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hw_solutions_2

# hw_solutions_2 - Homework Solutions 2(Libo Chapter 4 4.5(a...

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Homework Solutions # 2 (Liboff Chapter 4) 4.5 (a) Given in book. (b) | ϕ ψ | f = a | ϕ , where a = dx ψ * f . (c) f | ϕ ψ | f . (d) f | ϕ ψ | ψ = f | ϕ . 4.8 g ( x ) = 2 a n a n sin nπx a where a n = 2 a a 0 dx sin nπx a x ( x - a ) e ikx = 1 2 i 2 a a 0 dx ( x 2 - ax ) e i ( k + a ) x - e - i ( k + a ) x Now the tedious integration. Use I ( α ) = a 0 dx e iαx = 1 e iaα - 1 a 0 dx x e iαx = - i ∂I ∂α = 1 α 2 e iaα - 1 - ia α e iaα a 0 dx x 2 e iαx = - 2 I ∂α 2 = 2 i α 3 e iaα - 1 + 2 a α 2 e iaα - ia 2 α e iaα Afer a bunch of algebra I’m not going to reproduce here, we get a n = 2 a ( - 1) n e ika - 1 1 ( k + nπ/a ) 3 ) - 1 ( k + nπ/a ) 3 ) - ia ( - 1) n e ika 1 ( k + nπ/a ) 2 - 1 ( k - nπ/a ) 2 - a 2 i ( - 1) n e ika - 1 1 ( k + nπ/a ) 2 - 1 ( k - nπ/a ) 2 Phys 580 HW# 2 Solutions 1

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4.9 || ψ + ϕ || 2 = ψ + ϕ | ψ + ϕ = ( ψ | + ϕ | )( | ψ + | ϕ ) = ψ | ψ + ϕ | ψ + ψ | ϕ + ϕ | ϕ Orthogonality means ϕ | ψ = 0, so || ψ + ϕ || 2 = ψ | ψ + ϕ | ϕ = || ψ || 2 + || ϕ || 2 4.12 (a) i ( ˆ A ˆ B - ˆ B ˆ A ) = - i ( ˆ B ˆ A - ˆ A ˆ B ) = i ( ˆ A ˆ B - ˆ B ˆ A ), so Hermitian. (b) Same, except for no factor of i , so it’s anti-Hermitian ( ˆ O = - ˆ O ).
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