Modern Physics · Undergraduate Engineering

Particle in a One-Dimensional Infinite Potential Well

A particle is confined between two perfectly impenetrable walls separated by a distance a. This simple model shows how confinement leads to discrete energy levels and characteristic wave functions.

1. Physical picture: the particle in a box

Schematic of a one-dimensional infinite potential well, with infinite walls at x equals zero and x equals a, zero potential inside, and wave-function and probability-density curves for n equals 1, 2 and 3. — Conceptual schematic of the well and its first three stationary states. The top arrow illustrates the one-dimensional direction; it is not a trajectory for a stationary quantum eigenstate.

Inside the well

For \(0\lt x\lt a\), the potential energy is defined as \(V(x)=0\). The particle is free of potential-energy forces in this region in the idealized model.

At and outside the walls

The ideal walls have \(V(x)=\infty\) for \(x\leq 0\) and \(x\geq a\). The particle cannot exist beyond the walls, so the wave function must vanish at both boundaries.

Model assumption: “Free to move” means the potential is constant inside the box. In quantum mechanics, the confined particle is described by a wave function; a stationary state does not have a definite classical back-and-forth path.

2. Boundary conditions and allowed states

For a particle of mass m in the well, the time-independent Schrödinger equation inside the box is

\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi,\qquad 0\lt x\lt a \]

Impenetrable walls require \(\psi(0)=0\) and \(\psi(a)=0\). Only standing waves that fit an integer number of half-wavelengths inside the box satisfy both conditions:

\[ \psi_n(x)=\sqrt{\frac{2}{a}}\sin\!\left(\frac{n\pi x}{a}\right),\qquad n=1,2,3,\ldots \]

The associated quantized energies are

\[ E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}=\frac{n^2h^2}{8ma^2} \]

Here, \(\hbar=\frac{h}{2\pi}\). The energy is proportional to \(n^2\) and inversely proportional to the square of the box width, \(a^2\). The ground-state energy is non-zero: confinement does not permit a zero-energy stationary state.

3. Explore the first three stationary states

Each row pairs the wave function with its probability density. The two plots share the same horizontal coordinate and the probability-density baseline is aligned with the wave function’s \(\psi=0\) axis.

Marker speedSlow

The orange marker is a synchronized teaching aid that helps compare the same x-position in both plots. It is not a literal trajectory, and the probability density of a stationary eigenstate does not travel with the marker.

4. How to interpret the plots

Wave function \(\psi_n(x)\): may be positive or negative; its sign is an amplitude, not a negative probability.
Probability density \(|\psi_n(x)|^2\): is always non-negative. The probability of finding the particle between x and x + dx is approximately \(|\psi_n(x)|^2\,dx\).
Nodes: points where \(\psi_n(x)=0\). The state n has n − 1 internal nodes; all states also vanish at the two walls.
Ground state \(n=1\): one positive lobe and no internal node.
First excited state \(n=2\): two lobes of opposite wave-function sign and one internal node at a/2.
Second excited state \(n=3\): three lobes and two internal nodes at a/3 and \(\frac{2a}{3}\).

Important distinction: ψₙ(x) oscillates about its zero axis, while |ψₙ(x)|² sits on or above that same zero baseline. Squaring the wave function removes its sign and gives the spatial probability distribution.

5. Check your understanding

Why must the wave function be zero at x = 0 and x = a?
How does the energy of n = 3 compare with the ground-state energy E₁?
Where is the internal node of the first excited state?
Why can probability density never be negative even though the wave function can?

Show concise answers

The infinite walls exclude the particle from the outside region, requiring the boundary values \(\psi(0)=\psi(a)=0\).
\(E_3=3^2E_1=9E_1\).
At \(x=\frac{a}{2}\).
Because \(|\psi|^2\) is the squared magnitude of the wave function.