The Infinite Square Well

The infinite square well — also called a particle in a box — is the first exactly solvable quantum system. It is simple enough to work out by hand yet rich enough to introduce every major idea in bound-state quantum mechanics: energy quantization, stationary states, wave-function normalization, and the zero-point energy. Everything in this module builds on this foundation.

Defining the potential

Imagine a particle of mass $m$ that is perfectly free inside a one-dimensional region of length $L$ but faces an impenetrable wall at both ends. Formally, the potential energy is

\begin{cases} 0 & 0 \leq x \leq L \\ \infty & \text{otherwise.} \end{cases}$$ The infinite walls are an idealization: in practice any real potential well has finite depth, but the infinite well captures the essential physics while keeping the algebra clean. Setting the left wall at $x = 0$ and the right wall at $x = L$ is a standard convention; the physics does not depend on where you place the origin. ## What the Schrödinger equation asks For a time-independent problem — a particle with definite energy $E$ — the wave function $\psi(x)$ satisfies the **time-independent Schrödinger equation** (TISE): $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\,\psi(x) = E\,\psi(x).$$ Outside the well, $V = \infty$ and the equation can only be satisfied if $\psi = 0$ there: the particle cannot exist in a region where its potential energy would be infinite. This gives the **boundary conditions** $$\psi(0) = 0, \qquad \psi(L) = 0.$$ Inside the well, $V = 0$ and the TISE simplifies to a familiar second-order ordinary differential equation: $$\frac{d^2\psi}{dx^2} = -k^2\,\psi, \qquad k^2 \equiv \frac{2mE}{\hbar^2}.$$ Here $k$ has units of inverse length and plays the role of the wave number. Because $E$ must be positive for a nontrivial solution to exist (if $E \leq 0$ then $k^2 \leq 0$, the solutions inside the well are $\sinh$ and $\cosh$, and the only combination that satisfies both $\psi(0)=0$ and $\psi(L)=0$ is the trivial $\psi \equiv 0$), we take $k > 0$. ## The general solution inside the well The general solution of $\psi'' = -k^2 \psi$ is a linear combination of sine and cosine: $$\psi(x) = A\sin(kx) + B\cos(kx),$$ where $A$ and $B$ are constants. Applying the left boundary condition $\psi(0) = 0$ immediately forces $B = 0$, because $\cos(0) = 1 \neq 0$. The solution reduces to $$\psi(x) = A\sin(kx).$$ Now apply the right boundary condition $\psi(L) = 0$: $$A\sin(kL) = 0.$$ Since $A = 0$ would give the trivial (unphysical) solution $\psi \equiv 0$, we need $\sin(kL) = 0$, which requires $kL = n\pi$ for some positive integer $n$. This is the quantization condition. <Callout type="tip"> The integer $n$ is the **quantum number** of the state. Only discrete values of $k$ — and therefore only discrete values of energy — are allowed. This is why the energy spectrum is quantized. </Callout> ## Quantized wave numbers and energies Inserting $k_n = n\pi / L$ into the definition $k^2 = 2mE/\hbar^2$ and solving for $E$ gives the **energy eigenvalues**: $$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2mL^2}, \qquad n = 1, 2, 3, \ldots$$ Several features are immediately remarkable: - **Quantization**: only energies $E_1, E_2, E_3, \ldots$ are possible — there is no classical counterpart to this restriction. - **Zero-point energy**: the lowest allowed energy is $E_1 = \pi^2\hbar^2/(2mL^2) > 0$. The particle can never be at rest; this is a direct consequence of the uncertainty principle — a particle at rest would have $\Delta p = 0$, which the uncertainty relation $\Delta x \,\Delta p \geq \hbar/2$ forbids when the particle is confined to a finite region. - **Level spacing grows with $n$**: the gap between consecutive levels is $E_{n+1} - E_n = (2n+1)E_1$, so higher levels are more widely spaced. The corresponding wave functions are $$\psi_n(x) = A_n \sin\!\left(\frac{n\pi x}{L}\right), \qquad 0 \leq x \leq L.$$ The normalization constant $A_n$ is fixed by the requirement $\int_0^L |\psi_n(x)|^2\,dx = 1$, which gives $A_n = \sqrt{2/L}$. (The derivation uses the identity $\int_0^L \sin^2(n\pi x / L)\,dx = L/2$.) The normalized wave functions are therefore $$\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right).$$ ## Physical picture Each $\psi_n$ is a standing wave that fits exactly $n$ half-wavelengths inside the box: the wavelength is $\lambda_n = 2L/n$, so the wave number $k_n = 2\pi/\lambda_n = n\pi/L$ matches what we derived from the boundary conditions. The ground state ($n=1$) has one arch, the first excited state ($n=2$) has two arches with a node at the center, and so on. The number of nodes (zeros of $\psi$ in the interior) equals $n - 1$. Probability is distributed as $|\psi_n(x)|^2 = (2/L)\sin^2(n\pi x / L)$. For large $n$, this oscillates rapidly and its average over any macroscopic interval approaches $1/L$ — uniform across the box. This agreement with the classical prediction (the particle spends equal time everywhere when bouncing back and forth) is an instance of the **correspondence principle**: quantum mechanics reduces to classical mechanics in the appropriate limit. ## Summary The infinite square well sets up the template for all bound-state problems: 1. Write down the potential and identify the regions. 2. Solve the TISE in each region. 3. Apply boundary conditions — here, $\psi = 0$ at both walls. 4. Accept only solutions that are normalizable — here, $A \neq 0$ forces $kL = n\pi$. 5. Read off the energy eigenvalues and normalized eigenfunctions. In the next lesson we examine the boundary conditions more carefully and see how the continuity requirements differ when the well walls are finite rather than infinite.