Presentation on theme: "The Wavelike Properties of Particles"— Presentation transcript:
1 The Wavelike Properties of Particles Chapter 5The Wavelike Properties of Particles
2 The Wavelike Properties of Particles The de Broglie HypothesisMeasurements of Particles WavelengthsWave PacketsThe Probabilistic Interpretation of the Wave FunctionThe Uncertainty PrincipleSome Consequences of Uncertainty PrincipleWave-Particle Duality
3 The de Broglie Hypothesis As it was showed by Thomson the rays of a cathode tube can be deflected by electric and magnetic fields and therefore must consist of electrically charged particles.Thomson showed that all the particles have the same charge-to-mass ratio q/m. He also showed that particles with this charge-to-mass ratios can be obtained using any materials for the cathode, which means that these particles, now called electrons, are a fundamental constituent of matter.
4 The de Broglie Hypothesis Since light seems to have both wave and particle properties, it is natural to ask whether matter (electrons, protons) might also have both wave and particle characteristics.In 1924, a French physics student, Louis de Broglie, suggested this idea in his doctoral dissertation.For the wavelength of electron, de Broglie chose:λ = h/pf = E/hwhere E is the total energy, p is the momentum, andλ is called the de Broglie wavelength of the particle.
5 The de Broglie Hypothesis For photons these same equations results directly from Einstein’s quantization of radiation E = hf and equation for an energy of a photon with zero rest energy E = pc :E = pc = hf = hc/λUsing relativistic mechanics de Broglie demonstrated, that this equation can also be applied to particles with mass and used them to physical interpretation of Bohr’s hydrogen-like atom.
6 The de Broglie Wavelength Using de Broglie relation let’s find the wavelength of a 10-6g particle moving with a speed 10-6m/s:Since the wavelength found in this example is so small, much smaller than any possible apertures, diffraction or interference of such waves can not be observed.
7 The de Broglie Wavelength The situation are different for low energy electrons and other microscopic particles.Consider a particle with kinetic energy K. Its momentum is found fromIts wavelength is than
8 The de Broglie Wavelength If we multiply the numerator and denominator by c we obtain:Where mc2=0.511MeV for electrons, and K in electronvolts.
9 The de Broglie Wavelength We obtained the electron wavelength:Similarly, for proton (mc2 = 938 MeV for protons)
10 The de Broglie Wavelength For the molecules of a stationary gas at the absolute temperature T, the square average speed of the molecule v2 is determined by Maxwell’s LawThan the momentum of the molecule is:Knowing that the mass of He atom, for instance, is 6.7x10-24g, (kB=1.38x10-23J/K) we obtain for He wavelength:
11 The de Broglie Wavelength Similarly, for the molecule of hydrogen.and for the thermal neutronsThis calculations show, that for the accelerated electrons, for atoms of helium, hydrogen molecules under the room temperature, for thermal neutrons and other “slow” light particles de Broglie wavelength is on the same order as for soft X-rays. So, we can expect, that diffraction can be observed for this particles
12 Show that the wavelength of a nonrelativistic neutron is where Kn is the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a keV neutron?
13 Show that the wavelength of a nonrelativistic neutron is where Kn is the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a keV neutron?Kinetic energy, K, in this equation is in Joules(a)
14 Show that the wavelength of a nonrelativistic neutron is where Kn is the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a keV neutron?(b)
15 The nucleus of an atom is on the order of 10–14 m in diameter The nucleus of an atom is on the order of 10–14 m in diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus?
16 The nucleus of an atom is on the order of 10–14 m in diameter The nucleus of an atom is on the order of 10–14 m in diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus?(a)
17 The nucleus of an atom is on the order of 10–14 m in diameter The nucleus of an atom is on the order of 10–14 m in diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus?,(b)With its
18 Electron Interference and Diffraction The electron wave interference was discovered in 1927 by C.J. Davisson and L.H.Germer as they were studying electron scattering from a nickel target at the Bell Telephone Laboratories.After heating the target to remove an oxide coating that had accumulate after accidental break in the vacuum system, they found that the scattered electron intensity is a function of the scattered angle and show maxima and minima. Their target had crystallized during the heating, and by accident they had observed electron diffraction.Then Davisson and Germer prepared a target from a single crystal of nickel and investigated this phenomenon.
19 The Davisson-Germer experiment. Low energy electrons scattered at angle Φ from a nickel crystal are detected in an ionization chamber. The kinetic energy of electrons could be varied by changing the accelerating voltage on the electron gun.
20 Scattered intensity vs detector angle for 54-ev electrons Scattered intensity vs detector angle for 54-ev electrons. Polar plot of the data. The intensity at each angle is indicated by the distance of the point from the origin. Scattered angle Φ is plotted clockwise started at the vertical axis.
21 The same data plotted on a Cartesian graph The same data plotted on a Cartesian graph. The intensity scale are the same on the both graphs. In each plot there is maximum intensity at Φ=50º, as predicted for Bragg scattering of waves having wavelength λ = h/p.
22 Scattering of electron by crystal Scattering of electron by crystal. Electron waves are strongly scattered if the Bragg condition nλ = D SinΦ is met.
23 Test of the de Broglie formula λ = h/p Test of the de Broglie formula λ = h/p. The wavelength is computed from a plot of the diffraction data plotted against V0-1/2, where V0 is the accelerating voltage. The straight line is 1.226V0-1/2 nm as predicted from λ = h/(2mE)-1/2
24 Test of the de Broglie formula λ = h/p Test of the de Broglie formula λ = h/p. The wavelength is computed from a plot of the diffraction data plotted against V0-1/2, where V0 is the accelerating voltage. The straight line is 1.226V0-1/2 nm as predicted from λ = h/(2mE)-1/2
25 A series of a polar graphs of Davisson and Germer’s data at electron accelerating potential from 36 V to 68 V. Note the development of the peak at Φ = 50º to a maximum when V0 = 54 V.
26 Variation of the scattered electron intensity with wavelength for constant Φ. The incident beam in this case was 10º from the normal, the resulting diffraction causing the measured peaks to be slightly shifted from the positions computed from nλ = D Sin Φ.
27 Schematic arrangement used for producing a diffraction pattern from a polycrystalline aluminum target.
28 Diffraction pattern produced by x-rays of wavelength 0 Diffraction pattern produced by x-rays of wavelength nm and an aluminum foil target.
29 Diffraction pattern produced by 600-eV electrons and an aluminum foil target ( de Broigle wavelength of about 0.05 nm: )
30 Diffraction pattern produced by 600-eV electrons and an aluminum foil target ( de Broigle wavelength of about 0.05 nm: ) .
31 Diffraction pattern produced by 0 Diffraction pattern produced by eV neutrons (de Broglie wavelength of nm) and a target of polycrystalline copper. Note the similarity in the pattern produced by x-rays, electrons, and neutrons.
32 Diffraction pattern produced by 0 Diffraction pattern produced by eV neutrons (de Broglie wavelength of nm) and a target of polycrystalline copper. Note the similarity in the pattern produced by x-rays, electrons, and neutrons.
33 In the Davisson–Germer experiment, 54 In the Davisson–Germer experiment, 54.0-eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)
34 In the Davisson–Germer experiment, 54 In the Davisson–Germer experiment, 54.0-eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)
35 In the Davisson–Germer experiment, 54 In the Davisson–Germer experiment, 54.0-eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)
36 A photon has an energy equal to the kinetic energy of a particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of c ? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds?
37 A photon has an energy equal to the kinetic energy of a particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of c ? (c) What value does the ratio of the two wavelengths approach at high particle speeds? (d) At low particle speeds?For a particle:For a photon:
38 A photon has an energy equal to the kinetic energy of a particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of c ? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds?(a)(b)
39 becomes nearly equal to γ. and A photon has an energy equal to the kinetic energy of a particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of c ? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds?(c)Asbecomes nearly equal to γ.and,(d)
40 What is “waving”? For matter it is the probability of finding the particle that waves. Classical waves are the solution of the classical wave equationHarmonic waves of amplitude y0, frequency f and period T:where the angular frequency ω and the wave number k are defined byand the wave or phase velocity vp is given by
41 If the film were to be observed at various stages, such as after being struck by 28 electrons the pattern of individually exposed grains will be similar to shown here.
42 After exposure by about 1000 electrons the pattern will be similar to this.
43 And again for exposure of about 10,000 electrons we will obtained a pattern like this.
44 Two source interference pattern Two source interference pattern. If the sources are coherent and in phase, the waves from the sources interfere constructively at points for which the path difference dsinθ is an integer number of wavelength.
45 Grows of two-slits interference pattern Grows of two-slits interference pattern. The photo is an actual two-slit electron interference pattern in which the film was exposed to millions of electrons. The pattern is identical to that usually obtained with photons.
46 Using relativistic mechanics, de Broglie was able to derive the physical interpretation of Bohr’s quantization of the angular momentum of electron.He demonstrate that quantization of angular momentum of the electron in hydrogenlike atoms is equivalent to a standing wave condition:for n = integerThe idea of explaining discrete energy states in matter by standing waves thus seems quite promising.
47 Standing waves around the circumference of a circle Standing waves around the circumference of a circle. In this case the circle is 3λ in circumference. For example, if a steel ring had been suitable tapped with a hammer, the shape of the ring would oscillate between the extreme positions represented by the solid and broken lines.
48 Wave pulse moving along a string Wave pulse moving along a string. A pulse have a beginning and an end; i.e. it is localized, unlike a pure harmonic wave, which goes on forever in space and time.
49 Two waves of slightly different wavelength and frequency produced beats. (a) Shows y(x) at given instant for each of the two waves. The waves are in phase at the origin but because of the difference in wavelength, they become out of phase and then in phase again.
50 (b) The sum of these waves (b) The sum of these waves. The spatial extend of the group Δx is inversely proportional to the difference in wave numbers Δk, where k is related to the wavelength by k = 2π/λ.
51 F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)] BEATSConsider two waves of equal amplitude and nearly equal frequencies and wavelengths.The sum of the two waves is (superposition):F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)]
54 BEATS This is an equivalent of an harmonic wave whose amplitude is modulated byWe have formed wave packets of extend Δx and can imagine each wave packet representing a particle.
55 Δx Δp ≈ h - Uncertainty Principle NowThe particle is in the region Δx, the momentum of the particle in the range Δk:p = ћk → Δp = ћΔkΔx Δp ≈ h - Uncertainty PrincipleIn order to localize the particle within a region Δx, we need to relax the precision on the value of the momentum, Δp .
56 Gaussian-shaped wave packets y(x) and the corresponding Gaussian distributions of wave numbers A(k). (a) A narrow packet. (b) A wide packet. The standard deviations in each case are related by σxσk = ½.
57 Wave packet for which the group velocity is half of phase velocity Wave packet for which the group velocity is half of phase velocity. Water waves whose wavelengths are a few centimeters, but much less than the water depth, have this property. The arrow travels at the phase velocity, following a point of constant phase for the dominant wavelength. The cross at the center of the group travels at the group velocity.
58 A three-dimensional wave packet representing a particle moving along the x-axis. The dots indicate the position of classical particle. Note that the particle spreads out in the x and y directions. This spreading is due to dispersion , resulting from the fact that the phase velocity of the individual wave making up the packet depends on the wavelength of the waves.
59 “Seeing an electron” with a gamma-ray microscope.
60 Because of the size of the lens, the momentum of the scattered photons is uncertain by Δpx ≈psinθ = hsinθ/ λ. Thus the recoil momentum of the electron is also uncertain by at least this amount.
61 The position of the electron can not be resolved better than the width of the central maximum of the diffraction pattern Δx ≈ λ/sinθ. The product of the uncertainties Δpx Δx is therefore of the order of Planck’s constant h.
62 The Interpretation of the Wave Function Given that electrons have wave-like properties, it should be possible to produce standing electron waves. The energy is associated with the frequency of the standing wave, as E = hf, so standing waves imply quantized energies.The idea that discrete energy states in atom can be explained by standing waves led to the development by Erwin Schrödinger in 1926 mathematical theory known as quantum theory, quantum mechanics, or wave mechanics.In this theory electron is described by a wave function Ψ that obeys a wave equation called the Schrödinger equation.
63 The Interpretation of the Wave Function The form of the Schrödinger equation of a particular system depends on the forces acting on the particle, which are described by the potential energy functions associated with this forces.Schrödinger solved the standing wave problem for hydrogen atom, the simple harmonic oscillator, and other system of interest. He found that the allowed frequencies, combined with E=hf, resulted in the set of energy levels, found experimentally for the hydrogen atom.Quantum theory is the basis for our understanding of the modern world, from the inner working of the atomic nucleus to the radiation spectra of distant galaxies.
64 The Interpretation of the Wave Function The wave function for waves in a string is the string displacement y. The wave function for sound waves can be either the displacement of the air molecules, or the pressure P. The wave function of the electromagnetic waves is the electric field E and the magnetic field B.What is the wave function for the electron Ψ? The Schrödinger equation describes a single particle. The square of the wave function for a particle describes the probability density, which is the probability per unit volume, of finding the particle at a location.
65 The Interpretation of the Wave Function The probability of finding the particle in some volume element must also be proportional to the size of volume element dV.Thus, in one dimension, the probability of finding a particle in a region dx at the position x is Ψ2(x)dx. If we call this probability P(x)dx, where P(x) is the probability density, we haveP(x) = Ψ2(x)The probability of finding the particle in dx at point x1 or point x2 is the sum of separate probabilitiesP(x1)dx + P(x2)dx.If we have a particle at all the probability of finding a particle somewhere must be 1.
66 The Interpretation of the Wave Function Then, the sum of the probabilities over all the possible values of x must equal 1. That is,This equation is called the normalization condition. If Ψ is to satisfy the normalization condition, it must approach zero as x is approach infinity.
67 Probability Calculation for a Classical Particle It is known that a classical point particle moves back and forth with constant speed between two walls at x = 0 and x = 8cm. No additional information about of location of the particle is known.(a) What is the probability density P(x)?(b) What is the probability of finding the particle at x=2cm?(c) What is the probability of finding the particle between x=3.0 cm and x=3.4 cm?
68 A Particle in a BoxWe can illustrate many of important features of quantum physics by considering of simple problem of particle of mass m confined to a one-dimensional box of length L.This can be considered as a crude description of an electron confined within an atom, or a proton confined within a nucleus.According to the quantum theory, the particle is described by the wave function Ψ, whose square describes the probability of finding the particle in some region. Since we are assuming that the particle is indeed inside the box, the wave function must be zero everywhere outside the box: Ψ =0 for x≤0 and for x≥L.
69 A Particle in a BoxThe allowed wavelength for a particle in the box are those where the length L equals an integral number of half wavelengths.L = n( λn/2) n = 1,2,3,…….This is a standing wave condition for a particle in the box of length L.The total energy of the particle is its kinetic energyE = (1/2)mv2 = p2/2mSubstituting the de Broglie relation pn = h/λn,
70 A Particle in a Box where Then the standing wave condition λn= 2L/n gives the allowed energies:where
71 A Particle in a Box The equation gives the allowed energies for a particle in the box.This is the ground state energy for a particle in the box, which is the energy of the lowest state.
72 A Particle in a BoxThe condition that we used for the wave function in the boxΨ = 0 at x = 0 and x = Lis called the boundary condition.The boundary conditions in quantum theory lead to energy quantization.Note, that the lowest energy for a particle in the box is not zero. The result is a general feature of quantum theory.If a particle is confined to some region of space, the particle has a minimum kinetic energy, which is called zero-point energy. The smaller the region of space the particle is confined to, the greater its zero-point energy.
73 A Particle in a BoxIf an electron is confined (i.e., bond to an atom) in some energy state Ei, the electron can make a transition to another energy state Ef with the emission of photon. The frequency of the emitted photon is found from the conservation of the energyhf = Ei – EfThe wavelength of the photon is thenλ = c/f = hc/(Ei – Ef)
74 Standing Wave Function The amplitude of a vibrating string fixed at x=0 and x=L is given aswhere An is a constant and is the wave number.The wave function for a particle in a box are the same:Using , we have
75 Standing Wave Function The wave function can thus be writtenThe constant An is determined by normalization conditionThe result of evaluating the integral and solving for An is independent from n.The normalized wave function for a particle in a box are thus
76 Graph of energy vs. x for a particle in the box, that we also call an infinitely deep well. The set of allowed values for the particle’s total energy En is E1(n=1), 4E1(n=2), 9E1(n=3) …..
77 Wave functions Ψn(x) and probability densities Pn(x)= Ψn2(x) for n=1, 2, and 3 for the infinity square well potential.
78 Probability distribution for n=10 for the infinity square well potential. The dashed line is the classical probability density P=1/L, which is equal to the quantum mechanical distribution averaged over a region Δx containing several oscillations. A physical measurement with resolution Δx will yield the classical result if n is so large that Ψ2(x) has many oscillations in Δx.
79 Photon Emission by Particle in a Box An electron is in one dimensional box of length 0.1nm. (a) Find the ground state energy. (b) Find the energy in electron-volts of the five lowest states, and then sketch an energy level diagram. (c) Find the wavelength of the photon emitted for each transition from the state n=3 to a lower-energy state.
81 The probability of a particle being found in a specified region of a box. The particle in one-dimensional box of length L is in the ground state. Find the probability of finding the particle (a) anywhere in a region of length Δx = 0.01L, centered at x = ½L; (b) in the region 0<x<(1/4)L.
82 Expectation ValuesThe most that we can know about the position of the particle is the probability of measuring a certain value of this position x. If we measure the position for a large number of identical systems, we get a range of values corresponding to the probability distribution.The average value of x obtained from such measurements is called the expectation value and written ‹x›. The expectation value of x is the same as the average value of x that we would expect to obtain from a measurement of the position of a large number of particles with the same wave function Ψ(x).
83 Expectation ValuesSince Ψ2(x)dx is the probability of finding a particle in the region dx, the expectation value of x is:The expectation value of any function f(x) is given by:
84 Calculating expectation values Find (a) ‹ x › and (b) ‹ x2› for a particle in its ground state in a box of length L.
85 Complex NumbersA complex number has the form a+ib, with i2=-1 or i=√-1 – imaginary unit.a - real part; b – imaginary part; i – imaginary unit(a +ib) + (c +id) = (a+c) + i(b+d)m(a +ib) = ma + imb(a +ib) (c +id) = (ac - bd) + i(ad + bc)The absolute value of a + ib is denoted by │a+ib│ and is given by │a+ib│= √ a2 + b2
86 (a+ib)*∙ (a+ib) = (a-ib) (a+ib)=a2 + b2 Complex NumbersThe complex conjugate of a+ib is denoted by (a+ib)* and is given(a+ib)* = (a-ib)Then(a+ib)*∙ (a+ib) = (a-ib) (a+ib)=a2 + b2
87 Polar Form of Complex Numbers p = √ a2 + b2 = │a + ib│We can represent the number (a + ib) in the complex xy plane.Then the polar coordinatesa + ib ≡ p(cosφ +isinφ)Remembering the Euler formula:eiφ = (cosφ+isinφ)a + ib = p eiφImaginary axispbφaReal axisEuler Identities:eiφ = cosφ + isinφe-iφ = cosφ - isinφwhere i = √-1
88 Fourier TransformIn quantum mechanics, our basic function is the pure sinusoidal plane wave describing a free particle, given in equation:We are not interest here in how things behave in time, so we chose a convenient time of zero. Thus, our “building block” isNow we claim that any general, nonperiodic wave function ψ(x) can be expressed as a sum/integral of this building blocks over the continuum of wave numbers:
89 Fourier TransformThe amplitude A(k) of the plane wave is naturally a function of k, it tell us how much of each different wave number goes into the sum. Although we can’t pull it out of the integral, the equation can be solved for A(k). The result is:The proper name of for A(k) is the Fourier transform of the function ψ(x).
90 ψ(x) = 0 │x│> a Let use Euler identities: _eika = cosφ+isinφ eika- e-ika = 2isinkaAnd we can overwrite the equation for A(k):ψ(x) =
91 General Wave PacketsAny point in space can be described as a linear combination of unit vectors. The three unit vectors î, ĵ, and constitute a base that can generate any points in space.In similar way: given a periodic function, any value that the function can take, can be produced by the linear combination of a set of basic functions. The basic functions are the harmonic functions (sin or cos). The set of basic function is actually infinite.
92 The General Wave Packet A periodic function f(x) can be represented by the sum of harmonic waves:y(x,t) = Σ [Aicos(kix – ωit) + Bisin(kix – ωit)]Ai and Bi ≡ amplitudes of the waves with wave number ki and angular frequency ωi.For a function that is not periodic there is an equivalent approach called Fourier Transformation.
93 Fourier Transformation A function F(x) that is not periodic can be represented bya sum (integral) of functions of the typee±ika = Cosφ±iSinφIn math terms it called Fourier Transformation. Given a function F(x)wheref(kj) represents the amplitude of base function e-ikx used to represent F(x).
94 The Schrödinger Equation The wave equation governing the motion of electron and other particles with mass m, which is analogous to the classical wave equationwas found by Schrödinger in 1925 and is now known as the Schrödinger equation.
95 The Schrödinger Equation Like the classical wave equation, the Schrödinger equation is a partial differential equation in space and time.Like Newton’s laws of motion, the Schrödinger equation cannot be derived. It’s validity, like that of Newton’s laws, lies in its agreement with experiment.
96 We will start from classical description of the total energy of a particle: Schrödinger converted this equation into a wave equation by defining a wavefunction, Ψ. He multiplied each factor in energy equation with that wave function:
97 To incorporate the de Broglie wavelength of the particle he introduced the operator, ,which provides the square of the momentum when applied to a plane wave:If we apply the operator to that wavefunction:where k is the wavenumber, which equals 2π/λ.We now simple replace the p2 in equation for energy:
98 Time Independent Schrödinger Equation This equation is called time-independent Schrödinger equation.E is the total energy of the particle.The normalization condition now becomes∫ Ψ*(x)Ψ(x)dx = 1
99 A Solution to the Srödinger Equation Show that for a free particle of mass m moving in one dimension the functionis a solution of the time independent Srödinger Equation for any values of the constants A and B.
100 Energy Quantization in Different Systems The quantized energies of a system are generally determined by solving the Schrödinger equation for that system. The form of the Schrödinger equation depends on the potential energy of the particle.The potential energy for a one-dimensional box from x = 0 to x = L is shown in Figure. This potential energy function is called an infinity square-well potential, and is described by:U(x) = 0, 0<x<LU(x) = ∞, x<0 or x>L
101 ψ(0) = A sin (k0) + B cos (0x) = 0 + B A Particle in Infinity Square Well PotentialInside the box U(x) = 0, so the Schrödinger equation is written:where E = ħω is the energy of the particle, orwhere k2 = 2mE/ħ2The general solution of this equation can be written asψ(x) = A sin kx + B cos kxwhere A and B are constants. At x=0, we haveψ(0) = A sin (k0) + B cos (0x) = 0 + B
102 A Particle in Infinity Square Well Potential The boundary condition ψ(x)=0 at x=0 thus gives B=0 and equation becomesψ(x) = A sin kxWe received a sin wave with the wavelength λ related to wave number k in a usual way, λ = 2π/k. The boundary condition ψ(x) =0 at x=L givesψ(L) = A sin kL = 0This condition is satisfied if kL is any integer times π, orkn = nπ / LIf we will write the wave number k in terms of wavelength λ = 2π/k, we will receive the standing wave condition for particle in the box:nλ / 2 = L n = 1,2,3,……
103 A Particle in Infinity Square Well Potential Solving k2 = 2mE/ħ2 for E and using the standing wave condition k = nπ / L gives us the allowed energy values:whereFor each value n, there is a wave function ψn(x) given by
104 A Particle in Infinity Square Well Potential Compare with the equation we received for particle in the box, using the standing wave fitting with the constant An = √2/L determined by normalization:Although this problem seems artificial, actually it is useful for some physical problems, such as a neutron inside the nucleus.
105 A Particle in a Finite Square Well This potential energy function is described mathematically by: U(x)=V0, x<0U(x)=0, 0<x<LU(x)=V0, x>LHere we assume that 0 ≤E≤V0. Inside the well, U(x)=0, and the time independent Schrödinger equation is the same as for the infinite well
106 A Particle in a Finite Square Well orwhere k2 = 2mE/ħ2. The general solution isψ(x) = A sin kx + B cos kxbut in this case, ψ(x) is not required to be zero at x=0, so B is not zero.
107 A Particle in a Finite Square Well Outside the well, the time independent Schrödinger equation isorwhere
108 The Harmonic Oscillator More realistic than a particle in a box is the harmonic oscillator, which applies to an object of mass m on a spring of force constant k or to any systems undergoing small oscillations about a stable equilibrium. The potential energy function for a such oscillator is:where ω0 = √k/m=2πf is the angular frequency of the oscillator. Classically, the object oscillates between x = +A and x=-A. Its total energy iswhich can have any nonnegative value, including zero.
109 Potential energy function for a simple harmonic oscillator Potential energy function for a simple harmonic oscillator. Classically, the particle with energy E is confined between the “turning points” –A and +A.
110 The Harmonic Oscillator Classically, the probability of finding the particle in dx is proportional to the time spent in dx, which is dx/v. The speed of the particle can be obtained from the conservation of energy:The classical probability is thus
111 The Harmonic Oscillator The classical probability isAny values of the energy E is possible. The lowest energy is E=0, in which case the particle is in the rest at the origin. The Shrödinger equation for this problem is
112 The Harmonic Oscillator In quantum theory, the particle is represented by the wave function ψ(x), which is determined by solving the Schrödinger equation for this potential.Only certain values of E will lead to solution that are well behaved, i.e., which approach zero as x approach infinity. Normalizeable wave function ψn(x) occur only for discrete values of the energy En given bywhere f0=ω0/2π is the classical frequency of the oscillator.
113 The Harmonic Oscillator where f0=ω0/2π is the classical frequency of the oscillator. Thus, the ground-state energy is ½ħω and the exited energy levels are equally spaced by ħω.
115 Energy levels in the simple harmonic oscillator potential Energy levels in the simple harmonic oscillator potential. Transitions obeying the selection rule Δn=±1 are indicated by the arrows. Since the levels have equal spacing, the same energy ħω is emitted or absorbed in all allowed transitions. For this special potential, the frequency of emitted or absorbed photon equals the frequency of oscillation, as predicted by classical theory.
116 The Harmonic Oscillator Compare this with uneven spacing of the energy levels for the particle in a box. If a harmonic oscillator makes a transition from energy level n to the next lowest energy level (n-1), the frequency f of the photon emitted is given by hf = Ef – Ei. Applying this equation gives:The frequency f of the emitted photon is therefore equal to the classical frequency f0 of the oscillator.
117 Wave function for the ground state and the first two excited states of the simple harmonic oscillator potential, the states with n=0, 1, and 2.
118 Probability density for the simple harmonic oscillator plotted against the dimensionless value , for n=0, 1,and 2. The blue curves are the classical probability densities for the same energy, and the vertical lines indicate the classical turning points x = ±A
119 Molecules vibrate as harmonic oscillators Molecules vibrate as harmonic oscillators. Measuring vibration frequencies enables determination of force constants, bond strengths, and properties of solids.
120 Verify that , where α is a positive constant, is a solution of the Schrödinger equation for the harmonic oscillator
121 OperatorsAs we have seen, for a particle in a state of definite energy the probability distribution is independent of time. The expectation value of x is then given byIn general, the expectation value of any function f(x) is given by
122 OperatorsIf we know the momentum p of the particle as function of x, we can calculate the expectation value ‹p›. However, it is impossible in principle to find p as function of x since, according to uncertainty principle, both p and x can not be determined at the same time.To find ‹p› we need to know the distribution function for momentum. If we know Ψ(x), the distribution function can be found by Fourier analysis. It can be shown that ‹p› can be found from
123 Operators Similarly, ‹p2› can be found from Notice that in computing the expectation value the operator representing the physical quantity operates on Ψ(x), not on Ψ*(x). This is not important to the outcome when the operator is simply some function of x, but it is critical when the operator includes a differentiation, as in the case of momentum operator.
124 Expectation Values for p and p2 Find ‹p› and ‹p2› for the ground state wave function of the infinity square well.
125 The time-independent Schrödinger equation can then be written: In classical mechanics, the total energy written in terms of position and momentum variables is called the Hamiltonian functionIf we replace the momentum by the momentum operator pop and note that U = U(x), we obtain the Hamiltonian operator Hop:The time-independent Schrödinger equation can then be written:
126 The advantage of writing the Schrödinger equation in this formal way is that it allows for easy generalization to more complicated problems such as those with several particles moving in three dimensions.We simply write the total energy of the system in terms of position and momentum and replace the momentum variables by the appropriate operators to obtain the Hamiltonian operator for the system.
127 x component of momentum py y component of momentum pz SymbolPhysical quantityOperatorf(x)Any function of x (the position, x; the potential energy U(x), etc.pxx component of momentumpyy component of momentumpzz component of momentumEHamiltonian (time-independent)Hamiltonian (time-dependent)EkKinetic energyLzZ component of angular momentum
128 Minimum Energy of a Particle in a Box An important consequence of the uncertainty principle is that a particle confined to a finite space can not have zero kinetic energy.Let’s consider a one-dimensional box of length L. If we know that the particle is in the box, Δx is not larger than L. This implies that Δp is at least ħ/L. Let us take the standard deviation as a measure of Δp:
129 Minimum Energy of a Particle in a Box If the box is symmetric, p will be zero since the particle moves to the left as often as to the right. Thenand the average kinetic energy is:
130 Minimum Energy of a Particle in a Box The average kinetic energy of a particle in a box is:Thus, we see that the uncertainty principle indicate that the minimum energy of a particle in a box cannot be zero. This minimum energy is called zero-point energy.
131 The Hydrogen AtomThe energy of an electron of momentum p a distance r from a proton isIf we take for the order of magnitude of the position uncertainty Δx = r, we have:(Δp2) = p2 ≥ ћ2/r2The energy is then
132 The Hydrogen AtomThere is a radius rm at which E is minimum. Setting dE/dr = 0 yields rm and Em:rm came out to be exactly the radius of the first Bohr orbitThe ground state energy
133 The Hydrogen AtomThe potential energy of the electron-proton system varies inversely with separation distanceAs in the case of gravitational potential energy, the potential energy of the electron-proton system is chosen to be zero if the electron is an infinity distance from the proton. Then for all finite distances, the potential energy is negative.
134 The Hydrogen AtomLike the energies of a particle in a box and of a harmonic oscillator, the energy levels in the hydrogen atom are described by a quantum number n. The allowed energies of the hydrogen atom are given byEn = eV/n2,n = 1,2,3,……Energy-level diagram for the hydrogen atom. The energy of the ground state is eV. As n approaches ∞ the energy approaches 0.
135 Step PotentialConsider a particle of energy E moving in region in which the potential energy is the step functionU(x) = 0, x<0U(x) = V0, x>0What happened whena particle moving fromleft to right encountersthe step?The classical answer issimple: to the left of thestep, the particle moveswith a speed v = √2E/m
136 Step PotentialAt x =0, an impulsive force act on the particle. If the initial energy E is less than V0, the particle will be turned around and will then move to the left at its original speed; that is, the particle will be reflected by the step. If E is greater than V0, the particle will continue to move to the right but with reduced speed given byv = √2(E – U0)/m
137 Step PotentialWe can picture this classical problem as a ball rolling along a level surface and coming to a steep hill of height h given by mgh=V0.If the initial kinetic energy of the ball is less than mgh, the ball will roll part way up the hill and then back down and to the left along the lower surface at it original speed. If E is greater than mgh, the ball will roll up the hill and proceed to the right at a lesser speed.
138 The quantum mechanical result is similar when E is less than V0 The quantum mechanical result is similar when E is less than V0. If E<V0 the wave function does not go to zero at x=0 but rather decays exponentially. The wave penetrates slightly into the classically forbidden region x>0, but it is eventually completely reflected.
139 When the wavelength changes suddenly, part of Step PotentialThis problem is somewhat similar to that of total internal reflection in optics.For E>V0, the quantum mechanical result differs from the classical result. At x=0, the wavelength changes fromλ1=h/p1 = h/√2mEtoλ2=h/p2 = h/√2m(E-V0).When the wavelength changes suddenly, part ofthe wave is reflected and part of the wave is transmitted.
140 Reflection Coefficient Since a motion of an electron (or other particle) is governed by a wave equation, the electron sometimes will be transmitted and sometimes will be reflected.The probabilities of reflection and transmission can be calculated by solving the Schrödinger equation in each region of space and comparing the amplitudes of transmitted waves and reflected waves with that of the incident wave.
141 Reflection Coefficient This calculation and its result are similar to finding the fraction of light reflected from the air-glass interface. If R is the probability of reflection, called the reflection coefficient, this calculation gives:where k1 is the wave number for the incident wave and k2 is the wave number for the transmitted wave.
142 Transmission Coefficient The result is the same as the result in optics for the reflection of light at normal incidence from the boundary between two media having different indexes of refraction n.The probability of transmission T, called the transmission coefficient, can be calculated from the reflection coefficient, since the probability of transmission plus the probability of reflection must equal 1:T + R = 1In the quantum mechanics, a localized particle is represented by the wave packet, which has a maximum at the most probable position of the particle.
143 Time development of a one dimensional wave packet representing a particle incident on a step potential for E>V0. The position of a classical particle is indicated by the dot. Note that part of the packet is transmitted and part is reflected.
144 Reflection coefficient R and transmission coefficient T for a potential step V0 high versus energy E (in units V0).
145 A particle of energy E0 traveling in a region in which the potential energy is zero is incident on a potential barrier of height V0=0.2E0. Find the probability that the particle will be reflected.
146 Lets consider a rectangular potential barrier of height V0 and with a given by: U(x) = 0, x<0 U(x) = V0, 0<x<a U(x) = 0, x>a
147 Barrier PotentialWe consider a particle of energy E , which is slightly less than V0, that is incident on the barrier from the left. Classically, the particle would always be reflected. However, a wave incident from the left does not decrease immediately to zero at the barrier, but itwill instead decay exponentially in the classically forbidden region 0<x<a. Upon reaching the far wall of the barrier (x=a), the wave function must join smoothly to a sinusoidal wave function to the right of barrier.
148 The potentials and the Schrödinger equations for the three regions are as follows: Region I (x<0) V = 0, Region II (0<x<a) V = V0, Region III (x>a) V = 0,
149 Barrier PotentialIf we have a beam of particle incident from left, all with the same energy E<V0, the general solution of the wave equation are, following the example for a potential step,where k1 =√2mE/ħ and α = √2m(V0-E)/ħThis implies that there is some probability of the particle (which is represented by the wave function) being found on the far side of the barrier even though, classically, it should never pass through the barrier.
150 We assume that we have incident particles coming from the left moving along the +x direction. In this case the term Aeik1x in region I represents the incident particles. The term Be-ik1x represents the reflected particles moving in the –x direction. In region III there are no particles initially moving along the -x direction. Thus G=0, and the only term in region III is Feik1x. We summarize these wave functions:
151 Barrier Potential For the case in which the quantity αa = √2ma2(V0 – E)/ħ2is much greater than 1, the transmission coefficient is proportional to e-2αa, withα = √2m(V0 – E)/ħ2The probability of penetration of the barrier thus decreases exponentially with the barrier thickness a and with the square root of the relative barrier height (V0-E). This phenomenon is called barrier penetration or tunneling. The relative probability of its occurrence in any given situation is given by the transmission coefficient.
152 A wave packet representing a particle incident on two barriers of height just slightly greater than the energy of the particle. At each encounter, part of the packet is transmitted and part reflected, resulting in part of the packet being trapped between the barriers from same time.
153 A 30-eV electron is incident on a square barrier of height 40 eV A 30-eV electron is incident on a square barrier of height 40 eV. What is the probability that the electron will tunnel through the barrier if its width is (a) 1.0 nm?(b) 0.1nm?
154 The penetration of the barrier is not unique to quantum mechanics The penetration of the barrier is not unique to quantum mechanics. When light is totally reflected from the glass-air interface, the light wave can penetrate the air barrier if a second peace of glass is brought within a few wavelengths of the first, even when the angle of incidence in the first prism is greater than the critical angle. This effect can be demonstrated with a laser beam and two 45° prisms.
155 α- DecayThe theory of barrier penetration was used by George Gamov in 1928 to explain the enormous variation of the half-lives for α decay of radioactive nuclei.Potential well shown on the diagram for an α particle in a radioactive nucleus approximately describes a strong attractive force when r is less than the nuclear radius R. Outside the nucleus the strong nuclear force is negligible, and the potential is given by the Coulomb’s law, U(r) = +k(2e)(Ze)/r, where Ze is the nuclear charge and 2e is the charge of α particle.
156 α- DecayAn α-particle inside the nucleus oscillates back and forth, being reflected at the barrier at R. Because of its wave properties, when the α-particle hits the barrier there is a small chance that it will penetrate and appear outside the well at r = r0. The wave function is similar to that for a square barrier potential.
157 The probability that an α-particle will tunnel through the barrier is given by which is a very small number, i.e., the α particle is usually reflected. The number of times per second N that the α particle approaches the barrier is given bywhere v equals the particle’s speed inside the nucleus.The decay rate, or the probability per second that the nucleus will emit an α particle, which is also the reciprocal of the mean life time , is given by
158 The decay rate for emission of α particles from radioactive nuclei of Po212. The solid curve is the prediction of equationThe points are the experimental results.
159 Applications of Tunneling Nanotechnology refers to the design and application of devices having dimensions ranging from 1 to 100 nmNanotechnology uses the idea of trapping particles in potential wellsOne area of nanotechnology of interest to researchers is the quantum dotA quantum dot is a small region that is grown in a silicon crystal that acts as a potential wellNuclear fusionProtons can tunnel through the barrier caused by their mutual electrostatic repulsion
160 Resonant Tunneling Device Electrons travel in the gallium arsenide semiconductorThey strike the barrier of the quantum dot from the leftThe electrons can tunnel through the barrier and produce a current in the device
161 Scanning Tunneling Microscope An electrically conducting probe with a very sharp edge is brought near the surface to be studiedThe empty space between the tip and the surface represents the “barrier”The tip and the surface are two walls of the “potential well”
162 Scanning Tunneling Microscope The STM allows highly detailed images of surfaces with resolutions comparable to the size of a single atomAt right is the surface of graphite “viewed” with the STM
163 Scanning Tunneling Microscope The STM is very sensitive to the distance from the tip to the surfaceThis is the thickness of the barrierSTM has one very serious limitationIts operation is dependent on the electrical conductivity of the sample and the tipMost materials are not electrically conductive at their surfacesThe atomic force microscope (AFM) overcomes this limitation by tracking the sample surface maintaining a constant interatomic force between the atoms on the scanner tip and the sample’s surface atoms.
164 SUMMARY 1. Time-independent Schrödinger equation: 2.In the simple harmonic oscillator:the ground wave function is given:where A0 is the normalization constant and a=mω0/2ħ.3. In a finite square well of height V0, there are only a finite number of allowed energies.
165 4.Reflection and barrier penetration: SUMMARY4.Reflection and barrier penetration:When the potentials changes abruptly over a small distance, a particle may be reflected even though E>U(x). A particle may penetrate a region in which E<U(x). Reflection and penetration of electron waves are similar for those for other kinds of waves.