Free Particle Solutions

What is a free particle?

A free particle is one subject to no potential energy at all: $V(x) = 0$ everywhere. Simple as that sounds, its solutions teach you the essential vocabulary of quantum mechanics — plane waves, wave number, and de Broglie momentum.

The time-independent Schrödinger equation for V = 0

With $V = 0$ , the time-independent Schrödinger equation (TISE) in one dimension becomes

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

Rearranging:

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

For $E > 0$ define the wave number

k \equiv \frac{\sqrt{2mE}}{\hbar},

so the equation becomes $d^2\psi/dx^2 = -k^2\psi$ , whose general solution is

\psi_k(x) = A\,e^{ikx} + B\,e^{-ikx}.

Each term is a plane wave: $e^{ikx}$ travels in the $+x$ direction and $e^{-ikx}$ in the $-x$ direction.

Full (time-dependent) plane-wave solution

Appending the universal time factor $e^{-iEt/\hbar}$ to each stationary-state solution gives the full wave

\Psi_k(x,t) = A\,e^{i(kx - \omega t)},

where the angular frequency is $\omega = E/\hbar = \hbar k^2/(2m)$ . This is a traveling wave with phase velocity

v_\phi = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m}.

Notice that $v_\phi = p/(2m) = v_{\text{classical}}/2$ — the phase velocity is half the classical particle speed. This is not a paradox: a free particle is not a single plane wave but a wave packet, a superposition of many $k$ -values, and it is the group velocity $v_g = d\omega/dk = \hbar k/m = p/m$ that matches the classical speed.

Connection to de Broglie

Louis de Broglie (1924) proposed that every particle with momentum $p$ has an associated wavelength $\lambda = h/p$ . The wave number $k = 2\pi/\lambda$ therefore satisfies

k = \frac{p}{\hbar} = \frac{\sqrt{2mE}}{\hbar}.

This is precisely the $k$ we defined from the TISE — a beautiful internal consistency. Conversely, knowing $k$ tells you the momentum: $p = \hbar k$ .

Normalisation caveat

Strict plane waves $e^{ikx}$ are not normalisable over all of $\mathbb{R}$ : the integral $\int_{-\infty}^{\infty}|e^{ikx}|^2\,dx$ diverges. Physically, a free particle with a perfectly sharp momentum is infinitely spread out in space — a consequence of the uncertainty principle $\Delta x\,\Delta p \geq \hbar/2$ . Real particles are described by wave packets (Fourier superpositions of plane waves), which are normalisable. Plane waves are still the indispensable building blocks of those packets.

Try it

This is a numerical exercise — your code should return a number. Given a free electron with kinetic energy $K = 1.0\,\text{eV}$ , compute the wave number $k = p/\hbar$ in SI units ( $\text{m}^{-1}$ ). Use the de Broglie relation $p = \sqrt{2mK}$ .

Run your code to see the quantum state.