The Infinite Square Well

Consider a particle confined to a region $0 < x < a$ where the potential is zero, with infinite walls at $x = 0$ and $x = a$:

Outside the well, the wave function must be zero because the particle cannot exist where $V = \infty$. Inside the well ($0 < x < a$), the time-independent Schrödinger equation becomes:

The general solution is $\phi(x) = A\sin(kx) + B\cos(kx)$, where $k = \sqrt{2mE}/\hbar$. The boundary condition $\phi(0) = 0$ gives $B = 0$. The boundary condition $\phi(a) = 0$ requires $\sin(ka) = 0$, so $ka = n\pi$ for integer $n = 1, 2, 3, \dots$

The quantum number $n = 0$ is excluded because it would give $\phi(x) = 0$ everywhere, which is not normalizable. Negative values of $n$ give the same physical states as positive $n$ (up to an overall sign), so we restrict to $n \ge 1$.

The energy levels are quantized: $E_n = n^2 E_1$ where $E_1 = \pi^2\hbar^2/(2ma^2)$ is the ground state energy. The spacing between adjacent levels increases with $n$: $E_{n+1} - E_n = (2n+1)E_1$.

The eigenfunctions are orthonormal: $\int_0^a \phi_m^*(x)\,\phi_n(x)\,dx = \delta_{mn}$. This can be verified using trigonometric identities.

The eigenfunctions form a complete set: any reasonably well-behaved function $f(x)$ on $[0,a]$ can be expanded as a Fourier sine series: $f(x) = \sum_n c_n \phi_n(x)$. This is precisely the expansion we need to solve the time-dependent problem.

The $n$th eigenfunction has $n-1$ nodes (zeros) inside the well. The ground state ($n=1$) has no nodes, the first excited state ($n=2$) has one node at $x = a/2$, and so on. More nodes correspond to higher energy, a general feature of quantum mechanics.

Given an initial wave function $\psi(x,0)$, we expand it in the energy eigenbasis:

The expansion coefficients are found by Fourier's trick: multiply both sides by $\phi_m^*(x)$ and integrate, using orthonormality:

The full time-dependent solution attaches the phase factor $e^{-iE_n t/\hbar}$ to each term:

Although each stationary state has a time-independent probability density, a superposition does not. The probability density $|\psi(x,t)|^2$ can exhibit complex time-dependent behavior, including oscillations and revivals.

A particle in an infinite square well of width $a$ has initial wave function $\psi(x,0) = A\,x(a-x)$. Find the expansion coefficients, the probability of the ground state, and the expectation value of the energy.

Step 1 — Normalization. $\int_0^a x^2(a-x)^2\,dx = a^5/30$. Thus $A = \sqrt{30/a^5}$.

Step 2 — Expansion coefficients. $c_n = \int_0^a \sqrt{2/a}\,\sin(n\pi x/a) \cdot \sqrt{30/a^5}\,x(a-x)\,dx$. Using integration by parts twice, $c_n = \frac{2\sqrt{15}}{n^3\pi^3}[1-(-1)^n]$. Only odd $n$ contribute.

Step 3 — Ground state probability. $P(E_1) = |c_1|^2 = 960/\pi^6 \approx 0.9986$. The particle is almost certainly in the ground state because the parabolic initial state closely resembles $\sin(\pi x/a)$.

Step 4 — Energy expectation value. $\langle H\rangle = \sum_n |c_n|^2 E_n = \sum_{n\text{ odd}} \frac{960}{n^6\pi^6} \cdot \frac{n^2\pi^2\hbar^2}{2ma^2} = \frac{5\hbar^2}{ma^2}$.