The Free Particle

Introduction

The simplest problem in quantum mechanics is a particle on which no forces act: a free particle . There are no walls, no nuclei and no springs — just a particle moving through empty space. Solving this case shows what the Schrödinger equation does when nothing confines the particle, and the answer is striking: every positive energy is allowed. Quantisation, it turns out, is not automatic. It appears only when a particle is confined, which is the lesson of the particle-in-a-box problem that follows.

Core explanation

Setting up the equation. With V = 0 everywhere, the one-dimensional Hamiltonian contains only the kinetic term, and the Schrödinger equation becomes

−(ħ²/2m) d²ψ/dx² = Eψ

Rearranging gives d²ψ/dx² = −k²ψ, where k² = 2mE/ħ². We need a function whose second derivative is a negative constant times itself.

The solutions. Functions with this property are e^(ikx) and e^(−ikx), or equivalently sin kx and cos kx. The general solution is

ψ(x) = A e^(ikx) + B e^(−ikx)

and the energy is

E = ħ²k²/2m

Definite momentum. Apply the momentum operator p̂ₓ = −iħ d/dx to e^(ikx): the result is −iħ(ik)e^(ikx) = ħk e^(ikx). So e^(ikx) is an eigenfunction of momentum with eigenvalue +ħk, describing a particle moving in the positive x-direction. Similarly e^(−ikx) has momentum −ħk and describes motion to the left. Since p = ħk and k = 2π/λ, this is just the de Broglie relation p = h/λ, now derived rather than assumed. The energy is E = p²/2m, the classical kinetic energy.

Standing waves. The combinations cos kx and sin kx are superpositions of left- and right-moving waves with equal weight. They have definite energy but not definite momentum: a measurement would give +ħk or −ħk with equal probability.

No quantisation. Nothing restricts k. Any real value is acceptable, because the wavefunction stays finite everywhere and there are no boundaries to satisfy. The energy E = ħ²k²/2m therefore takes any value from zero upwards: the free particle has a continuum of energies.

Position is completely uncertain. For ψ = A e^(ikx), the probability density is ψ ² = A ², the same everywhere. A particle of perfectly defined momentum is equally likely to be found anywhere — the extreme case of the uncertainty principle. This also means the plane wave cannot be normalised over infinite space; real free particles are described by wave packets, superpositions of plane waves with a spread of k.

Formulae

ψ = A e^(ikx) + B e^(−ikx); k = (2mE)^½/ħ; E = ħ²k²/2m = p²/2m; p = ħk = h/λ; k = 2π/λ.

Step-by-step reasoning

To find the free-particle wavefunction for a given energy:

1. Set V = 0 and write the Schrödinger equation. 2. Rearrange into d²ψ/dx² = −k²ψ and identify k = (2mE)^½/ħ. 3. Write the general solution as a combination of e^(ikx) and e^(−ikx). 4. Choose A or B to describe motion in one direction, or combine them for a standing wave. 5. Note that no condition restricts k, so any E ≥ 0 is allowed.

Visual explanation

Plot the real part of e^(ikx): an endless cosine wave with wavelength 2π/k. Its modulus squared is a flat horizontal line. Raising the energy squeezes the wave to shorter wavelengths but leaves the flat probability line unchanged.

Real-world analogy

A free particle is like a wave on an infinitely long rope with no fixed ends. You can shake it at any frequency you like, and a wave of that wavelength will travel along it. Only when the rope is tied down at both ends, as on a guitar, are particular notes selected.

Real-world example

Electrons in an electron microscope, after acceleration, travel through vacuum as very nearly free particles. Their momentum is controlled by the accelerating voltage, giving de Broglie wavelengths of a few picometres at typical voltages — short enough to resolve individual columns of atoms in crystals.

Why?

Why are the free-particle energies continuous while atomic energies are discrete? Quantisation comes from boundary conditions: a confined wavefunction must fit its surroundings and vanish far away, which only certain wavelengths can do. With no confinement, there is nothing for the wave to fit, so every wavelength is acceptable.

Common misconception

"All quantum systems have discrete energy levels." Only bound states do. Free particles, and electrons ejected from atoms above the ionisation energy, have continuous energies — which is why an atomic spectrum shows a continuum beyond the series limit.

Worked example

Question: An electron moves freely with a kinetic energy of 1.0 eV (1.60 × 10⁻¹⁹ J). Find k and its de Broglie wavelength. (mₑ = 9.11 × 10⁻³¹ kg, ħ = 1.055 × 10⁻³⁴ J s)

Reasoning: k = (2mE)^½/ħ = (2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹)^½ / 1.055 × 10⁻³⁴ = (5.40 × 10⁻²⁵ kg m s⁻¹) / 1.055 × 10⁻³⁴ = 5.12 × 10⁹ m⁻¹. Then λ = 2π/k = 1.23 × 10⁻⁹ m.

Answer: k ≈ 5.1 × 10⁹ m⁻¹ and λ ≈ 1.2 nm.

Quick check

1. Is the wavefunction sin kx an eigenfunction of the momentum operator for a free particle? Answer: No; it is an equal mixture of momenta +ħk and −ħk, although it is an eigenfunction of the Hamiltonian with E = ħ²k²/2m.

Exam focus

Be able to solve the free-particle equation, show that e^(ikx) is a momentum eigenfunction with eigenvalue ħk, and explain why the energies form a continuum. Link p = ħk to the de Broglie relation.

Advanced insight

Because plane waves cannot be normalised, they are handled either by "box normalisation" — placing the particle in a very large box with periodic boundaries so that k is quantised in tiny steps — or by delta-function normalisation. The same device is used for electrons in metals, where the free-electron model treats conduction electrons as nearly free particles.

Summary

For a free particle V = 0 and the Schrödinger equation has plane-wave solutions e^(±ikx) with E = ħ²k²/2m. These are states of definite momentum ±ħk, reproducing the de Broglie relation. No boundary conditions restrict k, so energies form a continuum from zero. A plane wave is spread uniformly through space, illustrating the uncertainty principle at its extreme.

Practice questions

1. Show that e^(−ikx) is an eigenfunction of p̂ₓ and state its eigenvalue. Answer: −iħ d/dx e^(−ikx) = −iħ(−ik)e^(−ikx) = −ħk e^(−ikx), so the eigenvalue is −ħk. 2. By what factor does the energy of a free particle change if its wavelength is halved? Answer: k doubles, and E ∝ k², so the energy increases by a factor of four. 3. Why can the plane wave e^(ikx) not be normalised over all space? Answer: Its probability density A ² is constant everywhere, so its integral over infinite space diverges unless A = 0. 4. Explain why a spectrum of atomic hydrogen shows a continuum beyond the series limit. Answer: Above the ionisation energy the electron is free, and a free particle can have any kinetic energy, so absorption occurs at all higher frequencies.