hw_solutions_3 - Homework Solutions # 3 (Liboff Chapter 5)...

hw_solutions_3
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Unformatted text preview: Homework Solutions # 3 (Liboff Chapter 5) 5.2 (a) If we interpret ψ to be a probability amplitude for each neutron, R dx | ψ | 2 = 1. We get A 2 Z a dx ( x- a ) 2 x 2 = A 2 " x 5 5- ax 4 2 + a 2 x 3 3 # a = A 2 a 5 30 = 1 So A = q 30 /a 5 . If we interpret | ψ | 2 to be the neutron density, we should multiply A by √ 1000. Let’s stick with the first. (b) Since ψ is symmetric about x = a/ 2, half the neutrons must be between 0 and a/ 2. 500. (c) Find h 5 | ψ i : s 2 a A Z a dxx ( x- a )sin 5 πx a = s 60 a 6 " 2 x (5 π/a ) 2 sin 5 πx a + 2 (5 π/a ) 3- x 2 5 π/a ! cos 5 πx a- a (5 π/a ) 2 sin 5 πx a + ax 5 π/a cos 5 πx a # a = 8 √ 15 (5 π ) 3 For the probability, P (5) = |h 5 | ψ i| 2 = 6 . 4 × 10- 5 For 1000 particles, this still means 6 . 4 × 10- 2 . (d) Using ˆ H =- ¯ h 2 2 m ∂ 2 ∂x 2 , h E i =- ¯ h 2 A 2 2 m Z a dx ( x 2- ax ) ∂ 2 ∂x 2 ( x 2- ax ) =- 30¯ h 2 ma 5 a 3 3- a 3 2 !...
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