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PU-202 / L06 · lecture

The Infinite Square Well

~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable solving second-order linear differential equations with boundary conditions and understand the probabilistic interpretation of the wavefunction \(\psi(x)\).

Why this matters

The infinite square well is the simplest non-trivial quantum system that exhibits discrete energy levels.

While the free particle has a continuous spectrum, confining a particle to a finite region forces the wavefunction to satisfy specific boundary conditions. This constraint quantises the allowed energies, providing the first concrete example of the quantisation that defines quantum mechanics. It serves as the foundational model for understanding bound states, zero-point energy, and the structure of the Hilbert space formalism introduced in the next lecture.

The Physical Picture

Imagine a particle of mass \(m\) trapped inside a one-dimensional box of length \(L\). The walls of the box are impenetrable; if the particle were to touch them, the potential energy would become infinite. Consequently, the probability of finding the particle outside the region \(0 < x < L\) is exactly zero. Inside the box, the potential is zero, so the particle behaves like a free particle, but its wavefunction must vanish at the boundaries \(x=0\) and \(x=L\). This requirement forces the wave to "fit" into the box in specific ways, much like a string fixed at both ends can only vibrate at certain frequencies. These allowed modes correspond to discrete energy levels, and the lowest energy is not zero, reflecting the fact that a confined particle cannot be completely at rest.

Boundary Conditions and Quantisation

The time-independent Schrödinger equation inside the well (\(V=0\)) is \(-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi\). The general solution is a combination of sine and cosine terms. Applying the boundary condition \(\psi(0)=0\) eliminates the cosine term, leaving \(\psi(x) = A \sin(kx)\). The second boundary condition, \(\psi(L)=0\), requires \(\sin(kL)=0\). This implies that \(kL\) must be an integer multiple of \(\pi\), i.e., \(k_n = \frac{n\pi}{L}\) for \(n=1, 2, 3, \dots\). Note that \(n=0\) yields a trivial zero wavefunction and is discarded. The wave number \(k\) is related to energy by \(E = \frac{\hbar^2 k^2}{2m}\). Substituting \(k_n\) yields the discrete energy spectrum:

\[ E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \]

The corresponding normalised eigenfunctions are:

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

The factor \(\sqrt{2/L}\) ensures that \(\int_0^L |\psi_n(x)|^2 dx = 1\). The integer \(n\) is the quantum number, labelling the stationary states. The ground state (\(n=1\)) has no nodes inside the well, while the first excited state (\(n=2\)) has one node at \(x=L/2\).

Zero-Point Energy and Scaling

A striking feature of the infinite square well is that the minimum energy \(E_1\) is non-zero. This is a direct consequence of the Heisenberg uncertainty principle. Confining the particle to a region of size \(L\) implies an uncertainty in position \(\Delta x \sim L\). The uncertainty principle \(\Delta x \Delta p \geq \hbar/2\) then implies a minimum momentum uncertainty \(\Delta p \sim \hbar/L\). The kinetic energy, proportional to \(p^2\), therefore has a lower bound proportional to \(\hbar^2/L^2\). This "zero-point energy" is a purely quantum effect with no classical analogue; a classical particle in a box could have zero kinetic energy if it were at rest.

The energy levels scale as \(E_n \propto n^2\). This quadratic dependence arises because the wavelength \(\lambda_n = 2L/n\) decreases as \(n\) increases, leading to a higher wave number \(k_n\) and thus higher kinetic energy. The spacing between adjacent levels, \(\Delta E_n = E_{n+1} - E_n\), increases with \(n\). Specifically, \(\Delta E_n \propto (2n+1)\). This means that for large \(n\), the energy levels become more widely spaced, but the relative spacing \(\Delta E_n / E_n \approx 2/n\) decreases, recovering the classical continuum limit.

Probability Density and Nodes

The probability density \(|\psi_n(x)|^2\) determines where the particle is likely to be found. For the ground state, \(|\psi_1(x)|^2 = \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right)\), which is symmetric about \(x=L/2\) and peaks at the centre. For higher states, the probability density oscillates, with \(n-1\) nodes (points where the probability is zero) inside the well. These nodes are a direct consequence of the boundary conditions and the orthogonality of the eigenfunctions. The expectation value of position for any state \(\psi_n\) is \(\langle x \rangle = L/2\), reflecting the symmetry of the potential. However, the expectation value of momentum is zero, \(\langle p \rangle = 0\), because the particle is equally likely to be moving left or right in a stationary state.

The orthogonality of the eigenfunctions, \(\int_0^L \psi_m(x) \psi_n(x) dx = \delta_{mn}\), is a crucial property. It ensures that different energy states are linearly independent and form a complete basis for the Hilbert space of square-integrable functions on \([0, L]\). This completeness allows any arbitrary wavefunction \(\Psi(x)\) to be expanded as a superposition of the eigenstates \(\psi_n(x)\), with coefficients determined by the inner product \(\langle \psi_n | \Psi \rangle\).

The derivation

◆ DERIVATION EMBED · D-143 · not re-derived here

Bound States of the Infinite Square Well

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 En = n2π2ℏ2/(2mL2), n = 1, 2, 3, …, with orthonormal eigenfunctions ψn(x) = √(2/L) sin(nπx/L).

En = n2π2ℏ2 / (2mL2),    ψn(x) = √(2/L) sin(nπx/L),    n = 1, 2, 3, …
Open full derivation →
Tier 3 · watch out

A common misconception is that the particle "bounces" off the walls in the infinite square well. In reality, the wavefunction is zero at the walls, and the probability current is zero everywhere in a stationary state. The particle does not have a definite trajectory; rather, its state is a standing wave. The concept of "bouncing" is a classical heuristic that fails to capture the quantum nature of the system, particularly the absence of a well-defined position and momentum simultaneously. C

Worked example

Problem. An electron is confined in a one-dimensional infinite square well of width \(L = 1.0 \times 10^{-9} \mathrm{m}\). Calculate the ground state energy \(E_1\) in electronvolts (eV) and the first excited state energy \(E_2\).

  1. Identify the given values: \(m = 9.11 \times 10^{-31} \mathrm{kg}\), \(L = 1.0 \times 10^{-9} \mathrm{m}\), \(\hbar = 1.055 \times 10^{-34} \mathrm{J \cdot s}\).
  2. Use the energy formula \(E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}\) for \(n=1\).
  3. Calculate \(E_1 = \frac{(1)^2 \pi^2 (1.055 \times 10^{-34})^2}{2(9.11 \times 10^{-31})(1.0 \times 10^{-9})^2}\).
  4. Numerator: \(\pi^2 \approx 9.87\), \(\hbar^2 \approx 1.113 \times 10^{-68}\). Product \(\approx 1.10 \times 10^{-67}\).
  5. Denominator: \(2 \times 9.11 \times 10^{-31} \times 10^{-18} = 1.822 \times 10^{-48}\).
  6. \(E_1 \approx \frac{1.10 \times 10^{-67}}{1.822 \times 10^{-48}} \approx 6.04 \times 10^{-20} \mathrm{J}\).
  7. Convert to eV: \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\). \(E_1 \approx \frac{6.04 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.377 \mathrm{eV}\).
  8. Calculate \(E_2 = 4 E_1 \approx 4 \times 0.377 = 1.51 \mathrm{eV}\).
\[ E_1 \approx 0.38 \mathrm{eV}, \quad E_2 \approx 1.51 \mathrm{eV} \]

Units check. The formula yields Joules. Conversion to eV is consistent with the definition of the electronvolt. The scaling \(E_2 = 4E_1\) is correct for \(n^2\) dependence.

CHECK YOURSELF

Three quick tests

  1. What is the value of the wavefunction \(\psi_n(x)\) at the boundaries \(x=0\) and \(x=L\)?
    AnswerIt is zero at both boundaries, satisfying the Dirichlet boundary conditions.
  2. How does the energy \(E_n\) scale with the width \(L\) of the well?
    AnswerIt scales as \(1/L^2\). A smaller well leads to higher energy levels due to increased confinement.
  3. Why is the ground state energy non-zero?
    AnswerDue to the Heisenberg uncertainty principle. Confinement implies momentum uncertainty, resulting in a minimum kinetic energy.
PROBLEMS

Problem set for PU-202 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-143 · Threads: matter, waves, energy