Download Solutions to Modern Physics II: Quantum Mechanics Homework Set 2 and more Assignments Physics in PDF only on Docsity! PHY 373 Modern Physics II: Quantum Mechanics, Homework Set 2 Solutions Matthias Ihl 09/10/2007 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 2.1 (a) We have: 1 = ∫ +∞ −∞ dx|Ψ(x, t)|2 = e2Γt/~ ∫ +∞ −∞ dx|ψ(x)|2. (1) Since the second term is independent of t, the first term also has to be time- independent in order for the product to be 1. Therefore Γ = 0. (b) Since V,E are real, it follows that if ψ(x) is a solution to the Schrödinger eq. (2.5), ψ∗ also is a solution. Since eq. (2.5) is a linear equation, any linear combination of solutions is again a solution. More specifically, both Re(ψ)= (ψ + ψ∗)/2 and Im(ψ)= (ψ − ψ∗)/2i are solutions of eq. (2.5). For any complex solution, we can always consider two equivalent real ones. Therefore we can restrict ourselves to real ψ(x). (c) Noting that ∂2/∂(−x)2 = ∂2/∂x2, we observe that, if ψ(x) is a solution to (2.5), ψ(−x) also is a solution as long as V (x) = V (−x). Therefore the even and odd linear combinations ψ±(x) = ψ(x) ± ψ(−x) (2) are also solutions. But since ψ = 1 2 (ψ+ + ψ−), any solution can be expressed in terms of even and odd solutions. 1 2 Problem 2.5 (a) Using the orthonormality of the ψns, we get 1 = ∫ |Ψ|2dx = |A|2 ∫ (|ψ1|2 + |ψ1|2) = 2|A|2. (3) Thus, A = 1/ √ 2. (b) With proper normalization, we find Ψ(x, t) = 1√ 2 ( ψ1e −iE1t/~ + ψ2e −iE2t/~ ) = 1√ a eiωt ( sin( πx a ) + sin( 2πx a )e−3iωt ) , (4) |Ψ(x, t)|2 = 1 a ( sin2( πx a ) + sin2( 2πx a ) + 2 sin( πx a ) sin( 2πx a ) cos(3ωt) ) . (5) (c) Expectation value of x: 〈x〉 = ∫ x|Ψ(x, t)|2dx = a 2 − 16a 9π2 cos(3ωt), (6) after solving a lengthy integral. The amplitude is 32 9π2 a 2 < a 2 and the angular frequency is 3ω. (d) Expectation value of p (ze kveek vay): 〈p〉 = md〈x〉 dt = 16amω 3π2 sin(3ωt) = 8~ 3a sin(3ωt), (7) after plugging in ω = π 2~ 2ma2 . This has the correct units, since a appears in the denominator. (e) Expectation value of H : 〈H〉 = 1 2 (E1 + E2) = 5π2~2 4ma2 , (8) which is the average of E1 and E2. The probability for getting E1 or E2 is given by |c1|2 = 1/2 and |c2|2 = 1/2, resp.
