Perturbed Harmonic Oscillator
Author: R. Tobin

Shows the effect of an approximately delta-function perturbation on the energies and wavefunctions of the two lowest-lying states of a harmonic oscillator.  The slider controls the strength (and sign) of the potential.

\frac{V_0}{\hbar \omega}

Show wavefunctions

Coeff(n) = sqrt(2^(n) * n!)
H0(x) = 1
H1(x) = 2 * x
H2(x) = 4 * x^(2) - 2
H3(x) = 8 * x^(3) - 12 * x
H4(x) = 16 * x^(4) - 48 * x^(2) + 12
Phi0(x) = H0(x) * ℯ^((-x^(2)) / 2) / Coeff(0)
Phi1(x) = H1(x) * ℯ^((-x^(2)) / 2) / Coeff(1)
Phi2(x) = H2(x) * ℯ^((-x^(2)) / 2) / Coeff(2)
α = 0 (slider from -0.5 to 0.5, step 0.01)
w = 0.1
V2(x) = α / w * ℯ^(-(x / w)^(2)) / sqrt(π)
V(x) = 0.5 * x^(2) + V2(x)
E0(x) = 0.5 + α / sqrt(π)
E1(x) = 1.5
E2(x) = 2.5 + α / (2 * sqrt(π))
Phi4(x) = H4(x) * ℯ^((-x^(2)) / 2) / Coeff(4)
Scale = 0.8
Phi0u(x) = E0(x) + Scale * Phi0(x)
Phi2u(x) = E2(x) + Scale * Phi2(x)
a = Integral(V2, -0.5, 0.5)
b = false
Phi2p(x) = E2(x) + Scale * (Phi2(x) - α / sqrt(8 * π) * Phi0(x) + α / 2 * sqrt(3 / (16 * π)) * Phi4(x))
Phi0p(x) = E0(x) + Scale * (Phi0(x) + α / sqrt(8 * π) * Phi2(x) - α / 4 * sqrt(3 / (8 * π)) * Phi4(x))