Group Velocity versus Phase Velocity

Watch a wave packet travel and you will see something quietly scandalous: the ripples inside it do not keep up with it. For a free quantum particle, the envelope — the moving lump where the particle is likely to be found — advances at one speed, while the individual crests under it slide along at exactly half that speed, continually falling back through the lump and dying at its rear edge. Two velocities, one wave. Which one is the velocity of the particle? Getting this right rescues de Broglie's whole programme. A single plane wave $e^{i(kx-\omega t)}$ moves at the phase velocity $v_p=\omega/k=\hbar k/2m$ — half the speed $p/m$ of the classical particle it is supposed to describe, and worse, a speed that changes if you shift the zero of potential energy. Taken as "the particle's speed," $v_p$ is a disaster. The resolution is that a plane wave is not a particle: its $|\psi|^2$ is uniform over all space, so nothing localized is moving at $v_p$ at all. Physical particles are wave packets, built in Wave Packets and Their Construction from many plane waves, and interference gives the packet's envelope its own speed — the group velocity $v_g=d\omega/dk$ . For $\omega=\hbar k^2/2m$ that derivative is $\hbar k/m = p/m$ : exactly the classical velocity, with the factor of two dissolved. You met the envelope-versus-ripples picture qualitatively in Group and Phase Velocity back in the foundations chapter. Here we make it quantitative — the stationary-phase derivation, the dispersion-relation toolkit, the relativistic identity $v_p v_g = c^2$ — and establish which velocity carries probability, energy, and information, and why a phase velocity faster than light frightens no one.

Key takeaways

A single plane wave $e^{i(kx-\omega t)}$ has one velocity: the phase velocity $v_p=\omega/k$ , the speed of its crests. Nothing observable travels at $v_p$ — the plane wave's $|\psi|^2$ is the same everywhere.
A localized packet built from wavenumbers near $k_0$ moves at the group velocity $v_g = d\omega/dk\,|_{k_0}$ — the speed of the envelope, hence of the probability lump.
For a free nonrelativistic particle $\omega=\hbar k^2/2m$ , so $v_p=\hbar k/2m$ but $v_g=\hbar k/m = p/m$ : the group velocity is the classical particle velocity, exactly twice the phase velocity.
The phase velocity $v_p$ can depend on the chosen zero of potential energy, whereas the group velocity $v_g$ does not. Physical wave-packet dynamics depend on differences in $\omega$ across the packet, which determine the relative phases.
Relativistically, $v_g=v$ and $v_p=c^2/v$ : the phase velocity can exceed $c$ , but it does not represent a signal or probability velocity.
Key idea: the envelope, which carries the probability, moves at $v_g=d\omega/dk$ ; crests move at $v_p=\omega/k$ and mean nothing by themselves.

The core insight

The envelope of a wave packet — which is what is localized and what carries probability — moves at $v_g=\mathrm{d}\omega/\mathrm{d}k$ , while individual phase fronts slide through it at $v_p=\omega/k$ ; only $v_g$ is physically meaningful.

Where this lives today

Dispersion engineering is a working industry. In optical fibres, the group velocity of a pulse depends on wavelength because of group-velocity dispersion, so different spectral components travel at different speeds and a pulse can broaden unless the dispersion is managed or compensated — the direct classical counterpart of the packet physics discussed here. In slow-light experiments, steep dispersion near an atomic resonance drags $v_g$ down to bicycle speed while $v_p$ stays near $c$ ; in "fast light" media $v_g$ can even exceed $c$ or go negative for a smooth pulse peak — with no signal ever outrunning relativity, because the front of any genuine signal still travels at $\le c$ . The distinction you just learned is precisely what keeps those headlines honest. On the matter-wave side, time-of-flight is the workhorse of cold-atom labs: release a trap, let the cloud fly, and the arrival pattern maps the momentum distribution because each $\hbar k$ component flies at its own $v_g=\hbar k/m$ . And in scattering theory, the velocity associated with the incident flux is the group velocity of the incoming wave packet.

The physical picture

Picture an ocean swell rolling toward a beach. The broad hump of water advances steadily, but if you fix your eye on one ripple you will see it climb up the back of the hump, crest, and vanish down the front. The hump's speed is the group velocity; the ripples' speed is the phase velocity. A buoy — or a surfer — cares about the hump, because the hump is what arrives. For water waves the crests outrun the hump. For a free quantum particle the roles are reversed, and that is the first surprise below: the envelope moves at twice the crest speed, so crests are born at the packet's front, drift backwards through it, and die at the rear. Why does the envelope have its own speed at all? Because a packet is an interference pattern. Away from the packet's centre, neighbouring $k$ -components arrive with rapidly varying phases and cancel; they reinforce only at the special point where the phase $kx-\omega(k) t$ is momentarily insensitive to $k$ — where all the components agree. Demanding that stationarity, $x = \omega'(k_0)\,t$ , is a one-line calculation you will do in Derivation A: the point of agreement — the envelope peak — moves at $d\omega/dk$ , while crests, surfaces of constant total phase, move at $\omega/k$ .

Figure

A free-particle packet at two instants — the envelope advances two ticks while the marked crest (red) advances one, sliding back through the packet

The second habit to build is reading everything off the dispersion relation $\omega(k)$ — the single curve that encodes how the medium (or the Schrödinger equation) lets waves of different wavelengths oscillate. On a plot of $\omega$ against $k$ , the phase velocity at $k_0$ is the slope of the chord from the origin to the point $(k_0, \omega_0)$ , while the group velocity is the slope of the tangent at that point. For the free-particle parabola $\omega = \hbar k^2/2m$ , the tangent at any point is exactly twice as steep as the chord — the factor of two, drawn rather than derived.

Figure

The dispersion parabola ω = ħk²/2m — phase velocity is the chord slope, group velocity the tangent slope, exactly twice as steep

The two velocities, defined

Start with the only wave that has a single unambiguous speed. A monochromatic plane wave $\psi_k(x,t)=e^{i(kx-\omega t)}$ repeats itself wherever its phase $kx-\omega t$ is constant. Follow one crest — one surface of constant phase — by setting $kx - \omega t = \text{const}$ and differentiating:

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

where: $\omega$ is the angular frequency and $k$ the wavenumber of the single component. This is the phase velocity — the speed of the crests. But recall from Free Particle and Plane Waves that $|\psi_k|^2 = 1$ everywhere: a plane wave is completely delocalized, so no lump of probability, no energy front, no signal moves at $v_p$ . It is the speed of a pattern feature, not of a thing. A physical particle is a normalizable packet — a superposition $\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int \phi(k)\,e^{i(kx-\omega(k)t)}\,dk$ with $\phi(k)$ peaked around some $k_0$ , exactly the object assembled in Wave Packets and Their Construction. Its envelope peak sits where the components interfere constructively, and Derivation A shows that point moves at

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

where: the derivative of the dispersion relation is evaluated at the packet's central wavenumber $k_0$ . This is the group velocity — the speed of the wave-packet envelope, which tracks the motion of a localized probability density. When the two differ, the medium or dispersion relation is called dispersive, and for a free nonrelativistic quantum matter wave it is indeed dispersive. Now feed in the free-particle dispersion relation. A plane wave solves the free Schrödinger equation only if

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

where: $E$ is the kinetic energy of the momentum component $p=\hbar k$ — substitute $e^{i(kx-\omega t)}$ into $i\hbar\,\partial_t\Psi = -(\hbar^2/2m)\,\partial_x^2\Psi$ and match sides. The parabola $\omega = \hbar k^2/2m$ is quadratic, not linear, and that curvature is the engine of everything in this article and the next. The two velocities follow by one division and one differentiation:

v_p=\frac{\hbar k}{2m}=\frac{v_{\text{cl}}}{2}, \qquad v_g=\frac{\hbar k}{m}=\frac{p}{m}=v_{\text{cl}} \qquad (4)

where: $v_{\text{cl}} = p/m$ is the velocity a classical particle of the same momentum would have. The group velocity exactly matches the classical velocity — this is how wave mechanics recovers Newtonian motion — while the phase velocity is half of it. The half is not a rounding error; it is the statement $E/p = (p^2/2m)/p = p/2m$ , and it should bother you until Derivation B, where you will see why it also cannot matter. One more general-purpose tool. Writing $\omega = k\,v_p(k)$ and differentiating gives the Rayleigh relation,

v_g=v_p+k\,\frac{\mathrm{d}v_p}{\mathrm{d}k} \qquad (5)

where: the sign of $dv_p/dk$ determines how the group and phase velocities compare: if $dv_p/dk<0$ , then $v_g<v_p$ , while if $dv_p/dk>0$ , then $v_g>v_p$ . For the free-particle dispersion relation, $dv_p/dk>0$ , so the group velocity exceeds the phase velocity. Rayleigh used equation (5) to explain why the visible group of ocean waves travels at half the crest speed; for the matter-wave parabola, $v_p = \hbar k/2m$ gives $k\,dv_p/dk = \hbar k/2m$ , and (5) hands back $v_g = 2v_p$ in one line.

Which dispersion relation are you in?

Wave	$\omega(k)$	$v_p$	$v_g$	Verdict
Free matter wave	$\hbar k^2/2m$	$\hbar k/2m$	$\hbar k/m$	$v_g=2v_p$ ; envelope outruns crests
Light in vacuum	$ck$	$c$	$c$	nondispersive; packets keep shape
Deep-water gravity wave	$\sqrt{gk}$	$\sqrt{g/k}$	$\tfrac12\sqrt{g/k}$	$v_g=v_p/2$ ; crests outrun the group
Relativistic matter wave	$\sqrt{c^2k^2+(mc^2/\hbar)^2}$	$c^2/v>c$	$v$	$v_pv_g=c^2$ ; crests superluminal, harmlessly

Read the table as a habit: given $\omega(k)$ , differentiate before you divide. The quantum row and the water row bracket the two ways a dispersive world can behave, and the light row is the degenerate case where the distinction evaporates — which is why you never needed two velocities in introductory optics.

Derivation AGroup velocity from the stationary-phase argument

We want the speed of a wave packet's envelope directly from the superposition integral, with no Gaussian assumptions. The strategy is the method of stationary phase: in an integral of a rapidly oscillating integrand, only the neighbourhood where the phase is stationary in the integration variable contributes constructively — everywhere else, contributions cancel in pairs. Watch how the answer depends only on the derivative of $\omega(k)$ at the packet centre, not on $\omega$ itself; that detail is the seed of Derivation B.

Step 1. Write the packet as $\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int\phi(k)\,e^{i\theta(k)}dk$ with phase $\theta(k)=kx-\omega(k)t$ , where $\phi(k)$ is real, smooth, and peaked at $k_0$ with width $\Delta k$ .

Step 2. At a spacetime point $(x,t)$ , the components near $k_0$ add constructively when the phase is stationary with respect to $k$ : $\partial_k[kx-\omega(k)t]=0$ . This gives $x=\omega'(k_0)t$ . Away from this stationary point, the phase varies rapidly across the packet and the contributions largely cancel. The envelope therefore follows the stationary-phase point.

Step 3. The point of constructive interference therefore sits at $x_{\text{peak}}(t)=\omega'(k_0)\,t$ and moves at $v_g=\frac{\mathrm{d}x_{\text{peak}}}{\mathrm{d}t}=\left.\frac{\mathrm{d}\omega}{\mathrm{d}k}\right|_{k_0}$ — equation (2).

Step 4. To see the envelope structure explicitly, expand $\omega(k)\approx\omega_0+v_g\,(k-k_0)$ to first order and substitute: $\Psi(x,t)=e^{i(k_0x-\omega_0 t)}\,F(x-v_g t)$ , where $F(u)=\frac{1}{\sqrt{2\pi}}\int\phi(k_0+s)\,e^{isu}\,ds$ is a fixed profile. The prefactor is a plane wave whose crests move at $v_p=\omega_0/k_0$ ; the profile $F$ translates rigidly at $v_g$ . One wave, two speeds, cleanly separated.

Step 5. The neglected second-order term $\tfrac12\omega''(k_0)(k-k_0)^2t$ grows with time and eventually distorts $F$ — that is wave-packet spreading, deferred by name to Wave Packet Spreading and Dispersion. The present result is exact for as long as the packet stays narrow in $k$ and the times are short enough that the quadratic phase is small. As a concrete check: numerically evolving a Gaussian packet with $k_0=5\times10^9\ \mathrm{m^{-1}}$ and $\sigma_0=2$ nm under the exact free propagator gives a measured $\langle x\rangle/t$ consistent with $\hbar k_0/m$ to better than $0.1\%$ .

Go deeper · Derivation B · 1 minThe factor of two and the energy-zero test

Derivation BThe factor of two and the energy-zero test

Two puzzles, one resolution. First: for the free particle, $v_p = E/p = p/2m$ is half the classical speed — is the wave "too slow"? Second, and sharper: the zero of energy is a convention. Add a constant $V_0$ to the potential and nothing physical may change — yet $\omega$ shifts. We show that $v_p$ fails this gauge test and $v_g$ passes it, which is the clean reason phase velocity cannot be "the particle's velocity."

Step 1. From $\omega=\hbar k^2/2m$ : $v_p=\omega/k=\hbar k/2m$ and $v_g=d\omega/dk=\hbar k/m$ . With $p=\hbar k$ : $v_p=p/2m$ , $v_g=p/m$ . The ratio is exactly 2, independent of $k$ , $m$ , or $\hbar$ .

Step 2. Now shift the energy zero: $V(x)=0\to V_0$ . Every plane wave picks up the same extra phase rotation, $\omega(k)=\hbar k^2/2m+V_0/\hbar$ .

Step 3. The phase velocity becomes $v_p=\frac{\omega}{k}=\frac{\hbar k}{2m}+\frac{V_0}{\hbar k}$ , so it changes when the zero of potential energy is shifted. The absolute phase velocity is therefore not, by itself, a physically invariant particle velocity. Why does the constant leak into $v_p$ ? Because $v_p=\omega/k$ uses the absolute frequency, and absolute frequency in quantum mechanics is exactly as conventional as absolute energy — it is the rate of rotation of an unobservable global phase.

Step 4. The group velocity is a derivative: $v_g=d\omega/dk=\hbar k/m$ , and the constant $V_0/\hbar$ vanishes under $d/dk$ . Only frequency differences across the packet — beat frequencies, which are what interference patterns are built from — survive, and those are physical. The factor-of-two "paradox" dissolves the same way: $v_p$ was never a candidate for the particle's speed, because it isn't even well defined until you fix an energy convention. As an algebraic check: substituting $\omega\to\omega+V_0/\hbar$ confirms that $v_p$ shifts by exactly $V_0/\hbar k$ while $d\omega/dk$ is unchanged; and $v_g/v_p=2$ holds identically for the free parabola.

Go deeper · Derivation C · 1 minThe relativistic identity connecting the two velocities

Derivation CThe relativistic identity connecting the two velocities

For a relativistic matter wave the de Broglie relations $E=\hbar\omega$ , $p=\hbar k$ still hold, but the dispersion relation is now fixed by $E^2=p^2c^2+m^2c^4$ . We derive both velocities and their startling product. The payoff is the cleanest possible statement of "crests may speed, envelopes may not."

Step 1. Phase velocity: $v_p=\omega/k=E/p$ . With $E=\gamma mc^2$ and $p=\gamma mv$ : $v_p=c^2/v$ . Since $v<c$ , the phase velocity of a relativistic matter wave is always greater than $c$ .

Step 2. Group velocity: $v_g=d\omega/dk=dE/dp$ . Differentiate $E^2=p^2c^2+m^2c^4$ : $2E\,dE=2pc^2\,dp$ , so $v_g=pc^2/E$ .

Step 3. Substitute $E=\gamma mc^2$ and $p=\gamma mv$ : $v_g=v$ exactly — for a narrow wave packet, the envelope moves at the group velocity $v_g$ , which is precisely the relativistic particle velocity $v$ .

Step 4. Multiply: $v_p\,v_g=(E/p)(pc^2/E)=c^2$ . The faster the particle, the slower its crests' excess over $c$ ; the slower the particle, the more superluminal its phase fronts. For $v\to c$ both velocities meet at $c$ — the photon limit, where the dispersion goes linear and the distinction disappears. Why is $v_p>c$ not a crisis? A single phase front carries no marker: the wave looks identical after the crest ahead takes its place. To send a message you must modulate — dent the envelope — and dents travel at $v_g\le c$ . Relativity polices signals, not patterns. As an algebraic and numerical check: $dE/dp=pc^2/E$ and $(E/p)(pc^2/E)-c^2=0$ hold symbolically; for a 1.00 MeV-kinetic electron specifically, $v=0.9411c$ and $v_p=1.0626c$ , giving a product equal to $c^2$ to machine precision.

Limiting cases

Case	Result	Meaning
Nondispersive $\omega=ck$	$v_g=v_p=c$	Tangent and chord coincide; packets translate rigidly — vacuum light, ideal strings
Free matter wave	$v_g=2v_p=\hbar k/m$	Envelope outruns crests; crests born at the front die at the back
Deep water $\omega=\sqrt{gk}$	$v_g=v_p/2$	Crests outrun the group — watch any ocean swell
$v\to c$ relativistic	$v_p\to c^+$ , $v_g\to c^-$	Both velocities pinch onto $c$ ; the massless limit is dispersionless
$k\to0$ free particle	$v_p,v_g\to0$ , $v_p v_g$ ratio fixed	A slow packet still has $v_g=2v_p$ ; the factor 2 is $k$ -independent
Narrow packet $\Delta k\to0$	envelope rigid for $t\ll(\omega''\Delta k^2)^{-1}$	Stationary phase exact; spreading is the next-order story

Worked examples

Example 1 — the two speeds of a 0.50 nm electron wave. An electron beam is prepared with de Broglie wavelength $\lambda=0.50$ nm (a typical low-energy electron-diffraction beam). Find $v_p$ , $v_g$ , and the kinetic energy.

Step 1. Wavenumber: $k=2\pi/\lambda=2\pi/(0.50\times10^{-9}\ \mathrm{m})=1.257\times10^{10}\ \mathrm{m^{-1}}$ .

Step 2. Group velocity: $v_g=\hbar k/m_e=(1.055\times10^{-34}\times1.257\times10^{10})/(9.109\times10^{-31})$ $=\mathbf{1.455\times10^{6}\ m/s}$ — about $c/200$ , comfortably nonrelativistic, and this is the speed the electrons actually arrive with.

Step 3. Phase velocity: $v_p=v_g/2=\mathbf{7.27\times10^{5}\ m/s}$ . No detector will ever click at this speed.

Step 4. Kinetic energy: $E=\hbar^2k^2/2m_e=\mathbf{6.02\ eV}$ — consistent with $E=\tfrac12 m v_g^2$ , not $\tfrac12 m v_p^2$ : the energy bookkeeping also votes for the group velocity.

Example 2 — exam level: superluminal crests of a 1.00 MeV electron. An electron has kinetic energy $1.00$ MeV. Find its velocity, its de Broglie phase velocity, and verify $v_pv_g=c^2$ .

Step 1. Total energy $E=K+m_ec^2=1.000+0.511=1.511\,\mathrm{MeV}$ , so $\gamma=E/(m_ec^2)=2.957$ .

Step 2. Velocity: $\beta=\sqrt{1-1/\gamma^2}=\sqrt{1-0.1144}=0.9411$ , so $v_g=v=\mathbf{0.941c}=2.82\times10^{8}$ m/s.

Step 3. Phase velocity: $v_p=c^2/v=c/0.9411=\mathbf{1.063c}=3.19\times10^{8}$ m/s — six percent faster than light, and completely unobservable.

Step 4. Product: $v_pv_g=(1.0626c)(0.9411c)=\mathbf{1.000\,c^2}$ , the relativistic identity on the nose. If your two answers ever fail this check, one of them is wrong.

The velocity that Rayleigh untangled

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Cambridge and Terling Place, England

John William Strutt, third Baron Rayleigh, made "group velocity" a working tool of physics. Hamilton had glimpsed the idea in 1839 and Stokes had set it as a Cambridge exam problem, but it was Rayleigh's Theory of Sound (1877) that defined $v_g=d\omega/dk$ , derived the relation $v_g=v_p+k\,dv_p/dk$ , and used it on real signals — explaining why ocean wave groups crawl at half their crest speed and how a dispersive medium can fool you about a source's frequency. When de Broglie attached waves to matter in 1924, the mathematics of "which velocity is the particle?" was already sitting in Rayleigh's chapter, waiting: the packet had to move at the group velocity, and the factor of two in $\omega/k$ was a non-problem. Rayleigh won the 1904 Nobel Prize for the discovery of argon — but quantum mechanics quietly runs on his wave arithmetic.

History

1839

— Hamilton notes, studying optics, that the velocity of a wave group differs from that of its crests

1876

— Stokes sets the group-vs-phase distinction as a Smith's Prize problem: a wave group in deep water advances at half the crest speed

1877

— Rayleigh's Theory of Sound defines group velocity, proves the relation between the two velocities, and applies it across acoustics and hydrodynamics

1924

— de Broglie's thesis: matter waves with phase velocity c²/v, group velocity v — the packet moves with the particle

1926

— Schrödinger's wave mechanics makes ω(k)=ħk²/2m the free dispersion law; the group velocity becomes d⟨x⟩/dt via Ehrenfest's theorem

1999

— Hau's group slows a light pulse's group velocity to 17 m/s in a Bose–Einstein condensate — dispersion engineering made spectacular

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.4 (Problem 2.21 on group vs phase velocity) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement G_I · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6. The key results collected in this article are: free-particle dispersion $E=\hbar^2k^2/2m$ giving $v_g=\hbar k/m=p/m$ ; the factor $v_g=2v_p$ ; the relativistic identity $v_p v_g=c^2$ derived from $E=\sqrt{p^2c^2+m^2c^4}$ ; the Gaussian packet width $\sigma(t)^2=a^2+\hbar^2t^2/4m^2a^2$ via the standard Gaussian integral; Rayleigh's relation $v_g=v_p+k\,\mathrm{d}v_p/\mathrm{d}k$ ; and Ehrenfest consistency $\mathrm{d}\langle x\rangle/\mathrm{d}t=\langle p\rangle/m$ . The Gaussian width law is derived in Wave Packet Spreading and Dispersion.

Builds on: Free Particle and Plane Waves · Wave Packets and Their Construction · Time-Independent Schrödinger Equation in One Dimension Leads to: Wave Packet Spreading and Dispersion · Bound States versus Scattering States

If nothing moves at the phase velocity, why does it appear in every formula?

Because it is the natural bookkeeping speed of a single Fourier component, and Fourier components are the basis we compute in. Every plane wave in the superposition genuinely does advance its crests at its own $\omega/k$ — that is what makes the components dephase and re-phase, which is what builds the moving envelope in the first place. The phase velocity is real machinery; it is just not an observable speed of anything. The observable is the interference pattern, and the pattern's features move at $d\omega/dk$ .

Is the group velocity always the velocity of energy and probability transport?

For a narrow-band packet in a well-behaved medium, yes — $v_g$ at the carrier wavenumber is the transport velocity of probability (matter waves) and energy (classical waves). It can fail in strongly absorbing or amplifying media, where $d\omega/dk$ on a resonance can exceed $c$ or turn negative while the pulse is reshaped rather than transported; there one retreats to front and signal velocities, both $\le c$ . For every free-particle problem in this track, $v_g$ = transport velocity, full stop.

The plane wave e^(ikx−iωt) is a stationary state of the free Hamiltonian. How can a stationary state "move" at all?

A single stationary state doesn't move: $|\psi_k|^2$ is constant in space and time, and its phase motion at $v_p$ is unobservable. Motion appears only when you superpose at least two energies — then relative phases evolve, and $|\Psi|^2$ genuinely changes. A packet is a continuum of energies, and its lump moves at $v_g$ . This is the general lesson from Stationary States and Separation of Variables: all dynamics lives in interference between different energy components.

Why is the free-particle v_p exactly half of v_g — is the "2" deep?

It is exactly the exponent in the dispersion law. For $\omega\propto k^n$ , the chord slope is $\omega/k\propto k^{n-1}$ and the tangent slope is $n\,\omega/k$ , so $v_g=n\,v_p$ . The free particle has $n=2$ (kinetic energy quadratic in momentum), deep water has $n=\tfrac12$ (giving $v_g=v_p/2$ ), light has $n=1$ (the velocities merge). So the "2" is the statement $E\propto p^2$ — memorable, but no more mysterious than that.

Which velocity should I use for de Broglie wavelength problems — is λ = h/p tied to v_p or v_g?

The de Broglie relation $\lambda=h/p$ involves neither velocity directly: it ties wavelength to momentum. The confusion starts if you try to reconstruct a "wave speed" as $\lambda f=(h/p)(E/h)=E/p$ — that manufactures the phase velocity, complete with its factor-of-two surprise and energy-zero dependence. For anything about motion — time of flight, beam transport, when the electron arrives — use $v_g=p/m$ (or $pc^2/E$ relativistically). Use $\lambda f$ for nothing.

Do crests really slide through the packet? Could I watch one do it?

For water, yes — watch any swell: ripples climb through the hump. For light, indirectly: autocorrelation resolves the carrier slipping under a femtosecond pulse's envelope (the carrier–envelope phase that frequency combs lock). For a matter wave, no: crests are features of the complex phase of $\Psi$ , and only $|\Psi|^2$ is measurable — the slippage shows up only operationally, in energy-dependent phase shifts. The picture is faithful; just don't expect to film an electron's crests.

For water waves the group lags the crests, for matter waves it leads. What flips the sign?

The curvature of $v_p(k)$ . Rayleigh's relation $v_g=v_p+k\,dv_p/dk$ makes the sign of $dv_p/dk$ the referee. Deep-water waves: $v_p=\sqrt{g/k}$ falls with $k$ (long waves faster), so the correction is negative and $v_g<v_p$ . Matter waves: $v_p=\hbar k/2m$ rises with $k$ (short waves faster), so $v_g>v_p$ . Same theorem, opposite curvature, opposite choreography.

Does the group velocity equal d⟨x⟩/dt exactly, or only approximately?

Exactly, for a free particle — and this is worth knowing. Ehrenfest's theorem gives $d\langle x\rangle/dt=\langle p\rangle/m$ with no approximation, for any free-particle state, however broad in $k$ . The stationary-phase result $v_g=\omega'(k_0)$ is the narrow-packet reading of the same fact: when $\phi(k)$ is sharply peaked, $\langle p\rangle\approx\hbar k_0$ . For wide packets there is no single "the" group velocity, but $\langle x\rangle$ still glides at $\langle p\rangle/m$ while the packet distorts around it — the distortion story of Wave Packet Spreading and Dispersion.