The finite square well

A symmetric well has V(−x) = V(x), so the parity operator commutes with H and every bound state can be chosen even or odd. That is a theorem rather than a convenience: in one dimension bound states are non-degenerate, so there is no third option and no mixed case to worry about. The saving is real. Without parity you carry four constants, two inside and one in each tail, and four matching equations. With it, the interior is either A cos kx or A sin kx, the left tail is the mirror of the right, and a single wall at x = +a carries the whole condition. Even states have 0, 2, 4 nodes and odd states have 1, 3, 5, so the ground state is always the nodeless even one. Write that down before any root-finding starts and you already know the parity of every level you are about to find.

At a finite step both ψ and ψ′ are continuous; only an infinite singularity in V, a hard wall or a delta spike, can break ψ′. Take an even state: ψ = A cos kx for |x| ≤ a, and ψ = B e^(−κx) for x beyond a. Continuity of ψ gives A cos ka = B e^(−κa), and continuity of ψ′ gives −Ak sin ka = −κB e^(−κa). Divide the second by the first and both amplitudes cancel, leaving k tan ka = κ. That division is the whole trick, and it is why the quantisation condition is a statement about the logarithmic derivative ψ′/ψ: you never touch normalisation to find an energy, and you normalise afterwards only if you want a probability. Repeat the argument with the sine and you get −k cot ka = κ. Each condition still contains two unknowns, k and κ, but they are not independent.

The well ties k and κ together: k² + κ² = 2mV₀/ħ² for every bound state, because E cancels between the two definitions. Multiply by a², write z = ka and z₀ = (a/ħ)√(2mV₀), and the pair becomes z² + (κa)² = z₀², while the matching conditions become z tan z = √(z₀² − z²) and −z cot z = √(z₀² − z²). Now everything sits on one graph: the tangent and cotangent branches climb away from the z axis, a quarter circle of radius z₀ sweeps down to meet them, and every intersection is one bound state. The point of the rewrite is that z₀ is the only parameter left. Two wells with the same z₀, whatever their separate mass, width and depth, have the same spectrum in units of ħ²/2ma². An electron in a 1.0 nm well 5.0 eV deep has z₀ = 5.73.

The tangent branches begin at z = 0, π, 2π and the cotangent branches at π/2, 3π/2, so a new branch opens every π/2 and the circle catches one more each time z₀ grows past another half-multiple of π: N = floor(2z₀/π) + 1. Since the first branch starts at z = 0, N is never zero. Translate that threshold back into physics. Setting z₀ = (n − 1)π/2 gives V₀ = (n − 1)²π²ħ²/2mL² with L = 2a, which is exactly level n − 1 of the infinite well of the same width. So state n appears when V₀ passes Eₙ₋₁, not Eₙ, and the finite well always holds one more level than there are infinite-well levels below its rim. For the 5.0 eV, 1.0 nm well those are 0.376, 1.50 and 3.38 eV, three of them, and the finite well holds four: 0.27, 1.08, 2.38 and 4.06 eV above the floor.

Outside the well ψ falls as e^(−κ|x|) and |ψ|² as e^(−2κ|x|). The penetration depth 1/κ = ħ/√(2mE_b), the e-folding length of ψ, is set by how far the state sits below the rim, so the weaker the binding the wider the tail. Deeply bound states barely notice: the ground state of the 5.0 eV, 1.0 nm well has 1/κ = 0.090 nm and keeps 99% of its probability inside. The top state of the same well, bound by only 0.94 eV, has 1/κ = 0.20 nm and leaks 23% out. Push further and the tail takes over: a 0.20 eV well 0.40 nm wide binds by 0.033 eV, has 1/κ = 1.07 nm, and puts 70% of the electron outside the well that holds it. A useful summary is that the finite well behaves like an infinite well widened by one penetration depth at each wall. For the ground state of the 5.0 eV, 1.0 nm well, π/k = 1.176 nm against L + 2/κ = 1.180 nm, agreement to 0.3%, though the rule degrades as the binding weakens.

Let the well go shallow. For small z₀ the even condition gives z ≈ z₀ and κa ≈ z₀², so the binding energy tends to E_b = ħ²κ²/2m → 2mV₀²a²/ħ², quadratic in the product V₀a rather than linear in the depth. It is tiny, but it is not zero: a symmetric one-dimensional well binds however shallow it is. Three dimensions behave differently. For an s state let u(r) be r times the radial wavefunction; then u obeys the same one-dimensional equation with the extra condition u(0) = 0, which discards the entire even family. Only odd solutions survive, and those need z₀ past π/2, so a spherical well has a minimum depth, V₀ > π²ħ²/8μR². The deuteron sits right at that edge: with R ≈ 2.0 fm the threshold is 25.6 MeV, the fitted depth is about 36.6 MeV, and z₀ = 1.88 against 1.571. The result is a nucleus bound by 2.22 MeV with no excited state and 64% of its wavefunction outside the range of the force.