𝓗𝓍𝓎 = ½ (-∇² + ρ²).

ρ² = x² + y² and ħ = m = ω = 1.

Griffith's Eq. 2.71:

    Hₙ₊₁(ξ) = 2ξ Hₙ(ξ) - 2n Hₙ₋₁(ξ)




In 2D cartesian coordinates, the del operator is defined

    ∇f = [∂/∂x f, ∂/∂y f].


∇²f = ∇⋅∇f = ∇⋅[∂/∂x f, ∂/∂y f] 
    = ∂²/∂x² f + ∂²/∂y² f.

∴ ∇² = ∂²/∂²x + ∂²/∂²y.

Then,
    
    𝓗𝓍𝓎 = ½ (-(∂²/∂²x + ∂²/∂²y) + (x² + y²)).

𝓗𝓍𝓎 = ½ (-(∂²/∂²x + ∂²/∂²y) + (x² + y²))
    = ½ (x² - ∂²/∂²x + y² - ∂²/∂²y)
    = ½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y).

𝓗𝓍 + 𝓗𝓎 = ½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y).

∎

The Schrodinger Equation then reads,

    [½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y)] Ψ = (E𝓍 + E𝓎) Ψ.

Assuming a separable solution Ψ(x,y) = X(x) Y(y), with E = E𝓍 + E𝓎.

[½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y)] X(x) Y(y) = (E𝓍 + E𝓎) X(x) Y(y).

½(x² - ∂²/∂²x) X(x) Y(y) + ½(y² - ∂²/∂²y) X(x) Y(y) 
    = (E𝓍 + E𝓎) X(x) Y(y).

[1/X(x)] [½(x² - ∂²/∂²x) X(x)] + [1/Y(y)] [½(y² - ∂²/∂²y) Y(y)] 
    = (E𝓍 + E𝓎).

So, I have two differential equations,

    ½(x² - ∂²/∂²x) X(x) = E𝓍 X(x), and
    ½(y² - ∂²/∂²y) Y(y) = E𝓎 Y(y).

The solutions to these differential equations are the same as for the 1D harmonic oscillator. They have eigenvalues (n + 1/2), where ħ = ω = 1, with n = 0,1,2,... .

∴ the eigenvalues for the combined operator are n𝓍 + n𝓎 + 1.

The degeneracy is pretty obvious, just from counting the possibilities: there is n+1 degeneracy for each value of n = n𝓍 + n𝓎.

So,
    n     degeneracy
    ─────────────────
    0          1
    1          2
    2          3
    3          4
    4          5
    5          6


The Hermite polynomials help to generate the eigenstates of this:

Hₙ(x) = (-1)ⁿ exp(x²) d/dxⁿ exp(-x²/2) = (2x - d/dx)ⁿ * 1.

The first six polynomials are

H₀(x) = 1.
H₁(x) = 2x.
H₂(x) = 4x² - 2.
H₃(x) = 8x³ - 12x.
H₄(x) = 16x⁴ - 48x² + 12.
H₅(x) = 32x⁵ - 160x³ + 120x.
H₆(x) = 64x⁶ - 480x⁴ + 720x² - 120.


The wave functions involving these polynomials, with the unitizations given in the intro, are

    Ψₙₘ(x) = π^(-1/4) 1/√(2ⁿ n!) Hₙ(x) exp(-x²/2) 
                π^(-1/4) 1/√(2ᵐ m!) Hₘ(y) exp(-y²/2).

Ψₙₘ(x) = 1/√π 1/√(2ⁿ n! 2ᵐ m!) Hₙ(x) Hₘ(y) exp(-(x²/2 + y²/2)).

For n = {1,...,∞}, the lowest possible energy is 3, and this level has no degeneracy.

For additional levels, 




a𝓍 = 1/√2(x + ιp𝓍)
a𝓎 = 1/√2(y + ιp𝓎)
=======
The lowering operators are

    a𝓍 = 1/√2 (x + ιp𝓍) and
    a𝓎 = 1/√2 (y + ιp𝓎).

They are not hermitian, but x,y and p𝓍,p𝓎 are, so the raising operators are

    a𝓍᛭ = 1/√2 (x - ιp𝓍) and
    a𝓎᛭ = 1/√2 (y - ιp𝓎).

Applying these to the ground state ❙00❭, I can find the first six states, with normalization:

a𝓍᛭❙00❭ = 1/√2 (x - ιp𝓍)❙00❭ 
        = 1/√2 (x❙00❭ - ιp𝓍❙00❭)


    



b) & c)

I'm still working out the algebra, here. I will try to finish it as soon as I can, but I know I also have new work to do.


I finished much of this assignment, but need to get done faster in the future.