The particle in a box
Statement
Quantised energies as a consequence of confinement.
Why it matters
born-oppenheimer justifies solving an electronic Schrödinger equation at fixed nuclear positions; the particle in a box is the simplest possible such solvable case — a single particle confined to a region with zero potential inside and infinite potential outside — and it establishes, with minimal mathematics, the central quantum idea that confinement alone forces energy quantisation. That idea is reused conceptually by hydrogen-atom-solution's much harder exact solution, and directly, if qualitatively, by lcao-molecular-orbitals and huckel-theory's treatment of delocalised pi electrons, often modelled explicitly as particles confined to a one-dimensional box along the conjugated chain.
Hypotheses
Proof
Result
Reading. Confinement alone forces the particle's energy to take only a discrete set of values, growing as \(n^2\), rather than any continuous value; larger boxes give more closely spaced, less obviously quantised levels.
Scope. Exactly solvable in one dimension; readily extends to two or three dimensions by adding independent quantum numbers per dimension; the confined free-electron picture underlying huckel-theory's qualitative pi-system treatment is a direct, if approximate, application to real conjugated molecules.
Corollaries & converses
- The zero-point energy \(E_1=h^2/(8mL^2)\) is never zero, a direct manifestation of the Heisenberg uncertainty principle: perfectly confining a particle (\(\Delta x\) finite) forbids simultaneously having exactly zero momentum.
- Level spacing \(\Delta E=E_{n+1}-E_n\) grows with \(n\) and shrinks as \(L\) or \(m\) grows, explaining both why quantum confinement effects are most visible for light particles (electrons) in nanometre-scale boxes and why macroscopic "boxes" show levels so closely spaced as to appear classically continuous.
- huckel-theory's free-electron treatment of conjugated polyenes models delocalised pi electrons approximately as particles in a one-dimensional box along the chain, directly reusing this result's \(E_n\) formula to estimate absorption wavelengths.
Fails without
- Treat the box walls as finite rather than genuinely infinite (Hypotheses): the wavefunction is no longer forced to vanish exactly at the boundary, and a real evanescent tunnelling tail exists outside the box, an effect entirely absent from, and not describable by, the idealised infinite-well solution derived here.
- Include \(n=0\) as an allowed state: \(n=0\) forces \(\psi=0\) everywhere, describing no particle at all rather than a valid zero-energy bound state, directly contradicting both the normalisation requirement and the uncertainty-principle argument behind zero-point energy (Corollaries).
Common errors
- Including \(n=0\) as an allowed state; \(n=0\) gives \(\psi=0\) everywhere, so the ground state is \(n=1\), and the particle can never have exactly zero energy.
- Forgetting the normalisation constant \(\sqrt{2/L}\), or forgetting to normalise the wavefunction at all before using it to compute expectation values or probabilities.
- Assuming energy levels are evenly spaced, as for the harmonic oscillator; particle-in-a-box levels instead spread apart quadratically with \(n\), a qualitatively different spectrum.
- Treating the box walls as finite (soft) rather than genuinely infinite, forgetting that a finite well permits nonzero wavefunction amplitude with an evanescent tail outside the box — a distinct, more advanced case not treated by this idealised model.
Discussion
The particle in a box is among the very first problems solved once Schrödinger's wave equation was formulated in 1926, valued pedagogically precisely because its exact solution requires only elementary differential-equation and boundary-condition techniques, in sharp contrast to the far more involved hydrogen-atom-solution.
Extending to two or three dimensions by separation of variables gives independent quantum numbers along each axis; for a box with unequal side lengths, different combinations of quantum numbers can coincidentally share the same total energy, whereas a perfectly cubic box produces genuine, symmetry-required degeneracy — multiple distinct states sharing exactly the same energy, a concept reused extensively once atomic and molecular orbital degeneracies are discussed.
Common misconception: that a confined particle's lowest-energy state should be one of exactly zero energy, perfectly at rest. Zero-point energy (Corollaries) rules this out entirely — confinement itself is precisely what forces a nonzero minimum energy, an early, clean illustration of the uncertainty principle's real physical consequences.
Worked examples
Reading. Even the simplest possible confined-particle model gives energy gaps of a size directly comparable to visible/near-UV photon energies once the box length is nanometre-scale, exactly the length scale of a real conjugated pi system.
Scope. The identical \(E_n\) formula and gap calculation apply to any particle mass and box length, with the numerical outcome depending only on those two inputs.
Problems
- Compute \(E_1\), \(E_2\), and \(E_3\) for an electron in a \(2\,\text{nm}\) box, and comment on the spacing pattern.
Solution
\(E_n=n^2h^2/(8m_eL^2)\); with \(L=2\times10^{-9}\,\text{m}\), \(E_1\propto1\), \(E_2\propto4\), \(E_3\propto9\) (in units of \(E_1\)), so the levels are not evenly spaced — the gap \(E_2-E_1=3E_1\) is smaller than the gap \(E_3-E_2=5E_1\), spacing that grows with \(n\), unlike the evenly spaced harmonic oscillator. - Show that as \(L\to\infty\) at fixed \(n\), \(E_n\to0\), and interpret this physically.
Solution
\(E_n=n^2h^2/(8mL^2)\) decreases without bound as \(L\) grows for any fixed \(n\), approaching zero as \(L\to\infty\); physically, this recovers the classical free-particle limit, where an unconfined particle can have arbitrarily small (continuous) kinetic energy, consistent with quantisation being a direct consequence of confinement (Result) rather than an intrinsic property of the particle itself. - Compute the probability of finding the particle in the left half of the box (\(0\) to \(L/2\)) for the \(n=1\) state, and explain qualitatively why the \(n=2\) state gives the same result despite having a node at the centre.
Solution
For \(n=1\), \(\psi_1^2\) is symmetric about the box's midpoint (a single smooth hump peaking at \(L/2\)), so exactly half the probability lies in each half: \(P(0\to L/2)=0.5\). For \(n=2\), \(\psi_2^2\) has a node exactly at the centre but is still symmetric left-right about that centre, so the total probability again splits exactly \(0.5/0.5\) between the two halves, even though the detailed shape (two humps either side of the node) differs completely from the \(n=1\) case.