Superpositions in the Box

The energy eigenstates $\psi_n$ of the infinite square well are the simplest solutions to the Schrödinger equation, but a real particle is rarely in a single eigenstate. The most general state is a superposition — a linear combination of eigenstates — and the way that superposition evolves in time reveals beating patterns and oscillating probability densities that have no classical parallel.

Building a superposition

The normalized eigenstates of the infinite square well of width $L$ are

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

with energies $E_n = n^2 E_1$ where $E_1 = \pi^2\hbar^2/(2mL^2)$ is the ground-state energy.

Any normalizable wave function inside the well can be written as a superposition of these eigenstates. At time $t = 0$ the general state is

$\Psi(x, 0) = \sum_{n=1}^{\infty} c_n\,\psi_n(x),$

where the complex coefficients $c_n$ encode how much of each eigenstate is present. Normalization requires the coefficients to satisfy

$\sum_{n=1}^{\infty} |c_n|^2 = 1,$

because the $\psi_n$ form an orthonormal set: $\int_0^L \psi_m(x)\,\psi_n(x)\,dx = \delta_{mn}$ . Squaring and integrating $|\Psi|^2$ then reduces to the sum of $|c_n|^2$ by orthonormality, so the condition above is precisely the requirement that total probability equals one.

Time evolution

Each energy eigenstate carries its own time-dependent phase factor. Adding the phase $e^{-iE_n t/\hbar}$ to the $n$ -th term gives the full time-evolved state:

$\Psi(x,t) = \sum_{n=1}^{\infty} c_n\,\psi_n(x)\,e^{-iE_n t/\hbar}.$

This is a direct consequence of the time-dependent Schrödinger equation: each stationary state $\psi_n(x) e^{-iE_n t/\hbar}$ solves it individually, and the equation is linear, so any superposition also solves it.

Probability of measuring a given energy

The Born rule for energy measurements is beautifully simple: if the particle is in state $\Psi$ and you measure its energy, the probability of obtaining the specific eigenvalue $E_n$ is

$P(E_n) = |c_n|^2.$

The particle collapses into eigenstate $\psi_n$ upon that measurement. Notice that the time-dependent phase $e^{-iE_n t/\hbar}$ has modulus one, so $P(E_n) = |c_n|^2$ is independent of when you measure — the energy probabilities are frozen even as the spatial probability density oscillates.

A worked two-state example

Consider a particle prepared at $t = 0$ in the state

$\Psi(x,0) = \frac{3}{5}\,\psi_1(x) + \frac{4}{5}\,\psi_2(x).$

First, check normalization: $(3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1$ . Good — the coefficients satisfy $|c_1|^2 + |c_2|^2 = 1$ .

The energy probabilities are

$P(E_1) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36, \qquad P(E_2) = \left(\frac{4}{5}\right)^2 = \frac{16}{25} = 0.64.$

They add to one, as required.

The time-evolved state is

$\Psi(x,t) = \frac{3}{5}\,\psi_1(x)\,e^{-iE_1 t/\hbar} + \frac{4}{5}\,\psi_2(x)\,e^{-iE_2 t/\hbar}.$

The probability density is

+ 2\cdot\frac{3}{5}\cdot\frac{4}{5}\,\psi_1(x)\psi_2(x)\cos\!\left(\frac{(E_2-E_1)t}{\hbar}\right).$$ The cross term oscillates at the **beat frequency** $(E_2 - E_1)/\hbar = (4-1)E_1/\hbar = 3E_1/\hbar$, so the probability density sloshes back and forth with period $T = 2\pi\hbar/(3E_1)$. This is a quantum phenomenon with no classical analogue. ## Try it This is a numerical exercise — `return` a number. A particle is prepared in $\Psi = (3/5)\,\psi_1 + (4/5)\,\psi_2$. What is the probability of measuring energy $E_1$? <RunnableExample />