Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chemistry 532 Problem Set 8 Fall, 2012 Solutions 1. Evaluate the commutators [L̂i , x2 + y 2 + z 2 ] where i = x, y or z. [Hint: This problem can be done in two lines if you think.] Answer: [L̂i , x2 + y 2 + z 2 ] = [L̂i , r2 ] = 0 because L̂i is independent of r for all i. 2. Let ψ be an eigenfunction of L̂z . Show that < ψ|L̂x |ψ >=< ψ|L̂y |ψ >= 0. Answer: Let L̂z ψ = mh̄ψ We have L̂x = Then hψ|L̂x |ψi = 1 [L̂y , L̂z ] ih̄ 1 hψ|[L̂y , L̂z ]|ψi ih̄ 1 hψ|L̂y L̂z − L̂z L̂y |ψi ih̄ mh̄ = hψ|L̂y − L̂y |ψi = 0 ih̄ = We also have L̂y = Then hψ|L̂y |ψi = 1 [L̂z , L̂x ] ih̄ 1 hψ|[L̂z , L̂x ]|ψi ih̄ 1 hψ|L̂z L̂x − L̂x L̂z |ψi ih̄ mh̄ = hψ|L̂x − L̂x |ψi = 0 ih̄ = 1 3. Show that if any operator,  commutes with two components of the angular momentum, it commutes with the third. Answer: Let [Â, L̂x ] = [Â, L̂y ] = 0. We must show that [Â, L̂z ] = 0 as well. Now 1 [Â, [L̂z , L̂y ]] ih̄ [Â, L̂z ] = = 1 {ÂL̂x L̂y − ÂL̂y L̂x − L̂x L̂y  + L̂y L̂x Â} ih̄ 1 {ÂL̂x L̂y − ÂL̂y L̂x − ÂL̂x L̂y + ÂL̂y L̂x } = 0 ih̄ ~ The same proof holds for the other components of L. = 4. Show that the spherical harmonics are eigenfunctions of the operator L̂2x + L̂2y , and determine the eigenvalues. This problem should take about 4 lines of math. Answer: L̂2 = L̂2x + L̂2y + L̂2z L̂2x + L̂2y = L̂2 − L̂2z (L̂2x + L̂2y )Y`m = (L̂2 − L̂2z )Y`m = h̄2 [`(` + 1) − m2 ]Y`m The eigenvalues are h̄2 [`(` + 1) − m2 ]. 5. Calculate the first two non-vanishing terms in the expansion of the function ( F (x) = −1 −1 ≤ x ≤ 0 +1 0<x≤1 in Legendre polynomials. [See Table 6-3 on page 283 of the text.] Answer: ∞ X F (x) = A`0 P`0 (x) `0 =0 Z 1 −1 ∞ X F (x)P` (x) dx = A` 0 Z 1 `0 =0 = ∞ X A` 0 `0 =0 = A` 2`0 −1 P`0 (x)P` (x) dx 2 δ`,`0 +1 2 2` + 1 2` + 1 Z 1 A` = F (x)P` (x) dx 2 −1 2 Z 1 1Z 1 1 Z0 A0 = F (x) dx = [− dx + dx] = 0 2 −1 2 −1 0 Z 1 3 3Z 1 x dx = 3 x dx = A1 = 2 −1 2 0 Z 1 51 (3x2 − 1)F (x) dx = 0 A2 = 2 2 −1 7Z 1 3 5 3 A3 = x − x F (x) dx 2 −1 2 3 Z 1 3 5 3 =7 x − x dx 0 2 3 7 21 5 1 − =− = 2 12 2 8 Then 3 7 F (x) = P1 (x) − P3 (x) + . . . 2 8 6. Schmidt orthogonalize the set of functions {1, x, x2 , x3 , . . .} on the interval −1 ≤ x ≤ 1, and show that the first three members of the orthogonal set at proportional to the first three Legendre polynomials. Answer: v1 = 1 P0 (x) = 1 v2 = x + av1 hv1 |v2 i = Z 1 Z 1 x dx + a −1 dx = 0 −1 0 + 2a = 0 v2 = x a=0 P1 (x) = x v3 = x2 + bx + c = x2 + bv2 + cv1 hv1 |v3 i = c=− Z 1 2 x dx + c −1 Z 1 x3 dx + b −1 1 v3 = x − 3 Z 1 x2 dx = 0 −1 2 0+ b=0 3 2 dx = 0 −1 12 1 1Z 1 2 x dx = − =− 2 −1 23 3 hv2 |v3 i = Then Z 1 b=0 3 2 1 3 1 P2 (x) = (3x2 − 1) = x − = v3 2 2 3 2 3 7. Determine uncertainty relations between the three components of the linear momentum and the three components of the angular momentum of a system. Answer: 1 ∆px ∆Lx ≥ |h[p̂x , L̂x ]i| 2 # " ∂ ∂ ∂ 2 ,y −z =0 [p̂x , L̂x ] = −h̄ ∂x ∂z ∂y Then ∆px ∆Lx = 0 1 ∆px ∆Ly ≥ |h[p̂x , L̂y ]i| 2 # " ∂ ∂ ∂ 2 ,z −x [p̂x , L̂y ] = −h̄ ∂x ∂x ∂z 2 = h̄ ∂ ∂ ∂ ∂ x −x ∂x ∂z ∂z ∂x = h̄2 ! ∂ = ih̄p̂z ∂z Then ∆px ∆Ly ≥ h̄ |hpz i| 2 Next, ∂ ∂ ∂ ,x −y ∂x ∂y ∂x 2 [p̂x , L̂z ] = −h̄ = −h̄2 ! ∂ = −ih̄p̂y ∂y so that ∆px ∆Lz ≥ h̄ |hpy i| 2 In a similar way ∆py ∆Lx ≥ h̄ |hpz i| 2 ∆py ∆Ly = 0 ∆pz ∆Lx ≥ h̄ |hpy i| 2 ∆pz ∆Ly ≥ and ∆py ∆Lz ≥ h̄ |hpx i| 2 h̄ |hpx i| 2 ∆pz ∆Lz = 0 8. A particle of mass m is confined to move on the surface of a sphere of radius R. The state of the particle at time t = 0 is described by the wavefunction Ψ(θ, φ) = 1/2 1 3 [Y11 (θ, φ) + Y10 (θ, φ) + Y1−1 (θ, φ)] where Ylm (θ, φ) are the spherical harmonics. 4 (a) If Lz is measured for this system at time t = 0, what are the possible outcomes of the measurement and with what relative probabilities? Answer: Lz = −h̄, 0, h̄ each with probability P = 1 √ 3 !2 = 1 3 (b) If L2 is measured for this system at time t = 0, what are the possible outcomes of the measurement and with what relative probabilities? Answer: L2 = h̄2 1(1 + 1) = 2h̄2 P =1 (c) What is the average value of Lx over Ψ(θ, φ)? Answer: L̂+ = L̂x + iL̂y L̂− = L̂x − iL̂y so that 1 L̂x = (L̂+ + L̂− ) 2 1 1 hψ|L̂x |ψi = Y11 + Y10 + Y1−1 (L̂+ + L̂− ) Y11 + Y10 + Y1−1 3 2 √ √ √ √ h̄ = hY11 + Y10 + Y1−1 |( 2|Y11 i + 2|Y10 i + 2|Y10 i + 2|Y1−1 i) 6 √ h̄ √ 2 2h̄ = 4 2= 6 3 9. The quantum mechanical Hamiltonian operator for a two-dimensional particle of mass m confined to move on the perimeter of a ring of radius R is given by Ĥ = − h̄2 d2 2mR2 dφ2 √ (a) Show that the set of functions {1/ 2π exp(inφ)} form a normalized complete set of simultaneous eigenfunctions of the Hamiltonian and the z-component of the angular momentum. Impose boundary conditions on the set of wavefunctions to determine the possible values of the quantum number n. Answer: h̄ d inφ L̂z einφ = e = nh̄einφ i dφ inφ Ĥe h̄2 d2 inφ n2 h̄2 inφ =− e = e 2mR2 dφ2 2mR2 einφ = ein(φ+2π) = einφ ein2π 5 Then ein2π = 1 n = 0, ±1, ±2, . . . Z 2π 2π 1 e−inφ einφ dφ = =1 2π 0 2π (b) Suppose at time t = 0 the quantum state of the particle is described by the wavefunction ψ(φ) = N cos2 (φ) where N is the normalization. If the z-component of the angular momentum and the energy of the particle are measured simultaneously at time t = 0, what values can be obtained for each observable and with what probabilities? Answer: 2π Z 2π 6π 2 4 2 3φ N cos φ dφ = N =1 = N2 8 0 8 0 4 N= 3π 4 3π 1/2 cos2 φ = 1/2 4 6 cn = = 1/2 4 6 1/2 ∞ X 1 cn √ einφ 2π n=−∞ 1 Z 2π cos2 φe−inφ dφ π 0 1 Z 2π 1 iφ (e + e−iφ )2 e−inφ dφ π 0 4 1/2 2 1 1 = δn,2 + δn,−2 + δn,0 3 2 2 Then for Lz we can measure 2h̄ or −2h̄ with probabilities of 1/6 each or we can find Lz = 0 with probability 2/3. The energy levels are En = n2 h̄2 2mR2 We can obtain from an energy measurement 4h̄2 E= 2mR2 i.e. n = 2 with probability 1/3 and E = 0 with probability 2/3. 10. For the 1s ground state of the hydrogen atom, find (a) hri Answer: ψ1s (r) = 6 1 πa30 !1/2 e−r/a0 1 Z ∞ Z π Z 2π hri = 3 dr dθ dφ r2 sin θe−r/a0 re−r/a0 πa0 0 0 0 4π Z ∞ dr r3 e−2r/a0 = πa0 0 4 a30 = (b) h1/ri Answer: 1 r 3 = a0 2 = 4 a30 ! a20 4 ! = 1 a0 ! 4 Z∞ dr e−2r/a0 a30 0 = = (d) hT i Answer: ! = (c) h1/r2 i Answer: 1 r2 6a40 24 ! 4 Z∞ dr r e−2r/a0 a30 0 ! 4 a30 ! a0 2 = 2 a20 h̄2 ∂ 2 ∂ L̂2 h̄2 2 r + − ∇ =− 2µ 2µr2 ∂r ∂r 2µr2 L̂2 ψ1s (r) = 0 d2 2 d 1 d 2d r = + 2 2 r dr dr dr r dr Then ! T̂ ψ1s (r) = Now and Then d2 2 d + ψ1s (r) 2 dr r dr 1 d −r/a0 = − e−r/a0 e dr a0 1 d2 −r/a0 e = 2 e−r/a0 2 dr a0 h̄2 h̄2 2 −r/a0 − ∇e =− 2µ 2µ 7 ! 1 2 − e−r/a0 2 a0 ra0 and h̄2 h̄2 hT i = − hψ1s |ψ1s i + 2µa20 2µa0 2 2 h̄ h̄2 h̄ =− + = 2µa20 µa20 2µa20 1 r 11. An electron in a hydrogen atom occupies a 3dxy orbital. If Lz , the z-component of the angular momentum of the electron is measured, determine what values can be obtained and with what probabilities? Answer: !1/2 1 1 Ψ3dxy (r, θ, φ) = √ r2 e−r/3a0 sin2 θ sin 2φ 7 81 2π a0 Because h̄ ∂ L̂z = i ∂φ we need only focus on the φ part of the wavefunction; i.e. we take ψ(φ) = A sin 2φ Normalizing ψ we have A2 Z 2π 0 1 Z 4π 1 y 1 dφ sin2 2φ = A2 − sin 2y dy sin2 y = A2 2 2 2 4 0 4π 0 2 =A π=1 or 1 ψ(φ) = √ sin 2φ π √ The normalized eigenfunctions of L̂z are {1/ 2πeinφ }, n = 0, ±1, ±2, . . .. Then ∞ X 1 1 √ sin 2φ = cn √ einφ π 2π n=−∞ or 1 Z 2π −inφ cn = √ e sin 2φ dφ π 2 0 1 Z 2π −inφ 1 i2φ [e − e−i2φ ] dφ = √ e 2i π 2 0 1 Z 2π iφ(2−n) √ = [e − e−iφ(2+n) ] dφ 2iπ 2 0 2π √ [δn,2 − δn,−2 ] = 2πi 2 1 = √ [δn,2 − δn,−2 ] i 2 Then a measurement of Lz can find Lz = ±2h̄ each with a probability of 1/2. 8