Bound states: infinite and finite square wells

Where V is constant, the time-independent equation reads ψ″ = −(2m/ħ²)(E − V)ψ, a constant-coefficient second-order equation, so each flat region admits just two independent solutions. Where E exceeds V, ψ = A sin kx + B cos kx with k = √(2m(E − V))/ħ, and ψ curves back towards the axis. Where E falls short of V, ψ = C e^(κx) + D e^(−κx) with κ = √(2m(V − E))/ħ, and ψ curves away from it. Solving a square well means writing one such form per region and stitching them together. At a finite step ψ″ jumps by a finite amount, so ψ and ψ′ are both continuous; at an infinite step only ψ = 0 survives. In each outer region normalizability deletes the exponential that grows away from the well. A finite well then has four amplitudes tied by four homogeneous matching equations, and a non-zero solution exists only when their determinant vanishes: one equation for E, with discrete roots.

Put rigid walls at x = 0 and x = L. ψ(0) = 0 removes the cosine, and ψ(L) = 0 needs sin kL = 0, so kL = nπ: exactly n half-wavelengths fit, λₙ = 2L/n, as on a string fixed at both ends. The energy is all kinetic, E = ħ²k²/(2m), so Eₙ = n²π²ħ²/(2mL²) = n²h²/(8mL²). n = 0 gives ψ ≡ 0, no particle at all, and negative n only repeats states with a sign change. Normalizing, ∫₀ᴸ A² sin²(nπx/L) dx = A²L/2 = 1, so A = √(2/L). For an electron in a 1.00 nm box, E₁ = 0.376 eV, E₂ = 1.504 eV and E₃ = 3.384 eV; the 2 → 1 photon carries 1.128 eV, a wavelength of 1099 nm. Because E ∝ 1/(mL²), halving the width quadruples every level, and a proton in the same box sits 1836 times lower.

E₁ = h²/(8mL²) lies above the floor. A particle at rest on the bottom would need a ψ with no curvature that still vanishes at both walls, and only ψ ≡ 0 manages that. The uncertainty principle puts a number on it: in ψ₁, Δx = L√(1/12 − 1/(2π²)) = 0.181L and Δp = πħ/L, so ΔxΔp = 0.568ħ, just above ħ/2. Higher states are told apart by nodes. ψₙ has n − 1 interior zeros, and each added node forces extra curvature, which is extra kinetic energy; that ordering holds for any one-dimensional well, so a numerically found state is labelled by counting its zero crossings. At large n the correspondence principle takes over: |ψₙ|² oscillates so fast that its average over any fixed interval approaches the classical 1/L, and the fractional gap (2n + 1)/n² shrinks towards zero. A 1 × 10⁻¹⁵ kg dust grain in a 1 μm box has E₁ = 5.5 × 10⁻⁴¹ J, so room-temperature energies put it near n ~ 10¹⁰, where the ladder is indistinguishable from a continuum.

Lower the walls to V₀. Inside write ψ = sin(kx + φ); outside, ψ decays away from each wall. Matching ψ′/ψ at x = 0 gives k cot φ = κ, so tan φ = k/κ and sin φ = √(E/V₀): φ is the phase the wave already has on reaching the wall. Symmetry demands the same phase at x = L, so kL + φ = nπ − φ, which is kL = nπ − 2 arcsin(√(E/V₀)). Rigid walls have φ = 0 and return kL = nπ. Finite walls have φ positive, so kL falls short of nπ and every level lies below its rigid-wall partner: in a 1.00 nm, 2.0 eV well an electron has 0.229, 0.887 and 1.815 eV instead of 0.376, 1.504 and 3.384 eV. Odd n are the even-parity states, obeying k tan(kL/2) = κ; even n are odd-parity, obeying −k cot(kL/2) = κ. The equation is transcendental, so solve it by bisection or graphically. A quick estimate lets the wave fill a box L + 2δ wide: for the ground state that gives 0.376 eV/1.293² = 0.225 eV, within 2%, though it fails near the rim.

Outside the well ψ ∝ e^(−x/δ) with δ = ħ/√(2m(V₀ − E)). For the 0.229 eV ground state of the 2.0 eV well, V₀ − E = 1.771 eV and δ = 0.195 nm/√1.771 = 0.147 nm. With ψ = sin(kx + φ) inside, the normalization integral is exactly L/2 + δ, of which the two tails carry δ sin²φ. That gives Pₒᵤₜ = (E/V₀)(2δ/(L + 2δ)): 0.026 for the ground state, 0.120 for n = 2 and 0.432 for n = 3, so 43% of position measurements on that state find the electron inside the walls. The limits check the formula. V₀ → ∞ sends δ and Pₒᵤₜ to zero, the rigid box; E → V₀ sends δ to infinity and Pₒᵤₜ to 1, a state spreading out as it unbinds. Energy conservation survives: a measurement that finds the electron in the wall confines it to within about δ, which costs a kinetic energy of order ħ²/(2mδ²), and that is exactly V₀ − E.

State n stops being bound when its energy reaches the rim. There φ = π/2 and the condition becomes k₀L = (n − 1)π with k₀ = √(2mV₀)/ħ, so state n exists only once √(V₀/E₁∞) exceeds n − 1, where E₁∞ is the infinite-well ground energy for the same width. Hence N = ⌈√(V₀/E₁∞)⌉. In one dimension the first threshold is zero, so every attractive well, however shallow, binds one state. For the 1.00 nm, 2.0 eV well √(2.0/0.376) = 2.31 and N = 3, although only two infinite-well levels lie below 2.0 eV. The correspondence check is a phase-space count: N ≈ 2L√(2mV₀)/h is the classically allowed area, width L times momentum range 2√(2mV₀), divided by h. Raise m, L or V₀ and the well fills with states until the spectrum looks continuous. Three dimensions differ: for l = 0 the equation for rR(r) behaves like the odd-parity states alone, so a spherical well of radius a binds nothing unless V₀ exceeds π²ħ²/(8ma²).