mirror of
https://asciireactor.com/otho/phy-521.git
synced 2024-11-22 14:05:06 +00:00
128 lines
3.3 KiB
Plaintext
128 lines
3.3 KiB
Plaintext
|
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.
|