Modern Physics II: Quantum Mechanics, Homework Set 4 Solutions | PHY 373, Assignments of Physics

Download Modern Physics II: Quantum Mechanics, Homework Set 4 Solutions | PHY 373 and more Assignments Physics in PDF only on Docsity! PHY 373 Modern Physics II: Quantum Mechanics, Homework Set 4 Solutions Matthias Ihl 02/28/2007 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/teaching.html We will frequently work in God-given units c = ~ = 1. The casual reader may also want to set 1 = 2 = π = −1. 1 Problem 1 The wave function 〈x|ψ〉 = ψ(x) = Neikxe− x 2 2σ2 (1) desribes a wave packet traveling with momentum k in the positive x-direction. (I changed p → k to avoid confusion! k is arbitrary and has nothing to do with the momentum operator p̂ below.) (a) Normalization: 〈ψ|ψ〉 = ∫ +∞ −∞ dx〈ψ|x〉〈x|ψ〉 = ∫ +∞ −∞ dxψ∗(x)ψ(x) = ∫ +∞ −∞ dxN2e− x 2 σ2 = N2 √ πσ. 1 Therefore, N2 = 1√ πσ . (b) Expectation values: 〈x̂〉ψ = 〈ψ|x̂|ψ〉 = ∫ +∞ −∞ dx〈ψ|x̂|x〉〈x||ψ〉 = ∫ +∞ −∞ dx x|ψ(x)|2 = 0, 〈p̂〉ψ = 〈ψ|p̂|ψ〉 = ∫ +∞ −∞ dxψ∗(x)(−i~) ∂ ∂x ψ(x) = ∫ +∞ −∞ dx 1√ πσ (−i~)(− x σ2 + ik)e −x 2 σ2 = ~k. (c) Uncertainties: √ ∆x2 = √ 〈x2〉 = √ ∫ +∞ −∞ dx x2|ψ(x)|2 = σ√ 2 . √ ∆p2 = √ 〈p2〉 − ~2k2 = √ ( ∫ +∞ −∞ dx (−i~)2√ πσ (− x σ2 + ik)2e −x2 σ2 ) − ~2k2 = √ ~2k2 + ( ∫ +∞ −∞ dx ~2√ πσ5 x2e −x2 σ2 ) − ~2k2 = ~√ 2σ . (d) Probability to find particle on positive x-axis: ∫ +∞ 0 dx |ψ(x)|2 = 1 2 . (2) 2 Problem 2 (a) Normalization: We assume that 〈0|0〉 = 1 is properly normalized. Then 〈n|n〉 = 〈0| a n √ n! a†n√ n! |0〉 = 1 n! 〈0|ana†n|0〉 = 1 n! ( 〈0|an−1a†na|0〉 + 〈0|an−1[a, a†n]|0〉 ) = 1 n! 〈0|an−1[a, a†n]|0〉 = n n! 〈0|an−1a†n−1|0〉 = 1 (n− 1)!〈0|a n−1a†n−1|0〉. since ψ1(x) and ψ2(x) are orthonormal. (b) With ω = π 2~ 2ma2 , we have Ψ(x, t) = 1√ 2 √ 2 a ( sin( π a x)e−iωt + sin( 2π a x)e−i4ωt ) (8) Moreover, following Example 2.1 on page 29, |Ψ(x, t)|2 = 1 a ( sin2( π a x) + sin2( 2π a x) ) + 2 a sin( π a x) sin( 2π a x) cos(3ωt). (9) (c)Expectation value for x: 〈x〉 = ∫ a 0 dx x|Ψ(x, t)|2 = a 2 − 16a 9π2 cos(3ωt). (10) (d)Expectation value for p: 〈p〉 = ∫ a 0 dx Ψ∗(−i~ ∂ ∂x )Ψ(x, t) = 8~ 3a sin(3ωt). (11) (e)We want to find the energy eigenvalues of ψ1(x) and ψ2(x): 〈x|H|ψi〉 = Ei〈x|ψi〉. (12) With H = − ~2 2m ∂2 ∂x2 , the eigenvalues can be calculated to be E1 = ~ 2 2m π2 a2 , (13) E2 = ~ 2 2m 4π2 a2 . (14) The probabilities of finding either E1 or E2 are given by |〈ψ1|Ψ〉|2 = | ∫ a 0 dx ψ∗1(x, t)Ψ(x, t)|2 = 1 2 , (15) |〈ψ2|Ψ〉|2 = | ∫ a 0 dx ψ∗2(x, t)Ψ(x, t)|2 = 1 2 . (16) Expectation value of H : 〈H〉Ψ = ( − ~ 2 2m ) ∫ a 0 dx Ψ∗(x, t) ∂2 ∂x2 Ψ(x, t) = 5 2 ~ 2 2m π2 a2 = 1 2 E1 + 1 2 E2. (17) So the expectation value of H turns out to be the weighted sum of the energy eigenvalues, as expected. 4 Problem 4 (−i~ ∂ ∂x )n〈x|ψ〉 = ∫ +∞ −∞ dp (−i~ ∂ ∂x )n〈x|p〉〈p|ψ〉 = ∫ +∞ −∞ dp (−i~ ∂ ∂x )ne i ~ px〈p|ψ〉 = ∫ +∞ −∞ dp (−i~)n( ip ~ )n〈x|p〉〈p|ψ〉 = ∫ +∞ −∞ dp pn〈x|p〉〈p|ψ〉 = ∫ +∞ −∞ dp 〈x|p̂n|p〉〈p|ψ〉 = 〈x|p̂n|ψ〉.