hw_solutions_3

# hw_solutions_3 - Homework Solutions 3(Libo Chapter 5 5.2(a...

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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 0 dx ( x - a ) 2 x 2 = A 2 " x 5 5 - ax 4 2 + a 2 x 3 3 # a 0 = 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 0 dx x ( 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 0 = 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 0 dx ( x 2 - ax ) 2 ∂x 2 ( x 2 - ax ) = - 30¯ h 2 ma 5 a 3 3 - a 3 2 ! = h 2 ma 2 Note that this is just barely above E 1 = ¯ h 2 π 2 / 2 ma 2 . Phys 580 HW# 3 Solutions 1

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5.11 We start with the wavefunction ψ = q 2 a sin πx a , but with the wall now moved to x = 2 a , and the new energy eigenstates φ n = q 1
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