Download Solutions to Homework 1 - Quantum Mechanics | PHYS 324 and more Assignments Quantum Mechanics in PDF only on Docsity! Solution to Homework Set # 1 1. λ = h/p and p2/2m = KE = 3kBT/2. Therefore: p = √ 3mkBT = √ 3 × 1.67× 10−27kg × 1.38× 10−23J/K × 300K = 4.55× 10−24kg −m/s Then, λ = (6.626× 10−34J s)/(4.55× 10−24kg −m/s) = 1.46× 10−10 m. Note, this wavelength is comparable to the spacing between atoms in a solid, making thermal neutron beams a useful tool for studying the structure of materials. 2. (Problem 1.7 in the text) (a) ∫ ∞ −∞ ψ(x, 0) ∗ψ(x, 0)dx = 1 = A2 [ ∫ a 0 (x/a) 2dx + ∫ b a (b− x) 2/(b− a)2dx ] Changing variables to y = b− x in the second integral, 1 = A2 [ x3 3a2 ∣∣∣ a 0 + y3 3(b− a)2 ∣∣∣ b−a 0 ] = A2 [a 3 + b− a 3 ] = A2 × b 3 Therefore, A = √ 3/b. (b) ψ(x, 0) starts at 0 when x = 0 and increases linearly to a value of A when x = a. It then decreases linearly from a value of A at x = a to 0 when x = b. (c) The particle is most likely to be found where ψ(x, 0)∗ψ(x, 0) is largest because |ψ(x)|2 is the prob- ability density for finding the particle at x. From part (b), you see that ψ(x, 0)∗ψ(x, 0) is largest at x = a where it takes the value of A2 = 3/b. Therefore, at t = 0, the particle is most likely to found near x = a. (d) If P (x) is a probability density, then the probability of finding x to be between a and b is given by∫ b a P (x)dx. The probability of finding our particle to the left of a is then: P (x < a) = ∫ a −∞ ψ(x, 0)∗ψ(x, 0)dx = A2 ∫ a 0 (x/a)2dx = A2 x3 3a2 ∣∣∣ a 0 = 3 b × a 3 = a b When b = a, P (x < a) = 1 as it should because ψ(x, 0) = 0 for x > b. When b = 2a, ψ(x, 0) is symmetric about x = a, so we expect P (x < a) to be 1/2, in agreement with our result. (e) 〈x(t = 0)〉 = ∫ ∞ −∞ ψ(x, 0) ∗ x ψ(x, 0) dx: 〈x(t = 0)〉 = A2 [ ∫ a 0 x3 a2 dx + ∫ b a x(b− x)2 (b− a)2 dx ] = 3 b [ x4 4a2 ∣∣∣ a 0 + b2x2/2 − 2bx3/3 + x4/4 (b− a)2 ∣∣∣ b a ] = 3 b [a2 4 + 1 (b− a)2 [b2(b2 − a2) 2 − 2b(b 3 − a3) 3 + b4 − a4 4 ]] = 3 b [a2 4 + 1 12(b− a)2 [ b4 − 6b2a2 + 8ba3 − 3a4 ]] = 3 b [a2 4 + 1 12(b− a)2 [ (b− a)2(b2 + 2ba− 3a2) ]] = 3 b [a2 4 + b2 + 2ba 12 − a 2 4 ] = b+ 2a 4 1
