The finite well, split by parity

Write V(x) = −V₀ for |x| ≤ a and 0 outside. Then V(−x) = V(x), so the parity operator Π, defined by Πψ(x) = ψ(−x), commutes with Ĥ. Π is both Hermitian and unitary with Π² = 1, so its eigenvalues are +1 and −1, and Ĥ and Π can be diagonalised together. In one dimension you get something stronger than the usual permission to choose: bound states of a 1D Hamiltonian are non-degenerate, so if ψ is an eigenvector then Πψ is an eigenvector at the same E, hence Πψ = cψ with c² = 1. Every bound state is therefore even or odd with no choice in the matter, and the spectrum interleaves — even, odd, even, odd in ascending energy, the nth state carrying n − 1 nodes. What this buys is bookkeeping. A naive attack has four unknown coefficients and four matching equations at x = ±a. Fixing the parity first reduces the interior solution to cos kx or sin kx alone, kills the growing exponential outside by normalisability, and leaves a single condition at the single wall x = +a; the wall at −a is then satisfied automatically by reflection.

Inside the well E + V₀ is positive, so ψ″ = −k²ψ with k = √(2m(E + V₀))/ℏ and ψ oscillates. Outside, E is negative, so ψ″ = +κ²ψ with κ = √(−2mE)/ℏ; the growing branch exp(+κx) is not square-integrable beyond the wall, so only exp(−κx) survives — and it is that rejection, not the matching, that makes the spectrum discrete. Take the even sector: ψ = A cos(kx) for |x| ≤ a and ψ = C exp(−κ(x − a)) beyond it. Because V is finite everywhere, ψ and ψ′ are both continuous at x = a; the kink you tolerated at an infinite wall is gone. Dividing the two conditions removes A and C at a stroke: the interior logarithmic derivative at the wall is −k tan(ka), the exterior one is −κ, so k tan(ka) = κ. Repeat with ψ = A sin(kx) in the odd sector, where the interior value is +k cot(ka), and the condition is −k cot(ka) = κ. Both demand a positive κ, which is the reminder that a root of the squared equation with tan(ka) negative is an artefact, not a state.

Set z = ka and κa for the two reduced wavenumbers. The definitions give k² + κ² = 2mV₀/ℏ², so z² + (κa)² = z₀² with z₀ = a√(2mV₀)/ℏ, and the conditions become z tan z = κa and −z cot z = κa. Every physical input — mass, half-width, depth — is now packed into the single radius z₀, and the bound-state problem is the intersection of a quarter circle with a fixed set of branches that never move. Put numbers on it with ℏ²/2mₑ = 3.81 eV Ų, so z₀² = V₀a²/(3.81 eV Ų) for an electron. An electron in a = 2.00 Å at V₀ = 15.24 eV gives z₀ = 4.00. A GaAs quantum well 10.0 nm wide with a 0.300 eV offset and m* = 0.067mₑ has ℏ²/2m* = 56.9 eV Ų = 0.569 eV nm², so z₀ = 5.00 nm × √(0.300 eV/0.569 eV nm²) = 3.63. The two systems differ by a factor of 25 in size and 50 in depth, and in reduced form they are nearly the same problem: three bound states each.

The branches never move, so the count is geometry alone. z tan z rises from 0 to +∞ on every interval (nπ, nπ + π/2), and −z cot z does the same on every (nπ + π/2, (n + 1)π). The quarter circle starts at (0, z₀) and falls monotonically to (z₀, 0), so it crosses exactly once every branch whose left endpoint lies below z₀, and misses the rest. Those endpoints are 0, π/2, π, 3π/2, …, hence N = ⌊2z₀/π⌋ + 1, with parities alternating from even. For z₀ = 4.00 that is ⌊2.546⌋ + 1 = 3: even, odd, even. Two consequences. Because the first branch starts at z = 0, N is at least 1 for every positive z₀ — a symmetric well in one dimension always binds an even state however shallow. The three-dimensional spherical well does not: its l = 0 radial function obeys u(0) = 0, which is the odd condition, whose first branch starts at π/2, so a 3D well of z₀ below π/2 binds nothing. And at each threshold the new state arrives with E approaching 0, so κ approaches 0 and 1/κ diverges: it is bound, but spread far outside the well.

Do not hand F(z) = z tan z − √(z₀² − z²) to a root finder over a wide interval. F jumps from +∞ to −∞ at every z = π/2 + nπ, and a bracketing method reads that jump as a sign change and returns the pole, to machine precision, as a spurious root. Fix it twice over. Bracket by branch: the nth even root lies in (nπ, min(nπ + π/2, z₀)) and the nth odd root in (nπ + π/2, min((n + 1)π, z₀)). And clear the denominator first — multiply by cos z or sin z to get f(z) = z sin z − y cos z for even states and h(z) = z cos z + y sin z for odd, with y = √(z₀² − z²). Both are bounded and analytic. f gains no spurious zero, since f = 0 together with cos z = 0 would also need z sin z = 0; h does vanish trivially at z = 0, where sin kx is identically zero, so the odd bracket must start at π/2, never at 0. For z₀ = 4.00, f(0) = −4.000 and f(1.5) = +1.234, so scipy.optimize.brentq(f, 0, 1.5) returns z₁ = 1.2524. Cross-check by diagonalising the discretised Hamiltonian on a grid several decay lengths wider than the well; if the lowest eigenvalue drifts as you widen the box, the box is doing the binding, not the well.

Convert the root: Eₙ = −V₀(1 − zₙ²/z₀²), measured from the top of the well. At z₀ = 4.00 with V₀ = 15.24 eV, z₁ = 1.2524 gives E₁ = −13.75 eV, so the ground state sits 1.49 eV above the floor. Infinite walls of the same full width 4.00 Å would put it at π²ℏ²/(2mL²) = 2.35 eV, 57% higher, and κ says why: with κa = 3.799 the decay length is 1/κ = a/3.799 = 0.527 Å, so the state occupies about 4.00 + 2 × 0.527 = 5.05 Å rather than 4.00 Å. Finite walls lower every level because they let the state spread. Now the honesty. Vertical faces are a fiction: a real heterojunction is graded over two or three monolayers, and the effective mass differs across it. The correct interface condition is then not ψ′ continuous but BenDaniel-Duke — ψ and (1/m*)ψ′ continuous — which turns the even condition into k tan(ka) = (m_w/m_b)κ. With equal masses it collapses to what you derived; with a 30% mass step every root moves, and a subband spacing quoted to three figures without it is quoting the model, not the sample.