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79 lines
2.2 KiB
Plaintext
79 lines
2.2 KiB
Plaintext
Problem 1
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a) The wave function ψ(ξ) = ξᵖ exp(-ξ²/2) L(ξ²), should satisfy the boundary condition for either even or odd fuctions of ξ with p=0 or p=1, respectively, for any function L(u).
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Recall ξ = √(mω/ħ) x.
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dξ = √(mω/ħ) dx = α dx.
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ψ(ξ) = ξᵖ exp(-ξ²/2) L(ξ²).
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ψ'(ξ) = d/dξ (ξᵖ exp(-ξ²/2) L(ξ²))
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= (d/dξ ξᵖ) exp(-ξ²/2) L(ξ²)
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+ ξᵖ (d/dξ exp(-ξ²/2)) L(ξ²)
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+ ξᵖ exp(-ξ²/2) (d/dξ L(ξ²)).
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For an even state, ψ'(0) = 0 and p=0, so
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ψ'(ξ) = (d/dξ 1) exp(-ξ²/2) L(ξ²)
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+ 1 d/dξ exp(-ξ²/2) L(ξ²)
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+ 1 exp(-ξ²/2) d/dξ L(ξ²).
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ψ'(ξ) = - ξ exp(-ξ²/2) L(ξ²)
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+ exp(-ξ²/2) L'(ξ²).
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ψ'(0) = L'(0).
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This must be wrong, somehow... in this case the boundary condition only applies if L'(0) = 0.
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∎
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For an odd state, ψ(0) = 0 and p=1, so
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ψ(0) = 0ᵖ * ... = 0. This is trivial.
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∎
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b) Equation 2.72 from Griffiths:
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dHₙ/dξ = 2nHₙ₋₁(ξ).
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Substituting the state ψ(ξ), I can obtain the associated Laguerre differential equation
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uL'' + (p - ½ + 1 - u)L' + kL = 0, with u = ξ².
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A better and equivalent substitution uses
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ψ(x,t) = A exp(ι(kx - (ħk²/2m)t)).
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h = ξᵖ L(ξ²).
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u = ξ².
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ξᵖ = ξ²
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h = ξᵖ L(ξ²)
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ψ(x,t) = A exp( ι(kx - k²/4πm ξᵖ L(ξ²) t) )
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= A exp( ι(kx - k²/4πm ξᵖ L(u) t) ).
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This... is probably not the "h" you meant.
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h = ξᵖ L(ξ²).
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dHₙ/dξ = 2nHₙ₋₁(ξ).
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h'(ξ) = dh/dξ = pξᵖ⁻¹ L(ξ²) + ξᵖ L'(ξ²).
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h''(ξ) = pξᵖ⁻¹ L(ξ²) + ξᵖ L'(ξ²) = p(p-1)ξᵖ⁻² L(ξ²) + pξᵖ⁻¹ L'(ξ²) + pξᵖ⁻¹ L'(ξ²) + ξᵖ L''(ξ²).
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p(p-1)ξᵖ⁻² L(ξ²) + pξᵖ⁻¹ L'(ξ²) + pξᵖ⁻¹ L'(ξ²) + ξᵖ L''(ξ²).
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Hₙ in Griffiths maps to hₚ in the problem sheet, I'll assume.
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Hₙ = ξⁿ.
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