Why n must be a positive integer

Write Ĥ = −(ℏ²/2m) d²/dx² and it looks as though all the physics sits in the derivative. It does not. The same expression is symmetric on many different domains, and each choice — Dirichlet ψ(0) = ψ(L) = 0, Neumann ψ′(0) = ψ′(L) = 0, periodic, Robin — is a different self-adjoint operator with a different spectrum. The infinite-wall limit of the finite well is what selects the Dirichlet domain: as V₀ grows, the decay length 1/κ outside the well shrinks to zero and the wavefunction is pinned to zero at both walls. Only after that choice does the eigenvalue question have an answer, and the question is a linear-algebra one: for which E is the kernel of Ĥ − E bigger than the zero subspace? The differential equation supplies candidate functions; the domain decides which candidates are vectors of the space you are working in.

Take any ψ in the domain and compute ⟨ψ|Ĥ|ψ⟩ = −(ℏ²/2m) ∫ ψ* ψ″ dx. Integrate by parts once: the result is (ℏ²/2m) ∫ |ψ′|² dx minus a boundary term (ℏ²/2m)[ψ* ψ′] evaluated at 0 and L, and that boundary term vanishes because ψ itself vanishes at both walls. So the expectation of Ĥ is never negative for a state in the domain, and the whole spectrum lies in [0, ∞). Equality needs ∫|ψ′|² = 0, so ψ′ = 0 almost everywhere, so ψ is constant — and a constant vanishing at x = 0 is the zero function. Hence E is strictly positive, and the sinh-and-cosh algebra usually run at this point is one special case of a single line. Notice exactly where the boundary condition entered: on the Neumann domain the same boundary term also vanishes, and there the constant ψ = 1/√L is a genuine eigenvector with E = 0. Zero-point energy is a fact about these walls, not about this derivative.

With E positive set k = √(2mE)/ℏ, so ψ″ = −k²ψ and the solution space is span{sin kx, cos kx}, two-dimensional for every k. The wall at x = 0 acts on the vector: ψ(0) = B = 0 deletes the cosine and cuts the space to one dimension, spanned by sin kx. The wall at x = L cannot delete anything further without emptying the space, so it acts on the number instead: A sin(kL) = 0. That asymmetry is worth naming, because it settles two later results at once. Since only one dimension survives at any E, no two independent eigenvectors can share an eigenvalue: the one-dimensional box is non-degenerate, exactly as the Wronskian argument says. And since the second condition constrains k rather than ψ, it is the one that manufactures a spectrum.

For k = 5.12 × 10⁹ m⁻¹, which is E = 1.00 eV for an electron, ψ(x) = sin(kx) solves the differential equation perfectly and is smooth everywhere. It is rejected for exactly one reason: sin(kL) = −0.917 at L = 1.00 nm, so the only multiple of it that vanishes at the far wall is zero times it. And ψ ≡ 0 is not a low-probability state; it has ⟨ψ|ψ⟩ = 0, so the Born rule has no probability to distribute and no rescaling can normalise it. This is also what to compute numerically. Integrate outward from ψ(0) = 0 with ψ′(0) = 1, record the miss at the far wall, ψ(L) = sin(kL)/k, whose zeros are those of the shooting function F(E) = sin(kL) because 1/k never vanishes, and hand F to scipy.optimize.brentq on each energy bracket where it changes sign. The eigenvalues are the isolated zeros of F, and every energy between them is a legitimate solution of the ODE that is not a state.

sin(kL) = 0 gives kL = nπ for integer n, and the three cases are not symmetric. n = 0 gives k = 0 and ψ ≡ 0, the zero vector — not a particle sitting still but no state at all, and it is normalisability that rejects it, long before any uncertainty argument is reached. Negative n gives sin(−nπx/L) = −sin(nπx/L): the same one-dimensional subspace, and since states are rays in L²([0, L]) a factor of −1 is a global phase and unobservable. It also gives the same energy, because E depends on k². So the distinct labels are n = 1, 2, 3, …, each used once, and Eₙ = ℏ²kₙ²/(2m) = n²π²ℏ²/(2mL²). Normalisation, ∫ A² sin²(nπx/L) dx over [0, L] = A²L/2 = 1, then fixes A = √(2/L); it sets the length of the surviving ray and changes nothing about which energies exist.

Put numbers in. For an electron, E₁ = π²ℏ²/(2mL²) = 0.376 eV at L = 1.00 nm, so E₂ = 1.504 eV and the 2 → 1 photon carries 1.128 eV, a wavelength of 1.10 μm. Because E₁ ∝ 1/(mL²), the same box holding a proton gives E₁ = 0.376 eV ÷ 1836 = 2.05 × 10⁻⁴ eV, already well under the 0.026 eV of room-temperature thermal energy. Widen the box holding the electron to 1.00 cm and E₁ falls from 0.376 eV by a factor of 10¹⁴ to 3.8 × 10⁻¹⁵ eV: the ladder is still there, still exactly n², and completely unresolvable. Take the walls away altogether and there is no second boundary condition left to impose, so every k is allowed, the spectrum is the continuum [0, ∞), and the eigenfunctions exp(ikx) are δ-normalised rather than square-integrable. Quantisation is not a property of small things; it is what confinement does, and it fades smoothly as the confinement is relaxed.