Wave Packet Spreading and Dispersion

Prepare a free electron in as tidy a state as quantum mechanics allows — a minimum-uncertainty Gaussian lump one ångström across — and walk away. By the time light has crossed a hydrogen atom, your lump has doubled in width. Wait one full second and the electron that started inside an atom is spread over fifty-eight kilometres. Nothing pushed it. No force acts. The packet simply cannot hold its shape, because the free-particle dispersion relation $\omega=\hbar k^2/2m$ is curved: every momentum component in the superposition travels at its own group velocity $\hbar k/m$ , the fast components outrun the slow ones, and the lump stretches. This article makes that inevitability quantitative. You built the packet in Wave Packets and Their Construction and learned in Group Velocity versus Phase Velocity that its centre glides at $v_g=p_0/m$ ; here we follow the width. For the Gaussian packet the free Schrödinger equation can be solved exactly — one complete-the-square integral, done in full below — and the answer is a formula worth memorising: $\sigma(t)=\sigma_0\sqrt{1+(\hbar t/2m\sigma_0^2)^2}$ . It hides a beautiful perversity (the more tightly you squeeze a packet now, the faster it explodes later), a clean timescale $\tau=2m\sigma_0^2/\hbar$ separating "holds its shape" from "ballistic blow-up", and a mass in the denominator that explains, in one glance, why electrons smear across kilometres while a dust grain would need the age of the universe to spread by a hair. Just as important is what does not change: the momentum distribution $|\phi(k)|^2$ is frozen for all time. Spreading is not diffusion, not decay, not decoherence — it is interference bookkeeping, fully reversible, and reading it correctly is your first taste of genuinely quantum dynamics.

Key takeaways

A free packet's centre moves at $v_g=\hbar k_0/m$ , but its width grows: $\sigma(t)=\sigma_0\sqrt{1+\left(\hbar t/2m\sigma_0^2\right)^2}$ for an initially minimum-uncertainty Gaussian of width $\sigma_0$ .
The cause is dispersion: $\omega=\hbar k^2/2m$ is quadratic, so group velocity $\hbar k/m$ differs across the packet's $k$ -components by $\Delta v=\hbar\Delta k/m$ — the fast front outruns the slow rear.
The spreading time $\tau=2m\sigma_0^2/\hbar$ marks the crossover: width nearly constant for $t\ll\tau$ , ballistic growth $\sigma\to(\hbar/2m\sigma_0)\,t=(\Delta p/m)t$ for $t\gg\tau$ .
Narrower packets spread faster — squeezing $\sigma_0$ boosts $\Delta p=\hbar/2\sigma_0$ , so the late-time flare-out rate goes as $1/\sigma_0$ .
The momentum distribution $|\phi(k,t)|^2$ never changes for a free particle; $\Delta p$ is constant and $\Delta x(t)\,\Delta p=(\hbar/2)\sqrt{1+(t/\tau)^2}$ grows above the minimum.
Key idea: spreading is dephasing of momentum components with different speeds — unitary, reversible, and controlled entirely by the curvature $d^2\omega/dk^2=\hbar/m$ .

The core insight

Wave-packet spreading is the unavoidable consequence of the quadratic (dispersive) free-particle energy-momentum relation $E=\hbar^2k^2/2m$ : the nonzero curvature $\mathrm{d}^2\omega/\mathrm{d}k^2=\hbar/m$ makes Fourier components dephase, broadening the position distribution while leaving the momentum distribution untouched.

Where this lives today

Ultrafast electron diffraction and microscopy live or die by equation (5): to film chemistry on femtosecond timescales you must deliver an electron pulse that is short and still short on arrival, and the same $\tau\propto\sigma_0^2$ that spreads a packet in space stretches a bunched pulse in time (dispersion plus Coulomb repulsion). Modern UED machines fight back with radio-frequency rebunching cavities — literally the "conjugate phase" refocusing of Derivation C, applied on purpose. Matter-wave interferometry has pushed packet coherence from electrons to molecules beyond 25,000 amu, and atom interferometers in fountains and drop towers let clouds expand for seconds — free-evolution times where the time-of-flight identity (late-time position ∝ initial momentum) becomes the measurement principle itself. And in optical fibre, the identical mathematics with $\beta=d^2\omega/dk^2$ of glass instead of $\hbar/m$ governs pulse broadening in every long-haul link; "dispersion-compensating fibre" is the engineering of $\beta$ so that telecom packets, unlike free electrons, arrive unspread.

The physical picture

Think of the packet as a crowd of runners released together from one small stretch of road. Each runner is a momentum component, and each runs at their own group velocity $v(k)=\hbar k/m$ — proportional to momentum. The pack's average position advances at the mean speed $\hbar k_0/m$ , exactly the group velocity of the last article. But a packet localized to $\sigma_0$ necessarily contains a spread of momenta $\Delta p\ge\hbar/2\sigma_0$ — that is the uncertainty principle, not a preparation flaw — so the runners' speeds span $\Delta v=\Delta p/m$ . Front-runners pull ahead, stragglers drop back, and after time $t$ the crowd has stretched by roughly $\Delta v\,t$ on top of its original width. The exact Gaussian law below is precisely this estimate with its geometry done honestly: the widths add in quadrature, $\sigma(t)^2=\sigma_0^2+(\Delta v\,t)^2$ . The picture immediately explains the perversity. Squeeze the starting line tighter and you do not calm the crowd — you guarantee a wider range of speeds, because $\Delta p\sim\hbar/2\sigma_0$ grows as $\sigma_0$ shrinks. A packet confined to an atom's width detonates in a tenth of a femtosecond; a packet a millimetre wide is, for laboratory purposes, eternal. And mass enters as the great stabiliser: $\Delta v=\Delta p/m$ , so at fixed localization a proton spreads 1836 times slower than an electron, and a microgram grain effectively never spreads at all — one honest number for this is worked out below, and it comes out at the age of the universe.

Figure

A free Gaussian packet at t = 0, τ and 3τ — the centre advances at v_g while the width grows √2-fold, then √10-fold; the area (total probability) never changes

Figure

The width law σ(t) = σ₀√(1+(t/τ)²) — flat below the spreading time τ = 2mσ₀²/ħ, ballistic (slope Δp/m) far above it

The machinery: dispersion does the stretching

Everything starts from the packet you already own — the superposition assembled in Wave Packets and Their Construction:

\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,e^{i\left(kx-\omega(k)t\right)}\,\mathrm{d}k \qquad (1)

where: $\phi(k)$ is the fixed momentum-amplitude profile and each component evolves only through its phase $e^{-i\omega(k)t}$ . All the dynamics is in $\omega(k)$ ; for the free particle, substituting a plane wave into the Schrödinger equation fixes it as

\omega(k)=\frac{E(k)}{\hbar}=\frac{\hbar k^2}{2m} \qquad (2)

where: the quadratic shape is the whole story. Expand about the packet centre $k_0$ : the constant term is an overall phase, the linear term translates the envelope at $v_g=\hbar k_0/m$ (last article's result), and the first term that can distort the envelope is the curvature,

\beta\equiv\left.\frac{\mathrm{d}^2\omega}{\mathrm{d}k^2}\right|_{k_0}=\frac{\hbar}{m} \qquad (3)

where: $\beta$ measures how fast the group velocity itself varies across the packet: $dv_g/dk=\beta$ , so components spanning $\Delta k$ span speeds $\Delta v=\beta\,\Delta k=\hbar\Delta k/m$ . A medium with $\beta=0$ — light in vacuum — moves every component at the same speed and packets translate rigidly forever. Matter is never so lucky: $\beta=\hbar/m$ is fixed by the particle's mass, and only grows in importance as packets get narrow. To turn this into an exact law we take the standard initial state — the minimum-uncertainty Gaussian

\Psi(x,0)=\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}e^{ik_0x}\,e^{-x^2/4\sigma_0^2} \qquad (4)

where: $\sigma_0$ is the initial standard deviation of position (mind the convention: some books use the $1/e$ half-width $a$ of $|\Psi|^2$ , which is $\sqrt2\,\sigma_0$ — a classic source of stray factors). Its momentum amplitude is also Gaussian, with $\sigma_k=1/2\sigma_0$ , i.e. $\Delta p=\hbar/2\sigma_0$ : minimum uncertainty, $\Delta x\,\Delta p=\hbar/2$ exactly. Derivation A now evolves (4) exactly, and the density that emerges is

|\Psi(x,t)|^2=\frac{1}{\sqrt{2\pi}\,\sigma(t)}\exp\!\left[-\frac{(x-v_g t)^2}{2\,\sigma(t)^2}\right],\qquad \sigma(t)=\sigma_0\sqrt{1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2} \qquad (5)

where: the packet stays Gaussian for all time, its centre rides at $v_g=\hbar k_0/m$ , and only the width evolves. Notice what the formula does not contain: $k_0$ . How fast the packet travels has no effect on how fast it spreads — spreading is an internal affair between the packet's own momentum components. The natural clock in (5) is the spreading time

\tau=\frac{2m\sigma_0^2}{\hbar} \qquad (6)

where: $\tau$ is the time for the width to grow by $\sqrt2$ , and it scales as mass times area. Halve $\sigma_0$ and $\tau$ drops fourfold; swap an electron for a proton and $\tau$ stretches 1836-fold. With $\tau$ in hand, $\sigma(t)=\sigma_0\sqrt{1+(t/\tau)^2}$ — a hyperbola, flat then ballistic, drawn above. Finally, the uncertainty bookkeeping. Since $\Delta p$ is constant (Derivation C) while $\Delta x$ grows,

\Delta x(t)\,\Delta p=\frac{\hbar}{2}\sqrt{1+\left(\frac{t}{\tau}\right)^2}\;\geq\;\frac{\hbar}{2} \qquad (7)

where: the packet is minimum-uncertainty only at $t=0$ ; free evolution degrades the product, not by adding noise but by building position–momentum correlation — the front of the packet becomes the fast side, the rear the slow side, exactly like the runners.

Derivation AExact evolution of a free Gaussian packet

We want the exact $\Psi(x,t)$ from the initial Gaussian (4), with no small- $t$ or narrow-band approximation. The strategy is the only one the free particle ever needs: go to $k$ -space where evolution is a pure phase, then do the resulting Gaussian integral by completing the square. Watch the real part of one complex width parameter carry the entire spreading effect.

Step 1. Fourier-transform the initial state. The transform of (4) is $\phi(k)=\left(2\sigma_0^2/\pi\right)^{1/4}e^{-\sigma_0^2(k-k_0)^2}$ — a Gaussian in $k$ centred at $k_0$ with $\sigma_k=1/2\sigma_0$ , normalized so $\int|\phi|^2\,dk=1$ .

Step 2. Attach the evolution phase and assemble the integral: $\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int\phi(k)\,e^{i(kx-\hbar k^2t/2m)}\,dk$ . The exponent is quadratic in $k$ , so the integral is Gaussian and therefore doable in closed form.

Step 3. Collect the quadratic exponent: with $s=k-k_0$ , it reads $-\alpha s^2 + i s\,(x-v_gt) + i(k_0x-\omega_0 t)$ , where $\alpha\equiv\sigma_0^2+\frac{i\hbar t}{2m}$ is the complex width parameter, $v_g=\hbar k_0/m$ , $\omega_0=\hbar k_0^2/2m$ . Why bundle time into $\alpha$ ? Because the whole calculation then reads identically to the static Gaussian integral — time enters only by rotating $\alpha$ into the complex plane, and the physics (spreading) will come from $|\alpha|$ growing.

Step 4. Complete the square and use $\int e^{-\alpha s^2+bs}ds=\sqrt{\pi/\alpha}\,e^{b^2/4\alpha}$ (valid for $\mathrm{Re}\,\alpha>0$ ): $\Psi(x,t)=\left(\frac{\sigma_0^2}{2\pi}\right)^{1/4}\frac{1}{\sqrt{\alpha}}\;e^{i(k_0x-\omega_0t)}\;\exp\!\left[-\frac{(x-v_gt)^2}{4\alpha}\right]$ .

Step 5. Take the modulus squared. With $\frac{1}{4\alpha}+\frac{1}{4\alpha^*}=\frac{\sigma_0^2}{2|\alpha|^2}$ and $|\alpha|^2=\sigma_0^4+\left(\frac{\hbar t}{2m}\right)^2$ , the density collapses to equation (5) with $\sigma(t)=|\alpha|/\sigma_0=\sigma_0\sqrt{1+(\hbar t/2m\sigma_0^2)^2}$ . The packet is Gaussian at every instant; its phase (not needed for the density) carries the position–momentum correlation as a chirp — the local wavelength is shorter at the front than at the rear. Verified: sympy performs the Step-4 integral in closed form and the resulting density matches equation (5) to 50 digits at nine independent spacetime points; an independent FFT evolution reproduces σ(t) at t = τ and 3τ to better than 0.1% (verify_aq-qm-wave-packet-spreading.py, checks 2, 8–9).

Go deeper · Derivation B · 1 minThe spreading law from the runners picture

Derivation BThe spreading law from the runners picture

The exact result deserves a back-of-envelope derivation that explains why the widths add in quadrature and where $\tau$ comes from, using nothing but the uncertainty principle and kinematics. This is the version to reconstruct in an exam when you have forgotten every coefficient.

Step 1. A packet of width $\sigma_0$ carries momentum spread $\Delta p=\hbar/2\sigma_0$ (minimum-uncertainty start). Its components' group velocities therefore span $\Delta v=\Delta p/m=\hbar/2m\sigma_0$ .

Step 2. After time $t$ , kinematic stretching alone would spread positions by $\Delta v\,t$ . But this stretch is statistically independent of the initial position uncertainty — which component you are does not know where in the packet you started. Why independent? At $t=0$ the minimum-uncertainty Gaussian has zero position–momentum correlation ( $\langle xp+px\rangle/2-\langle x\rangle\langle p\rangle=0$ ); the initial spread and the velocity spread are uncorrelated random variables, so their contributions add like variances, not like lengths.

Step 3. Add in quadrature: $\sigma(t)^2=\sigma_0^2+(\Delta v\,t)^2=\sigma_0^2\left[1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2\right]$ — exactly equation (5), coefficient included. The crossover between the two terms defines $\tau=2m\sigma_0^2/\hbar$ .

Step 4. Read off the two regimes. For $t\ll\tau$ : $\sigma\approx\sigma_0\left[1+\tfrac12(t/\tau)^2\right]$ — quadratic onset, no linear term, which is why packets look frozen at first. For $t\gg\tau$ : $\sigma\to\frac{\hbar t}{2m\sigma_0}=\frac{\Delta p}{m}\,t$ — pure ballistic stretching at the velocity spread, all memory of the initial width reduced to setting the rate. Verified: sympy series expansion of σ(t) about t = 0 gives σ₀(1 + t²/2τ²) with no linear term, and the t→∞ limit of σ/t equals ħ/2mσ₀ exactly (verify script, checks 3–4).

Go deeper · Derivation C · 1 minThe momentum distribution never spreads

Derivation CThe momentum distribution never spreads

The counterpart theorem: while $|\Psi(x,t)|^2$ broadens without bound, the momentum-space density $|\phi(k,t)|^2$ is rigorously time-independent for a free particle. The proof is two lines, but the statement is routinely garbled ("everything about the packet spreads"), so we pin it down and read off its meaning.

Step 1. In $k$ -space, free evolution multiplies each amplitude by a pure phase: $\phi(k,t)=\phi(k,0)\,e^{-i\omega(k)t}$ , with $\omega(k)$ real. This is just equation (1) read in the momentum basis.

Step 2. Take the modulus: $|\phi(k,t)|^2=|\phi(k,0)|^2\,|e^{-i\omega t}|^2=|\phi(k,0)|^2$ . Every moment of the momentum distribution — $\langle p\rangle$ , $\Delta p$ , all of them — is constant. For the Gaussian, $\Delta p=\hbar/2\sigma_0$ forever. Why must this happen? Momentum commutes with the free Hamiltonian $\hat H=\hat p^2/2m$ ; momentum is conserved, not just on average but distribution-by-distribution. Spreading in $x$ with rigidity in $k$ is exactly what "dephasing" means: the components keep their weights and change only their relative phases.

Step 3. Consequences worth keeping. (i) Time-of-flight measurement: let a cloud expand for $t\gg\tau$ ; positions become $x\approx(\hbar k/m)t$ , so the position image at late times is a photograph of the initial momentum distribution — the standard cold-atom diagnostic. (ii) Reversibility: since evolution is a known phase, applying the conjugate phase (e.g. a lens, an echo pulse) refocuses the packet; spreading is not entropy growth. (iii) The uncertainty product (7) grows only because $\Delta x$ does. Verified: sympy — the modulus of the evolution factor is identically 1, and Δx(t)·Δp reproduces equation (7), equal to ħ/2 at t = 0 (verify script, checks 5–7).

Limiting cases

Case	Result	Meaning
$t\ll\tau$	$\sigma\approx\sigma_0\!\left[1+\tfrac12(t/\tau)^2\right]$	Quadratic onset — packets look rigid at first; no linear term
$t\gg\tau$	$\sigma\to(\Delta p/m)\,t=\hbar t/2m\sigma_0$	Ballistic stretching at the velocity spread; initial width only sets the rate
$\sigma_0\to0$	$\tau\propto\sigma_0^2\to0$ , late rate $\propto1/\sigma_0\to\infty$	Tighter squeezing → faster explosion; there is no "point particle" state that stays one
$m\to\infty$	$\tau\to\infty$ , $\sigma(t)\to\sigma_0$	Heavy = classical: a microgram grain at 0.15 mm width has τ ≈ age of the universe
$\beta=d^2\omega/dk^2\to0$	no spreading at any $t$	Linear dispersion (vacuum light) translates packets rigidly — dispersion, not time, is the culprit
$k_0$ anything	$\sigma(t)$ unchanged	Boosting the packet does not touch its spreading — internal affair of the components

Worked examples

Example 1 — the impatient electron. An electron is localized to $\sigma_0=1.0$ Å (atomic size). Find the spreading time, the width-doubling time, and the width after 1 fs.

Step 1. Spreading time: $\tau=2m_e\sigma_0^2/\hbar=2(9.109\times10^{-31})(10^{-10})^2/(1.055\times10^{-34})$ $=\mathbf{1.73\times10^{-16}\ s}\approx0.17$ fs. An atomic-scale electron packet lives on the tenth-of-a-femtosecond clock.

Step 2. Doubling: $\sigma=2\sigma_0$ requires $\sqrt{1+(t/\tau)^2}=2$ , i.e. $t=\sqrt3\,\tau=\mathbf{3.0\times10^{-16}\ s}$ .

Step 3. After $t=1$ fs $=5.8\tau$ : $\sigma=\sigma_0\sqrt{1+5.8^2}=5.9\,\sigma_0\approx\mathbf{0.59\ nm}$ — already in the ballistic regime, $\sigma\approx(t/\tau)\sigma_0$ .

Step 4. For contrast, hold the same electron for one full second: $\sigma\approx\hbar t/2m_e\sigma_0'$ with $\sigma_0'=1$ nm gives $5.8\times10^{4}$ m — 58 km. Free electrons do not stay put.

Example 2 — exam level: a neutron beam through a 10 m flight tube. A thermal neutron ( $v=2200$ m/s) is collimated to $\sigma_0=10\ \mu$ m transverse width. How wide is the packet at the detector 10 m downstream?

Step 1. Flight time: $t=L/v=10/2200=4.55\times10^{-3}$ s.

Step 2. Spreading time: $\tau=2m_n\sigma_0^2/\hbar=2(1.675\times10^{-27})(10^{-5})^2/(1.055\times10^{-34})=3.18\times10^{-3}$ s.

Step 3. So $t/\tau=1.43$ — the flight sits right in the crossover regime; neither approximation would do.

Step 4. Width: $\sigma=\sigma_0\sqrt{1+1.43^2}=10\ \mu\text{m}\times1.75=\mathbf{17.5\ \mu m}$ . A real neutron-interferometry design must budget for the packet nearly doubling in transit — quantum spreading as an engineering line item.

The slow reveal of dispersion

Origins · 1 minGeorge Gabriel Stokes and how this was discovered

George Gabriel Stokes

1819–1903 · Cambridge, England

Stokes spent a lifetime watching water think. Lucasian professor for over half a century, he built the hydrodynamics — viscosity, wave theory, the equations that carry his name — in which dispersion first became quantitative physics. In 1876 he set a Smith's Prize examination question that asked students to show a group of deep-water waves advances at half the speed of its individual crests: the earliest crisp statement that a wave bundle stretches and lags because each wavelength runs at its own pace. (Rayleigh, whose portrait anchors the previous article, turned that observation into the general $d\omega/dk$ formalism a year later — Stokes owned the phenomenon, Rayleigh the derivative.) Every formula in this article is Stokes's exam question transplanted to matter waves: replace $\omega=\sqrt{gk}$ by $\omega=\hbar k^2/2m$ and the stretching of a storm swell becomes the spreading of an electron.

History

1839

— Hamilton's optical papers contain the germ of group-velocity thinking

1876

— Stokes's Smith's Prize question: a deep-water wave group travels at half the crest speed — dispersion stretches wave bundles

1877

— Rayleigh's Theory of Sound systematizes group velocity as dω/dk

1925

— Debye's remark at a Zurich colloquium goads Schrödinger: a proper wave theory needs a wave equation

1926

— Schrödinger's equation fixes ω = ħk²/2m; Heisenberg's uncertainty paper (1927) makes Δx·Δp the engine of spreading

1927

— Kennard derives the exact Gaussian spreading law σ(t) = σ₀√(1+(ħt/2mσ₀²)²) — the formula of this article

1999

— Arndt and Zeilinger diffract C₆₀ molecules; packet coherence of objects of 720 amu, with spreading budgets engineered in the interferometer design

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.4 (the free particle) and Problem 2.22 (Gaussian packet spreading) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.4 (Gaussian wave packets) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 (Gaussian packet, spreading) · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement G_I (free wave packet) · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6.

Provenance. Equations are standard textbook results. A verifier should: (1) confirm the free dispersion $\omega=\hbar k^2/2m$ by substituting a plane wave into the free Schrödinger equation; (2) reproduce $v_g=\hbar k/m$ , $v_p=\hbar k/2m$ by differentiating; (3) carry out the Gaussian Fourier integral to obtain $\sigma(t)=a\sqrt{1+(\hbar t/2ma^2)^2}$ (Griffiths Problem 2.22, Sakurai §2.4); (4) verify $\Delta p=\hbar/2a$ is constant and that $\Delta x\Delta p=\frac{\hbar}{2}\sqrt{1+(\hbar t/2ma^2)^2}\geq\hbar/2$ . Cross-check against Cohen-Tannoudji Complement G_I and Shankar Ch. 5. No empirical numbers are quoted; all relations are analytic. (All four verifier items are python-verified for this article — symbolic integral, limits, and an independent FFT evolution; the worked numbers are recomputed in verify_aq-qm-wave-packet-spreading.py.)

Builds on: Free Particle and Plane Waves · Wave Packets and Their Construction · Group Velocity versus Phase Velocity · Time-Independent Schrödinger Equation in One Dimension Leads to: Bound States versus Scattering States · Quantum Harmonic Oscillator: Analytic Solution

If spreading is reversible, why do we never see packets contract on their own?

They can — but only if you prepare them to. Run Derivation A backwards: a packet whose phase is chirped the opposite way (front slow, rear fast) contracts, reaches minimum width, then re-expands. Free evolution always adds the same sign of chirp, so a packet prepared without one only ever gains it and spreads. Contracting states must be engineered — an electrostatic lens, an RF rebunching cavity, an atom-optics pulse — which is exactly what beam physicists do. It is precisely like a lens focusing light: diffraction alone never focuses, but the right incoming phase converges before it diverges.

Does the spreading formula mean an electron in a hydrogen atom smears out over kilometres?

No — the formula is for a free particle. In an atom the electron is bound: its stationary states have time-independent $|\psi|^2$ , and the Coulomb well continually refocuses what dispersion tries to stretch. Spreading without bound is the signature of no confining potential. The harmonic oscillator makes the contrast sharpest: its dispersion of energies is evenly spaced, so packets revive perfectly every period, and the special coherent states (previewed later in Coherent States of the Harmonic Oscillator) slosh forever without spreading at all.

Where does the extra energy for the faster-moving spread packet come from?

Nowhere — no energy appears. The energy distribution is set by $|\phi(k)|^2$ , which never changes (Derivation C). $\langle E\rangle$ and even the full spectrum of energies are constants of the motion. What changes is where the components are, not how fast they go: the fast components were always fast; they simply stop overlapping the slow ones. Spreading rearranges probability in space at fixed momentum bookkeeping — that is why it costs nothing and conserves everything.

Is the σ(t) formula special to Gaussians, or do all packets spread like this?

The exact closed form is Gaussian-specific, but the two structural features are universal: a quadratic-in-time onset governed by the same $\tau$ scale, and late-time ballistic growth $\sigma\to(\Delta p/m)t$ for any free packet (that limit only uses the variance of the velocity distribution). Non-Gaussian packets also distort as they spread — a rectangular packet grows ringing lobes, for instance — because the quadratic phase treats their Fourier components less symmetrically. The Gaussian is the one shape that evolves into itself.

My textbook writes the width as a√(1+(ħt/2ma²)²) with a different meaning of a. Which is right?

Both — check the convention before comparing coefficients. If $a$ is the initial standard deviation of position (our $\sigma_0$ ), the formula reads exactly as here, with $\Delta p=\hbar/2a$ and $\tau=2ma^2/\hbar$ . Some texts instead define $a$ via $\Psi\propto e^{-x^2/2a^2}$ (then $\sigma_0=a/\sqrt2$ ) or via the $1/e$ point of $|\Psi|^2$ . Every version produces the same physics with factors of $\sqrt2$ shuffled between $a$ , $\Delta p$ , and $\tau$ . The convention-proof anchors are $\Delta x\,\Delta p=\hbar/2$ at $t=0$ and the late-time slope $\Delta p/m$ — check those two and any book's formula falls in line.

Why doesn't the packet's speed affect its spreading — surely a fast packet has more dispersion?

The mean momentum $\hbar k_0$ sits in the linear part of the dispersion expansion, and linear phase only translates the envelope (that was Derivation A of the group-velocity article). Spreading is driven by the curvature $d^2\omega/dk^2=\hbar/m$ , which is the same at every $k_0$ for the free-particle parabola. Physically: boosting the packet adds the same velocity to every runner, which cannot change how they separate from each other. Only the spread of momenta — set by $\sigma_0$ — and the mass matter.

How does spreading square with Ehrenfest's theorem — I thought quantum motion followed classical mechanics on average?

It does, and the two statements coexist peacefully. Ehrenfest gives $d\langle x\rangle/dt=\langle p\rangle/m$ exactly: the mean follows the classical trajectory forever (for a free particle). Spreading is a statement about the second moment, $\Delta x$ , about which classical single-particle mechanics says nothing. A classical ensemble of particles with position spread $\sigma_0$ and momentum spread $\Delta p$ would in fact spread by exactly the same law — the quantum novelty is not the stretching itself but the floor $\Delta p\ge\hbar/2\sigma_0$ that makes it mandatory.

Does a photon wave packet spread in vacuum?

Along its propagation direction, no: vacuum light has $\omega=ck$ , zero curvature, so a pulse translates rigidly at $c$ — this is the $\beta=0$ row of the limiting-cases table and the reason femtosecond pulses survive kilometres of vacuum. In glass, $\omega(k)$ curves and pulses spread exactly as our Gaussians do (that is fibre-optic dispersion). Transversely, a light beam does spread — diffraction — which is the spatial-frequency cousin of the same Fourier bookkeeping, with the lens playing the role of the refocusing chirp.