Boundary Conditions in the Well

The infinite square well

The infinite square well — often called a particle in a box — is the simplest system in which a particle is confined by quantum mechanics. It consists of a potential energy that is zero inside an interval of length $L$ and infinite everywhere outside:

V(x) = \begin{cases} 0, & 0 \le x \le L, \\ \infty, & \text{otherwise.} \end{cases}

A particle with finite energy cannot penetrate a region where $V = \infty$ , so the wave function must vanish in those regions. Outside the well,

\psi(x) = 0 \quad \text{for } x < 0 \text{ and } x > L.

Why the wave function must be continuous

One of the fundamental requirements of quantum mechanics is that the wave function $\psi(x)$ be continuous everywhere. The reason is the Schrödinger equation itself: an abrupt jump in $\psi$ would make the first derivative $d\psi/dx$ contain a delta-function spike, and the second derivative $d^2\psi/dx^2$ blow up even more violently. But the equation ties $d^2\psi/dx^2$ to $(V-E)\psi$ , which stays finite wherever $V$ is finite — so no such jump in $\psi$ is allowed. (At an infinitely steep wall, where $V \to \infty$ , only the slope $d\psi/dx$ is permitted to jump; $\psi$ itself must still join on continuously.)

Combining continuity with the fact that $\psi = 0$ outside the well, continuity at the boundaries forces

\psi(0) = 0 \quad \text{and} \quad \psi(L) = 0.

These two equations are the boundary conditions for the infinite square well.

General solution inside the well

Inside the well ( $0 \le x \le L$ ) the potential is zero, so the time-independent Schrödinger equation reduces to

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi.

Rearranging gives $d^2\psi/dx^2 = -k^2\psi$ with $k^2 = 2mE/\hbar^2$ , whose general solution is

\psi(x) = A\sin(kx) + B\cos(kx).

Now apply the boundary conditions one at a time.

At $x = 0$ :

\psi(0) = A\sin(0) + B\cos(0) = B = 0.

So $B = 0$ and only the sine term survives:

\psi(x) = A\sin(kx).

At $x = L$ :

\psi(L) = A\sin(kL) = 0.

Because $A \neq 0$ (a zero wave function everywhere would not be normalizable), the sine itself must vanish:

\sin(kL) = 0.

This is satisfied when $kL = n\pi$ for any positive integer $n = 1, 2, 3, \ldots$ , giving

k_n = \frac{n\pi}{L}.

Negative integers produce the same set of functions (just with a sign flip absorbed into $A$ ), and $n = 0$ gives $\psi = 0$ identically, which is not normalizable.

What the boundary conditions have achieved

Starting from a completely general oscillatory solution, two algebraic constraints — $\psi(0) = 0$ and $\psi(L) = 0$ — have forced:

The cosine term to vanish entirely ( $B = 0$ ).
The wave number $k$ to take only the discrete set of values $k_n = n\pi/L$ .

The stationary-state wave functions are therefore

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

The amplitude $A$ is fixed by normalization in the next steps of the derivation. These functions are orthogonal and form a complete basis for square-integrable functions on $[0, L]$ — any allowed state of the particle can be written as a superposition of them.