Quantum Chemistry Exam 2 Solutions

Transcription

1 Chemistry 46 Fall 17 Dr. Jean M. Standard November 8, 17 Name KEY Quantum Chemistry Exam Solutions 1.) ( points) Answer the following questions by selecting the correct answer from the choices provided. a.) Which of the statements below are true about the Variation Principle and Interleaving Theorem and their consequences? 1) The energy expectation value of the nth excited state of a system is always greater than or equal to the exact energy eigenvalue of the nth excited state. ) The ground state energy expectation value of a wavefunction may never be equal to the exact ground state energy eigenvalue of the system. 3) When the basis set size is increased, the energy expectation value for a given state decreases monotonically toward an asymptote. 4) The Variation Principle and Interleaving Theorem are valid only for properly normalized wavefunctions. 5) Both (1) and (). 6) Both (1) and (3). 7) Both (1) and (4). b.) Which of the following statements are false about quantum mechanical angular momentum? 1) When ˆ z is applied to the spherical harmonic function Y 3 ( θ,φ), the eigenvalue is!. ) For each value of the angular momentum quantum number l, there are l +1 values of the magnetic quantum number m. 4) When ˆ is applied to the spherical harmonic function Y 31 θ,φ ( ), the eigenvalue is 1!. 5) Both (1) and (3). 6) Both (1) and (4). 3) The operators ˆ x and ˆ y share a common set of eigenfunctions.

2 1. Continued c.) Which of the following statements about the Heisenberg Uncertainty Principle and its consequences are true? 1) The position and momentum (both in the same direction) of a particle may never be measured simultaneously and exactly. ) The position of a particle in the x and y directions may be measured simultaneously and exactly. 3) In order for the momentum of system to be measured exactly, the system must be in a state corresponding to an eigenfunction of the momentum operator. 4) The uncertainty product for two observable properties is related to the commutator of the two operators corresponding to the observables. 5) All of the above. 6) None of the above. d.) Which of the following statements are false about the quantum mechanical harmonic oscillator? 1) The expectation value of the position, x, equals for all eigenfunctions ψ v ( x). ) The probability density exhibits zero amplitude for values of x beyond the classical turning points of the system. 3) Doubling the value of the force constant k increases the harmonic frequency ν by. 4) The eigenfunction ψ ( x) α π 1/ / possesses two nodes. 5) All of the above. 6) None of the above. ( 4αx ) e α x /

3 3.) ( points) Briefly explain and discuss the significance of each of the following topics or concepts. a.) Basis set A basis set corresponds to a set of functions used to represent an approximate wavefunction. The basis set representation of an approximate wavefunction ψ approx has the form ψ approx N c i φ i, i1 where c i are linear coefficients (often obtained via the linear variation method) and φ i are the fixed set of N functions chosen as the basis set. In the limit that N, the representation is exact if the functions φ i form a complete set. b.) Quantum mechanical uncertainty The quantum mechanical uncertainty ΔA for an observable A is defined in terms of the expectation values, ΔA A A 1/. The uncertainty is a purely quantum mechanical effect that arises as a result of the wave-like nature of matter. The quantum mechanical uncertainty is closely related to the statistical definition of the standard deviation. It provides a measure of the spread in the distribution of the observable A. c.) Franck-Condon Principle The Franck-Condon Principle applies to electronic transitions in molecular systems. Because electronic motion is very fast compared to nuclear motion, the Franck-Condon Principle states that any electronic transition must occur vertically on a potential energy diagram. The Franck-Condon Principle also leads to the idea that the overlap integral S nm is related to the intensity I of the electronic transition, I S nm. The overlap integral Snm is defined as S nm ψ g* e n ( x) ψ m ( x) dx, Here, the function ψ g n ( x) corresponds to the ground state vibrational wavefunction with quantum number e n, the function ψ m ( x) corresponds to the excited state vibrational wavefunction with quantum number m, and x is the bond displacement, x r r e.

4 4.) continued d.) Morse oscillator The Morse oscillator provides a model for the potential energy of a vibrating molecule. The form of the Morse oscillator potential is ( ) D e 1 e ax V x, where D e is the dissociation energy, a is a constant, and x is the bond displacement. A graph of the Morse oscillator potential for a typical bond (O-H in this case) is shown below. Note that unlike the harmonic oscillator model of molecular vibrations, which is valid only at low energies, the Morse oscillator demonstrates the correct dissociation behavior and so is valid for all energies.

5 5 3.) ( points) The ground state wavefunction of the particle in a box is given by ψ 1 ( x) sin π x, x. The wavefunction is equal to zero outside this range. a.) Using the ground state particle in a box wavefunction, calculate the uncertainty in position, Δx. The uncertainty Δx is defined as Δx [ x x ] 1/. The expectation values x and x must be evaluated. The expectation value x for the ground state wavefunction is x ψ 1 x ψ 1 ψ 1 ψ 1 ψ 1 x ψ 1, since the particle in a box wavefunctions are normalized. Evaluating the expectation value, x ψ 1 x ψ 1 x sin π x x sin π x dx x sin π x dx. From the integral table, x sin bx dx x cosbx b + x sinbx b + cosbx b 3. Substituting, x x sin π x dx x cos π x π + x sin π x π + cos π x x 3. π x

6 6 3 a.) Continued Evaluating at the limits yields, x x π cosπ + 3 π sinπ + 3 π 3 cosπ ( π 1 ) π π 43 π 3 8 π. 3 π ( ) + 3 ( π 3 1) 3 ( π 3 1 ) 3 π 3 cos The expectation value x for the ground state wavefunction is x ψ 1 x ψ 1 ψ 1 ψ 1 ψ 1 x ψ 1, since the particle in a box wavefunctions are normalized. Evaluating the expectation value, we have x ψ 1 x ψ 1 sin π x x sin π x dx x xsin π x dx. From the integral table, x sin bx dx x 4 xsinbx 4b cosbx 8b. Substituting, x xsin π x dx x 4 xsin π x 4 π cos π x 8 π x x.

7 7 3 a.) Continued Evaluating at the limits yields, x 4 4 4π sinπ 8π cosπ 4π ( ) 4 8π + 8π ( 8π 1 ) ( 8π 1 ) 8π cos x 4. The expectation values may now be inserted into the uncertainty expression for x, Δx x x 1/ π 8 π 1/ Δx π 8 π or Δx /, This result makes sense because the uncertainty in position cannot be larger than, since the particle is excluded from the region outside the box by the infinite potential, and an uncertainty corresponding to not quite half of the width of the box seems reasonable for a measure of the spread of the position for the ground state.

8 8 3.) continued b.) Using the ground state particle in a box wavefunction, calculate the uncertainty in momentum Δp x. The uncertainty Δp x is defined as Δp x [ p x p x ] 1/. The expectation values p x and px must be evaluated. The expectation value p x state wavefunction is for the ground p x ψ 1 p ˆ x ψ 1 ψ 1 p ˆ x ψ 1, ψ 1 ψ 1 since the particle in a box wavefunctions are normalized. Substituting into the expectation value, p x p x ψ 1 p ˆ x ψ 1 sin π x! d dx! sin π x dx sin π x d dx sin π x dx. Evaluating the second derivative, d dx sin π x π sin π x. Substituting, p x d! sin π x dx sin π x dx π! 3 sin π x dx. From the integral table, sin bx dx x sinbx. 4b

9 9 3 b.) Continued Substituting, the expectation value becomes, p x π! 3 sin π x dx π! 3 sin π x x 4 π x x. Evaluating at the limits, p x p x π! 3 π! 3 π! 3 π!. sin( π ) 4π 4π sin ( ) ( 4π ) ( 4π ) The expectation value p x for the ground state wavefunction is p x ψ 1 ˆp x ψ 1 ψ 1 ψ 1 ψ 1 ˆp x ψ 1, since the particle in a box wavefunctions are normalized. Substituting into the expectation value, we have p x ψ 1 ˆp x ψ 1 sin π x i! d dx sin π x dx p x i! sin π x d dx sin π x dx. Evaluating the first derivative, d dx sin π x π cos π x.

10 1 3 b.) Continued Substituting, p x i! i! π sin π x d dx sin π x dx sin π x cos π x dx. From the integral table, sin bx cosbx dx sin bx b. Substituting, the expectation value becomes, p x i! i! sin π x d dx sin π x dx sin π x π x x. Evaluating at the limits, p x i! i! p x. π sin ( π ) π ( ) π ( ) π sin ( ) These expectation values may now be inserted into the uncertainty expression for p x, 1/ Δp x p x p x π! 1/ ( ) Δp x π!.

11 11 1.) Continued c.) Compare your calculated uncertainty product Δx Δp x to the result required by the Heisenberg Uncertainty Principle. Does your calculated uncertainty product make sense? Explain. The uncertainty product can be calculated for the ground state of the particle in a box, Δp x Δx π! π 8 π 1 3 4! π 8 π π 4 Δp x Δx 1.166!. 1/ 1/ Since the Heisenberg Uncertainty Principle states that Δp x Δx!, the result of 1.166! does make sense because it is larger than the minimum uncertainty product,.5!.

12 1 4.) ( points) Consider the differential equation d dx + C x + D x E ψ (x), where C, D, and E are real constants, and the range of x is < x. a.) Obtain an equation for the asymptotic solution in the limit that x. Show that a solution of the form ψ(x) e ±βx satisfies the asymptotic equation, where β is positive and real. Determine the relationship between β and the constant E. As x, the middle two terms in the differential equation, C x asymptotic form of the differential equation is and D x, go to zero. Thus, the d dx E ψ(x). The solution can be written in the form of an exponential. Assume a solution of the form ψ(x) e ±βx, where β is a positive constant. The derivatives of the function are ψ ʹ(x) ± β e ±βx, ψ ʹʹ(x) β e ±βx. Substituting the second derivative into the asymptotic equation, we have d dx E ψ(x) β e ±βx E e ±βx ( β E) e ±βx. Dividing both sides of the equation by e ±βx yields, β E. Therefore, the exponent is given by β E, or β E. The asymptotic solution thus has the form ψ(x) e ± E x.

13 13 4.) continued b.) Explain whether all or part of the asymptotic solutions corresponds to an acceptable wavefunction in the limit that x. The asymptotic solution is ψ(x) e ± E x. In the limit that x, the asymptotic solution has the behavior that e E x and e + E x. Since an acceptable wavefunction must remain finite, the solution with positive exponent must be thrown out. Thus, the only valid asymptotic solution is the one with negative exponent, ψ(x) e E x. c.) Write the full solution to the differential equation as the asymptotic part times a remainder function, g(x). Obtain a differential equation for the remainder g(x). To get the remainder, we write the full solution as the asymptotic part times the remainder. In this case, ψ(x) e βx g(x), where the function g(x) is the remainder function and β E. Calculating the derivatives, ( ), ψ ʹ(x) β e βx g(x) + e βx g ʹ(x) e βx βg(x) + g ʹ(x) ψ ʹʹ(x) β e βx [ βg(x) + g ʹ(x) ] + e βx βg ʹ(x) + e βx g ʹʹ(x) βg ʹ(x) + β g(x). [ g ʹʹ(x) ] Substituting these expressions into the original differential equation, we have d dx + C x + D x E ψ (x), or ʹʹ ψ (x) + C x ψ (x) + D ψ (x) Eψ (x), x e βx g ʹʹ(x) βg ʹ(x) + β g(x) e βx g ʹʹ(x) βg ʹ(x) + β + + C x e βx g(x) + C x + D x E D x e βx g(x) Ee βx g(x) g(x).

14 14 4 c.) continued Dividing both sides of the equation by the factor e βx, the equation for the remainder function g(x) is g ʹʹ(x) βg ʹ(x) + β + C x + D x E g(x). Using the relation that β E, we can eliminate two terms on the left, giving g ʹʹ(x) βg ʹ(x) + C x + D x g(x). d.) Express the remainder g(x) as an infinite power series in x. Use this to obtain a recursion relation for the coefficients of the power series. Next, we assume a power series solution for the remainder, Evaluating the derivatives, g(x) g ʹ(x) a k x k. a k k x k 1, k1 g ʹʹ(x) a k k( k 1) x k. k Next, the summations can be rewritten so that they start at, g ʹ(x) a k k x k 1, g ʹʹ(x) a k+1 k( k +1) x k 1. The derivatives are then substituted into the differential equation for the remainder determined in part (c), g ʹʹ(x) βg ʹ(x) + C x + D x g(x), a k+1 k( k +1) x k 1 β a k k x k 1 + C x + D x a k x k.

15 15 4 d.) continued Incorporating the powers of x from C x and D x into the last summation and splitting it into two parts yields a k+1 k( k +1) x k 1 β a k k x k 1 + a k+1 k( k +1) x k 1 β a k k x k 1 + C x C x + D x a k x k + a k x k D x a k x k a k+1 k( k +1) x k 1 β a k k x k 1 + C a k x k + D a k x k 1. The only summation that does not include the factor x k 1 is the third one. At this point you get stuck because there is no way to combine them into a single summation. That is because the D term should have had a derivative in it (i.e., the term in the differential equation should have been D d x dx instead of just D x ). Then all the summations can be written with the same power of x and a recursion formula may be developed.