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Chapter 8 Quantum Theory: Techniques and Applications Using Quantum Mechanics on Simple Systems , Phys. Chem. 2nd Ed T. Engel, P. Reid (Ch15) Objectives • Using the postulates to understand the particle in the box (1-D, 2-D and 3-D) Outline 1. 2. 3. 4. The Free Particle The Particle in a One-Dimensional Box Two- and Three-Dimensional Boxes Using the Postulates to Understand the Particle in the Box and Vice Versa 8.1 The Free Particle • For a free particle in a one-dimensional space on which no forces are acting, the Schrödinger equation is ℏ 2 d 2ψ ( x ) − = − Eψ ( x ) 2 2 m dx ψ (x ) • is a function that can be differentiated twice to return to the same function ( 2 mE / ℏ )x + ikx 2π 2 mE ( x ) = A e = A e ψ + + where k = = 2 λ ℏ ψ − (x ) = A e −i (2 mE / ℏ )x = A e −ikx +i + 2 2 − − 8.1 The Free Particle • If x is restricted to the interval − L ≤ x ≤ L then the probability of finding the particle in an interval of length dx can be calculated. P(x )dx = ψ * ( x )ψ ( x )dx L ∫ψ * (x)ψ (x )dx −L = A+ A+ e −ikx e + ikx dx L A+ A+ ∫ e −ikx e + ikx dx −L = dx 2L 8.2 The Particle in a OneOne-Dimensional Box • When consider particle confined to a box in 1D, the potential is V(x) =0, (for 0<x <a) V(x) =∞, (for x <0, or x ≥a) 8.2 The Particle in a OneOne-Dimensional Box • Consider the boundary condition satisfying 1-D, ψ (0) = ψ (a ) = 0 • The acceptable wave functions must have the form of nπx ψ n = A sin , for n = 1, 2,3, 4... a • Thus the normalized eigenfunctions are ψ n (x ) = 2 nπx sin a a Physical Chemistry Fundamentals: Figure 8.3 Fig. 8.3 The first five normalized wavefunctions of a particle in a box. Each wavefunction is a standing wave, and successive functions possess one more half wave and a correspondingly shorter wavelength. 8.2 The Particle in a OneOne-Dimensional Box • Energies for the Particle in a Box k 2ℏ 2 Ek = 2m ka = nπ (n = 1,2,...) n 2π 2 ℏ 2 n 2 h 2 En = = (n = 1,2,...) 2 2 2 ma 8ma Zero - point energy of a particle in a box : h2 E1 = 8ma 2 - the lowest, irremovable energy Physical Chemistry Fundamentals: Figure 8.2 Fig. 8.2The allowed energy levels for a particle in a box. Note that the energy levels increase as n2, and that their separation increases as the quantum number increases. Example 1 From the formula given for the energy levels for the particle in the box, En = h 2 n 2 / 8ma 2 for n = 1, 2, 3, 4… , we can see that the spacing between adjacent levels increases with n. This appears to indicate that the energy spectrum does not become continuous for large n, which must be the case for the quantum mechanical result to be identical to the classical result in the high-energy limit. Example 1 A better way to look at the spacing between levels is to form the ratio (En+1 − En ) / En . By forming this ratio, we see that ∆E / E becomes a smaller fraction of the energy as n → ∞ . This shows that the energy spectrum becomes continuous for large n. Solution We have, En+1 − En = En ( [ ] h 2 ( n +1) 2 − n 2 / 8 ma 2 h 2 n 2 / 8 ma 2 ) 2n + 1 = n2 which approaches zero as n → ∞ . Both the level spacing and the energy increase with n, but the energy increases faster (as n2), making the energy spectrum appear to be continuous as n→∞ •The correspondence principle states that classical mechanics emerges from quantum mechanics as high quantum numbers are reached. Physical Chemistry Fundamentals: Figure 8.4 The probability density for a particle in a box with length a (or L) is 2 a nπx a ψ 2 ( x) = sin 2 (8.8) Fig. 8.4 (a) The first two wavefunctions, (b) the corresponding probability distributions, and (c) a representation of the probability distribution in terms of the darkness of shading. Example 8.1 Using the particle in a box solutions (p.292) • What is the probability, P, of locating the electron between x = 0 (the left-hand end of a box) and x = 0.2 nm in its lowest energy state in a cbox of length 1.0 nm? • Method The value of Ψ2dx is the probability of finding the particle in the small region dx located at x; therefore, the total probability of finding the particle in the specified region is the integral of Ψ2dx over that region. The wavefunction of the particle is given in eqn 8.4b with n = 1. • Answer The probability of finding the particle in a region between x = 0 and x = l is 2l nπ x l 1 2π nl P = ∫ψ dx = ∫ sin 2 dx = − sin a0 a a 2nπ a 0 l 2 n ψ n ( x) = 2 nπ x sin a a (8.4b) • Set n = 1 and l = 0.2 nm, which gives P = 0.05. The result corresponds to a chance of 1 in 20 of finding the particle in the region. As n becomes infinite, the sine term, which is multiplied by 1/n, makes no contribution to P and the classical result, P = l/a, is obtained. • Test 8.4 Calculate the probability that a particle in the state with n = 1 will be found between x = 0.25a and x = 0.75a in a box of length a (with x = 0 at the left-hand end of the box). • Correct Answer: 0.82 A particle in a box – Key points • (a) The energies of a particle constrained to move in a finite region of space are quantized; • (b) The energies and wavefunctions for a particle moving in a box are labelled by quantum numbers. The wavefunctions of a particle constrained to move in a one-dimensional box are mutually orthogonal sine functions with the same amplitude but different wavelengths. • The zero point energy is the lowest, irremovable energy of a particle in a box. • The correspondence principle states that classical mechanics emerges from quantum mechanics as high quantum numbers are reached. Physical Chemistry Fundamentals: Figure 8.5 Fig. 8.5 A twodimensional square well. The particle is confined to the plane bounded by impenetrable walls. As soon as it touches the walls, its potential energy rises to infinity. Motion in two dimensions - Separation of variables Physical Chemistry Fundamentals: Figure 8.6 Fig. 8.6 The wavefunctions for a particle confined to a rectangular surface depicted as contours of equal amplitude. (a) n1 = 1,n2 = 1, the state of lowest energy, (b) n1 = 1,n2 = 2, (c) n1 = 2,n2 = 1, and (d) n1 = 2, n2 = 2. Figure 8.7 Degeneracy L1 = L2 = L Fig. 8.7 The wavefunctions for a particle confined to a square surface. Note that one wavefunction can be converted into the other by a rotation of the box by 90°. The two functions correspond to the same energy. Degeneracy and symmetry are closely related. Motion in two and more dimensions - Key points • (a) The separation of variables technique can be used to solve the Schrodinger equation in multiple dimensions. The energies of a particle constrained to move in two or three dimensions are quantized. • (b) Degeneracy occurs when different wavefunctions correspond to the same energy. Many of the states of a particle in a square or cubic box are degenerate. 8.3 TwoTwo- and ThreeThree-Dimensional Boxes • • 1-D box is useful model system as it allows focus to be on quantum mechanics instead of mathematics. For 3-D box, the potential energy is V ( x, y , z ) = 0 for 0 < x < a, 0 < y < b, 0 < z < c; V ( x, y , z ) = ∞ otherwise • Inside the box, the Schrödinger equation can be written as h2 ∂ 2 ∂2 ∂2 2 + 2 + 2 ψ (x , y , z ) = Eψ ( x, y , z ) − 2m ∂ x ∂y ∂z 8.3 TwoTwo- and ThreeThree-Dimensional Boxes • The total energy eigenfunctions have the form nxπx nyπy nzπz ψnxnynz ( x, y, z) = Nsin sin sin a b c • And the total energy has the form 2 2 2 n h nx nz y E= 2 + 2 + 2 8m a b c 2 8.4 Using the Postulates to Understand the Particle in the Box and Vice Versa Postulate 1 The state of a quantum mechanical system is completely specified by a wave function ψ ( x , t ) . The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 is given by ψ * ( x0 , t )ψ (x0 , t )dx . • This postulate states that all information obtained about the system is contained in the wave function. Example 88-2 Consider the function ψ ( x ) = c sin (πx / a ) + d sin (2πx / a) a. Is ψ (x ) an acceptable wave function for the particle in the box? b. Is ψ (x ) an eigenfunction of the total energy operator? c. Is ψ (x ) normalized? Solution a. If ψ (x ) is to be an acceptable wave function, it must satisfy the boundary conditions ψ (x ) =0 at x=0 and x=a. Its first and second derivatives must ψ (x ) also be well-behaved functions between x=0 and x=a. This is the case for ψ (x ) . We conclude that ψ (xψ) (x) = c sin (πx / a ) + d sin (2πx / a ) is an acceptable wave function for the particle in the box. Solution b. Although ψ ( x ) = c sin (πx / a ) + d sin (2πx / a ) may be an acceptable wave function, it need not be an eigenfunction of a given operator. To see if ψ (x ) is ψ (x ) an eigenfunction of the total energy operator, the operator is applied to the function: 2 (x ) 2 2 ℏ2 d ψ 2 π x ℏ π πx 2πx πx − c sin ( a ) + d sin c sin + 4d sin = 2 2 2 m dx a 2ma a a The result of this operation is not ψ (x ) multiplied by a constant. Therefore, ψ (x) is not an eigenfunction of the total energy operator. Solution c. To see if ψ (x ) is normalized, the following integral is evaluated: 2 πxψ (x ) 2πx c sin ∫0 a + d sin a dx a x πx 2πx 2 π 2 2πx ψ (x ) = ∫ c * c sin + d * d sin + ( cd * + c * d ) sin sin dx a a a a 0 a 2 2 πx 2 πx 2πx 2 2πx = ∫ c sin dx + ∫ d sin dx + ∫ (cd * +c * d ) sin sin dx a a a a 0 0 0 a a a Solution Using the standard integral ∫ sin 2bydy = y / 2 − (1/ 4b ) sin by and recognizing that the third integral is zero 2 2 ψ2(x 2 π)x 2 2πx c sin + d sin dx ∫ a a 0 2 a a (sin 2π − sin 0 ) 2a a (sin 4π − sin 0 ) a 2 2 = c ψ−(x ) + d − = c + d 2 2 4π 8π 2 a ( ) Solution Therefore, ψ (x ) is not normalized, but the function ψ (x ) 2 πx 2π c sin + d sin a a a is normalized for the condition that c 2 + d 2 = 1 ψ (x ) Note that a superposition wave function has a more complicated dependence on time than does an eigenfunction of the total energy operator. Solution For instance, ψ (x ) for the wave function under consideration is given by 2 −iE / h πx −iE ψ (x ) ψ ( x, t ) = ce sin + ce a a 1 ψ (x ) 2 /h 2πx sin ≠ ψ ( x ) f (t ) a This wave function cannot be written as a product of a function of x and a function of t. Therefore, it is not a standing wave and does not describe a state whose properties are, in general, independent of time. 8.4 Using the Postulates to Understand the Particle in the Box and Vice Versa • Acceptable Wave Functions for the Particle in a Box Example 88-3 What is the probability, P, of finding the particle in the central third of the box if it is in its ground state? Solution For the ground state, ψ 1 (x ) = 2 / a sin (πx / a ) . From the postulate, P is the sum of all the probabilities of finding the particle in intervals of width dx within the central third of the box. This probability is given by the integral 2 P= a πx ∫a / 3sin a dx 2a/ 3 2 Solution Solving this integral, 2 a a P= − a 6 4π 2π 4π − sin sin 3 3 = 0.609 Although we cannot predict the outcome of a single measurement, we can predict that for 60.9% of a large number of individual measurements, the particle is found in the central third of the box. 8.4 Using the Postulates to Understand the Particle in the Box and Vice Versa Postulate 3 In any single measurement of the observable that corresponds to the operator  , the only values that will ever be measured are the eigenvalues of that operator. 8.4 Using the Postulates to Understand the Particle in the Box and Vice Versa Postulate 4 If the system is in a state described by the wave function ψ (x, t ) , and the value of the observable a is measured once each on many identically prepared systems, the average value of all of these measurements is given by ∞ a = ˆ ψ (x , t )dx * ( x , t ) A ψ ∫ −∞ ∞ ∫ψ * (x, t )ψ (x, t )dx −∞ 8.4 Using the Postulates to Understand the Particle in the Box and Vice Versa • Expectation Values for E, p, and x for a Superposition Wave Function Example 88-4 Assume that a particle is confined to a box of length a, and that the system wave function is ψ ( x ) = 2 / a sin (πx / a ) a. Is this state an eigenfunction of the position operator? b. Calculate the average value of the position that would be obtained for a large number of measurements. Explain your result. Example 88-4 a. The position operator xˆ = x . Because , xψ (x ) = x 2 / a sin (πx / a ) ≠ cψ (x ) where c is a constant, the wave function is not an eigenfunction of the position operator. Example 88-4 b. The expectation value is calculated using the fourth postulate: x = 2 2 πx 2 πx πx sin x sin dx = xsin dx ∫ ∫ a 0 a a 0 a a a a Using the standard integral a 2 x 2 cos 2bx x sin 2bx ∫0 x (sin bx ) dx = 4 − 8b 2 − 4b Example 88-4 We have a 2πx 2πx x sin 2 cos a2 a 2 x 2 a 2 a 2 a a x = − − = − 2 − 0 + 2 = 2 a4 π a 4 8π π 8π 2 4 8 a a 0 The average position is midway in the box. This is exactly what we would expect, because the particle is equally likely to be in each half of the box.