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128 lines
3.3 KiB
Plaintext
128 lines
3.3 KiB
Plaintext
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𝓗𝓍𝓎 = ½ (-∇² + ρ²).
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ρ² = x² + y² and ħ = m = ω = 1.
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Griffith's Eq. 2.71:
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Hₙ₊₁(ξ) = 2ξ Hₙ(ξ) - 2n Hₙ₋₁(ξ)
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In 2D cartesian coordinates, the del operator is defined
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∇f = [∂/∂x f, ∂/∂y f].
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∇²f = ∇⋅∇f = ∇⋅[∂/∂x f, ∂/∂y f]
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= ∂²/∂x² f + ∂²/∂y² f.
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∴ ∇² = ∂²/∂²x + ∂²/∂²y.
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Then,
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𝓗𝓍𝓎 = ½ (-(∂²/∂²x + ∂²/∂²y) + (x² + y²)).
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𝓗𝓍𝓎 = ½ (-(∂²/∂²x + ∂²/∂²y) + (x² + y²))
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= ½ (x² - ∂²/∂²x + y² - ∂²/∂²y)
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= ½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y).
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𝓗𝓍 + 𝓗𝓎 = ½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y).
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∎
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The Schrodinger Equation then reads,
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[½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y)] Ψ = (E𝓍 + E𝓎) Ψ.
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Assuming a separable solution Ψ(x,y) = X(x) Y(y), with E = E𝓍 + E𝓎.
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[½(x² - ∂²/∂²x) + ½(y² - ∂²/∂²y)] X(x) Y(y) = (E𝓍 + E𝓎) X(x) Y(y).
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½(x² - ∂²/∂²x) X(x) Y(y) + ½(y² - ∂²/∂²y) X(x) Y(y)
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= (E𝓍 + E𝓎) X(x) Y(y).
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[1/X(x)] [½(x² - ∂²/∂²x) X(x)] + [1/Y(y)] [½(y² - ∂²/∂²y) Y(y)]
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= (E𝓍 + E𝓎).
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So, I have two differential equations,
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½(x² - ∂²/∂²x) X(x) = E𝓍 X(x), and
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½(y² - ∂²/∂²y) Y(y) = E𝓎 Y(y).
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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,... .
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∴ the eigenvalues for the combined operator are n𝓍 + n𝓎 + 1.
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The degeneracy is pretty obvious, just from counting the possibilities: there is n+1 degeneracy for each value of n = n𝓍 + n𝓎.
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So,
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n degeneracy
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─────────────────
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0 1
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1 2
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2 3
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3 4
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4 5
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5 6
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The Hermite polynomials help to generate the eigenstates of this:
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Hₙ(x) = (-1)ⁿ exp(x²) d/dxⁿ exp(-x²/2) = (2x - d/dx)ⁿ * 1.
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The first six polynomials are
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H₀(x) = 1.
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H₁(x) = 2x.
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H₂(x) = 4x² - 2.
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H₃(x) = 8x³ - 12x.
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H₄(x) = 16x⁴ - 48x² + 12.
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H₅(x) = 32x⁵ - 160x³ + 120x.
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H₆(x) = 64x⁶ - 480x⁴ + 720x² - 120.
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The wave functions involving these polynomials, with the unitizations given in the intro, are
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Ψₙₘ(x) = π^(-1/4) 1/√(2ⁿ n!) Hₙ(x) exp(-x²/2)
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π^(-1/4) 1/√(2ᵐ m!) Hₘ(y) exp(-y²/2).
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Ψₙₘ(x) = 1/√π 1/√(2ⁿ n! 2ᵐ m!) Hₙ(x) Hₘ(y) exp(-(x²/2 + y²/2)).
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For n = {1,...,∞}, the lowest possible energy is 3, and this level has no degeneracy.
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For additional levels,
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a𝓍 = 1/√2(x + ιp𝓍)
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a𝓎 = 1/√2(y + ιp𝓎)
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=======
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The lowering operators are
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a𝓍 = 1/√2 (x + ιp𝓍) and
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a𝓎 = 1/√2 (y + ιp𝓎).
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They are not hermitian, but x,y and p𝓍,p𝓎 are, so the raising operators are
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a𝓍᛭ = 1/√2 (x - ιp𝓍) and
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a𝓎᛭ = 1/√2 (y - ιp𝓎).
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Applying these to the ground state ❙00❭, I can find the first six states, with normalization:
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a𝓍᛭❙00❭ = 1/√2 (x - ιp𝓍)❙00❭
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= 1/√2 (x❙00❭ - ιp𝓍❙00❭)
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b) & c)
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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.
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I finished much of this assignment, but need to get done faster in the future.
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