Wave packets

A pure harmonic wave $A\cos(kx-\omega t)$ has one exact wavenumber $k$ , but it fills all of space and time — it never "arrives" or "is here." A real pulse of light, sound, or water is confined to a finite region, so it cannot be a single harmonic wave. It must be a superposition of many harmonic waves of slightly different wavenumbers, chosen so that they reinforce inside a small region and cancel outside it. That localized bump is a wave packet: a carrier oscillation at the central wavenumber $k_0$ , riding inside a slowly varying envelope that gives the packet its size and position. This article continues directly from the two-wave beats picture: instead of two waves, superpose a whole band of wavenumbers clustered around $k_0$ , and ask how the packet's spatial width $\Delta x$ relates to the spread of wavenumbers $\Delta k$ used to build it. That inverse relation, $\Delta x\,\Delta k \gtrsim 1$ , is the deliverable — the classical seed of the Heisenberg uncertainty principle met later.

The core insight

A wave packet is a superposition of harmonic waves over a narrow band of wavenumbers $\Delta k$ centred on $k_0$ , producing a localized envelope of width $\Delta x$ obeying $\Delta x\,\Delta k \gtrsim 1$ — narrower packets always demand a broader spread of wavenumbers.

The core idea

A packet is many harmonic waves added together. Superpose waves with wavenumbers spread around a central value $k_0$ , each carrying amplitude $A(k)$ :

y(x,t) = \int A(k)\,\cos\big(kx-\omega(k)t\big)\,dk,

where: $A(k)$ is the amplitude given to wavenumber $k$ (appreciable only over a band of width $\Delta k$ around $k_0$ ), and $\omega(k)$ is that wave's frequency. Near $x=0$ the phases $kx$ for different $k$ stay close together and the waves add up constructively; far from $x=0$ the phases spread out and average to zero. The result is a bump — an envelope — riding on a rapid oscillation at $k_0$ , exactly the beat pattern seen with two waves, now built from a continuous band.

Width and bandwidth trade off. A packet squeezed into a small $\Delta x$ needs a wide $\Delta k$ to build it; a packet spread over a large $\Delta x$ can be built from a narrow $\Delta k$ . This qualitative bound holds for any kind of wave — water, sound, or light:

\Delta x \, \Delta k \gtrsim 1.

where: $\Delta x$ is the spatial extent of the localized bump and $\Delta k$ is the spread of wavenumbers used to build it (both order-of-magnitude, not precisely defined widths). The relation later becomes, via $p=\hbar k$ , the seed of $\Delta x\,\Delta p \gtrsim \hbar$ — a classical wave fact here, not yet a quantum statement.

The two-wave packet is the simplest case. Two waves of wavenumbers $k_0\pm\Delta k/2$ and frequencies $\omega_0\pm\Delta\omega/2$ , equal amplitude $A$ , combine (by the same sum-to-product algebra as beats) into

y(x,t) = 2A\cos\!\left(\frac{\Delta k}{2}x - \frac{\Delta\omega}{2}t\right)\cos(k_0x-\omega_0 t),

a fast carrier $\cos(k_0x-\omega_0t)$ modulated by a slow envelope depending on $x,t$ only through $\Delta k\,x-\Delta\omega\,t$ . The envelope's spatial period is $\sim 2\pi/\Delta k$ : squeeze the constructive bump into a smaller $\Delta x$ and $\Delta k$ must grow — the seed of the boxed relation above.

Figure

Three curves showing that adding more waves around a central wavenumber narrows the envelope while the carrier oscillation stays the same

Derivation AWidth times bandwidth from a band of three waves

The goal is to show, using only the sum-to-product identity a student already owns, that adding waves over a band of wavenumbers $\delta k$ around a centre $k_0$ produces a localized envelope whose first zero sits at a distance $x$ that shrinks as the band $\delta k$ widens — the concrete arithmetic behind the qualitative bound $\Delta x\,\Delta k\gtrsim 1$ , using three waves as the simplest band with more than two members.

Step 1. Write the sum of three equal-amplitude waves at $t=0$ , with wavenumbers $k_0-\delta k$ , $k_0$ , $k_0+\delta k$ :

y(x,0) = A\cos\big((k_0-\delta k)x\big) + A\cos(k_0x) + A\cos\big((k_0+\delta k)x\big).

This is the $t=0$ snapshot of the band-sum $\int A(k)\cos(kx-\omega t)\,dk$ with just three sample points in $k$ , chosen symmetrically about $k_0$ .

Step 2. Combine the outer two terms with the sum-to-product identity $\cos P+\cos Q = 2\cos\big(\tfrac{P-Q}{2}\big)\cos\big(\tfrac{P+Q}{2}\big)$ :

A\cos\big((k_0-\delta k)x\big)+A\cos\big((k_0+\delta k)x\big) = 2A\cos(\delta k\,x)\cos(k_0x).

The two outer waves alone already show a carrier $\cos(k_0x)$ times a slow modulation $\cos(\delta k\,x)$ .

Step 3. Add back the middle wave $A\cos(k_0x)$ and factor out the shared carrier:

y(x,0) = \big[1+2\cos(\delta k\,x)\big]A\cos(k_0x) = E(x)\,A\cos(k_0x),

where the envelope is $E(x) = 1+2\cos(\delta k\,x)$ . For $\delta k\ll k_0$ , $E(x)$ varies slowly compared with the rapid carrier $\cos(k_0x)$ — exactly the carrier-times-envelope structure of the two-wave beat case, now built from three waves.

Step 4. Find where the envelope first vanishes. Setting $E(x)=0$ gives $\cos(\delta k\,x) = -\tfrac12$ , so the smallest positive solution is $\delta k\,x = 2\pi/3$ , i.e.

x = \frac{2\pi}{3\,\delta k}.

This first zero marks the edge of the central bump — a rough measure of the packet's half-width. As $\delta k$ grows, $x$ shrinks in direct proportion: doubling the wavenumber spread halves the location of the first zero, which is precisely the inverse relationship packed into $\Delta x\,\Delta k\gtrsim 1$ . Adding still more waves across the same band sharpens this crude triple-humped shape into the single smooth central peak of a true packet, with overall width $\Delta x\sim 2\pi/\Delta k$ .

Worked example

An optical pulse has a wavenumber spread of $\Delta k = 2.0\times10^{4}\ \text{m}^{-1}$ . Estimate the pulse's minimum spatial length $\Delta x$ , and, given it travels at $c=3\times10^8\ \text{m/s}$ , its minimum duration $\Delta t$ .

Step 1. Apply the width–bandwidth bound $\Delta x\,\Delta k\gtrsim 1$ , solved for $\Delta x$ :

\Delta x \gtrsim \frac{1}{\Delta k} = \frac{1}{2.0\times10^{4}} = 5.0\times10^{-5}\ \text{m} = 50\ \mu\text{m}.

A pulse built from waves spread over $\Delta k = 2.0\times10^4\ \text{m}^{-1}$ cannot be shorter than about $50\ \mu\text{m}$ , no matter how the amplitudes $A(k)$ are chosen.

Step 2. Convert the spatial length to a time using the pulse's travel speed, $\Delta t \sim \Delta x/c$ :

\Delta t \sim \frac{5.0\times10^{-5}}{3\times10^{8}} \approx 1.7\times10^{-13}\ \text{s} \approx 0.17\ \text{ps}.

The pulse cannot be shorter than about $50\ \mu\text{m}$ in space, or about $0.17\ \text{ps}$ in time. Shorter pulses always need a broader spread of wavenumbers — the reason ultrafast laser pulses inherently carry a wide range of colours.

Common traps

Thinking a wave packet has one exact wavelength; a localized packet is never monochromatic — it is always a superposition over a band of $k$ 's, however narrow that band is.
Treating $\Delta x\,\Delta k\gtrsim 1$ as an exact equation with a fixed constant; it is an order-of-magnitude statement whose precise constant depends on how $\Delta x,\Delta k$ are defined (e.g. standard deviations for a Gaussian packet).
Confusing the envelope (slow modulation, carries the packet's shape and location) with the carrier (rapid oscillation at $k_0,\omega_0$ ) — in a dispersive medium these move at different speeds, the subject of the next article on group velocity.
Forgetting that the waves being added must stay in phase (coherent superposition); randomly-phased waves smear out and never produce a clean, well-defined localized packet.

Ask the tutor: if the jump from two waves making beats to many waves making a packet feels abrupt, ask the AI tutor to build the packet up one wave at a time, or to explain why $\Delta x\,\Delta k\gtrsim 1$ is not an exact equality.

Why can't a wave packet ever have just one exact wavelength?

A single exact wavenumber $k$ means a pure harmonic wave $A\cos(kx-\omega t)$ , and that wave has the same amplitude at every $x$ , for all time — it is not confined anywhere, so it is not "localized." Localization requires interference: waves of different $k$ must cancel outside a small region and reinforce inside it, and cancellation is only possible if there is more than one $k$ present. So confinement in space always costs you a spread in wavenumber, however small that spread might be.

Is Δx Δk ≳ 1 the same as the Heisenberg uncertainty principle?

Not yet — at this stage it is a purely classical statement about waves, true for water ripples, sound pulses, or light, with no quantum mechanics involved. It becomes the seed of the uncertainty principle only later, when matter is treated as a wave and the relation $p=\hbar k$ (from de Broglie's hypothesis) turns a wavenumber spread $\Delta k$ into a momentum spread $\Delta p$ , giving $\Delta x\,\Delta p\gtrsim\hbar$ . Keep the two ideas separate for now: here it is geometry of superposition, not a limit on measurement.

Why does adding more waves make the envelope look like a single clean bump instead of the bumpy shape from just two or three waves?

Two or three waves interfere constructively at $x=0$ but also partially reconstruct at other points, giving the secondary bumps you can see in the three-wave case (its envelope has more than one zero). As you pack in more wavenumbers across the same band $\Delta k$ , those secondary reinforcements get washed out because far more phases have to line up simultaneously away from $x=0$ , which almost never happens once the band is densely filled. In the limit of a continuous integral over $k$ , only the central peak survives cleanly, and its width settles to the value set by $\Delta x\sim 1/\Delta k$ .

Wave packets

The core idea

Worked example

Common traps

Before this

Where this leads