phy-520/hw7.motes
2020-12-23 16:28:58 -05:00

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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.