Graphical Solution — 1D Symmetric Well
z·tan(z) even
−z·cot(z) odd
√(z₀²−z²)
Parameters
V₀
5.00 eV
z₀
—
Bound states
—
V₀ (well depth)
—
Mode
1D
3D Spherical Well — ℓ = 0 states
In 3D the radial equation for ℓ=0 maps to the odd-parity 1D condition:
−z·cot(z) = √(z₀²−z²)
This requires z₀ > π/2, i.e.
V₀ > — eV
—