Chemistry 532 Problem Set 7 Spring 2012 Solutions

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1 Chemistry 53 Problem Set 7 Spring 01 Solutions 1. The study of the time-independent Schrödinger equation for a one-dimensional particle subject to the potential function leads to the differential equation V (x) = κ x d u dz 1 4 u + α z u = 0 where z is a dimensionless coordinate related to x. (a) Show that an asymptotic solution to the differential equation is given by u(z) = e z/ As z d u dz 1 4 u = 0 u = e sz u = se sz u = s e sz ( s 1 ) e sz = 0 4 Because u 0 as z, A = 0 and s = ± 1 u = Ae z/ + Be z/ u = Be z/ (b) Write u(z) = f(z)e z/ and show that the differential equation satisfied by f is given by z d f dz z df dz + αf = 0 1

2 u(z) = f(z)e z/ u (z) = f (z)e z/ 1 f(z)e z/ u (z) = f (z)e z/ 1 f (z)e z/ 1 f (z)e z/ f(z)e z/ = e z/ [ f (z) f (z) f(z) ] e z/ [ f (z) f (z) f(z) 1 4 f(z) + α z f(z) ] = 0 or f (z) f (z) + α z f(z) = 0 z d f dz z df dz + αf(z) = 0 (c) Use the power series method to solve the differential equation for f(z) by finding a recursion relation between the expansion coefficients. f(z) = c n z n f (z) = n=0 n=1 nc n z n 1 f (z) = n(n 1)c n z n n= n(n 1)c n z n 1 nc n z n + α c n z n n=1 n=0 n=0 In the first sum we let k = n 1 so that n = k + 1. or z n [n(n + 1)c n+1 nc n + αc n ] = 0 k=0 c n+1 c n = n α n(n + 1)

3 . For the n=1 state of the harmonic oscillator, find the most likely position(s) of the particle. ( ) 4α P = ψ 1/ ψ = x e αx with α = mω/ h. dp dx = ( ) 4α 1/ [xe αx αx 3 e αx ] or = 0 at x = x max x max = x 3 maxα x max = ± 1 ( ) 1/ h = ± α mω 3. Find the eigenfunctions and eigenvalues of the Hamiltonian for a particle of mass m moving in a potential defined by { x < 0 V (x) = 1 kx x 0 [Hint: Draw the potential and think about the boundary conditions. This problem requires no algebra.] Ĥ = h d m dx + 1 mω x for 0 x <. This is the same Hamiltonian operator as found in the harmonic oscillator problem. Unlike the harmonic oscillator, this system has boundary conditions lim ψ(x) = ψ(0) = 0. x These boundary conditions are obeyed by the odd-n solutions to the harmonic oscillator. ( E n = n + 1 ) hω for ψ n (x) = ( ( ) α 1/4 e αx n!)1/ / H n n ( αx) n = 1, 3, 5,... 3

4 4. A three dimensional harmonic oscillator of mass m, has potential energy function V (x, y, z) = 1 k xx + 1 k yy + 1 k zz where k x, k y and k z are the force constants. Use the known solutions to the Schrödinger equation for the one-dimensional harmonic oscillator to find the eigenvalues and eigenfunctions of the Hamiltonian operator for the three dimensional harmonic oscillator. Ĥ = h m x + 1 k xx h m y + 1 k yy h m z + 1 k zz where where = ĥ1(x) + ĥ(y) + ĥ3(z) d ĥ 1 (x) = h m dx + 1 k xx d ĥ (y) = h m dy + 1 k yy d ĥ 3 (z) = h m dz + 1 k zz Ψ nx,ny,nz (x, y, z) = ψ nx (x)φ ny (y)λ nz (z) ĥ 1 ψ nx = E nx ψ nx ĥ φ ny = E ny φ ny ĥ 3 λ nz = E nz λ nz so that ψ, φ and λ solve the one-dimensional Schrödinger equation for the harmonic oscillator, and the eigenvalues are given by with E nx,ny,nz = ( n x + 1 ) ( hω x + n y + 1 ) ( hω y + n z + 1 ) hω z ω x = k x m ω y = k y m ω z = k z m 5. A quantum system in a stationary state is represented by the coordinate space wavefunction ψ(x) = Ne ax4 where N is the normalization constant. 4

5 (a) Choosing the zero of potential energy to be V (0) = 0, find the system s potential energy function, V (x) and its associated energy level E. h m d ψ + V (x)ψ = Eψ dx ψ(x) = Ne ax4 ψ (x) = 4ax 3 Ne ax4 ψ (x) = 1ax Ne ax4 + 16a x 6 Ne ax4 h m [16a x 6 1ax ]ψ + V ψ = eψ 6a m h x 8 h a x 6 m + V = E If we set V (0) = 0 when E = 0 V (x) = 8 h a x 6 m 6 h ax m (b) Sketch V (x). To sketch, we set h = m = a = 1. the function to sketch is We have the following limits The zeros of V (x) occur at or V (x) = 8x 6 6x lim V (x) = 8x6 x ± lim V (x) = 6x x 0 x (8x 4 6) = 0 ( ) 3 1/4 ( 3 1/4 x = ±0, ±, ±i 4 4) 5

6 with 4 real roots V (x) x (c) Determine whether ψ(x) represents the ground state of the system. Be sure to explain your answer. The wavefunction has no nodes and must be the ground state wavefunction of the system. 6. Suppose that a harmonic oscillator of mass m and natural frequency ω at time t = 0 is in a non-stationary state given by ( ) 1 1/ Ψ(x, t = 0) = {φ 0 (x) + 3φ 1 (x)} 10 where {φ n (x)} are the complete set of normalized Hamiltonian eigenfunctions for the oscillator. (a) Find Ψ/ t at time t = 0. Ψ t = 1 i hĥψ = 1 1 i h (Ĥφ 0 + 3Ĥφ 1) 10 = 1 ( 1 1 i h hωφ ) hωφ 1 10 = ω ( 1 i 10 φ ) φ 6

7 (b) If the energy of the particle is measured at time t = 0, what possible energies can be obtained and with what probabilities? E = hω, 3 hω P ( hω/) = 1 10 P (3 hω/) = 9 10 (c) What is the average energy of the particle at time t = 0? E = 1 hω hω 10 = 7 5 hω (d) If the position of the particle is measured at time t = 0, what possible positions can be measured and with what relative probabilities? P (x)dx = Ψ(x) dx = 1 ) ( ) α 1/4 e αx 4α 10 ( / 3 1/4 + 3 xe / αx dx (e) Suppose the position of the particle is measured and found to be x = 0 at time t. If the energy is measured at time t instantaneously after the position measurement, what energies can be found? δ(x) = c n φ n n c n = δ(x) φ n (x) dx = φ n (0) Note that c n = 0 for all odd n. We can then measure any E n = (n + 1/) hω for n even. 7. A harmonic oscillator of mass m and natural frequency ω is in its ground Hamiltonian eigenstate. If the momentum of the oscillator is measured, what possible outcomes can be obtained and with what probabilities? ( ) α 1/4 ψ 0 (x) = e αx / φ 0 (p) = 1 h ψ 0 (x)e ipx/ h dx = 1 ( ) α 1/4 e αx / e ipx/ h dx h 7

8 = 1 ( ) α 1/4 ( h α ) 1/ ) exp ( (p/ h) 4α ) = 1 ( ) α 1/4 exp ( p α h α h Any momentum p can be measured in the range < p < with probability P (p)dp = φ 0 (p) dp = 1 ( ) ) α 1/ exp ( p α h α h dp 8. Use creation and annihilation operators to find x p for the n th state of a harmonic oscillator. Use the usual definitions x = [< x > < x > ] 1/ and Compare your result with p = [< p > < p > ] 1/ x p 1 < [ˆx, ˆp] > ( ) 1/ h ˆx = (â + â ) mω = ˆp = i ( ) 1/ mω h (â â ) x = ψ n ˆx ψ n ( ) 1/ h ψ n â + â ψ = 0 mω p = ψ n ˆp ψ n ) 1/ ( hmω = i ψ n â â ψ = 0 x = p = 0 ˆx = h mω [â + (â ) + ââ + â â] = h mω [â + (â ) + ˆN + 1] 8

9 and so that so that ˆp = mω h [â + (â ) ââ â â] = mω h [ ˆN + 1 â (â ) ] x = h (n + 1) mω p = mω h (n + 1) ( x) ( p) = h mω = h (n + 1) 4 mω h (n + 1) x p = h ( (n + 1) = h n + 1 ) Now x p 1 [ˆx, ˆp] = h The inequality is satisfied in general, and we have an equality for n = Show that < ˆT >=< V > for the n th state of the harmonic oscillator. Verify that this result is consistent with the virial theorem [see problem number 9 from problem set 6]. From problem 8 (this problem set) x = h (n + 1) mω p = mω h (n + 1) ˆT = 1 m p = hω (n + 1) 4 V = 1 mω x = hω 4 (n + 1) The virial theorem states that T = V V = x d dx V 9

10 Now or x d dx V = x d dx = mω x 1 mω x T = mω x = V T = V 10. At time t = 0 a harmonic oscillator of mass m and natural frequency ω is in a nonstationary state given by ψ = Axφ where the set {φ n } are the normalized Hamiltonian eigenfunctions for the harmonic oscillator and A is the normalization factor for the wavefunction. If the energy of the oscillator is measured at time t = 0, use creation and annihilation operators to determine what energies can be obtained and with what probabilities. A φ x φ = A h mω φ (â + â ) φ = A h mω φ ˆN â + (â ) φ = = A 5 h mω = 1 ( ) mω 1/ A = 5 h ( ) mω 1/ xφ = c n φ n 5 h n=0 ( ) mω 1/ c n = φ n x φ 5 h ( ) mω 1/ ( ) 1/ h φ n â + â φ 5 h mω = 1 5 φ n ( φ φ 3 ) [see the previous problem] ( ) 1/ ( ) 3 1/ = δ n,1 + δ n,3 5 5 if the energy is measured, we can find E 1 = 3/ hω with probability /5 and E = 7/ hω with probability 3/5. 10

11 11. Let â be an operator and â be its adjoint, such that ââ + â â = 1. Let ψ be an eigenfunction of â â with eigenvalue n ; i.e. â âψ n = nψ n Show that âψ n is an eigenfunction of â â with eigenvalue 1 n. Let ˆN = â â and Now But so that ˆNψ n = nψ n ˆNâ = â ââ â â + ââ = 1 â â = 1 ââ ˆNâ = (1 ââ )â = â(1 â â) = â(1 ˆN) ˆN(âψ n ) = â(1 ˆN)ψ n = â(1 n)ψ n = (1 n)(âψ n ) (âψ n ) is an eigenfunction of ˆN with eigenvalue (1 n). 11