Group and Phase Velocity

Watch any wave closely and a quiet scandal appears: the thing you would point to as "the wave" — the hump, the pulse, the place where the action is — does not move at the same speed as the ripples inside it. Ocean surfers know it: paddle after a set of swells and individual crests keep sliding forward through the group, being born at the back and dying at the front. Now attach a wave to an electron, as de Broglie taught us to, and the scandal becomes a crisis. A single matter wave $e^{i(kx-\omega t)}$ has crests that move at $\hbar k/2m$ — half the speed of the particle it supposedly describes. Worse, the relativistic version has crests that outrun light. If the wave is the particle, which speed is the particle's? This article untangles the two velocities at the heart of the answer. The phase velocity $v_p = \omega/k$ is the speed of one crest of one pure sinusoid. The group velocity $v_g = d\omega/dk$ is the speed of the envelope — the modulation that says where the wave's energy, and in quantum mechanics its probability, actually sits. We will build the distinction the honest way, out of the simplest possible packet: two waves beating against each other. That one construction, plus the free particle's dispersion relation $\omega = \hbar k^2/2m$ , delivers the punchline that saves the whole matter-wave programme: the envelope moves at exactly $p/m$ , the classical velocity. This is the concept-register introduction. The full quantitative machinery — the stationary-phase proof, the energy-zero subtlety, the relativistic identity — lives one chapter ahead in Group Velocity versus Phase Velocity.

⚡ Key Takeaways

A single sinusoid $e^{i(kx-\omega t)}$ has one speed, the phase velocity $v_p = \omega/k$ — the speed of its crests. Nothing localized travels at $v_p$ , because a pure plane wave is the same everywhere.
Superpose just two waves with nearby $(k, \omega)$ and an envelope appears. The envelope moves at $\Delta\omega/\Delta k$ — the discrete ancestor of the group velocity $v_g = d\omega/dk$ .
For a free quantum particle $\omega = \hbar k^2/2m$ , so $v_p = \hbar k/2m$ but $v_g = \hbar k/m = p/m$ : the envelope moves at exactly the classical velocity, and the factor-of-two crest speed is harmless.
Whether crests outrun the group or trail it depends only on the dispersion relation: light in vacuum has $v_g = v_p$ , matter waves have $v_g = 2v_p$ , deep-water waves have $v_g = v_p/2$ .
Detectors, energy, information, probability — everything observable rides the envelope at $v_g$ . Phase velocity, even when it exceeds $c$ , carries no signal.
Key idea: a localized particle is a packet of plane waves; its observable motion is the group velocity $v_g = d\omega/dk = p/m$ of the envelope, not the phase velocity $v_p = \omega/k$ of the underlying oscillations.

💡 The Core Insight

A localized particle is a packet of plane waves; its observable motion is the group velocity $v_g = d\omega/dk = p/m$ of the envelope, not the phase velocity $v_p = \omega/k$ of the underlying oscillations.

The physical picture

Take two tuning forks a few hertz apart and strike them together. You hear the average pitch — the carrier — swelling and fading at the difference frequency: beats. Now give the sound a direction of travel and ask where the loud spots go. The individual pressure crests move at one speed; the loud-soft pattern that organizes them moves at another. That pattern is the envelope, and its speed is set not by $\omega$ or $k$ separately but by how $\omega$ changes with $k$ .

Figure

Two waves make a beat — the envelope is a new object with its own speed

The figure is the whole subject in one picture. Two pure waves, each with a perfectly definite speed, add up to something with two speeds: a fast internal wiggle (the carrier, moving at essentially the average phase velocity) and a slow breathing pattern (the envelope, moving at the ratio of the differences, $\Delta\omega/\Delta k$ ). Neither wave alone has an envelope. The envelope is an interference object — it exists only because the components slide in and out of step, and it therefore moves at the speed at which the agreement point between the components moves. Which of the two speeds wins depends entirely on the medium's dispersion relation $\omega(k)$ — the rule that tells each wavelength what frequency, and hence what crest speed, it gets. Three worlds cover almost everything you will meet:

Figure

Three dispersion worlds — where the envelope beats the crests, ties them, or trails them

Matter waves sit in the middle panel: the envelope — the particle — genuinely outruns its own crests, each crest surfacing at the rear of the packet, sweeping forward through it at half the packet's speed relative to nothing observable, and dissolving at the front. Once you accept that only the envelope is stuff, all the paradoxes below become bookkeeping.

The two speeds, defined

Start with the object that has only one speed. A single sinusoidal component of wavenumber $k$ and angular frequency $\omega$ has crests wherever its phase $kx - \omega t$ is a multiple of $2\pi$ ; following one crest means holding that phase constant, and differentiating $kx - \omega t = \text{const}$ gives the crest's velocity.

v_p = \frac{\omega}{k} \qquad (1)

where: $v_p$ — the phase velocity, the speed of a single crest of a single pure sinusoid; $\omega$ — its angular frequency; $k$ — its wavenumber. A pure sinusoid fills all space with identical crests, so $v_p$ is the speed of a pattern, not of any localized thing — there is no lump whose arrival you could time. A localized object needs many components, and the moment there are many, a second velocity is born. For a superposition sharply concentrated near some central wavenumber $k_0$ , the envelope translates at the rate at which frequency changes across the participating wavenumbers.

v_g = \left.\frac{d\omega}{dk}\right|_{k_0} \qquad (2)

where: $v_g$ — the group velocity, the speed of the interference envelope; $k_0$ — the central wavenumber of the packet. Geometrically, $v_p$ is the slope of the chord from the origin to the point $(k, \omega)$ on the dispersion curve, while $v_g$ is the slope of the tangent there. The two agree only when the curve is a straight line through the origin — the dispersionless case. Which formula answers which question:

you ask	use	why
how fast does one crest of one component slide	$v_p = \omega/k$	phase is a per-component property
how fast does the lump of probability travel	$v_g = d\omega/dk$	the envelope is an interference object
how fast does a detector see the particle arrive	$v_g$	detectors respond to $\lvert\psi\rvert^2$ , which lives in the envelope
does the packet hold its shape	curvature $d^2\omega/dk^2$	dispersion beyond first order — see Wave Packet Spreading and Dispersion

Derivation AThe beat of two waves — an envelope is born with its own velocity

We derive the group velocity from the simplest packet that exists: exactly two plane waves with neighbouring wavenumbers and frequencies. The strategy is a single trigonometric factorization — writing the sum as a slow modulation multiplying a fast carrier — and then reading each factor's speed directly off its phase. This discrete route is worth owning because nothing is hidden in it: no integrals, no approximations, and the continuum formula $v_g = d\omega/dk$ appears as the obvious limit. Watch one subtlety: for the quadratic dispersion of matter waves the result is not merely approximate but exact, with the envelope moving at the mean of the two group velocities.

Step 1. Superpose two equal-amplitude waves, labelling them symmetrically about the mean: $k_{1,2} = \bar{k} \pm \tfrac{1}{2}\Delta k$ and $\omega_{1,2} = \bar{\omega} \pm \tfrac{1}{2}\Delta\omega$ :

\psi(x,t) = e^{i(k_1 x - \omega_1 t)} + e^{i(k_2 x - \omega_2 t)}

Why symmetric labels: they make the factorization drop out in one line — the average rides the carrier, the difference rides the envelope.

Step 2. Factor out the mean phase. The two exponentials share $e^{i(\bar{k}x - \bar{\omega}t)}$ , and what remains is a cosine of half the difference phase:

\psi(x,t) = 2\cos\!\left(\tfrac{1}{2}\left(\Delta k\, x - \Delta\omega\, t\right)\right)\, e^{i(\bar{k}x - \bar{\omega}t)}

Why this is the whole physics: the sum of two waves with one speed each has become a product of two travelling patterns with different speeds — the carrier and the envelope are now separate factors you can track separately.

Step 3. Read off the two velocities from the two phases. The carrier factor repeats when $\bar{k}x - \bar{\omega}t$ is constant, so its crests move at $\bar{\omega}/\bar{k} \approx v_p$ . The cosine envelope holds its value when $\Delta k\,x - \Delta\omega\,t$ is constant, so every hump and every node of the modulation moves at

v_{\text{env}} = \frac{\Delta\omega}{\Delta k}

Step 4. Take the continuum limit. Let the two wavenumbers approach each other, $\Delta k \to 0$ : the difference quotient becomes the derivative, $v_{\text{env}} \to d\omega/dk = v_g$ . A realistic packet is a continuous superposition over a narrow band of $k$ , and every neighbouring pair within the band performs exactly this beat. Why the quadratic case is exact: for $\omega = \hbar k^2/2m$ the difference quotient evaluates to $\Delta\omega/\Delta k = \hbar(k_1 + k_2)/2m$ — precisely the group velocity at the mean wavenumber, with no error at any finite $\Delta k$ . *Verified: the factorization, the envelope speed, and the quadratic-dispersion exactness are checked symbolically in this article's verification script; the general stationary-phase proof for continuous packets is owned by Group Velocity versus Phase Velocity.*

What the Schrödinger equation says

The beat construction works for any waves. To make it quantum mechanics we need the dispersion relation that matter waves obey, and that is not a choice — it is dictated by the wave equation itself, the law introduced in The Time-Dependent Schrödinger Equation. Insert a trial plane wave $\psi = e^{i(kx - \omega t)}$ into the free equation $i\hbar\,\partial_t\psi = -\tfrac{\hbar^2}{2m}\,\partial_x^2\psi$ and each derivative just multiplies: the left side produces $\hbar\omega$ , the right side $\hbar^2k^2/2m$ . The wave solves the equation only if

\omega(k) = \frac{\hbar k^2}{2m} \qquad (3)

where: $\omega(k)$ — the frequency assigned to wavenumber $k$ ; $m$ — the particle's mass. Multiply by $\hbar$ and this is just $E = p^2/2m$ wearing wave clothing (with $E = \hbar\omega$ from Planck and $p = \hbar k$ from de Broglie Matter Waves). The crucial feature is the square: $\omega$ grows faster than $k$ , the curve bends upward, and chord and tangent slopes must differ. Matter waves are irreducibly dispersive.

Derivation BThe free particle's two speeds — and the factor of two that saves de Broglie

We compute both velocities for a free quantum particle directly from the dispersion relation (3) and confirm the claim the whole chapter has been building toward: the envelope of a matter-wave packet moves at exactly the classical particle velocity. The calculation is two derivatives long; the payoff is conceptual. Keep your eye on the factor of two — it is not an error to be explained away but the signature of a curved dispersion relation, and it lands on the harmless velocity.

Step 1. Phase velocity — divide by $k$ :

v_p = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m} = \frac{v_{\text{classical}}}{2}

Why this looked like a disaster: taken as "the particle's speed," it is off by half. But no measurement rides a single crest — a pure plane wave is featureless, the same everywhere, with nothing localized to time.

Step 2. Group velocity — differentiate:

v_g = \frac{d\omega}{dk} = \frac{\hbar k}{m} = \frac{p}{m} = v_{\text{classical}}

Why this is the rescue: the probability lump — the only thing a detector ever registers — moves at exactly the Newtonian velocity. The wave description and the particle description agree about everything observable.

Step 3. Note the consistency with averages. A packet built symmetrically around $k_0$ carries mean momentum $\langle p\rangle = \hbar k_0$ , so $v_g = \hbar k_0/m = \langle p\rangle/m$ — which is precisely the statement $d\langle x\rangle/dt = \langle p\rangle/m$ derived as an operator theorem in Momentum Operator. The beat picture and the Ehrenfest picture are two views of one fact. Why we stop here: the phase velocity has a further embarrassment — it changes if you shift the zero of potential energy, while $v_g$ does not. That energy-zero test, and the stationary-phase machinery behind it, are worked in full in Group Velocity versus Phase Velocity. *Verified: both derivatives, the plane-wave substitution, and $v_g = 2v_p$ are checked symbolically; the 100 eV electron numbers below recompute them concretely.*

One relativistic sentence completes the story, because it is where the subject began: for a relativistic matter wave the phase velocity is $c^2/v$ — faster than light — while the group velocity works out to exactly the particle's velocity $v$ , and the elegant identity $v_p\,v_g = c^2$ ties them together. Nothing observable moves at $v_p$ , so relativity is untouched. The derivation from $E^2 = (pc)^2 + (mc^2)^2$ , and the numbers for a real relativistic electron, are owned by Group Velocity versus Phase Velocity.

When the crests outrun the group

Quantum mechanics gives the envelope the lead. It is worth seeing one world where the ranking flips, both because you can watch it from any beach and because it kills the temptation to think $v_g > v_p$ is a law of nature.

Derivation CDeep water — the world where the envelope trails its own crests

We compute both velocities for gravity waves on deep water, dispersion relation $\omega = \sqrt{gk}$ , and find the opposite ordering from matter waves: the group travels at half the crest speed. The point of the exercise is structural — the ratio $v_g/v_p$ is a pure readout of how $\omega$ scales with $k$ , and a square-root law is the mirror image of a square law. This is also a two-line derivation with a real-world payoff: it is how oceanographers time the arrival of a distant storm's swell.

Step 1. Phase velocity:

v_p = \frac{\omega}{k} = \sqrt{\frac{g}{k}}

Why long swells are fast: $v_p$ grows as $k$ shrinks — hundred-metre storm swells travel much faster than wind chop, which is why the sea's longest wavelengths arrive first as a storm's early messenger.

Step 2. Group velocity:

v_g = \frac{d\omega}{dk} = \frac{1}{2}\sqrt{\frac{g}{k}} = \frac{v_p}{2}

Why the halving: differentiating $k^{1/2}$ pulls down the exponent $\tfrac{1}{2}$ , exactly as differentiating the matter wave's $k^2$ pulled down a 2. Power law in, velocity ratio out: $\omega \propto k^s$ gives $v_g = s\,v_p$ in general — $s = 2$ for matter waves, $s = 1$ for light, $s = \tfrac{1}{2}$ for deep water.

Step 3. Watch it on the water. A group of swells arrives; the crest you track appears at the front of the group, slides backward relative to the group as the group advances, and vanishes at the rear — the exact time-reverse of the matter-wave movie, where crests are born at the back and die at the front. Verified: both derivatives checked symbolically; the 100 m swell numbers in Worked example 2 recompute them with g = 9.81 m/s².

Limiting cases

case	result	meaning
$\omega \propto k$ (light in vacuum, ideal string)	$v_g = v_p$	straight-line dispersion: chord and tangent coincide, packets translate rigidly, no beats between the speeds
$\omega \propto k^2$ (free matter wave)	$v_g = 2v_p$	the envelope outruns its crests; the particle moves classically at $p/m$ while crests loiter at half speed
$\omega \propto \sqrt{k}$ (deep-water gravity waves)	$v_g = v_p/2$	the mirror world: crests outrun the group — watch any ocean swell set do it
$\Delta k \to 0$ (two-wave beat)	$\Delta\omega/\Delta k \to d\omega/dk$	the discrete beat velocity converges to the derivative; for quadratic $\omega(k)$ it already equals $v_g(\bar{k})$ exactly
relativistic matter wave	$v_p = c^2/v > c$ , $v_g = v < c$	crests outrun light, probability does not; derivation owned by Group Velocity versus Phase Velocity

⚠️ One wave, one velocity

The instinct that "the wave's speed" is a single number is the root error of this whole topic. It is true only for straight-line dispersion, $\omega \propto k$ . The moment $\omega(k)$ curves, a wave has a family of velocities — crest speed $v_p$ , envelope speed $v_g$ , and (beyond this article) front and signal speeds — and they genuinely differ: by a factor of 2 either way in the two cases above. Asking "how fast is the electron's wave?" without saying which feature of the wave is like asking how fast a marching band is without saying whether you mean the drummer's hands or the band down the street. Every observable rides the envelope.

Worked examples

Example 1 — the two speeds of a 100 eV electron beam (confidence builder). A cathode-ray beam accelerates electrons through 100 V, the beam used in Probability Current and the Continuity Equation. Find $k$ , $\lambda$ , and both velocities.

Step 1. Classical speed from the kinetic energy: $v = \sqrt{2E/m} = \sqrt{2(100)(1.602\times10^{-19})/(9.109\times10^{-31})} = \mathbf{5.93\times10^{6}\ \text{m/s}}$ — about 2% of $c$ , safely nonrelativistic.

Step 2. Wavenumber and wavelength: $k = mv/\hbar = (9.109\times10^{-31})(5.93\times10^{6})/(1.055\times10^{-34}) = \mathbf{5.12\times10^{10}\ \text{m}^{-1}}$ , so $\lambda = 2\pi/k = \mathbf{0.123\ \text{nm}}$ — atomic-lattice scale, which is why such beams diffract off crystals.

Step 3. Group velocity: $v_g = \hbar k/m = \mathbf{5.93\times10^{6}\ \text{m/s}}$ — identical to Step 1's classical speed, as the derivation promised. The beam's arrival time at a detector is set by this number.

Step 4. Phase velocity: $v_p = \hbar k/2m = \mathbf{2.97\times10^{6}\ \text{m/s}}$ . No detector clicks at this speed. Sanity check: $\tfrac{1}{2}mv_g^2 = 100$ eV on the nose — the energy bookkeeping votes for the group velocity too.

Example 2 — timing a storm by its swell (exam level, classical contrast). A weather buoy reports 100-metre-wavelength swell arriving from a storm 1000 km offshore. When did the storm generate it — and what error would you make by timing the crests?

Step 1. Wavenumber: $k = 2\pi/\lambda = 2\pi/100 = \mathbf{0.0628\ \text{m}^{-1}}$ .

Step 2. Frequency and period from deep-water dispersion: $\omega = \sqrt{gk} = \sqrt{(9.81)(0.0628)} = \mathbf{0.785\ \text{s}^{-1}}$ , so the period is $T = 2\pi/\omega = \mathbf{8.0\ \text{s}}$ — a classic long-period forerunner swell.

Step 3. The two speeds: $v_p = \omega/k = \mathbf{12.5\ \text{m/s}}$ for individual crests, but the swell energy travels at $v_g = v_p/2 = \mathbf{6.25\ \text{m/s}}$ .

Step 4. Travel time: $t = (10^6\ \text{m})/(6.25\ \text{m/s}) = 1.6\times10^{5}\ \text{s} = \mathbf{44.4\ \text{hours}}$ . Timing crests instead would give 22.2 hours — a factor-of-two error, the deep-water mirror of the matter wave's factor of two. Surf forecasters live by the group velocity for exactly the reason quantum mechanics does: energy arrives with the envelope.

William Rowan Hamilton

1805–1865 · Trinity College Dublin and Dunsink Observatory

The group-velocity idea is a generation older than the name. In 1839, in two short papers to the Royal Irish Academy on "the propagation of vibration in crystallized media," Hamilton analyzed how a disturbance built from many vibrations travels, and found that the speed of the vibration pattern differs from the speed of propagation of the disturbance as a whole — the first clear separation of what we now call phase and group velocity. It was a characteristic Hamilton move: five years earlier his mathematical optics had predicted conical refraction, a wholly unanticipated behaviour of light in biaxial crystals, confirmed within months by his colleague Humphrey Lloyd and celebrated across Europe. But the vibration papers sank almost without trace; the distinction had to be rediscovered by Stokes, who set it as a Cambridge exam problem, and codified by Rayleigh in his Theory of Sound. Hamilton never knew that the derivative he isolated would, ninety years later, be the exact statement rescuing de Broglie's matter waves — the particle rides at $d\omega/dk$ . His wave work here is distinct from the mechanics that bears his name; that story belongs to the mechanics track.

⏳ History

1839

— Hamilton's Royal Irish Academy papers separate the speed of a vibration pattern from the speed of the travelling disturbance

1876

— Stokes sets the group-velocity distinction as a Smith's Prize examination problem at Cambridge

1877

— Rayleigh's Theory of Sound defines the group velocity dω/dk and applies it to real dispersive media

1924

— de Broglie's thesis attaches waves to matter; the superluminal phase velocity makes the group-velocity resolution essential

1926

— Schrödinger's equation fixes the matter-wave dispersion ω = ħk²/2m, making v_g = p/m a theorem

1999

— Hau's group slows a light pulse's group velocity to 17 m/s in an ultracold atomic gas, bicycle speed

Where this lives today

Group-velocity engineering is a working technology. Optical fibres carry data as pulse envelopes, and the fibre's group-velocity dispersion — the curvature of $\omega(k)$ , the same quantity that spreads quantum packets in Wave Packet Spreading and Dispersion — is what smears adjacent bits into each other; long-haul links interleave fibre segments with opposite dispersion so the smearing of one is undone by the next. At the exotic end, Lene Hau's 1999 experiment used electromagnetically induced transparency in an ultracold sodium cloud to drag light's group velocity down to 17 m/s, and follow-ups parked pulses entirely — the crests still oscillate at optical frequency while the envelope, the only thing carrying the pulse's identity, crawls or stops. In condensed-matter physics the concept is the definition of how electrons move: a Bloch electron in a crystal is a wave packet of lattice waves, and its transport velocity is the group velocity $v_g = \hbar^{-1}\,\partial E/\partial k$ read directly off the band structure that photoemission experiments map. When a band flattens, $v_g \to 0$ — electrons stall, interactions dominate, and flat-band materials like magic-angle twisted bilayer graphene become superconductors. Every one of those statements is this article's tangent-versus-chord distinction, applied.

Check yourself

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., Problem 2.43 and §2.4 (free particle and group velocity) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 (wave packets, Gaussian spreading) · Cohen-Tannoudji, Quantum Mechanics, Vol. 1, Complement G_I (free wave packet) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.4 · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6

Provenance. Upgraded in place from the AllQFT base article at Step-2 (concept register; the base's stationary-phase sketch, relativistic derivation, and Gaussian-spreading formula are not reproduced here — they are owned by Group Velocity versus Phase Velocity and Wave Packet Spreading and Dispersion per chapter ownership, and are recalled by title). A verifier should: (1) confirm the two-wave factorization and Δω/Δk envelope speed; (2) substitute e^i(kx−ωt) into the free TDSE to recover ω = ħk²/2m (Griffiths §2.4); (3) recompute v_p = ħk/2m, v_g = ħk/m and the 100 eV electron numbers (v_g = 5.93×10⁶ m/s, λ = 0.123 nm); (4) recompute the deep-water results v_g = v_p/2 and the 100 m swell numbers (v_p = 12.5 m/s, v_g = 6.25 m/s, 44.4 h from 1000 km); (5) check the quadratic-dispersion beat exactness Δω/Δk = ħ(k₁+k₂)/2m. All are asserted in this article's verification script.

Builds on: de Broglie Matter Waves · Wave–Particle Duality · The Wavefunction

Leads to: The Time-Dependent Schrödinger Equation · Born's Statistical Interpretation · Probability Density and Normalization

If nothing moves at the phase velocity, why does quantum mechanics keep it around at all?

Because the packet is built from components, and each component's phase evolves at its own v_p = ω/k. You cannot compute the envelope's motion, its spreading, or any interference effect without tracking those individual phases — the group velocity itself is defined by how they differ across the band. v_p is indispensable bookkeeping for the parts even though it is not the velocity of the whole.

Which velocity does a real detector actually measure?

The group velocity, always. A detector responds to probability density arriving at its location — the envelope. Time-of-flight measurements on electrons, neutrons, and molecules all confirm arrival at v_g = p/m. There is no experiment whose outcome depends on v_p directly; it enters only through phase differences between components, which is again group-velocity physics.

Why exactly a factor of two for matter waves — is something deep hiding in it?

It is the exponent of the dispersion law, nothing more mysterious. For any power law ω ∝ kˢ, differentiating gives v_g = s·v_p. Matter waves have s = 2 because kinetic energy is quadratic in momentum, so v_g = 2v_p; deep water has s = ½, so v_g = v_p/2. The "deep" content is that only ω-differences are observable — the absolute frequency, like the zero of energy, is convention. That subtlety is worked through in Group Velocity versus Phase Velocity.

Can the group velocity ever exceed c?

Yes — in anomalously dispersive optical media the envelope peak of a smooth pulse can move faster than c, or even backward. No signal does, because such media only reshape the pulse: the earliest possible arrival of new information (the front) still travels at c. For free matter waves the issue never arises: v_g = p/m < c always. The lesson stands: when velocities multiply, ask which feature of the wave each one describes.

A wave packet contains many wavenumbers. The group velocity at which k should I use?

The central one, k₀, provided the packet is narrow in k — then all components share nearly the same dω/dk and the envelope translates rigidly at v_g(k₀). If the packet is broad in k, different parts of it move at different group velocities and the very notion of "the packet's speed" blurs as the packet spreads. That regime is the subject of Wave Packet Spreading and Dispersion.

How does v_g = p/m connect to Ehrenfest's theorem?

They are the same statement in two languages. The beat/envelope argument says the interference pattern peaks where components agree in phase, and that point moves at dω/dk = ħk₀/m. The operator argument in Momentum Operator proves d⟨x⟩/dt = ⟨p⟩/m for any state, no packet approximation needed. For a narrow packet ⟨p⟩ = ħk₀, and the two results coincide exactly.

Do sound waves in air have different phase and group velocities?

Essentially no — over the audible range, air is very nearly non-dispersive: ω ≈ (343 m/s)·k, a straight line, so v_g ≈ v_p. That is why music arrives intact: if air had matter-wave-strength dispersion, a chord's frequencies would arrive at different times and melt into noise. Dispersion is the exception you must earn — vacuum light has none, air sound almost none, matter waves always.