Setting Up the Particle in a One-Dimensional Box

Introduction

The free particle showed that an unconfined particle can have any energy. What happens when we trap it? The particle in a one-dimensional box is the simplest model of confinement and the first bound-state problem solved exactly in almost every course on quantum chemistry. It is simple enough to solve in a few lines, yet it reveals quantised energy levels, zero-point energy and nodes — features shared by electrons in atoms, molecules and nanocrystals. This page sets the problem up carefully; the next pages solve it.

Core explanation

The model. A particle of mass m moves along the x-axis between two walls at x = 0 and x = L. Inside the box it feels no force, so its potential energy is constant and we choose it to be zero. The walls are perfectly impenetrable, which we represent by an infinite potential energy:

V(x) = 0 for 0 < x < L V(x) = ∞ for x ≤ 0 and x ≥ L

This potential is called an infinite square well . Its graph looks like a flat-bottomed trench with vertical sides of unlimited height.

Outside the box. Where V is infinite, the Schrödinger equation −(ħ²/2m)ψ″ + Vψ = Eψ can only be satisfied with finite energy E if ψ = 0. A non-zero ψ would give an infinite term Vψ that nothing else could balance. Physically, the particle has zero probability of being found in the walls or beyond them. So

ψ(x) = 0 for x ≤ 0 and x ≥ L

Inside the box. With V = 0 the equation is identical to the free-particle equation:

−(ħ²/2m) d²ψ/dx² = Eψ, or d²ψ/dx² = −k²ψ with k = (2mE)^½/ħ

For the box problem it is most convenient to write the general solution with real functions:

ψ(x) = A sin kx + B cos kx

where A and B are constants still to be determined. At this stage k — and hence E — could be anything, exactly as for a free particle.

What changes the picture. The difference from the free particle is that the inside and outside solutions must join up. An acceptable wavefunction must be continuous, so ψ inside must fall to zero at both walls, matching the zero outside. These boundary conditions , applied on the next page, remove the cos term and allow only certain values of k. That is where quantisation enters.

Why study a box? Real potentials never have infinite walls, but the model captures the essential physics of confinement. Electrons delocalised along a conjugated chain, electrons in a semiconductor quantum dot and gas molecules in a container are all approximated usefully as particles in boxes.

Formulae

V = 0 for 0 < x < L; V = ∞ elsewhere. Inside: d²ψ/dx² = −k²ψ with k² = 2mE/ħ². General solution: ψ = A sin kx + B cos kx. Outside: ψ = 0.

Step-by-step reasoning

To set up any confined-particle problem:

1. Draw the potential energy V(x) and identify the regions where it takes different forms. 2. In regions of infinite potential, set ψ = 0. 3. In each finite region, write the Schrödinger equation with the appropriate V. 4. Write the general solution in each region, leaving constants undetermined. 5. List the conditions — continuity at boundaries and normalisation — that will fix the constants.

Visual explanation

Sketch a horizontal axis with vertical lines rising to the top of the page at x = 0 and x = L. The floor between them is at V = 0. Every allowed wavefunction will be drawn inside this trench, pinned to zero where it meets each wall, like a string tied at both ends.

Real-world analogy

A bead threaded on a straight wire between two solid stoppers is a classical version of the box: it slides freely between them but can never pass them. Quantum mechanics adds that the bead behaves as a wave, and that wave must fit between the stoppers.

Real-world example

The π electrons of hexa-1,3,5-triene are spread along the carbon chain but are held within it by the attraction of the nuclei at its ends. Treating the chain as a one-dimensional box of length roughly equal to the chain gives a surprisingly good first estimate of the wavelength of light the molecule absorbs.

Why?

Why can we set V = 0 inside the box rather than some other value? Only differences in potential energy have physical meaning. Choosing zero as the floor simply sets the energy scale; a different constant would shift every energy level by the same amount without changing the spacing or the wavefunctions.

Common misconception

"The particle can leak a little way into the walls." For an infinite well it cannot: ψ is exactly zero at and beyond the walls. Penetration into a barrier happens only when the walls have finite height, which leads to tunnelling.

Worked example

Question: Show that ψ = A sin kx + B cos kx satisfies the Schrödinger equation inside the box, and find E in terms of k.

Reasoning: d²ψ/dx² = −Ak² sin kx − Bk² cos kx = −k²ψ. Substituting into −(ħ²/2m)ψ″ = Eψ gives (ħ²k²/2m)ψ = Eψ.

Answer: The function satisfies the equation for any A and B, with E = ħ²k²/2m.

Quick check

1. What is the value of the wavefunction at every point outside an infinite one-dimensional box, and why? Answer: It is zero, because an infinite potential energy multiplied by any non-zero ψ could not be balanced by a finite energy E.

Exam focus

Draw and label the potential, state V inside and outside, write the Schrödinger equation inside the box and give the general solution. Examiners expect you to say explicitly why ψ = 0 outside and that continuity will impose conditions at x = 0 and x = L.

Advanced insight

The derivative of ψ is discontinuous at the walls of an infinite well — the wavefunction meets the wall at an angle. This is allowed only because the potential jump is infinite; for any finite potential, both ψ and dψ/dx must be continuous, and the solutions outside decay exponentially instead of vanishing.

Summary

The particle in a one-dimensional box has V = 0 for 0 < x < L and infinite walls elsewhere. Outside the box ψ = 0. Inside, the Schrödinger equation is the free-particle equation, with general solution ψ = A sin kx + B cos kx and E = ħ²k²/2m. Continuity at the walls will restrict the allowed k, producing quantisation.

Practice questions

1. Write the potential energy function for a particle in a box of length L. Answer: V(x) = 0 for 0 < x < L and V(x) = ∞ for x ≤ 0 and x ≥ L. 2. What is k in terms of E for the particle inside the box? Answer: k = (2mE)^½/ħ. 3. Why is the general solution written with sin and cos rather than complex exponentials? Answer: Both forms are equivalent, but sin kx is already zero at x = 0, which makes applying the boundary conditions simpler. 4. Give two chemical systems that can be modelled as particles in a box. Answer: π electrons in a conjugated polyene chain and electrons confined in a semiconductor quantum dot.