Result

The particle in a box

T-070Home CU-301Threads quantum

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

The potential energy is exactly zero inside a fixed region of length \(L\) and infinite outside it.An idealisation: no real potential is a perfectly sharp, infinite well, but it is an excellent approximation for strongly confined systems and is exactly solvable, making it the standard first teaching example in quantum chemistry. The wavefunction is continuous and vanishes exactly at the box boundaries, \(\psi(0)=\psi(L)=0\).This boundary condition is required because the wavefunction must be finite and continuous everywhere, and cannot penetrate a genuinely infinite potential barrier. Only a single, non-interacting particle is considered.For multiple particles in a box, Pauli exclusion (for fermions) must additionally be imposed on top of the single-particle energy levels derived here, filling levels from the bottom up exactly as in the aufbau-type filling of atomic orbitals.

Proof

\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \]

Inside the box, \(V=0\), so the time-independent Schrödinger equation reduces to this simple second-order differential equation. A

\[ \psi(x) = A\sin(kx) + B\cos(kx), \qquad k^2 = \frac{2mE}{\hbar^2} \]

The general solution to Step 1 is a sinusoidal combination, with wavenumber \(k\) related to the energy. A

\[ \psi(0)=0 \Rightarrow B=0; \qquad \psi(L)=0 \Rightarrow \sin(kL)=0 \Rightarrow kL=n\pi,\ n=1,2,3,\ldots \]

Applying both boundary conditions (Hypotheses) forces \(B=0\) and restricts \(k\) to a discrete set of allowed values; \(n=0\) is excluded since it gives \(\psi=0\) everywhere — no particle at all. A

\[ E_n = \frac{n^2h^2}{8mL^2} \]

Substituting the allowed \(k=n\pi/L\) back into \(k^2=2mE/\hbar^2\) and solving for \(E\) gives the quantised energy levels, growing as \(n^2\). A

\[ \psi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right) \]

Normalising via \(\int_0^L|\psi_n|^2\,dx=1\) fixes the constant \(A\), giving the complete, normalised wavefunction for each allowed level. B

Result

\[ E_n = \frac{n^2h^2}{8mL^2}, \ n=1,2,3,\ldots \qquad \psi_n(x)=\sqrt{\tfrac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right) \]

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

\[ L=1\,\text{nm}, \ m=m_e: \quad E_1 = \frac{h^2}{8m_eL^2} \]

Substituting Planck's constant and the electron mass gives \(E_1\approx6\times10^{-20}\,\text{J}\approx0.38\,\text{eV}\), noticeably larger than typical room-temperature thermal energy \(kT\approx0.025\,\text{eV}\), illustrating that nanoscale confinement produces genuinely significant quantisation at ordinary temperature. A

\[ \Delta E = E_2-E_1 = \frac{3h^2}{8mL^2} \]

The energy gap for an \(n=1\to n=2\) transition follows directly from the Result; for a conjugated chain modelled qualitatively as a box, this same gap sets the wavelength of light absorbed, via \(\Delta E=hc/\lambda\), connecting this result directly to uv-vis-electronic. A

\[ E_1(1\,\text{nm electron}) \approx 0.38\,\text{eV}; \qquad \Delta E_{1\to2} = 3E_1 \]

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.

Further example. An electron is confined in a one-dimensional box of length \( L = 0.50 \, \text{nm} \). Calculate the wavelength \( \lambda \) of the photon emitted when the electron transitions from the \( n=2 \) state to the \( n=1 \) state. Use \( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_e = 9.109 \times 10^{-31} \, \text{kg} \), and \( c = 2.998 \times 10^8 \, \text{m/s} \).

\[ \Delta E = E_2 - E_1 = \frac{4h^2}{8m_eL^2} - \frac{h^2}{8m_eL^2} = \frac{3h^2}{8m_eL^2} \]

The energy difference between quantum states is the difference of their individual energies. Since \( E_n \propto n^2 \), the gap for \( 2 \to 1 \) is \( (2^2 - 1^2)E_1 = 3E_1 \). A

\[ \Delta E = \frac{3(6.626 \times 10^{-34})^2}{8(9.109 \times 10^{-31})(0.50 \times 10^{-9})^2} \]

Substitute the given constants and the box length \( L = 0.50 \, \text{nm} = 5.0 \times 10^{-10} \, \text{m} \) into the expression derived in Step 1. A

\[ \Delta E = \frac{3(4.390 \times 10^{-67})}{8(9.109 \times 10^{-31})(2.50 \times 10^{-19})} = \frac{1.317 \times 10^{-66}}{1.822 \times 10^{-48}} \approx 7.23 \times 10^{-19} \, \text{J} \]

Perform the arithmetic: the numerator is \( 3h^2 \) and the denominator is \( 8m_eL^2 \). The result is the energy of the emitted photon in Joules. A

\[ \lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(2.998 \times 10^8)}{7.23 \times 10^{-19}} \]

Use the photon energy-wavelength relation \( E = hc/\lambda \) to solve for \( \lambda \). This connects the quantised energy gap to the observable spectral line. A

\[ \lambda = \frac{1.986 \times 10^{-25}}{7.23 \times 10^{-19}} \approx 2.75 \times 10^{-7} \, \text{m} = 275 \, \text{nm} \]

Calculate the final wavelength. The value falls in the ultraviolet region, consistent with the high energy density of a nanoscale confinement. A

\[ \lambda \approx 275 \, \text{nm} \]

Reading. A transition in a 0.5 nm box emits UV light. This demonstrates that smaller boxes (stronger confinement) lead to larger energy gaps and shorter wavelengths, a key principle in nanotechnology and quantum dots.

Scope. This result holds for a particle in a 1D infinite potential well. It assumes the electron is the only particle and ignores electron-electron interactions, which are significant in real multi-electron systems but negligible for a single-particle model.

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.
Calculate the energy of the \(n=3\) state for an electron in a 1.0 nm box. Express your answer in Joules and eV.
Solution
Using \(E_n = \frac{n^2 h^2}{8 m_e L^2}\): \(L = 1.0 \times 10^{-9}\,\text{m}\). \(E_1 = \frac{h^2}{8 m_e L^2} \approx 6.02 \times 10^{-20}\,\text{J}\) (from previous examples, \(E_1\) for 1 nm is approx 0.376 eV). Let's calculate precisely: \(E_1 = \frac{(6.626 \times 10^{-34})^2}{8 (9.109 \times 10^{-31}) (10^{-9})^2} = \frac{4.390 \times 10^{-67}}{7.287 \times 10^{-48}} = 6.024 \times 10^{-20}\,\text{J}\). In eV: \(6.024 \times 10^{-20} / 1.602 \times 10^{-19} = 0.376\,\text{eV}\). For \(n=3\), \(E_3 = 9 E_1\). \(E_3 = 9 \times 6.024 \times 10^{-20}\,\text{J} = 5.42 \times 10^{-19}\,\text{J}\). \(E_3 = 9 \times 0.376\,\text{eV} = 3.38\,\text{eV}\).
An electron is confined in a 1D box. The energy difference between the \(n=1\) and \(n=2\) states is measured to be \(2.55\,\text{eV}\). Calculate the length of the box \(L\).
Solution
The energy gap is \(\Delta E = E_2 - E_1 = 3 E_1 = \frac{3 h^2}{8 m_e L^2}\). Rearranging for \(L\): \(L^2 = \frac{3 h^2}{8 m_e \Delta E}\) \(L = \sqrt{\frac{3 h^2}{8 m_e \Delta E}}\) Given \(\Delta E = 2.55\,\text{eV} = 2.55 \times 1.602 \times 10^{-19}\,\text{J} = 4.085 \times 10^{-19}\,\text{J}\). \(L = \sqrt{\frac{3 (6.626 \times 10^{-34})^2}{8 (9.109 \times 10^{-31}) (4.085 \times 10^{-19})}}\) Numerator: \(3 \times 4.390 \times 10^{-67} = 1.317 \times 10^{-66}\). Denominator: \(8 \times 9.109 \times 10^{-31} \times 4.085 \times 10^{-19} = 297.6 \times 10^{-50} = 2.976 \times 10^{-48}\). \(L^2 = \frac{1.317 \times 10^{-66}}{2.976 \times 10^{-48}} = 0.4425 \times 10^{-18} = 4.425 \times 10^{-19}\,\text{m}^2\). \(L = \sqrt{4.425 \times 10^{-19}} = \sqrt{44.25 \times 10^{-20}} \approx 6.65 \times 10^{-10}\,\text{m} = 0.665\,\text{nm}\).