Boundary Conditions and Quantisation

Introduction

Inside a one-dimensional box, the Schrödinger equation has the same solutions as for a free particle, and on its own it allows any energy. Yet a confined particle turns out to have only certain energies. The ingredient that makes the difference is the boundary conditions — the requirement that the wavefunction join smoothly onto its surroundings. This page shows exactly how imposing ψ = 0 at the two walls picks out a discrete set of waves, and why quantisation is a direct consequence of confinement rather than an extra assumption.

Core explanation

The starting point. Inside the box the general solution is ψ(x) = A sin kx + B cos kx with E = ħ²k²/2m. Outside, ψ = 0. An acceptable wavefunction must be continuous, so the inside solution must equal zero at x = 0 and at x = L.

The first wall, x = 0. Substituting gives ψ(0) = A sin 0 + B cos 0 = B. For ψ(0) = 0 we need B = 0. The cosine part is eliminated, leaving

ψ(x) = A sin kx

The second wall, x = L. Now ψ(L) = A sin kL = 0. One way to satisfy this is A = 0, but then ψ is zero everywhere and there is no particle at all — not an acceptable wavefunction. So instead sin kL must be zero. The sine function is zero only when its argument is a whole-number multiple of π:

kL = nπ, so k = nπ/L

Which values of n? If n = 0, then k = 0 and ψ = 0 everywhere, which is again rejected. Negative values of n give sin(−nπx/L) = −sin(nπx/L), which is the same function multiplied by −1; a wavefunction and its negative describe the same physical state because ψ ² is unchanged. The distinct allowed states are therefore labelled by

n = 1, 2, 3, …

Only certain wavelengths fit. Since k = 2π/λ, the condition kL = nπ is equivalent to

L = nλ/2

Only waves that fit a whole number of half-wavelengths into the box are allowed. The longest allowed wavelength is 2L (n = 1), then L (n = 2), then 2L/3 (n = 3), and so on.

Energy is quantised. Substituting k = nπ/L into E = ħ²k²/2m gives E = n²π²ħ²/2mL², which is usually written with h rather than ħ as Eₙ = n²h²/8mL². Because n is restricted to integers, the energy is restricted to a discrete ladder of values. Quantisation has emerged naturally from the boundary conditions.

The general principle. Every bound system in quantum mechanics is quantised for the same reason: the wavefunction must satisfy conditions at the edges of the region the particle occupies, or must vanish at great distances. For atoms, the requirement that ψ stays finite and dies away far from the nucleus gives the quantum numbers n, l and mₗ.

Formulae

ψ(0) = 0 gives B = 0. ψ(L) = 0 gives kL = nπ. Allowed k = nπ/L, λ = 2L/n, n = 1, 2, 3, … Energy Eₙ = n²h²/8mL².

Step-by-step reasoning

To derive quantisation from boundary conditions:

1. Write the general solution inside the region. 2. Apply the condition at the first boundary to eliminate one constant. 3. Apply the condition at the second boundary; reject the trivial solution ψ = 0. 4. Solve the resulting equation for the allowed values of k. 5. Substitute into the energy expression to obtain the allowed energies.

Visual explanation

Draw the box and fit waves between the walls. The first fits half a wavelength — a single hump. The second fits a full wavelength, one hump up and one down. The third fits one and a half wavelengths. A wave with, say, 1.3 half-wavelengths would end partway up the wall, breaking continuity, and so is forbidden.

Real-world analogy

A guitar string is fixed at the nut and the bridge. When plucked, it vibrates only at a fundamental frequency and its overtones, because only waves with nodes at both fixed ends can persist. The particle in a box obeys the same rule, with the wavefunction playing the part of the string.

Real-world example

Microwave ovens use a metal cavity in which only standing electromagnetic waves that fit between the walls can build up. The pattern of these allowed modes gives hot and cold spots, which is why many ovens include a turntable. The quantised modes arise from boundary conditions in just the way box states do.

Why?

Why must ψ be continuous at the walls? The wavefunction determines the probability density, and a sudden jump in ψ would make its second derivative, and therefore the kinetic energy, infinite. Since ψ is exactly zero outside, continuity demands ψ = 0 at each wall.

Common misconception

"Quantisation is imposed on quantum systems by assumption, as Bohr did." In the Schrödinger approach nothing is assumed about allowed energies. Quantisation follows automatically from requiring an acceptable wavefunction that meets the boundary conditions.

Worked example

Question: For an electron in a box of length 0.50 nm, find the allowed de Broglie wavelengths for the three lowest states.

Reasoning: λ = 2L/n. For n = 1, λ = 2 × 0.50 = 1.00 nm; for n = 2, λ = 0.50 nm; for n = 3, λ = 1.00/3 = 0.33 nm.

Answer: 1.00 nm, 0.50 nm and 0.33 nm; each fits a whole number of half-wavelengths into the box.

Quick check

1. Why is the state with n = 0 not allowed for a particle in a box? Answer: It gives ψ = sin 0 = 0 everywhere, so there would be no probability of finding the particle anywhere in the box.

Exam focus

Show each step: B = 0 from the first wall, sin kL = 0 from the second, rejection of A = 0, then kL = nπ. State clearly why n starts at 1 and why negative n adds nothing new. Link kL = nπ to L = nλ/2.

Advanced insight

If the box is replaced by a ring of circumference L, the boundary condition becomes periodic, ψ(x + L) = ψ(x), rather than ψ = 0. This allows e^(ikx) travelling waves with kL = 2πn, including n = 0 and negative n as distinct states, giving a different pattern of levels and degeneracies.

Summary

Continuity requires ψ = 0 at both walls of the box. The condition at x = 0 removes the cosine term; the condition at x = L forces sin kL = 0, so kL = nπ with n = 1, 2, 3, … Only waves fitting a whole number of half-wavelengths are allowed, and the energies Eₙ = n²h²/8mL² form a discrete ladder. Quantisation is a consequence of confinement.

Practice questions

1. What condition does ψ(0) = 0 impose on the general solution A sin kx + B cos kx? Answer: B = 0, so the solution reduces to ψ = A sin kx. 2. Show that kL = nπ is equivalent to L = nλ/2. Answer: k = 2π/λ, so (2π/λ)L = nπ, which rearranges to L = nλ/2. 3. Why do n = 2 and n = −2 describe the same state? Answer: sin(−2πx/L) = −sin(2πx/L); the wavefunctions differ only by a factor of −1, so ψ ² and every observable are identical. 4. What is the longest de Broglie wavelength allowed for a particle in a box of length L? Answer: 2L, for the n = 1 state, in which half a wavelength fits between the walls.