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).

    Recall ξ = √(mω/ħ) x.

    dξ = √(mω/ħ) dx = α dx.

    ψ(ξ) = ξᵖ exp(-ξ²/2) L(ξ²).

    ψ'(ξ) = d/dξ (ξᵖ exp(-ξ²/2) L(ξ²))
          =  (d/dξ ξᵖ) exp(-ξ²/2) L(ξ²)
            + ξᵖ (d/dξ exp(-ξ²/2)) L(ξ²)
            + ξᵖ exp(-ξ²/2) (d/dξ L(ξ²)).

    For an even state, ψ'(0) = 0 and p=0, so
    
    ψ'(ξ) = (d/dξ 1) exp(-ξ²/2) L(ξ²)
            + 1 d/dξ exp(-ξ²/2) L(ξ²)
            + 1 exp(-ξ²/2) d/dξ L(ξ²).
    
    ψ'(ξ) = - ξ exp(-ξ²/2) L(ξ²)
            + exp(-ξ²/2) L'(ξ²).
    
    ψ'(0) = L'(0).
    
    This must be wrong, somehow... in this case the boundary condition only applies if L'(0) = 0.
    
    ∎
    
    For an odd state, ψ(0) = 0 and p=1, so
    
    ψ(0) = 0ᵖ * ... = 0. This is trivial.
    
    ∎
    

b) Equation 2.72 from Griffiths:

    dHₙ/dξ = 2nHₙ₋₁(ξ).

Substituting the state ψ(ξ), I can obtain the associated Laguerre differential equation

    uL'' + (p - ½ + 1 - u)L' + kL = 0, with u = ξ².

A better and equivalent substitution uses
    ψ(x,t) = A exp(ι(kx - (ħk²/2m)t)).

    h = ξᵖ L(ξ²).

    u = ξ².

    ξᵖ  = ξ² 

    h = ξᵖ L(ξ²)

    ψ(x,t) = A exp( ι(kx - k²/4πm ξᵖ L(ξ²) t) )
           = A exp( ι(kx - k²/4πm ξᵖ L(u) t) ).

    This... is probably not the "h" you meant.

    h = ξᵖ L(ξ²).

    dHₙ/dξ = 2nHₙ₋₁(ξ).

    h'(ξ) = dh/dξ = pξᵖ⁻¹ L(ξ²) + ξᵖ L'(ξ²).

    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).

    p(p-1)ξᵖ⁻² L(ξ²) + pξᵖ⁻¹ L'(ξ²) + pξᵖ⁻¹ L'(ξ²) + ξᵖ L''(ξ²).    

    Hₙ in Griffiths maps to hₚ in the problem sheet, I'll assume.

    Hₙ = ξⁿ.