Midterm Exam Solutions - Fall 2011 - Quantum Mechanics | PHYSICS 115A, Exams of Quantum Mechanics

Download Midterm Exam Solutions - Fall 2011 - Quantum Mechanics | PHYSICS 115A and more Exams Quantum Mechanics in PDF only on Docsity! Physics 115A Fall 11 November 2, 2011 T. Tomboulis MIDTERM EXAM SOLUTIONS 1. [26 points ] A particle in a 1-dimensional infinite square well, 0 ≤ x ≤ L, is described at t = 0 by the wave-function Ψ(x, 0) = C [ 1√ 3 ψ3(x) + √ 3ψ5(x) ] , where ψn(x) are the infinite square well energy eigenfunctions. 1. Determine C so that Ψ(x, 0) is properly normalized. 2. What is the wave-function Ψ(x, t) at a later time t? 3. What is the probability that an energy measurement at some time t yields the value E1? What is the probability that it yields the value E5? 4. Calculate the average energy < H >. 5. What is < x2 >at time t? Note: You do not have to actually evaluate any integrals occurring in your answer. 6. Are there any times t > 0 at which < x2 > equals its value at t = 0? If yes, what are these values of t? We have ψ(x, 0) = ∑ n cn ψn with c3 = C 1√ 3 , c5 = C √ 3 , cn = 0 for n 6= 3, 5. 1. Using orthogonality ∫ dx ψ∗m(x)ψn(x) = δmn , we have 1 = ∫ ∞ −∞ dx ψ∗(x, t)ψ(x, t) = ∑ n |cn|2 = C2 [ 1 3 + 3 ] = C2 10 3 . 1 Hence C = √ 3 10 . 2. Ψ(x, t) = √ 3 10 [ 1√ 3 ψ3(x) e −iE3th̄ + √ 3ψ5(x) e −iE5th̄ ] . 3. For E1: c1 = 0, so Probability = 0. For E5: c5 = 3√ 10 , so Probability = 9 10 . 4. Again by orthogonality and since Ĥψn(x) = Enψ(x), < H >= ∑ n En |cn|2 = 1 10 E3 + 9 10 E5 , where En are the infinite square well energy levels. 5. < x2 > (t) = ∫ ∞ −∞ x2 |Ψ(x, t)|2 dx So: < x2 > (t) = 1 10 ∫ L 0 x2 ψ3(x) 2 dx+ 9 10 ∫ L 0 x2 ψ5(x) 2 dx + 3 5 ( ∫ L 0 x2 ψ3(x)ψ5(x) dx ) cos [ (E5 − E3)t/h̄ ] (1) 6. Yes, whenever the cos factor in (1) is unity, i.e. (E5 − E3)t/h̄ = 2π k , integer k i.e. at times t = T given by T = 2πh̄ (E5 − E3) k = mL2 4πh̄ k , k = 1, 2, . . . using En = π2h̄2n2 2mL2 . 2