Bound States of the Infinite Square Well

Derivation

Bound States of the Infinite Square Well

D-143 Home PU-202 Threads energy · waves · matter Depends on Separation of Variables and Stationary States, Hermitian Operators: Real Eigenvalues and Orthogonal Eigenstates

Statement

For a particle of mass m confined to the one-dimensional region 0 < x < L by a potential that is zero inside and infinite outside, the time-independent Schrödinger equation together with the Dirichlet boundary conditions ψ(0) = ψ(L) = 0 admits normalisable solutions only for a discrete set of energies E_n = n²π²ℏ²/(2mL²), n = 1, 2, 3, …, with orthonormal eigenfunctions ψ_n(x) = √(2/L) sin(nπx/L).

Why it matters

The infinite square well is the simplest quantum system with a genuinely bound spectrum, and it is the prototype for every confinement problem: quantum dots, nuclei modelled as boxes, electrons in a conjugated molecule, and the density of states in a metal all inherit its n² ladder and its half-wavelength counting. It shows, with almost no algebra, that quantisation is a consequence of boundary conditions imposed on a wave equation — not an extra postulate bolted onto classical mechanics.

It is also the reference against which harder wells are read. The finite well, the harmonic oscillator, and periodic lattices are all understood partly by contrast with this exactly solvable limit, so its eigenfunctions and energies must be at your fingertips.

Assumptions

The potential is exactly zero inside and strictly infinite outside.If the walls are finite, the wavefunction leaks (tunnels) into the classically forbidden region, energies drop below the n²π²ℏ²/(2mL²) values, and only finitely many bound states survive.

Motion is one-dimensional and non-relativistic.In three dimensions the spectrum becomes a sum over independent quantum numbers and degeneracies appear; at relativistic speeds the kinetic term ℏ²k²/2m must be replaced by the Dirac/Klein–Gordon dispersion.

The wavefunction is continuous everywhere, including at the walls.Continuity is what forces ψ to vanish at the boundary. Its derivative, by contrast, is allowed to jump here because the potential step is infinite — dropping the continuity-of-ψ requirement would destroy the quantisation condition entirely.

The Hamiltonian is self-adjoint on the domain of functions vanishing at both walls.The boundary conditions are not decoration: they define the domain on which Ĥ = −(ℏ²/2m) d²/dx² is Hermitian, guaranteeing a real spectrum and a complete orthonormal eigenbasis. A different boundary condition (e.g. periodic) is a different, equally valid, self-adjoint problem with a different spectrum.

Derivation

−(ℏ²/2m) d²ψ/dx² + V(x) ψ = E ψ

Time-independent Schrödinger equation for a stationary state of energy E (assumed prior result). A

d²ψ/dx² = −(2mE/ℏ²) ψ ≡ −k² ψ, k ≡ √(2mE)/ℏ

Inside the well V = 0, so the equation reduces to a constant-coefficient ODE. The definition k² = 2mE/ℏ² anticipates E > 0, justified in step 8. A

ψ(x) = A sin(kx) + B cos(kx)

General solution of a second-order linear ODE with real negative eigenvalue: two independent oscillatory modes with constants A, B fixed by boundary conditions. A

ψ(x) = 0 for x ≤ 0 and x ≥ L

An infinite potential makes any nonzero amplitude carry infinite energy, so the state cannot occupy the walls or the exterior. Continuity of ψ then forces the interior solution to match zero at each wall. A

ψ(0) = A sin 0 + B cos 0 = B = 0

Impose the left boundary condition; cos(0) = 1 kills the cosine branch. A

ψ(L) = A sin(kL) = 0 ⟹ sin(kL) = 0

Impose the right boundary condition. Taking A = 0 gives the trivial (unnormalisable, zero-probability) solution, so a nonzero state requires the sine to vanish. A

kL = nπ, n = 1, 2, 3, … ⟹ k_n = nπ/L

The zeros of sine are integer multiples of π. n = 0 gives ψ ≡ 0 (excluded); negative n only flip the sign of ψ and add nothing new, so the labels are the positive integers. A

E_n = ℏ²k_n²/(2m) = n²π²ℏ²/(2mL²)

Invert the definition k² = 2mE/ℏ² from step 2 and substitute the allowed k_n. Every E_n > 0, retroactively justifying the real k assumed in step 2 (no bound states exist at or below the well floor). B

∫₀^L |ψ_n|² dx = |A|² ∫₀^L sin²(nπx/L) dx = |A|² · L/2 = 1

Born rule: total probability must be one. Use sin²θ = (1 − cos 2θ)/2; the cosine integrates to zero over an integer number of half-periods, leaving L/2. B

A = √(2/L) ⟹ ψ_n(x) = √(2/L) sin(nπx/L)

Solve for the modulus of A; the overall phase is physically irrelevant, so choose A real and positive. A

∫₀^L ψ_m(x) ψ_n(x) dx = (2/L) ∫₀^L sin(mπx/L) sin(nπx/L) dx = δ_mn

Product-to-sum: sina sinb = ½[cos(a−b) − cos(a+b)]. For m ≠ n both cosines integrate to zero over [0,L]; for m = n the first term gives L/2. Orthonormality is also guaranteed a priori because these are eigenfunctions of a Hermitian operator with distinct eigenvalues. C

Result

E_n = n²π²ℏ² / (2mL²), ψ_n(x) = √(2/L) sin(nπx/L), n = 1, 2, 3, …

Reading. The allowed energies form a ladder spaced as the squares of the integers, with a nonzero ground state E₁ = π²ℏ²/(2mL²) — the zero-point energy demanded by confinement and the uncertainty principle. The n-th eigenfunction is a standing sine wave fitting exactly n half-wavelengths between the walls, with n − 1 interior nodes; adjacent states are orthogonal.

Units check. ℏ² carries (J·s)² = J²s²; dividing by mL² (kg·m²) gives J²s²/(kg·m²). Since J = kg·m²·s⁻², one factor of J cancels kg·m²·s⁻², leaving J — an energy, as required. The amplitude √(2/L) has units m^−1/2, so |ψ|² is a probability per unit length (m⁻¹) and ∫|ψ|²dx is dimensionless.

Limiting cases

Wide box / heavy particle (L → ∞ or m → ∞): level spacing E_n+1 − E_n ∝ 1/(mL²) → 0, the spectrum becomes effectively continuous, and the classical free particle is recovered (correspondence principle).
High quantum number (n ≫ 1): the probability density (2/L) sin²(nπx/L) oscillates so rapidly that its local average, 1/L, matches the uniform classical distribution of a particle bouncing between walls.
Fractional spacing: (E_n+1 − E_n)/E_n = (2n+1)/n² → 0 as n grows: the ladder looks smooth from far up even though absolute gaps widen.
Ground state: setting n = 1 gives the irreducible confinement energy; it cannot be lowered without widening the box, a direct manifestation of Δx Δp ≳ ℏ/2 with Δx ~ L.

Breaks when

Walls are finite. Any real potential is finite, so ψ penetrates the barriers, the sine no longer vanishes exactly at the edges, and the transcendental matching condition replaces kL = nπ. Energies are pushed below the ideal values and the number of bound states becomes finite.
Relativistic or strongly-bound regime. When E_n approaches mc² (very small L or very light particle), the non-relativistic kinetic relation E = ℏ²k²/2m fails and one must use the relativistic dispersion; the n² law is lost.
Interactions or many particles. The single-particle eigenbasis stops being the whole story once particles interact or fermion antisymmetry couples the coordinates; the box levels then act only as an unperturbed starting point.
Time-dependent or moving walls. If L = L(t), the states are no longer stationary; energy is not conserved and one must solve the time-dependent equation (quantum analogue of an adiabatic or sudden expansion).

Failure modes

Including n = 0. Students list a "ground state" at E = 0; but n = 0 makes ψ ≡ 0 everywhere, which is not a physical state. The lowest state is n = 1.
Forgetting to normalise. Writing ψ_n = sin(nπx/L) without the √(2/L) prefactor gives wrong probabilities and wrong expectation values.
Using the wrong well width in k. For a well running from −L/2 to +L/2 the eigenfunctions split into even cosines and odd sines; blindly reusing sin(nπx/L) mislabels parity and misplaces nodes.
Demanding continuous ψ′ at the walls. The derivative is discontinuous here because the potential is infinite; imposing derivative continuity yields no solutions.
Squaring the energy formula wrong. Treating the spacing as linear in n (like the harmonic oscillator) instead of quadratic; the box goes as n², not n.
Mismatched units for ℏ. Mixing ℏ = 1.055×10⁻³⁴ J·s with electron-volt energies without converting, giving answers off by 1.6×10⁻¹⁹.

Discussion

The physics of the box is entirely in its boundary conditions. The differential equation inside is that of a free particle, whose spectrum is continuous; it is the demand that ψ vanish at two points a distance L apart that selects a countable set of standing waves. This is the same mechanism that quantises a vibrating string's overtones — the quantum numbers here are literally the harmonic numbers of a string of length L, with de Broglie wavelength λ_n = 2L/n. Quantum mechanics adds only the identification of wave frequency with energy through E = ℏ²k²/2m.

The eigenfunctions form a complete orthonormal set, which is why the box is so useful: any well-behaved state vanishing at the walls can be expanded as ψ(x) = Σ_n c_n ψ_n(x) with c_n = ∫₀^L ψ_n ψ dx. This is nothing but a Fourier sine series, and |c_n|² is the probability of measuring energy E_n. Time evolution is then trivial: each coefficient picks up a phase e^−iE_nt/ℏ, so the box is the cleanest laboratory for watching wavepackets disperse and revive.

Parity and nodes carry physical content. The ground state has no interior node and is symmetric about the centre; each successive state adds one node and alternates symmetry about x = L/2. More nodes mean larger curvature, hence larger ⟨p²⟩ and higher energy — the variational intuition that "wigglier means costlier" is exact here. The nonzero ground-state energy is the confinement energy: pinning a particle to a region of size L forces a momentum spread of order ℏ/L and hence a kinetic energy of order ℏ²/(2mL²), which is precisely E₁ up to the factor π².

More deeply, the choice ψ(0) = ψ(L) = 0 is one self-adjoint extension of the free-particle kinetic operator on the interval; other extensions (periodic, antiperiodic, Robin) give physically different, equally consistent spectra. This is the finite-dimensional shadow of a general fact: the Hamiltonian is not defined until its domain is, and the boundary data are that domain. The reality of the spectrum and the completeness of the eigenbasis used in the expansion above are guaranteed by the self-adjointness proven for Hermitian operators, tying this concrete problem to the general spectral theorem.

Common misconceptions. The walls do not "reflect" a classical particle that still has definite position; the eigenstate is a genuine standing wave with no net current and a fixed energy. And the box is not "empty except at the walls" — the potential is zero inside precisely so that all the interesting structure lives in the interior wavefunction, not in the (idealised, unphysical) walls themselves.

Worked examples

Electron in a 0.50 nm box — ground-state energy.

A crude model of an electron confined to a single atom-sized region. Take m = m_e = 9.109×10⁻³¹ kg, L = 0.50 nm = 5.0×10⁻¹⁰ m, n = 1. A

E₁ = π²ℏ²/(2m_eL²)

Set n = 1 in the result. Symbols first, numbers next. A

E₁ = (9.8696)(1.0546×10⁻³⁴)² / [2(9.109×10⁻³¹)(5.0×10⁻¹⁰)²]

Substitute ℏ = 1.0546×10⁻³⁴ J·s and π² = 9.8696. Numerator = 9.8696 × 1.1122×10⁻⁶⁸ = 1.0977×10⁻⁶⁷ J²s². Denominator = 2 × 9.109×10⁻³¹ × 2.5×10⁻¹⁹ = 4.555×10⁻⁴⁹ kg·m². B

E₁ = 2.41×10⁻¹⁹ J ≈ 1.5 eV

Reading. An electron squeezed into half a nanometre has a zero-point energy of about 1.5 electron-volts — comparable to atomic and molecular binding energies, which is why the box is a serviceable back-of-envelope model for valence electrons.

Units check. J²s²/(kg·m²) = J, then divided by 1.602×10⁻¹⁹ J/eV gives eV.

Photon emitted in the n = 3 → n = 2 transition of the same electron.

Find the wavelength of light released when the electron drops between levels. Use E₁ = 2.41×10⁻¹⁹ J from Example 1. A

ΔE = E₃ − E₂ = (3² − 2²) E₁ = 5 E₁

Energies scale as n², so the gap is (9 − 4)E₁. Symbols before numbers. A

ΔE = 5 × 2.41×10⁻¹⁹ J = 1.205×10⁻¹⁸ J

Insert the ground-state value. A

λ = hc/ΔE = (6.626×10⁻³⁴)(2.998×10⁸) / 1.205×10⁻¹⁸

Planck relation E = hc/λ for the emitted photon. Numerator = 1.986×10⁻²⁵ J·m. B

λ = 1.65×10⁻⁷ m ≈ 165 nm

Reading. The transition lands in the ultraviolet — again the right order of magnitude for electronic transitions, confirming the box captures the essential energy scale even though real atoms need the Coulomb potential for accuracy.

Units check. (J·s)(m·s⁻¹)/J = m, a wavelength.

Problems

Show that the expectation value of position in any eigenstate is ⟨x⟩ = L/2, and explain why this is obvious from symmetry.
Solution
⟨x⟩ = (2/L) ∫₀^L x sin²(nπx/L) dx. Using sin²θ = (1−cos2θ)/2, the integral splits into (2/L)[∫x/2 dx − ∫(x/2)cos(2nπx/L) dx]. The first term gives (2/L)(L²/4) = L/2. The second, integrated by parts, vanishes because cos(2nπ) = 1 and sin(2nπ) = 0 at both limits. Hence ⟨x⟩ = L/2. It is obvious because |ψ_n|² is symmetric about the centre x = L/2 for every n.
Compute ⟨x²⟩ for the n-th state and hence the position uncertainty Δx. Evaluate Δx for the ground state.
Solution
Standard result: ⟨x²⟩ = L²[1/3 − 1/(2n²π²)]. Then (Δx)² = ⟨x²⟩ − ⟨x⟩² = L²[1/3 − 1/(2n²π²) − 1/4] = L²[1/12 − 1/(2n²π²)]. For n = 1: 1/12 − 1/(2π²) = 0.08333 − 0.05066 = 0.03267, so Δx = L√0.03267 = 0.181L. Reassuringly less than L: the particle is fairly well localised near the centre.
Show ⟨p⟩ = 0 and ⟨p²⟩ = (nπ ℏ/L)² in any eigenstate, and verify the Heisenberg product Δx Δp for n = 1.
Solution
⟨p⟩ = −iℏ(2/L)∫₀^L sin(nπx/L)·(nπ/L)cos(nπx/L) dx = 0, since the integrand is (proportional to) d/dx[sin²] and sin vanishes at both ends. For ⟨p²⟩: since Ĥ = p̂²/2m inside the well, ⟨p²⟩ = 2mE_n = 2m·n²π²ℏ²/(2mL²) = (nπℏ/L)². So Δp = nπℏ/L. For n = 1: Δx Δp = (0.181L)(πℏ/L) = 0.568 ℏ > ℏ/2 = 0.5 ℏ. The uncertainty bound is satisfied, and the ground state comes close to saturating it.
A proton (m = 1.673×10⁻²⁷ kg) is confined to a nucleus modelled as a box of width L = 1.0×10⁻¹⁴ m (10 fm). Find E₁ in MeV.
Solution
E₁ = π²ℏ²/(2mL²) = 9.8696 × (1.0546×10⁻³⁴)² / [2 × 1.673×10⁻²⁷ × (1.0×10⁻¹⁴)²]. Numerator = 9.8696 × 1.1122×10⁻⁶⁸ = 1.0977×10⁻⁶⁷. Denominator = 2 × 1.673×10⁻²⁷ × 1.0×10⁻²⁸ = 3.346×10⁻⁵⁵. E₁ = 3.28×10⁻¹³ J = 3.28×10⁻¹³/1.602×10⁻¹³ MeV = 2.0 MeV. The MeV scale is exactly why nuclear binding energies dwarf atomic ones — confinement to femtometres, not nanometres.
An electron sits in the n = 2 state of a box of width L. Find the probability of locating it in the left quarter, 0 ≤ x ≤ L/4.
Solution
P = (2/L) ∫₀^L/4 sin²(2πx/L) dx. With sin²θ = (1−cos4πx/L)/2: P = (2/L)[x/2 − (L/8π)sin(4πx/L)]₀^L/4 = (2/L)[L/8 − (L/8π)sin(π)]. Since sin(π) = 0, P = (2/L)(L/8) = 1/4 = 0.25. Exactly a quarter — for n = 2 the density has a full period in [0,L/2], so each quarter carries a symmetric equal share here. (Contrast n = 1, where the same integral gives ≈ 0.091, less than a quarter because the ground-state density is depleted near the wall.)