Particle in a One-Dimensional Box, and Quantum Tunnelling

By the end you should be able to: Derive the energy eigenvalues and normalised wave functions for a particle in a one-dimensional infinite potential well, calculate the allowed energies for a given width, and describe quantum tunnelling qualitatively.

This is the one problem in the syllabus you can solve completely, and it earns its place because quantisation emerges from it rather than being assumed. Bohr had to postulate discrete orbits; here the discrete energies fall out of a differential equation and a boundary condition.

The problem

A particle of mass $m$ is confined to $0 < x < L$ by walls of infinite potential:

V(x) = \begin{cases} 0 & 0 < x < L \\ \infty & \text{elsewhere}\end{cases}

Infinite potential means zero probability of being outside, so

\psi = 0 \quad \text{for } x \le 0 \text{ and } x \ge L

and by continuity, $\psi(0) = \psi(L) = 0$ . These are the boundary conditions, and they do all the work.

Solving inside the box

Inside, $V = 0$ , so the time independent equation reads

\frac{d^{2}\psi}{dx^{2}} + \frac{2mE}{\hbar^{2}}\psi = 0

Write $k^{2} = \dfrac{2mE}{\hbar^{2}}$ , giving

\frac{d^{2}\psi}{dx^{2}} + k^{2}\psi = 0

the simple harmonic form, with general solution

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

Applying the boundary conditions

At $x = 0$ :

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

So $B = 0$ and $\psi = A\sin kx$ . The cosine is eliminated because it does not vanish at the origin.

At $x = L$ :

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

Now $A \neq 0$ , or the wave function would vanish everywhere and describe no particle. Therefore

\sin kL = 0 \quad\Rightarrow\quad kL = n\pi, \qquad n = 1, 2, 3, \dots

k = \frac{n\pi}{L}

Note $n \neq 0$ . Taking $n = 0$ gives $k = 0$ and $\psi \equiv 0$ — no particle. This single exclusion is what produces the zero point energy below.

The energy eigenvalues

Substitute back into $k^{2} = 2mE/\hbar^{2}$ :

E_n = \frac{\hbar^{2}k^{2}}{2m} = \frac{\hbar^{2}}{2m}\frac{n^{2}\pi^{2}}{L^{2}}

Using $\hbar = h/2\pi$ :

\boxed{E_n = \frac{n^{2}h^{2}}{8mL^{2}}, \qquad n = 1,2,3,\dots}

Three consequences:

The energy is quantised. Only these values occur. Nothing was postulated; the boundary condition at $x = L$ produced it.
Energies go as $n^{2}$ , so levels get further apart as you go up: $E_1 : E_2 : E_3 = 1 : 4 : 9$ .
The lowest energy is not zero. $E_1 = h^{2}/8mL^{2}$ is the zero point energy. A confined particle can never be at rest — which is the uncertainty principle again: localising it within $L$ forces a momentum spread, hence kinetic energy. Confinement costs energy, and the narrower the box the higher the cost, since $E \propto 1/L^{2}$ .

The normalised wave function

With $\psi_n(x) = A\sin\dfrac{n\pi x}{L}$ , normalisation requires

\int_{0}^{L}A^{2}\sin^{2}\!\frac{n\pi x}{L}\,dx = 1

Using $\sin^{2}\theta = \tfrac12(1-\cos 2\theta)$ , the integral of the cosine over a whole number of half-periods vanishes and

A^{2}\cdot\frac{L}{2} = 1 \quad\Rightarrow\quad A = \sqrt{\frac{2}{L}}

\boxed{\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}}

Note $A$ is the same for every $n$ — the half-period cancellation works for all of them.

Reading the solutions

$\psi_1$ has no node between the walls; the particle is most likely found at the centre.
$\psi_2$ has a node at $x = L/2$ . The particle is never found at the midpoint, though it is found on both sides. Classically absurd, quantum mechanically routine.
$\psi_n$ has $n-1$ interior nodes.
As $n$ grows, the peaks crowd together and the density approaches the uniform classical distribution — the correspondence principle.

Quantum tunnelling — qualitative

The syllabus asks for a qualitative treatment, so no transmission coefficient is derived here.

Replace an infinite wall with a barrier of finite height $V_0$ and finite width, and send in a particle with $E < V_0$ . Classically it must reflect: it has not the energy to climb the barrier.

Quantum mechanically the wave function does not stop dead at the barrier. Inside, where $E < V$ , the equation gives a real exponential rather than an oscillation, so $\psi$ decays through the barrier rather than vanishing at it. If the barrier is thin enough, $\psi$ is still non-zero on the far side — and a non-zero $\psi$ means a non-zero probability of finding the particle there.

The particle can appear on the other side without ever having had enough energy to be on top. This is quantum tunnelling.

The transmission probability falls off exponentially with barrier width and with $\sqrt{V_0 - E}$ . That steep dependence is why tunnelling is invisible for everyday objects and decisive for electrons.

Where it matters:

Alpha decay — the alpha particle tunnels out of the nucleus.
The tunnel diode — Module 4, where tunnelling produces a negative resistance region.
The scanning tunnelling microscope — tunnelling current between tip and surface varies so sharply with distance that individual atoms are resolved.
Flash memory — charge tunnels onto and off the floating gate.
Leakage in small transistors — as gate oxides thin below a few nanometres, electrons tunnel through them, and the resulting leakage current is one of the limits on shrinking silicon devices.