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The Classical Free Particle Advanced Physical Chemistry Chemistry 5350 The simplest classical system imaginable is a particle with no forces acting on it: Q UANTUM T HEORY: T ECHNIQUES AND A PPLICATIONS In one dimension, the motion of a free particle can be described as: F = ma = m d2 x =0 dt2 This differential equation can be solved to obtain Professor Angelo R. Rossi x = x0 + v 0 t http://homepages.uconn.edu/rossi Department of Chemistry, Room CHMT215 The University of Connecticut The initial position x0 and the initial velocity v0 arise from constants of integration. Fall Semester 2013 To give them explicit values, the boundary conditions of the problem, the initial position and velocity must be known. [email protected] Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 The Quantum Mechanical Free Particle Free Particle Approach to the Wave Function The wave nature of the electron has been clearly shown in experiments like the photoelelectric effect The connection to the Schrödinger equation can be made by examining wave and particle expressions for energy: particle ⇐= p2 2m E wave =⇒ 2 hν Asserting the equivalence of these two expressions for energy p̂2 → −~2 What is the nature of the wave for an electron? The wave is the wave function for the electron. Starting with the expression for a traveling wave in one dimension, the connection can be made to the Schrodinger equation. This process makes use of the de Broglie relationship between wavelength and momentum and the Planck relationship between frequency and energy. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm ∂2 ∂x2 Ê → i~ ∂ ∂t and putting in the quantum mechanical operators for both brings us to the Shrodinger equation. The time-dependent Schrödinger equation takes the form − 3 Fall Semester 2013 ~2 ∂ 2 Ψ(x, t) ∂Ψ(x, t) = i~ 2m ∂x2 ∂t Last Updated: October 20, 2013 at 8:16pm 4 Free Particle Approach to the Wave Function The Time-Independent Wave Function for a Free Particle This gives a plane wave solution of the form The time-independent solutions for a free-particle wave function are Ψ(x, t) = Aei 2π −iωt λ The solutions are plane waves with one moving to the right (positive x direction), and the other moving to the left (negative x direction) Using the de Broglie relation gives 2πp p p 2π = = h = = k; λ h ~ 2π The general solutions are p (electron momentum) = ~k ψk (x) = A+ eikx + A− e−ikx The energy is given by Using the Planck relationship Ek = ω× ψ − (x) = A− e−ikx ψ + (x) = A+ e+ikx = Aeikx−iωt ~ ~ω E = = ; ~ ~ ~ E (electron energy) = ~ω p2 k 2 ~2 = 2m 2m The wave functions and energies (eigenfunctions and eigenvalues of Ĥ) are labeled with the index k. The wave function Ψ(x, t) = Aeikx−iωt is a complex function which can be expanded in the form k= Ψ(x, t) = A cos(kx − ωt) + iA sin(kx − ωt) 2π = λ r 2mEk ~2 Either the real or imaginary part of this function could be appropriate for a given application. In general, one is interested in particles which are free within some kind of boundary but have the boundary conditions set by a potential. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 5 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm Probability of Determining the Position of a Free Particle The Particle in a One-Dimensional Box A plane wave is not localized, and it is not possible to speak of its position as in the same way one can for a free classical particle. The particle in a box consists of a particle of mass m confined between two walls at x = 0 and x = a. 6 However, the probability of finding the particle in an interval of length dx can be calculated The free-particle wave functions cannot be normalized over the interval −∞ < x < +∞, but if x is restricted to the interval −L < x < +L • • A− A+ e−ikx eikx dx dx ψ ∗ (x)ψ(x)dx = = P (x)dx = R L RL ∗ (x)ψ(x)dx −ikx eikx dx 2L ψ A A e − + −L −L P (x)dx is independent of x. This result states that the particle is equally likely to be anywhere in the interval, which is equivalent to saying that nothing is known about the position of the particle. The wave function must be zero at the walls, and the solution for the wave function yields just sine waves. The idealized situation of a particle in a box with infinitely high walls is an application of the Schrödinger equation which yields some insights into particle confinement. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 7 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 8 The Particle in a One-Dimensional Box: Potential and Schrödinger Equation The Particle in a 1-D Box: Boundary Conditions The impenetrable walls of the particle in the box are modeled by making the potential energy infinite outside of a region of width a: Outside of the box, where the potential is infinite, the second derivative would be infinite if ψ(x) were not zero for all values of x outside the box. V (x) = 0 for a > x > 0 V (x) = ∞ for x ≥ a and x ≤ 0 ψ(0) = ψ(a) = 0 Moreover, inside the box where V (x) = 0, the wave function must be continuous. How does this change in potential affect the eigenfunctions that were obtained for the free particle? The above statements are boundary conditions that any well-behaved wave function for a one-dimensional box must satisfy. The Schrödinger equation is written in the following form: − Fall Semester 2013 The solution to Schrödinger equation for a particle inside the box is identical to that for a free particle. ~2 d2 ψ(x) + V (x)ψ(x) = Eψ(x) 2m dx2 Last Updated: October 20, 2013 at 8:16pm 9 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm The Particle in a 1-D Box: Boundary Conditions The Particle in a 1-D Box: Acceptable Wave Functions For ease in applying the boundary conditions, the solution of the Schrödinger equation in the box is written as Therefore, we conclude that ψn (x) = A sin ψ(x) = A sin kx + B cos kx Now apply the boundary conditions nπx a 10 The requirement that ka = nπ with n being an integer will turn out to have important consequences for the energy spectrum of a particle in a box. Acceptable wave functions for the particle in a box must have the form: ψ(0) = 0 + B = 0 ψ(a) = A sin ka = 0 The first condition can only be satisfied by the condition that B = 0. ψn (x) = A sin The second condition can be satisfied if either A = 0 or if ka = nπ with n being an integer. nπx a , for n = 1, 2, 3, 4, . . . Each different value of n corresponds to a different eigenfunction. Setting A to zero would mean the wave function will always be zero which is unacceptable because then there is no particle in the interval. But there IS a particle in the interval a > x > 0 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 11 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 12 The Particle in a 1-D Box: Normalized Wave Functions The Particle in a 1-D Box: Eigenvalues and Eigenfunctions The constant A can be determined by normalization by realizing that ψ ∗ (x)ψ(x)dx represents the probability of finding the particle in the interval dx centered at x Ra Ra ∗ dx = 1 ψ (x)ψ(x)dx = A∗ A 0 sin2 nπx a 0 What are the eigenvalues that go with these eigenfunctions? ~ d − 2m 2 A= 2 ; a ψn (x) = q 2 a sin ~ d − 2m dx2 2 nπx 2 q 2 a sin Last Updated: October 20, 2013 at 8:16pm = En ψn (x) nπx a = ~2 2m nπ 2 a q 2 a sin nπx a the following result for the energy (En ) is obtained a En = Fall Semester 2013 n (x) dx2 Applying the total energy operator to the eigenfunctions will give back the eigenfunction multiplied by the energy eigenvalue: because the probability of finding the particle somewhere in the entire interval is one. q 2ψ 13 The Particle in a 1-D Box vs The Free Particle Fall Semester 2013 ~2 2m nπ 2 a = h2 n 2 , 8ma2 = En ψn (x) for n = 1, 2, 3, . . . Last Updated: October 20, 2013 at 8:16pm 14 The Particle in a 1-D Box: Energies and Wave Functions An important difference between the energies of a particle in a box and the energies of a free particle is that the energy for a particle in a box can take on only discrete values. The lowest four energy levels for the particle in the box are shown in the adjacent figure and are superimposed on an energy versus distance diagram. The energy of a particle in the box is quantized, and n is a quantum number. The eigenfunctions are also shown on this diagram. Another important result is that the lowest allowed energy is greater than zero. The variation of the wave function with time is a standing wave which turns out to be a general result for a stationary state. The particle in the box has a nonzero minimum energy known as zero point energy. A standing wave has nodes that are at fixed distances as a function of time but move in time to produce a traveling wave. Energies and Wave Functions Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 15 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 16 The Particle in a 1-D Box: Probability Density The Particle in a 1-D Box: Calculating Energies What is the ground state energy for an electron that is confined to a potential well with a width of 0.2 nm? Using the formula for the allowed energies for a particle in a box, The probability of finding the electron outside the box is zero. The probability of finding the electron within an interval dx in the box depends both on the position and on the wave function. En = Although | ψ(x) |2 can be zero at nodal positions, | ψ(x) |2 dx is never zero for a finite interval dx inside the box. E1 = Last Updated: October 20, 2013 at 8:16pm for n = 1, 2, 3, . . . (6.626 x 10−34 J s)2 (1)2 = 1.506 x 10−18 J/e− 8 (9.109 x 10−31 kg)(0.2 x 10−9 m)2 E1 = (1.506 x 10−18 J/e− )(6.022 x 1023 mol−1 ) × Probability Density Fall Semester 2013 h2 n 2 , 8ma2 17 Fall Semester 2013 1 kJ = 907 kJ mol−1 103 J Last Updated: October 20, 2013 at 8:16pm The Particle in a 1-D Box: Calculating Probabilities The Particle in a 1-D Box: Calculating Expectation Values What is the probability of finding the particle in the box in a small region of space? The average value of x (< x >) for a particle in any of the eigenstates of a one-dimensional box is Z Z nπx a 2 a ∗ x sin2 dx = =< x > < x >n = ψn (x)xψn (x)dx = a 0 a 2 For a particle in a one-dimensional box, the ground-state wave function is given as ψ(x) = r 2 πx sin( ) a a What is the probability that the particle is in the middle third of the box? Thus, the particle is in the middle of box for all values of the quantum number n which is reasonable. The average value of x2 in the nth eigenstate ( < x2 >n ) is Z Z 2 a 2 2 nπx < x2 >n = ψn∗ (x)x2 ψn (x)dx = x sin dx a 0 a a π 2 a2 < x2 > n = −2 4 2πn 3 The probability density at point x in the interval dx is given by πx 2 sin2 dx a a P (x)dx = ψ ∗ (x)ψ(x)dx = P P Fall Semester 2013 a 2a ≤x≤ 3 3 a 2a ≤x≤ 3 3 = 2 a = 2 a Z 2a 3 a 3 1 a x− sin 2 4π sin2 πx dx a 2πx a 2a 3 a 3 = 0.609 Last Updated: October 20, 2013 at 8:16pm 18 19 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 20 The Particle in a 1-D Box: Classical Limit and the Correspondence Principle The Particle in a 1-D Box: Classical Limit and the Correspondence Principle The Correspondence Principle states that quantum mechanical predictions approach the predictions of classical mechanics as the quantum number approaches infinity. Examine the spacing between levels by forming a ratio: Looking at the equation below shows the dependence of the total energy eigenvalues on the quantum number n: En = En+1 − En = En ! = 2n + 1 n2 which approaches zero as n → ∞. h2 n2 , for n = 1, 2, 3, . . . 8ma2 Both the level spacing and energy increase with n, but the energy increases faster (≈ n2 ) making the energy spectrum appear to be continuous as n → ∞. However, it is not immediately obvious that the energy spectrum will become continuous in the classical limit of very large n because the spacing between adjacent energy levels increases with increasing n. Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 h2 [(n+1)2 −n2 ] 8ma2 h2 n 2 8ma2 21 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm The Particle in a Box: 2-D and 3-D Systems The Particle in a Box: 2-D and 3-D Systems The potential energy of a three-dimensional box is given by: This differential is solved by assuming that ψ(x, y, z) has the form V (x, y, z) = 0 for 0 < x < a; 0 < y < b; V (x, y, z) = ∞ otherwise 0<z<c ψ(x, y, z) = X(x)Y (y)Z(z) As for the one-dimensional box, the amplitude of the eigenfunctions of the total energy is zero outside the box. which is the product of three functions, each of which depends on only one of the variables. This assumption is valid Inside the three-dimensional box, the Schrödinger equation can be written − Fall Semester 2013 ~2 2m ∂2 ∂2 ∂2 + + ∂x2 ∂y 2 ∂z 2 22 • • because V (x, y, z) is independent of x, y, and z inside the box. because the potential is of the form ψ(x, y, z) = Eψ(x, y, z) Last Updated: October 20, 2013 at 8:16pm V (x, y, z) = Vx (x)+Vy (y)+Vz (z) 23 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 24 The Particle in a Box: 2-D and 3-D Systems The Particle in a Box: 2-D and 3-D Systems Substituting ψ(x, y, z) = X(x)Y (y)Z(z) into the Schrödinger equation for three-dimensional systems gives Each of the equations has the same form as the equation that was solved for the one-dimensional problem: 1 d2 X(x) 1 d2 Y (y) 1 d2 Z(z) ~2 + + = EX(x)Y (y)Z(z) − 2m X(x) dx2 Y (y) dy 2 Z(z) dz 2 ψnx ,ny ,nz (x, y, z) = N sin The total energy has the form: The energy E has independent contributions from the three coordinates: E = Ex + Ey + Ez which allows the original differential equation in three variables to be reduced to three differential equations, each with one variable: − nx πx ny πy nz πz sin sin a b c E= h2 8m n2x n2y n2z + 2 + 2 a2 b c ~2 d2 X(x) ~2 d2 Y (y) ~2 d2 Z(z) = Ex X(x); − = Ey Y (y); − = Ez Z(z) 2 2 2m dx 2m dy 2m dz 2 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 25 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm The Particle in a Box: 2-D and 3-D Systems The Particle in a Box: 2-D and 3-D Systems Because this is a three-dimensional problem, the eigenfunctions depend on three quantum numbers. If a = b = c, the energy can be written as For a particle in a two-dimensional box, the total energy eigenfunctions are given by E= ψnx ny (x, y) = N sin h2 n2 + n2y + n2z 8ma2 x ny 2 1 1 nx πx ny πy sin a b The total energy in terms of nx , ny , a, b is given by: Now, more than one set of the three quantum numbers can have the same energy: nx 1 2 1 26 nz 1 1 2 E= h2 8m n2x n2y + 2 a2 b In this case, the energy level is degenerate, and the number of states that have the same energy level is the degeneracy of the level. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 27 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 28 The Particle in a Box: 2-D and 3-D Systems The Classical Harmonic Oscillator To understand vibrations in molecules, it is important to understand the quantum mechanical treatment of a harmonic oscillator. As background, it is necessary to review the classical treatment of harmonic oscillator. The simplest example of a harmonic oscillator is a mass connected to a wall by means of an idealized spring in the absence of gravity. Contour plots of several eigenfunctions are shown here. The x and y directions of the box lie along the horizontal and vertical directions, respectively. The amplitude has been displayed as a gradation in colors with regions of positive and negative amplitude noted. As shown in the adjacent figure, the displacement of the mass is given by the x coordinate, and the origin of the coordinate system is taken at the equilibrium postion. The mass oscillates about its equilibrium postion, and the motion is said to be harmonic if the force F of the spring is directly proportional to the displacement, x. F = −kx The negative sign comes from the fact that the force F is opposite to the displacement. The constant k is referred to as a force constant and is small for a weak spring but large for a stiff spring. Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 29 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm The Classical Harmonic Oscillator Simple Harmonic Motion Since force is expressed as mass times accleeration, the equation of motion in the x directions is Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke’s Law. 30 The motion is sinusoidal in time and demonstrates a single resonant frequency d2 x + kx = 0 dt2 The general solution of this differential equation is x(t) = A sin ωt + B cos ωt q k where ω = is the angular frequency in radians per m second. Since q ω = 2πν, the resonant frequency is given as ν = 1 2π A mass on a spring will trace out a sinusoidal pattern as a function of time, as will any object vibrating in simple harmonic motion. k . m For example, while walking in a straight line at constant speed while carrying the vibrating mass, the mass will trace out a sinusoidal path in space as well as time. If the spring is initially stretched so that the mass is a position x0 and the velocity is zero, then let go, the time course of the motion is represented by x(t) = x0 cos ωt Since F = dV = −kx, integration of this equation yields dx V = 12 x2 which is the equation of a parabola as shown in the adjacent figure. Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 31 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 32 Quantum Harmonic Oscillator: Schrödinger Equation Eigenvalues of the Harmonic Oscillator The Schrödinger equation for a harmonic oscillator may be obtained by using the classical spring potential r 1 k 1 ω= (angular frequency) V (x) = kx2 = mω 2 x2 ; 2 2 m Substituting the function ψ(x) = Ce−α into the Schrödinger equation and imposing boundary conditions leads to the ground state energy for the quantum harmonic oscillator: As we saw previously, the Schrödinger equation with this form of the potential is − ~2 d2 ψ(x) 1 + mω 2 x2 ψ(x) = Eψ(x) 2m dx2 2 E0 = The imposition of boundary conditions leads to a discrete energy spectrum. x2 2 The lowest state accessible to the system has a non-zero energy, i.e. the zero point energy which is also true for the particle in the box. This is a Gaussian function which satisfies the requirement that the wave funcion approach zero at infinity which allows the wave function to be normalized. Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 1 ~ω = hν 2 2 The general solution to the Schrödinger equation produces a sequence of evenly spaced energy levels characterized by a quantum number, n. Since the second derivative of the wave function must give back the square of x plus a constant times the original function, the following form is suggested: ψ(x) = Ce−α x2 2 33 Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 Eigenfunctions of the Harmonic Oscillator Eigenfunctions of the Harmonic Oscillator The wave functions for the quantum harmonic oscillator contain the Gaussian form which allows them to satisfy the necessary boundary conditions at infinity. The Hermite polynomials for the first four levels of the harmonic oscillator are as follows: The wave functions for the first two energy levels are given by: ψ0 = where α = km ~2 12 α 41 π e− αx2 2 ; ψ1 = 4α3 π 1 4 x e− H0 (ξ) = 1; An even function is a function that satisfies f (x) = f (−x), and an odd function is a function that satisfies f (x) = −f (−x) In general, the wave functions of the harmonic oscillator are given by αx2 2 ; Nν = 1 1 (2ν ν!) 2 αx2 Since e− 2 is the wave functions for the harmonic oscillator is even, the even-odd character is determined by the Hermite polynomials. α 41 Thus, the harmonic oscillator wave functions are even when ν is even and odd when ν is odd making it easy to evaluate integrals. π 1 and the Hν (α 2 x) are called Hermite Polynomials Fall Semester 2013 H3 (ξ) = 8ξ 3 − 12ξ A property of the Hermite polynomials is that Hν (ξ) is an even function if ν is even and an odd function if ν is odd. . 1 H2 (ξ) = 4ξ 2 − 2; where ξ is a dummy variable. αx2 2 The wave functions have the shape of a Gaussian probability function. ψν = Nν Hν (α 2 x)e− H1 (ξ) = 2ξ; 34 Last Updated: October 20, 2013 at 8:16pm 35 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 36 Quantum Harmonic Oscillator: Eigenfunctions and Probabilities Probability Distributions for the Quantum Oscillator The solution of the Schrödinger equation for the quantum harmonic oscillator gives the wave functions, the energy levels as well as the probability distributions for the quantum states of the oscillator. The square of the wavefunction gives the probability of finding the oscillator at a particular value of x. There is a finite probability that the oscillator will be found outside the potential well as indicated by the smooth curve which is forbidden in classical physics. The most probable value of position for the lower states of a quantum oscillator is very different from the classical harmonic oscillator where it spends more time near the end of its motion. As the quantum number increases, the probabability distribution becomes more like that of the classical oscillator The normalized wavefunctions for the first four energy states gives are displayed on the left above. The probability of finding the oscillator at any given value of x is the square of the wavefunction, and these probabilities are shown at right above. The wavefunctions for higher n have more ”humps” within the potential well, and this corresponds to a shorter wavelength, a higher momentum according to the de Broglie equation, and therefore higher energy. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 37 Comparison of Classical and Quantum Probabilities for the Harmonic Oscillator Fall Semester 2013 The tendency to approach classical behavior for high quantum numbers is called the correspondence principle. For the ground state of the quantum harmonic oscillator, the correspondence principle seems far-fetched, since the classical and quantum predictions for the most probable location are in total contradiction. If the equilibrium position for the oscillator is taken to be x = 0, then the quantum oscillator predicts that for the ground state, the oscillator will spend most of its time near that center point. On the other hand, the classical spends very little time there because it has the maximum speed as it passes through x = 0. It will spend most of its time near the end points of its oscillation where the velocity is small. In going to higher and higher states of the quantum oscillator by increasing n, the most probable location shifts to the edges of the well. There is still the undulation of the probability which is characteristic of the wave solution, but at least the overall trend of the probability begins to look more like the classical probability which is shown by the dashed line. A comparison between the classical and quantum probability of finding the object which is oscillating at a given distance, x, from the equilibrium position can be made. For the classical case, the probability is greatest at the ends of the motion since it is moving more slowly and comes to rest instantaneously at the extremes of the motion. The relative probability of finding it in any interval, dx, is just the inverse of its average velocity over that interval. For the quantum mechanical case the probability of finding the oscillator in an interval, dx, is the square of the wavefunction, and that is very different for the lower energy states. In the diagram for the ground state (n = 0), the maximum probability is at the equilibrium point x = 0. Last Updated: October 20, 2013 at 8:16pm 38 The Correspondence Principle and the Quantum Oscillator The harmonic oscillator is a good example of how different quantum and classical results can be. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 39 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 40 Quantum Harmonic Oscillator: Zero-Point energy Quantum Harmonic Oscillator and the Uncertainty Principle If ∆x and ∆p are the uncertainty in position and momentum, respectively, the ground state energy for the quantum harmonic oscillator The most surprising difference for the quantum case is the zero-point vibration for the n = 0 ground state. E0 = This implies that molecules are not completely at rest, even at absolute zero temperature. can be shown to be the minimum energy allowed by the uncertainty principle. The quantum harmonic oscillator has implications far beyond the simple model presented here. This is a very significant physical result because it tells us that the energy of a system described by a harmonic oscillator potential cannot have zero energy. Physical systems such as atoms in a solid lattice or polyatomic molecules in the gas phase cannot have zero energy even at absolute zero temperature. It is the foundation for the understanding of complex modes of vibration in larger molecules, the motion of atoms in a solid lattice, the theory of heat capacity, and other properties of molecules and solids. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm ~ω = ∆x∆p 2 The energy of the ground vibrational state is often referred to as zero point vibration. The zero point energy is sufficient to prevent liquid helium-4 from freezing at atmospheric pressure, no matter how low the temperature. 41 Characteristics of Vibrational Motion in Molecules Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 42 Two Masses Connected by a Spring: Harmonic Motion When two masses are connected by a spring, there are two equations of motion: The discussion of the free particle and the particle in the box was useful for understanding how translational motion in various potentials is described in the context of wave-particle duality. d2 x1 = k(x2 − x1 − l0 ) dt2 d2 x2 = k(x2 − x1 − l0 ) dt2 In applying quantum mechanics to molecules, two other types of motion that molecules can undergo are vibration and rotation. m1 The simplest vibrational motion that can be imagined is that which occurs in a diatomic molecule. where l0 is the equilibrium length of the spring. The force on m1 is equal and opposite to the force on m2 as required by Newton’s third law. The oscillatory motion depends only on the relative coordinate x = x2 − x1 − l0 which yields d2 x = −k dt2 m2 1 1 + m1 m2 x The mass term can be taken to be the reciprocal of the rem2 duced mass µ defined by µ = mm1+m . and 1 The equation for harmonic motion also applies to mass m1 connected to mass m2 as shown in the figure above which represents an idealized diatomic molecule. µ d2 x dt2 2 + kx = 0 This equation is the same form as the previous equation for the harmonic oscillator with m replaced by the reduced mass and x is now the relative coordinate. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 43 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 44 Schrödinger Equation for a Vibrating Diatomic Molecule Vibrational Frequencies in Diatomic Molecules The following table provides transition frequencies and calculated force constants for the n = 0 to n = 1 vibrational level for diatomic molecules. A diatomic molecule vibrates somewhat like two masses on a spring with a potential energy that depends upon the square of the displacement from equilibrium. The wave-particle of mass µ vibrating around its equilibrium position is described by a set of wave functions ψn (x) Molecule Frequency/(1013 Hz) Force Constant (N/m) HF 8.72 970 HCl 8.66 480 HBr 7.68 410 HI 6.69 320 CO 6.42 1860 NO 5.63 1530 To find these wave functions and corresponding allowed energies of vibrational motion, the following Schrödinger equation must be solved: − ~2 d2 ψn (x) 2µ dx2 kx2 + 2 ψn (x) = En ψn (x) The energy levels of the quantum harmonic oscillator are quantized at equally spaced values and given by En = (n + 1 2 )~ω = (n + 1 2 )hν For a diatomic molecule, the natural frequency is of the form ω= s k µ ; ν = 1 2π s k µ m m where the reduced mass (µ) is given by µ = m 1+m2 1 2 The form of the frequency is the same as that for the classical simple harmonic oscillator. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 45 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm Comparison of Harmonic and More Realistic Potentials Characteristics of Vibrational Motion in Molecules The adjacent figure shows the potential energy, V (x), as a function of bond length, x for a diatomic molecule. xe represents the equilibrium bond length. The existence of a stable chemical bond implies that a minimum energy exists at some internuclear distance called the bond length 46 The configuration of atoms is dynamic rather than static, and so the chemical bond should be thought as having springs rather than a rigid bar connecting two atoms. The zero of energy is chosen to be the bottom of the potential. The red curve depicts a realistic potential in which the molecule dissociates for large values of x. Thermal energy increases the vibrational amplitude of their atom about their equilibrium positions. The orange curve shows a harmonic potential, V (x) = 21 kx2 which is a good approximation to the realistic potential near the bottom of the well. However, for T ≈ 300 K, only the lowest one or two vibrational levels are occupied for most molecules. In real systems, energy spacings are equal only for the lowest levels where the potential is a good approximation harmonic potential. Therefore, it is a good approximation to employ the functional form V (x) = 12 kx2 near the equilibrium bond length: The potential becomes steeply repulsive at short distances and levels out at large distances. The anharmonic terms which appear in the potential for a diatomic molecule are useful for mapping the detailed potential of such systems. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 47 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 48 The Classical Rigid Rotor Diatomic Molecule as a Classical Rigid Rotor In the adjacent figure, a particle is rotating around a fixed axis and has angular momentum and kinetic energy. The kinetic energy of a particle T in circular motion about a fixed point is expressed in terms of angular velocity or angular momentum L2 1 T = Iω 2 = 2 2I A model of a rigid diatomic molecule, i.e. with the bond distance fixed, provides a good example of a rigid rotor. The two masses rotate around their center of mass, satisfying the condition r1 m1 = r2 m2 . Because there is no potential energy, the classical Hamiltonian of a rigid rotor is just the kinetic energy. V (x, y, z) = 0 The rotational kinetic energy is where I = mr2 is the moment of inertia, L = Iω is the angular momentum, and ω is the angular velocity. In the expression for the rotational kinetic energy, the moment of inertia is analogous to mass, and angular velocity ω is analogous to linear velocity v for linear motion. T = where I is the moment of inertia of the diatomic molecule I = m1 r12 + m2 r22 = Last Updated: October 20, 2013 at 8:16pm m1 m2 R2 = µR02 m1 + m2 0 The rotational kinetic energy may also be written in terms of the angular momentum T = Fall Semester 2013 1 2 Iω 2 49 Fall Semester 2013 L2 L2 = 2I 2µR02 Last Updated: October 20, 2013 at 8:16pm Quantum Mechanical Motion in 2-D: Rotation on a Ring Quantum Mechanical Rotation in 2-D: Spherical Coordinates The quantum mechanical operator for a rigid rotor is given as ∂2 ∂2 ~2 ∂ 2 ~2 + + Ĥ = − ∇2 = − 2µ 2µ ∂x2 ∂y 2 ∂z 2 ~2 ∂ 2 ∂2 ∂2 − + + ψ = Eψ 2µ ∂x2 ∂y 2 ∂z 2 The relationship between Cartesian coordinates (x, y, z) and spherical coordinates (r, θ, φ) is given below: 50 In discussing rotation, it is more convenient to use spherical coordinates because they reflect they symmetry of the system being considered. The differential volume element in this coordinate system is dτ = r2 sin θdrdθdφ Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 51 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 52 Quantum Mechanical Rotation in 2-D: Schrödinger Equation Quantum Mechanical Rotation in 2-D: Boundary Conditions In spherical coordinates, the Laplacian operator ∇2 is given by ∂ 1 ∂ 1 ∂ 1 ∂ ∂2 + ∇2 = 2 R2 + 2 2 sin θ R ∂R ∂R ∂θ R sin θ ∂φ2 R2 sin θ ∂θ To obtain the solutions of the Schrödinger equation that describe the physical problem, it is necessary to introduce the boundary condition Φ(φ + 2π) = Φ(φ) This condition states that there is no way to distinguish the particle that has rotated n times around the circle from one that has rotated n + 1 times. Applying the single-valued condition For R fixed at R0 and θ fixed in a plane at 90◦ , the Schrödinger takes the form: ~2 d2 Φ(φ) − = EΦ(φ) 2µR02 dφ2 eiml [φ+2π] = e[iml φ] Thus, This equation has the same form as the Schrödinger equation for a free particle with two linearly independent solutions Using Euler’s relation cos 2πml + i sin 2πml = 1 Φ+ (φ) = A(+φ) eiml φ and Φ− (φ) = A(−φ) e−iml φ Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 e2πiml = 1 To satisfy this condition, ml = 0, ±1, ±2, ±3, . . . 53 Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 Quantum Mechanical Rotation in 2-D: Eigenfunctions Quantum Mechanical Rotation in 2-D: Eigenvalues The motivation for using the subscript l will become clear when rotation in two- and three-dimensions is considered. The normalization constant is Z 2π Φ∗ml (φ)Φml (φ)dφ = 1 Putting the eigenfunctions back into the Schrödinger equation allows the corresponding eigenvalues Eml to be calculated E ml = 54 ~2 m2l ~2 m2l = for ml = 0, ±1, ±2, ±3, . . . 2 2µR0 2I 0 (A+φ ) 2 Z 2π e −iml φ iml φ e A(+φ) A(−φ) Fall Semester 2013 dφ = (A+φ ) 0 2 Z States with +ml and −ml have the same energy even though the wave functions corresponding to these states are orthogonal to each other. 2π dφ = 1 Energy levels with ml 6= 0 are twofold degenerate. 0 1 =√ 2π 1 =√ 2π Last Updated: October 20, 2013 at 8:16pm 55 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 56 Quantum Mechanical Rotation in 3-D: Motion on a Sphere Quantum Mechanical Rotation in 3-D: Motion on a Sphere In the case just considered, the motion was constrained to two dimensions. The Schrödinger equation with zero potential energy in three-dimensions is given as ∂2 ~2 1 ∂ 1 1 ∂ ∂ 2 ∂ − + R + sin θ ψ = Eψ 2µ R2 ∂R ∂R ∂θ R2 sin2 θ ∂φ2 R2 sin θ ∂θ Now imagine the more familiar case of a diatomic molecule freely rotating in three-dimensional space. Again, we transform to the center of mass coordinate system, and the rotational motion is transformed to the motion of a particle on the surface of a sphere of radius R0 . Since the two masses of the rigid rotor are at fixed distances from the origin, R is constant, and we can ignore the derivatives with respect to R: ~2 1 ∂2 1 ∂ ∂ − + sin θ Y (θ, φ) = E Y (θ, φ) 2I sin2 θ ∂φ2 sin θ ∂θ ∂θ The bond length is assumed to remain constant as the molecule rotates giving rise to the term rigid rotor. The kinetic and potential operators are are expressed in spherical coordinates. The potential energy V (R, φ, θ) is zero because there is no hindrance to rotation. Y (θ, φ) is the product of two functions Y (θ, φ) = Θ(θ)Φ(φ) Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 57 Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 Quantum Mechanical Rotation in 3-D: Motion on a Sphere Eigenfunctions and Eigenvalues of a Particle on a Sphere One obtains two separate differential equations: Two quantum numbers l and ml which arise in the solution of this eigenvalue equations on the previous m slide and are represented by the function Yl l (θ, φ) and are called spherical harmonics. 1 d2 Φ(φ) = −m2l Φ(φ) dφ2 1 d dΘ(θ) sin θ sin θ + β sin2 θ = m2l Θ(θ) dθ dθ The first several spherical harmonics are given in the table below: Y00 = β= 1 1 (4π) 2 1 3 2 Y10 = 4π cos θ 1 3 2 1 sin θeiφ Y1 = 8π 1 3 2 Y1−1 = 8π sin θe−iφ 2µR02 E ~2 To ensure Y (θ, φ) remains single-values and finite, the following conditions must be met: ml The eigenfunctions Yl β = l(l + 1), for l = 0, 1, 2, 3, . . . ml (θ, φ) = so that El = ml = −l, −(l − 1), −(l − 2), . . . , 0, . . . , (l − 2), (l − 1), +l Last Updated: October 20, 2013 at 8:16pm Y20 = Y21 = Y2−1 = Y22 = 5 16π 1 1 15 2 (3 cos2 θ − 1) 2 sin θ cos θeiφ 8π 1 15 2 sin θ cos θe−iφ 8π 1 15 2 2 θe2iφ sin 32π (θ, φ) satisfy the eigenvalue equation ĤYl and Fall Semester 2013 58 ~2 m l(l + 1)Yl l (θ, φ) for l = 0, 1, 2, 3, . . . 2I ~2 l(l + 1) for l = 0, 1, 2, 3, . . . 2I The numerical factor in front of these functions ensures that they are normalized over the intervals 0 ≤ θ ≤ π and 0 ≤ φ2π. 59 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 60 Plots of p and d Linear Combinations of the Spherical Harmonics Review of Classical Angular Momentum Angular momentum is a vector and has components in the x, y, and z directions. r 3 1 px = √ Y11 + Y1−1 = sin θ cos φ 4π 2 To develop the quantum mechanical operators for angular momentum in the x, y, and z directions, we need to review the classical expressions for angular momentum in three dimensions. r 3 1 1 Y1 − Y1−1 = sin θ sin φ py = √ 4π 2i r 3 pz = Y10 = cos θ 4π r 5 3 cos2 θ − 1 dz2 = Y20 = 16π r 15 1 dxz = √ Y21 + Y2−1 sin θ cos θ cos φ 4π 2 r 15 1 1 Y2 − Y2−1 dyz = √ sin θ cos θ sin φ 4π 2i r 15 1 dx2 −y2 = √ Y22 + Y2−2 sin2 θ cos 2φ 16π 2 r 15 1 2 Y2 − +Y2−2 dxy = √ sin2 θ sin 2φ 16π 2i Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm A particle rotating around a fixed axis has an angular momentum and kinetic energy. L = Iω = mvr ω is the angular velocity. 61 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm Review of Classical Angular Momentum Review of Classical Angular Momentum In classical mechanics, the angular momentum rotating about a fixed point is represented by the vector L in the direction perpendicular to the plane of motion. If a mass m is rotating about a fixed point with linear velocity v , the angular momentum L is given by the cross product of the radius r and the linear momentum p The vectors r and p may be expressed in terms of their components and unit vectors i , j , and k pointing along the x, y, and z axes. 62 r = xii + yjj + zkk p = px i + py j + pz k The cross product of r and p may be calculated from a determinant: i j k L = r × p = x y z px py pz L = r × mvv = r × p The cross product of two vectors r and p is a vector of magnitude |rr | |pp| sin θ where θ is the angle between r and p . Thus, the angular momentum vector for the circular motion in the previous figure points up. L = (ypz − zpy )ii + (zpx − xpz )jj + (xpy − ypx )kk Lx = ypz − zpy Ly = zpx − xpz Lz = xpy − ypx Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 63 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 64 Quantum Mechanical Angular Momentum Review of Classical Angular Momentum The quantum mechanical operators for the angular momentum are obtained by replacing the components of momentum with their corresponding quantum mechanical operators. Specifically, The square of the angular momentum is given by the dot product of L with itself: L • L = L2 = L2x + L2y + L2z pˆx −→ −i~ ∂ ; ∂x pˆy −→ −i~ ∂ ; ∂y pˆz −→ −i~ The square of the angular momentum is a scalar. ∂ ∂z If no torque acts on a particle, its angular momentum is conserved, i.e. is a constant. This substitution yields: In classical mechanics, all possible values of L and E are permitted. ∂ ∂ Lˆx = −i~ y −z ∂z ∂y ∂ ∂ ˆ Ly = −i~ z −x ∂x ∂z ∂ ∂ −y L̂z = −i~ x ∂y ∂x Last Updated: October 20, 2013 at 8:16pm Fall Semester 2013 65 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm Angular Momentum: Quantum Mechanical Operators Quantum Mechanical Angular Momentum Operators The operator for the square of the angular momentum is given by: It is readily shown that Lˆx and Lˆy , Lˆy and Lˆz , Lˆx and Lˆz do not commute with each other. Lˆx , Lˆy , and Lˆz all commute with Lˆ2 . Lˆ2 = |L̂|2 = L̂ • L̂ = Lˆ2x + Lˆ2y + Lˆ2z Therefore, we can measure precisely the square of the total angular momentum, and one, but only one of its components. It is often more convenient to use the angular momentum operators in spherical coordinates: ∂ ∂ Lˆx = i~ sin φ + cot θ cos φ ∂θ ∂φ ∂ ∂ Lˆy = i~ − cos φ + cot θ sin φ ∂θ ∂φ Lˆ2 = −~2 Fall Semester 2013 L̂z = −~ 1 ∂ sin θ ∂θ ∂ ∂φ sin θ ∂ ∂θ + 1 ∂2 sin2 θ ∂φ2 Last Updated: October 20, 2013 at 8:16pm 66 Thus, if the magnitude of the total angular momentum q √ L| = L2 = L2x + L2y + L2z |L is measured, and Lz is measured, it is not possible to measure Lx or Ly precisely. The eigenfunction of Lˆ2 is also an eigenfunction of Lˆz but is not an eigenfunction of Lx or Ly which is an essential difference between classical and quantum mechanical systems. 67 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 68 Eigenvalues of Angular Momentum Operators Magnetic Quantum Numbers Since Lˆ2 and Lˆz commute, it is possible to construct an function that is an eigenfunction of both operators. Operating on the spherical harmonics with Lˆz , Lˆz Y ml (θ, φ) = ml ~ Y ml (θ, φ) For a classical rigid rotor, l thus, for the quantum mechanical operator ml = −l, −l + 1, . . . , 0, . . . , l − 1, +l Lˆ2 = 2I Ĥ and Lˆ2 Ylml (θ, φ) = l(l + 1)~2 Ylml (θ, φ) yields the following eigenvalues for the z component Lz of the angular momentum Lz = ml ~; l = 0, 1, 2, . . . According to this eigenvalue equation, the total angular momentum for a rigid rotor can have only the values L2 = l(l + 1)~2 The possible orientations of the momentum vector L for l = 1 with respect to a particular direction are shown in the adjacent figure. l = 0, 1, 2, . . . Last Updated: October 20, 2013 at 8:16pm 69 Magnitude of Quantum Mechanical Angular Momentum Since the x and y components of the angular momentum are indeterminant, the vector can be rotated about the z axis to lie anywhere on the conical surface. Since the Lx and Ly components are unknown, L can be described only as being on the surface of a cone as shown in the adjacent figure. The magnitude of the orbital angular momen1 tum L is [l(l + 1)] 2 ~. and the maximum component of Lz in a particular direction is l~. Thus, the magnitude of the angular momentum L is greater than its z component so that the angular momentum vector cannot point in the direction of an applied magnetic field or along a unique axis. Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm ml = −l, −l+1, . . . , 0, . . . , l−1, +l ml is referred to as the magnetic quantum number. l is referred to as the angular momentum quantum number. Fall Semester 2013 l with L2 = 2IT 71 Fall Semester 2013 Last Updated: October 20, 2013 at 8:16pm 70