momentum operators : (Energy operator)ψ = (energy) ×ψ (Operator correspond ing to an observable)ψ = (value of observable) ×ψ Specificat ioon of operators : x̂ = x × , h d p̂ x = i dx [7.29] Ex.7.6 Determining the value of an observable (p.269) • What is the linear momentum of a particle described by the wavefunction in eqn 7.22: with (a) B = 0, (b) A = 0? k 2 h2 ψ = Ae ikx + Be −ikx , E = • • 2m Method Operate on ψ with the operator corresponding to linear momentum eqn 7.29: Answer: p̂ x = (a ) B = 0 : h d i dx h d ψ h deikx h pˆ xψ = = A = A × ikeikx = kh Aeikx = khψ i dx i dx i p x = +kh (a ) A = 0 : h d ψ h de−ikx h pˆ xψ = = B = B × (−ik )e −ikx = − kh Be−ikx = −khψ i dx i dx i p x = −kh • • In (a) the particle is travelling to the right (positive x ) but in (b) it is travelling to the left (negative x ). Test 8.6 The operator for the angular momentum of a particle travelling in a circle in the xy -plane is ˆ d , where φ is its angular position. What is the angular momentum lz = (h / i ) φ of a particle described by the wavefunction e−2iφ? • Correct Answer: l z = −2h (c) The construction of operators (cont’d, p.270) 1 2 V̂ = kx × 2 1 h d h d h2 d 2 Ê k = =− 2 m i dx i dx 2m dx 2 (7.30) (7.31) 2 2 h d Hamiltonia n operator : Ĥ = Ê k + Vˆ = − + Vˆ (7.32) 2 2m dx Physical Chemistry Fundamentals: Figure 7.25 Fig. 7.25 Even if a wavefunction does not have the form of a periodic wave, it is still possible to infer from it the average kinetic energy of a particle by noting its average curvature. This illustration shows two wavefunctions: the sharply curved function corresponds to a higher kinetic energy than the less sharply curved function. (p.270) Physical Chemistry Fundamentals: Figure 7.26 Fig. 7.26 The observed kinetic energy of a particle is an average of contributions from the entire space covered by the wavefunction. Sharply curved regions contribute a high kinetic energy to the average; slightly curved regions contribute only a small kinetic energy. Physical Chemistry Fundamentals: Figure 7.27 Fig.7.27 The wavefunction of a particle in a potential decreasing towards the right and hence subjected to a constant force to the right. Only the real part of the wavefunction is shown, the imaginary part is similar, but displaced to the right. (p.271) (d) Hermitian operators • All the quantum mechanical operators that correspond to observables have a very special mathematical property: they are ‘hermitian’. An hermitian operator is one for which the following relation is true: { ˆ ψ j d τ = ψ j∗Ω Hermiticity : ∫ψ Ω ∫ ˆ ψ i dτ ∗ i ∫ ∞ −∞ ∞ ψ xψ j dτ = ∫ ψ j xψ dτ = ∗ i ∗ i −∞ } ∗ {∫ ψ xψ dτ } ∞ −∞ ∗ j [7.33] ∗ i J8.2 The hermiticity of the linear momentum operator : ∫ ∞ −∞ ∗ i ψ pˆ xψ j dx = {∫ ψ pˆ ψ dx} ∞ −∞ ∗ j ∗ x i integration by parts : dg df f dx = fg − g ∫ dx ∫ dxdx Test 7.7 Confirm that the operator d 2 /dx 2 is hermitian. The reality of eigenvalues • Two properties of Hermitian operators: their eigenvalues are real, and their eigenfunctions are ‘orthogonal’. All observables have real values, and so are represented by hermitian operators. • Justification 8.3 The reality of eigenvalues • For a wavefunction ψ that is normalized and is an eigenfunction of an hermitian operator with eigenvalue ω, ∗ ˆ ∗ ∗ Ω d = d = ψ ψ τ ψ ωψ τ ω ψ ∫ ∫ ∫ ψdτ = ω { } ∗ ˆ ˆ ψdτ = ω ω = ∫ψ Ωψ dτ = ∫ψ ∗ Ω ∗ ∗ hermiticity • ω* = ω confirms that ω is real. The Eigenfunctions of a Quantum Mechanical Operator Are Orthogonal • • Orthogonality is a concept of vector space. 3-D Cartesian coordinate space is defined by x• y = x • z = y • z = 0 • In function space, the analogous expression that defines orthogonality (正交 ) is J7.4 Orthogonality of wavefunctions (p.272) • Two different functions ψi and ψj are orthogonal if the integral (over all space) of their product is zero: Orthogonal ity : ∗ ψ i ∫ ψ j dτ = 0 for i ≠ j (7.34) • Wavefunctions corresponding to different eigenvalues of an hermitian operators are orthogonal. sin( a − b ) x sin( a + b) x − + constant (a 2 ≠ b 2 ) 2( a − b ) 2(a + b) cos(α ± β ) = cos α cos β m sin α sin β − 2 sin α sin β = cos(α + β ) − cos(α − β ) ∫ sin ax sin bxdx = ∫ 2π 0 sin x sin 2 xdx = 0 (a = 1, b = 2) • Test 8.8 Confirm that the functions sin x and sin 3x are mutually orthogonal. ∫ 2π 0 sin x sin 3 xdx = 0 Example Show graphically that sin x and cos 3x are orthogonal functions. Also show graphically that ∞ ∫ (sin mx)(sin nx)dx ≠ 0 for n = m = 1 −∞ Solution The functions are shown in the following graphs. The vertical axes have been offset to avoid overlap and the horizontal line indicates the zero for each plot. Because the functions are periodic, we can draw conclusions about their behaviour in an infinite interval by considering their behaviour in any interval that is an integral multiple of the period. Solution Solution The integral of these functions equals the sum of the areas between the curves and the zero line. Areas above and below the line contribute with positive and negative signs, respectively, ∞ and indicate that ∫−∞ (sin mx )(cos 3x )dx = 0 and ∫ (sin x )(sin x )dx > 0 . By similar means, we could show that any two functions of the type sin mx and sin nx or cos mx and cos nx are orthogonal unless n=m. Are the functions cos mx and sin mx(m=n) orthogonal? ∞ −∞ Physical Chemistry Fundamentals: Figure 7.28 Fig. 7.28 The integral of the function f(x) = sin x sin 2x is equal to the area (tinted) below the green curve, and is zero, as can be inferred by symmetry. The function—and the value of the integral— repeats itself for all replications of the section between 0 and 2π, so the integral from −∞ to ∞ is zero. 7.5 The information in a wavefunction (e) Superpositions and expectation values ψ = A(e ikx + e − ikx ) = 2 A cos kx (eqn.7.22 with A = B) • When the wavefunction of a particle is not an eigenfunction of an operator, the property to which the operator corresponds does not have a definite value. • The momentum not completely indefinite - cosine wavefunction is a linear combination, or sum, of eikx and e−ikx - definite momentum states. • The total wavefunction is a superposition of more than one wavefunction: Linear combination of basis functions • Any wavefunction as a linear combination of eigenfunctions of an operator - Suppose the wavefunction is known to be a superposition of many different linear momentum eigenfunctions • • • • (7.36) where the ck are numerical (possibly complex) coefficients and the ψk correspond to different momentum states. The functions ψk form a complete set in the sense that any arbitrary function can be expressed as a linear combination of them. 1. When the momentum is measured, in a single observation one of the eigenvalues corresponding to the ψk that contribute to the superposition will be found. 2. The probability of measuring a particular eigenvalue in a series of observations is proportional to the square modulus (|ck|2) of the corresponding coefficient in the linear combination. 3. The average value of a large number of observations is given by the expectation value, <Ω> , of the operator corresponding to the observable of interest. The Eigenfunctions of a Quantum Mechanical Operator Form a Complete Set • • The eigenfunctions of a quantum mechanical operator form a complete set. This means that any well-behaved wave function, f (x) can be expanded in the eigenfunctions of any of the quantum mechanical operators. The expectation value of an operator • The expectation value of an operator is defined as (7.37) (for normalized wavefunctions) An expectation value is the weighted average of a large number of observations of a property. J7.5 The expectation value of an operator ˆ with eigenvalue ω : Ω ˆ ψ = ωψ If ψ is an eigenfunct ion of Ω • the expectation value is • A wavefunction is not an eigenfunction of the operator of interest: can be written as a linear combination of eigenfunctions. • e.g. the wavefunction is the sum of two eigenfunctions <Ω> = |c1|2ω1 + |c2|2ω2 Example 7.7 Calculating an expectation value • Calculate the average value of the distance of an electron from the nucleus in the hydrogen atom in its state of lowest energy. • Method The average radius is the expectation value of the operator corresponding to the distance from the nucleus, x r. To evaluate <r>, we need the normalized wavefunction (Ex. 7.4): 1/ 2 1 ψ = 3 e − r / a0 (E x.7.4) πa0 Bohr radius a 0 = 52.9pm ˆ ψ dτ Ω = ψ ∗Ω [7.37] ∫ • Test 7.9 Evaluate the root mean square distance, <r2>1/2, of the electron from the nucleus in the hydrogen atom. • Correct Answer: 31/2a0 = 91.6 pm Mean kinetic energy of a particle in one dimension • The mean kinetic energy of a particle in one dimension is the expectation value of the operator given in eqn 7.31. • The kinetic energy is a kind of average over the curvature of the wavefunction: a large contribution to the observed value from regions where the wavefunction is sharply curved (d2ψ/dx2 is large) and the wavefunction itself is large (ψ* is large). 7.6 The uncertainty principle • Key points: The uncertainty principle restricts the precision with which complementary observables may be specified and measured. Complementary observables are observables for which the corresponding operators do not commute. • If the wavefunction is Aeikx , then the particle it describes has a definite state of linear momentum, namely travelling to the right with momentum px = +kh/2π . However, the position of the particle described by this wavefunction is completely unpredictable. In other words, if the momentum is specified precisely, it is impossible to predict the location of the particle. This statement is one-half of a special case of the Heisenberg uncertainty principle: • It is impossible to specify simultaneously, with arbitrary precision, both the momentum and the position of a particle. • The other half: if we know the position of a particle exactly, then we can say nothing about its momentum. Physical Chemistry Fundamentals: Figure 7.29 A wavefunction as a superposition of eigenfunction Fig. 7.29 The wavefunction for a particle at a welldefined location is a sharply spiked function that has zero amplitude everywhere except at the particle ’s position. Physical Chemistry Fundamentals: Figure 7.30 Fig. 7.30 The wavefunction for a particle with an ill-defined location can be regarded as the superposition of several wavefunctions of definite wavelength that interfere constructively in one place but destructively elsewhere. As more waves are used in the superposition (as given by the numbers attached to the curves), the location becomes more precise at the expense of uncertainty in the particle’s momentum. An infinite number of waves is needed to construct the wavefunction of a perfectly localized particle. Physical Chemistry Fundamentals: Table 7.2 Example 7.8 Using the uncertainty principle • Suppose the speed of a projectile of mass 1.0 g is known to within 1 µm s−1. Calculate the minimum uncertainty in its position. • Method Estimate ∆p from m ∆v, where ∆v is the uncertainty in the speed; then use eqn 7.39a to estimate the minimum uncertainty in position, ∆ q. • Answer The minimum uncertainty in position is (1 J = 1 kg m2 s−2). The uncertainty is completely negligible for all practical purposes concerning macroscopic objects. However, if the mass is that of an electron, then the same uncertainty in speed implies an uncertainty in position far larger than the diameter of an atom (the analogous calculation gives ∆q = 60 m); so the concept of a trajectory, the simultaneous possession of a precise position and momentum, is untenable. • Test 7.10 Estimate the minimum uncertainty in the speed of an electron in a onedimensional region of length 2a0. • Correct Answer: 547 km s−1 Complementary observables • General Heisenberg uncertainty principle: It applies to any pair of observables called complementary observables, which are defined in terms of the properties of their operators. Specifically, two observables Ω1 and Ω2 are complementary if (7.40) • When the effect of two operators depends on their order, they do not commute. • The commutator of the two operators is defined as (7.41) • The commutator of the operators for position and linear momentum is J 7.6 The commutator of position and momentum • Show that the operators for position and momentum do not commute (and hence are complementary observables) h xˆpˆ xψ − pˆ x xˆψ = − ψ = ihψ i → (7.42) • The Heisenberg uncertainty principle in the most general form - For any two pairs of observables, Ω1 and Ω2, the uncertainties (to be precise, the root mean square deviations of their values from the mean) in simultaneous determinations are related by: Postulates of Quantum Mechanics • Postulate (假定)1: All information that can be obtained about the state of a mechanical (physical) system is contained in a wave function Ψ, which is a continuous, finite, and single-valued function of time and of the coordinates of the particles of the system. (Mortimer 3e) Postulate 1 of Quantum Mechanics (cont) • This postulate implies that there is a one-to-one relationship between the state of the system and a wave function. • i.e., each possible state corresponds to one wave function, and each possible wave function corresponds to one state. • The terms “state function” and “wave function” are often used interchangeably. Information about values of mechanical variables such as energy and momentum must be obtained from the wave function, instead of from values of coordinates and velocities as in classical mechanics. • The 4th postulate will provide the method for obtaining this information. Postulate 2 • The wave function Ψ obeys the timedependent Schrodinger equation ∂ Ψ Hˆ Ψ = ih ∂t (1) where Ĥ is the Hamiltonian operator of the system. Postulate 2 (cont’d) • The time-independent Schrodinger equation can be derived from the time-dependent equation by assuming that the wave function is a product of a coordinate factor and a time factor: Ψ(q, t) = ψ(q)η(t) (2) where q stands for all of the coordinates of the particles in the system and where the coordinate wave function ψ satisfies the time-independent Schrodinger equation. • Not all wave functions consist of the two factors in Eq.(2), but all wave functions must obey the time-dependent Schrodinger equation. Postulate 3 Mathematical Operators and Mechanical Variables • There is a linear hermitian mathematical operator in one-to-one correspondence with every mechanical variable. • This postulate states that for each operator there is one and only one variable, and • for each variable there is one and only one mathematical operator. • A mathematical operator is a symbol that stands for performing one or more mathematical operations. • Usually denote an operator by a letter with a caret (︿) over it. Postulate 4 Expectation Values • (a) If a mechanical variable A is measured without experimental error, the only possible outcomes of the measurement are the eigenvalues of the operator  that corresponds to A. • (b) The expectation value for the error-free measurement of a mechanical variable A if given by the formula A = ∗ ˆ Ψ ∫ AΨdτ ∫ Ψ Ψdτ ∗ = Ψ Aˆ Ψ Ψ Ψ where  is the operator corresponding to the variable A, and where ψ=ψ(q, t) is the wave function corresponding to the state of the system at the time of the measurement. Postulate 5. Measurements and the Determination of the State of a System • Immediately after an error-free measurement of the mechanical variable A in which the outcome was the eigenvalue aj , the state of the system corresponds to a wave function that is an eigenfunction of  with eigenvalue equal to aj . Postulate 5 (cont’d) • This postulate says very little about the state of the system prior to a single measurement of the variable A, because the act of measurement can change the state of the system. • How a measurement can change the state of a system? Consider the determination of the position of a particle by the scattering of electromagnetic radiation: • When an airplane reflects a radar wave, the effect on the airplane is negligible because of the large mass of the airplane. • When an object of small mass such as an electron scatters ultraviolet light or X-rays, the effect is not negligible. Sum 7.7 The postulates of quantum mechanics (p.279) • The wavefunction. All dynamical information is contained in the wavefunction ψ for the system, which is a mathematical function found by solving the Schrödinger equation for the system. In one dimension: • The Born interpretation. If the wavefunction of a particle has the value ψ at some point r, then the probability of finding the particle in an infinitesimal volume dτ = dxd ydz at that point is proportional to |ψ|2dτ. • Acceptable wavefunctions. An acceptable wavefunction must be continuous, have a continuous first derivative, be single-valued, and be square-integrable. • Observables. Observables, Ω, are represented by operators,Ω̂ , built from position and momentum operators of the form • • or, more generally, from operators that satisfy the commutation relation [xˆ, pˆ x ] = ih • The Heisenberg uncertainty principle. It is impossible to specify simultaneously, with arbitrary precision, both the momentum and the position of a particle and, more generally, any pair of observable with operators that do not commute.