University Physics V · Particle in a Box · 5.3

Why n Must Be a Positive Integer

Energy is not quantised because small things come in lumps. The differential equation is solved at every E; what fails almost everywhere is the demand that a nonzero vector satisfy both walls at once. Run that argument here and you can run it on any confined system.

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Solving Ĥ|ψ⟩ = E|ψ⟩ is not solving a differential equation; it is asking which E leave a nonzero vector inside the operator's domain. For infinite walls on [0, L] that domain is the Dirichlet one: twice-differentiable functions with ψ(0) = ψ(L) = 0. Two facts then do all the work.

First, integrating ⟨ψ|Ĥ|ψ⟩ by parts leaves (ℏ²/2m)∫|ψ′|² dx once the boundary term dies on the walls, so every eigenvalue is positive and E ≤ 0 is gone before a single sine is written. Second, at positive E the solution space span{sin kx, cos kx} is two-dimensional; ψ(0) = 0 deletes the cosine and leaves one dimension, so ψ(L) = 0 can no longer constrain the vector and constrains the number instead: A sin(kL) = 0 with A ≠ 0, hence kL = nπ. The spectrum Eₙ = n²π²ℏ²/(2mL²) is what confinement costs, and the derivation is honest about its own price — walls of infinite height, so ψ′ jumps there and no material supplies such a potential; strictly one dimension, so no degeneracy; and a claim only about which energies carry normalisable states, not about a particle rattling between two walls.

Simple definition: Quantisation here means that the boundary-value problem −(ℏ²/2m)ψ″ = Eψ with ψ(0) = ψ(L) = 0 admits a solution other than ψ ≡ 0 only at the isolated energies Eₙ = n²π²ℏ²/(2mL²), with n = 1, 2, 3, …
Example: For an electron in L = 1.00 nm, E₁ = 0.376 eV and E₂ = 1.504 eV. A trial energy of 1.00 eV between them is not forbidden by the equation; it is simply the case where the only function meeting both walls is the zero vector.

The eigenvalue problem and its domainĤψ = −(ℏ²/2m) ψ″ = Eψ, ψ(0) = ψ(L) = 0

The two conditions are part of the operator, not decoration around it: change them and you change Ĥ, its spectrum, and its eigenvectors.

ψ lives in L²([0, L]); ℏ = 1.055 × 10⁻³⁴ J s; m in kg, L in m, E in J.

Positivity of Ĥ on the Dirichlet domain⟨ψ|Ĥ|ψ⟩ = (ℏ²/2m) ∫₀ᴸ |ψ′|² dx ≥ 0

Every eigenvalue is positive, and equality needs ψ′ ≡ 0, hence ψ constant, hence ψ ≡ 0 — so E ≤ 0 dies without any case-by-case algebra.

Integration by parts; the boundary term (ℏ²/2m)[ψ*ψ′] vanishes at both walls because ψ does.

The quantisation conditionA sin(kL) = 0 with A ≠ 0 ⇒ kₙ = nπ/L, n = 1, 2, 3, …

The ODE is solved by every k. Only ψ(L) = 0 discards all but a discrete set, so quantisation is a statement about the walls.

k = √(2mE)/ℏ in m⁻¹; n is a pure number; A = 0 would return the zero vector, which has no norm.

The spectrum of the infinite boxEₙ = n²π²ℏ²/(2mL²) = n²h²/(8mL²) = n²E₁

One number fixes the whole ladder, and the gaps Eₙ₊₁ − Eₙ = (2n + 1)E₁ widen with n — the opposite of the oscillator's equal rungs.

E₁ = 0.376 eV for an electron at L = 1.00 nm, and E₁ ∝ 1/(mL²), so E₁ = 0.376 eV ÷ (L/nm)².

The surviving eigenvector, normalisedψₙ(x) = √(2/L) sin(nπx/L), ⟨m|n⟩ = δₘₙ

Normalisation only sets the length of the surviving ray; it never adds or removes an eigenvalue, which is why it comes last.

Fixed by ∫|ψₙ|² dx = 1 over [0, L]; ψₙ has units m⁻¹⁄² and n − 1 interior nodes.

Which integers label distinct statesψ₀ ≡ 0 and ψ₋ₙ = −ψₙ, the same ray

n = 0 gives the zero vector, not a zero-energy state, and −n relabels a state already counted, so the spectrum is indexed once by n ≥ 1.

A state is a ray: |ψ⟩ and c|ψ⟩ with c ≠ 0 are the same physical state.

The problem is the operator plus its domain

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.

Positivity rules out E ≤ 0 before any sine appears

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.

The two walls do two different jobs

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.

Nontriviality is the entire content of quantisation

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 as a shooting function F(E) = sin(kL), and hand F to scipy.optimize.brentq between successive sign changes. 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.

Which integers survive, and why the list starts at 1

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.

How big the gaps are, and when they stop mattering

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 to 1.00 cm and E₁ falls by a further 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Trial wavenumber kL/π1.63

Box length L1.0 nm

Drag the trial until the arrow at the right wall collapses to nothing: it vanishes only at kL/π = 1, 2, 3 and 4, never between. Then stretch L and watch the same integers survive while the energy readout falls as 1/L².

TRIAL kL / π1.63

WALL MISS sin(kL)-0.918

TRIAL E / E₁2.66

TRIAL E, ELECTRON0.999 eV

Live interpretationTRIAL kL / π: 1.63. WALL MISS sin(kL): −0.918. TRIAL E / E₁: 2.66. TRIAL E, ELECTRON: 0.999 eV

Catch the common trap

Explain before calculating.

Misconception check

The usual reason given for excluding n = 0 is that a particle at rest would violate the uncertainty principle. That reaches the right answer by the wrong route, and it will mislead you the moment the boundary condition changes. n = 0 is excluded because it returns ψ ≡ 0, a vector of zero norm — not a state with no energy, but no state at all — and the rejection is made by normalisability, before Δx Δp is ever written down. Uncertainty explains the size of E₁, its ℏ²/(mL²) scaling, not its existence. The counter-case settles it: swap the Dirichlet walls for Neumann ones, ψ′(0) = ψ′(L) = 0, and the constant ψ = 1/√L is normalisable, is an eigenvector with E = 0, and carries Δp = 0 alongside Δx = L/(2√3). Nothing breaks, because on an interval p̂ is symmetric but not self-adjoint and the Robertson bound was never available to argue with. Zero-point energy belongs to the boundary condition.

In solving −(ℏ²/2m)ψ″ = Eψ on [0, L] with ψ(0) = ψ(L) = 0 you reach A sin(kL) = 0. Which reasoning forces kₙ = nπ/L with n = 1, 2, 3, …?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is confined to a one-dimensional box of length L = 0.500 nm with infinite walls. Find E₁ and E₃ in eV, and the wavelength of the photon emitted in the 3 → 1 transition.

Eₙ = n²π²ℏ²/(2mL²). With m = 9.109 × 10⁻³¹ kg and L = 5.00 × 10⁻¹⁰ m, the denominator is 2mL² = 4.555 × 10⁻⁴⁹ kg m².
The numerator is π²ℏ² = 9.8696 × (1.0546 × 10⁻³⁴ J s)² = 1.0976 × 10⁻⁶⁷, so E₁ = 1.0976 × 10⁻⁶⁷ ÷ 4.555 × 10⁻⁴⁹ = 2.410 × 10⁻¹⁹ J = 1.504 eV.
The n² law does the rest: E₃ = 9E₁ = 13.54 eV, and the emitted quantum carries E₃ − E₁ = 8E₁ = 12.03 eV.
λ = hc/ΔE = 1239.8 eV nm ÷ 12.03 eV = 103 nm.

AnswerE₁ = 1.504 eV, E₃ = 13.54 eV and λ = 103 nm, in the vacuum ultraviolet. Halving the box from 1.00 nm quadrupled every level, since Eₙ ∝ 1/L², and quadrupled every gap with it.

MediumFor an electron in a box of length L = 1.00 nm, test whether E = 1.00 eV is an eigenvalue by shooting: integrate ψ outward from ψ(0) = 0 with ψ′(0) = 1 and evaluate the miss at the far wall. Then name the two eigenvalues that bracket it.

k = √(2mE)/ℏ. With E = 1.00 eV = 1.6022 × 10⁻¹⁹ J, 2mE = 2.919 × 10⁻⁴⁹ and its root is 5.403 × 10⁻²⁵, so k = 5.403 × 10⁻²⁵ ÷ 1.0546 × 10⁻³⁴ = 5.123 × 10⁹ m⁻¹.
ψ″ = −k²ψ with ψ(0) = 0 and ψ′(0) = 1 integrates exactly to ψ(x) = sin(kx)/k, so the miss at the far wall is ψ(L) = sin(kL)/k. The factor 1/k never vanishes for positive k, so take the shooting function to be F(E) = sin(kL).
kL = 5.123 × 10⁹ × 1.00 × 10⁻⁹ = 5.123 rad, and sin(5.123) = −0.917. The miss is not zero, so at this energy the only function satisfying both walls is ψ ≡ 0 — a perfectly good solution of the ODE that is not a state.
F vanishes only when kL = nπ. Since π = 3.1416 and 2π = 6.2832 straddle 5.123, the trial lies between the first and second eigenvalues.
At L = 1.00 nm, Eₙ = n² × 0.376 eV, so those are E₁ = 0.376 eV and E₂ = 1.504 eV. Feeding F to scipy.optimize.brentq on [1.00, 2.00] eV, where it runs from −0.917 to +0.820, returns 1.504 eV.

AnswerF(1.00 eV) = sin(5.123) = −0.917 ≠ 0, so 1.00 eV is not an eigenvalue. It sits between E₁ = 0.376 eV and E₂ = 1.504 eV, and a root finder on F over [1.00, 2.00] eV returns 1.504 eV.

HardA student proposes a bound state at E = −0.100 eV for an electron in the same 1.00 nm box. (a) Show that the Dirichlet conditions force ψ ≡ 0. (b) Rule out every E ≤ 0 in one line, without cases. (c) Name a boundary condition that does admit E = 0, and give its eigenvector.

Negative E: put E = −ℏ²κ²/(2m), so ψ″ = +κ²ψ and ψ = A sinh(κx) + B cosh(κx). With |E| = 0.100 eV, 2m|E| = 2.919 × 10⁻⁵⁰ and κ = 1.620 × 10⁹ m⁻¹, so κL = 1.620.
ψ(0) = B = 0 leaves ψ = A sinh(κx), and ψ(L) = A sinh(1.620) = 2.428 A. Since sinh is strictly positive away from the origin, ψ(L) = 0 forces A = 0 and ψ ≡ 0. At E = 0 the same collapse happens with ψ = a + bx: the two walls give a = 0 and b = 0.
One line instead of two cases: for any ψ in the Dirichlet domain, ⟨ψ|Ĥ|ψ⟩ = (ℏ²/2m)∫|ψ′|² dx once ψ(0) = ψ(L) = 0 kills the boundary term, so the expectation of Ĥ is never negative and no negative eigenvalue can exist at any κ.
Equality would need ψ′ ≡ 0, hence ψ constant, hence — with ψ(0) = 0 — the zero vector. The spectrum is therefore strictly positive, starting at E₁ = 0.376 eV with nothing beneath it.
Neumann walls, ψ′(0) = ψ′(L) = 0, kill the same boundary term, but there the constant ψ = 1/√L = 3.16 × 10⁴ m⁻¹⁄² is admissible and satisfies Ĥψ = 0. Same differential operator, different domain, different spectrum.

Answersinh(1.620) = 2.428 ≠ 0 forces A = 0, so nothing exists at −0.100 eV; positivity of Ĥ on the Dirichlet domain removes every E ≤ 0 at once. Neumann walls do admit E = 0, with the constant eigenvector ψ = 1/√L.