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Plane waves as idealized states, wave packets as real particles, phase velocity $\omega/k$ and group velocity $d\omega/dk$, and the spreading of a packet in time.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to build wave packets from plane waves, distinguish phase from group velocity, and compute how fast a free packet spreads.
You know the de Broglie relations $p = \hbar k$ and $E = \hbar\omega$, the Gaussian wave packet from the lesson on uncertainty, and from the last two lessons that stationary states evolve by a phase $e^{-iEt/\hbar}$. You may also know, from waves, that a sum of waves of slightly different wavelengths makes beats. This lesson applies those ideas to a particle with no forces on it.
| Term | What it means |
|---|---|
| Plane wave | $e^{i(kx - \omega t)}$, a stationary state of the free particle with definite momentum $\hbar k$. |
| Dispersion relation | The relation between $\omega$ and $k$; for a free particle $\omega = \hbar k^2/2m$. |
| Wave packet | A superposition of plane waves with a spread of $k$, localized in space and normalizable. |
| Phase velocity | $v_p = \omega/k$, the speed of individual crests. |
| Group velocity | $v_g = d\omega/dk$, the speed of the packet's envelope and of the particle. |
| Dispersion | The spreading of a packet because its components travel at different speeds. |
| Spreading time | $\tau = 2m\sigma_0^2/\hbar$, the time after which a Gaussian packet's width grows significantly. |
With $V = 0$ the time-independent Schrödinger equation is $-\dfrac{\hbar^2}{2m}\psi'' = E\psi$, solved by $\psi = e^{ikx}$ with $E = \hbar^2k^2/2m$. Attaching the time dependence gives the plane wave
$$\Psi_k(x, t) = e^{i(kx - \omega t)}, \qquad \omega = \frac{\hbar k^2}{2m}.$$
It has definite momentum $\hbar k$ and definite energy, but $|\Psi_k|^2 = 1$ everywhere, so $\int|\Psi_k|^2dx$ is infinite: it cannot be normalized and does not describe a physical particle. It is, however, a building block. Adding plane waves with a spread of wave numbers gives a wave packet,
$$\Psi(x, t) = \frac{1}{\sqrt{2\pi}}\int\phi(k)\,e^{i(kx - \omega t)}\,dk,$$
localized where the components add in phase and normalizable when $\phi(k)$ is. Two speeds now appear. Individual crests move at the phase velocity $v_p = \omega/k = \hbar k/2m$. The envelope — the region where the particle is likely to be found — moves at the group velocity
$$v_g = \frac{d\omega}{dk} = \frac{\hbar k}{m} = \frac{p}{m},$$
exactly the classical velocity. So $v_g = 2v_p$ for matter waves: crests are born at the back of the packet, run forward through it and die out at the front. And because different components travel at different speeds, the packet spreads: a Gaussian of initial width $\sigma_0$ has $\sigma(t) = \sigma_0\sqrt{1 + (t/\tau)^2}$ with $\tau = 2m\sigma_0^2/\hbar$.
Another way: picture
Picture a group of runners who start bunched together, each at a slightly different pace. The bunch as a whole advances at the average pace, but it stretches out as the fast runners pull ahead and the slow ones fall back. A wave packet is that bunch; its components are the runners, each moving at its own phase speed. A tighter start needs a wider range of paces — the uncertainty principle — so the tighter the packet, the faster it stretches.
Another way: steps
Checks. The group velocity must equal $p/m$. The phase velocity of a matter wave is half that, not the particle's speed. When $\omega \propto k$, as for light in vacuum, the two velocities agree and packets do not spread. And the spreading time scales as mass times width squared: an electron localized to an atom spreads in attoseconds, while a grain of dust localized to a micrometer would take far longer than the age of the universe.
Take a packet whose components are concentrated near $k_0$ and expand the dispersion relation: $\omega(k) \approx \omega_0 + v_g(k - k_0)$ with $v_g = d\omega/dk$ at $k_0$. Then
$$\Psi(x, t) \approx e^{i(k_0x - \omega_0t)}\int\phi(k)\,e^{i(k - k_0)(x - v_gt)}\,dk.$$
The integral depends on $x$ and $t$ only through $x - v_gt$: it is an envelope of fixed shape sliding along at $v_g$. In front of it sits a carrier wave $e^{i(k_0x - \omega_0t)}$ moving at $\omega_0/k_0$, the phase velocity. The next term in the expansion, proportional to $d^2\omega/dk^2 = \hbar/m$, is what makes the envelope change shape: it is the source of spreading, and it vanishes only for linear dispersion.
The same argument explains why information and energy travel at the group velocity in every wave system, from signals in optical fibers to ripples on a pond, where the individual crests visibly outrun the group and vanish at its leading edge.
A plane wave cannot be normalized, so it cannot be the state of a real particle. Yet physicists use plane waves constantly, in scattering theory, in solid-state physics and in particle beams. The reason is that a packet many wavelengths long behaves, over any region small compared with its length, exactly like a plane wave. An electron beam from a microscope's gun has a momentum spread of a few parts in a million, so its packets extend over millions of wavelengths; treating each electron as a plane wave of definite momentum is an excellent approximation.
When a calculation with plane waves gives an infinite or ambiguous answer, go back to packets: the infinity is a sign that the idealization has been pushed too far. The formal tool for handling plane waves cleanly, normalization to a Dirac delta function, $\langle k|k'\rangle = \delta(k - k')$, appears in the formalism unit.
The spreading time $\tau = 2m\sigma_0^2/\hbar$ grows with the mass and with the square of the width. For an electron packet $1$ nm wide, $\tau = 17$ fs; for one $0.1$ nm wide, about the size of an atom, $\tau = 0.17$ fs, and the packet doubles in width in about $0.3$ fs. For a buckminsterfullerene molecule of $720$ atomic mass units, localized to $1$ nm, $\tau = 23$ ns. For a speck of dust of $10^{-12}$ kg localized to $1$ μm, $\tau$ is about $600$ years; for anything larger it exceeds the age of the universe.
That is why electrons behave so wavelike and everyday objects never seem to spread. The two regimes are not governed by different laws: the same formula, with different masses, gives both. Experiments that interfere ever heavier molecules, now past $25{,}000$ atomic mass units, push the frontier between them, showing that quantum spreading applies to large molecules too when they are isolated from their surroundings.
Light in vacuum has $\omega = ck$, a straight line, so its phase and group velocities are both $c$ and a light pulse keeps its shape. In glass the dispersion relation curves, because the refractive index depends on wavelength, and a short pulse spreads, a limit on how fast data can be sent through optical fiber. Engineers design fibers and add dispersion-compensating sections so that pulses stay sharp over thousands of kilometers.
Matter waves always disperse, because their $\omega$ goes as $k^2$. For a relativistic particle, $\hbar\omega = \sqrt{(\hbar kc)^2 + (mc^2)^2}$, and one finds $v_pv_g = c^2$: the phase velocity exceeds $c$ while the group velocity stays below it. No signal or energy travels faster than light, because both travel at the group velocity; the crests carry nothing that could be used to send a message.
The 2023 Nobel Prize in Physics went to Pierre Agostini, Ferenc Krausz and Anne L'Huillier for methods that produce pulses of light lasting a few hundred attoseconds, $10^{-18}$ s. Such short pulses are needed because electrons move and spread on that time scale. An electron wave packet localized to an atom, about $0.1$ nm across, has a spreading time $\tau = 2m\sigma_0^2/\hbar$ of about $0.17$ fs, or $170$ attoseconds, and it doubles in width in about $300$ attoseconds.
To photograph that motion, the probe must be shorter than the motion itself, just as a camera needs a shutter faster than a moving car. With attosecond pulses, researchers have timed how long it takes an electron to leave an atom after absorbing a photon, watched electrons redistribute in molecules before the nuclei can move, and studied how quickly a semiconductor switches from insulating to conducting — measurements that could guide the design of faster electronics. The calculation of packet spreading in this lesson sets the time scale all of them must beat.
A transmission electron microscope accelerates electrons to energies of $100$ to $300$ keV and uses them to image individual atoms. Each electron is a wave packet, and its group velocity is the electron's speed; its central wavelength, a few picometers, sets the ultimate resolution.
The microscope works well only if the packets are nearly plane waves, with a narrow spread of energies. Electron sources are rated by energy spread: a cold field-emission gun, with a spread of about $0.3$ eV, gives packets long and coherent enough for the finest images, while a heated tungsten filament, with a spread of a couple of electron volts, blurs fine detail because different parts of each packet focus at slightly different depths. Microscope designers call this chromatic aberration, and correcting it is one of the main reasons modern instruments cost millions of dollars. The same trade-off between spatial localization and spread of wave numbers governs both the packets and the images.
It is natural to think that a matter wave's crests move with the particle. They do not. For $\omega = \hbar k^2/2m$, the crests move at $v_p = \hbar k/2m$, half the particle's speed. What moves with the particle is the envelope of the packet, at the group velocity $d\omega/dk = \hbar k/m = p/m$. The two agree only when $\omega$ is proportional to $k$, as for light in vacuum.
A related error is to treat a plane wave as a real particle state. A plane wave has perfectly definite momentum and is spread uniformly over all space, so it cannot be normalized. Real particles are packets with some spread of momenta, and so they always spread in time, faster the more tightly they start.
An electron moves at $2.0 \times 10^{6}$ m/s. Find its momentum.
$p = mv = 9.109 \times 10^{-31} \times 2.0 \times 10^{6} = 1.82 \times 10^{-24}\ \text{kg m/s}$
Nonrelativistic, since $v \ll c$.
Find its wave number.
$k = \dfrac{p}{\hbar} = \dfrac{1.82 \times 10^{-24}}{1.0546 \times 10^{-34}} = 1.73 \times 10^{10}\ \text{m}^{-1}$
From the de Broglie relation.
Find its angular frequency.
$\omega = \dfrac{\hbar k^2}{2m} = \dfrac{1.0546 \times 10^{-34} \times (1.73 \times 10^{10})^2}{2 \times 9.109 \times 10^{-31}} = 1.73 \times 10^{16}\ \text{rad/s}$
The kinetic energy divided by $\hbar$.
Find the phase velocity.
$v_p = \dfrac{\omega}{k} = \dfrac{1.73 \times 10^{16}}{1.73 \times 10^{10}} = 1.0 \times 10^{6}\ \text{m/s}$
Half the particle's speed.
Find the group velocity.
$v_g = \dfrac{\hbar k}{m} = \dfrac{1.0546 \times 10^{-34} \times 1.73 \times 10^{10}}{9.109 \times 10^{-31}} = 2.0 \times 10^{6}\ \text{m/s}$
Equal to the particle's speed, as it must be.
Choose a Gaussian distribution of wave numbers.
$\phi(k) \propto e^{-(k - k_0)^2/4\sigma_k^2}$
Centered on $k_0$ with spread $\sigma_k$.
Write the packet at $t = 0$.
$\Psi(x, 0) \propto \int e^{-(k - k_0)^2/4\sigma_k^2}e^{ikx}\,dk$
A superposition of plane waves.
Shift the variable of integration.
$q = k - k_0: \quad \Psi(x, 0) \propto e^{ik_0x}\int e^{-q^2/4\sigma_k^2}e^{iqx}\,dq$
The carrier wave $e^{ik_0x}$ comes out front.
Use the Fourier transform of a Gaussian.
$\int e^{-q^2/4\sigma_k^2}e^{iqx}\,dq \propto e^{-\sigma_k^2x^2}$
The transform of a Gaussian is a Gaussian.
Read off the width in position.
$|\Psi|^2 \propto e^{-2\sigma_k^2x^2} = e^{-x^2/2\sigma_x^2} \quad\Rightarrow\quad \sigma_x = \dfrac{1}{2\sigma_k}$
Match to a normal distribution of variance $\sigma_x^2$.
Convert to momentum.
$\sigma_p = \hbar\sigma_k \quad\Rightarrow\quad \sigma_x\sigma_p = \dfrac{\hbar}{2}$
A Gaussian packet is a minimum-uncertainty state at $t = 0$.
An electron packet starts with width $\sigma_0 = 1.0$ nm. Write its width later.
$\sigma(t) = \sigma_0\sqrt{1 + (t/\tau)^2}$
The standard result for a free Gaussian packet.
Find the spreading time.
$\tau = \dfrac{2m\sigma_0^2}{\hbar} = \dfrac{2 \times 9.109 \times 10^{-31} \times (10^{-9})^2}{1.0546 \times 10^{-34}} = 1.73 \times 10^{-14}\ \text{s}$
About $17$ fs.
Find the width after $100$ fs.
$\sigma = 1.0\sqrt{1 + (100/17.3)^2} = 1.0\sqrt{1 + 33.4} = 5.87\ \text{nm}$
Almost six times wider.
Find the momentum spread.
$\sigma_p = \dfrac{\hbar}{2\sigma_0} = \dfrac{1.0546 \times 10^{-34}}{2 \times 10^{-9}} = 5.27 \times 10^{-26}\ \text{kg m/s}$
A minimum-uncertainty packet at the start.
Convert it to a velocity spread.
$\sigma_v = \dfrac{\sigma_p}{m} = \dfrac{5.27 \times 10^{-26}}{9.109 \times 10^{-31}} = 5.79 \times 10^{4}\ \text{m/s}$
Components differ in speed by tens of kilometers per second.
Check the long-time growth.
$\sigma_vt = 5.79 \times 10^{4} \times 10^{-13} = 5.79 \times 10^{-9}\ \text{m}$
At late times the width is just the velocity spread times the time.
Compare the two answers.
$5.87\ \text{nm} \approx 5.79\ \text{nm}$
They agree because $100$ fs is well past $\tau$.
Write the group velocity.
$v_g = \hbar k_0/m$
The packet moves at $d\omega/dk$.
Substitute the values.
$v_g = 1.1577 \times 10^{-4} \times 10^{10}$
With $k_0$ in m⁻¹.
Evaluate the speed.
An electron wave packet moves at $36 \times 10^{5}$ m/s. At what speed do the crests of its matter wave move?
Complete the worked solution: waves obey the dispersion relation $\omega = 4\,k^2$ (in consistent units). At $k = 8$, find $\omega$, the phase velocity and the group velocity.
Evaluate the angular frequency.
$\omega = 4 \times 8^2 =$ a
Substitute into the dispersion relation.
Divide by the wave number.
$v_p = \omega/k =$ b
The speed of the crests.
Differentiate the dispersion relation.
$v_g = 2 \times 4 \times 8 =$ d
The speed of the packet, twice the phase velocity.
For a free particle of mass $m$ described by plane waves $e^{i(kx - \omega t)}$, match each quantity to its expression.
| $\hbar k$ | $\hbar^2k^2/2m$ | $\hbar k/2m$ | $\hbar k/m$ | |
|---|---|---|---|---|
| the momentum | ||||
| the energy | ||||
| the phase velocity | ||||
| the group velocity |
For an electron, $\hbar^2/2m = 0.0381$ eV nm². Fill in the energy, in eV, of a plane wave with each wave number.
| energy (eV) | |
|---|---|
| $k = 2$ nm⁻¹ | |
| $k = 4$ nm⁻¹ | |
| $k = 6$ nm⁻¹ |
An electron wave packet is built from plane waves with wave numbers centered on $k_0 = 7$ nm⁻¹. How fast does the packet move, in km/s? Use $\hbar/m_e = 1.1577 \times 10^{-4}$ m²/s.
Answer: km/s
An electron is prepared as a Gaussian wave packet of width $\sigma_0 = 0.5$ nm. How wide is it $20$ fs later, in nm? Use $m = 9.109 \times 10^{-31}$ kg and $\hbar = 1.0546 \times 10^{-34}$ J s.
Answer: nm
In attosecond experiments, a short laser pulse frees an electron from an atom as a wave packet about $\sigma_0 = 0.1$ nm wide. How long does it take for the packet's width to double, in fs?
Answer: fs
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For an electron, $\hbar^2/2m = 0.0381$ eV nm². Fill in the energy, in eV, of a plane wave with each wave number.
| energy (eV) | |
|---|---|
| $k = 4$ nm⁻¹ | |
| $k = 8$ nm⁻¹ | |
| $k = 12$ nm⁻¹ |
You can describe a free particle as a wave packet. Explain to someone why the crests of an electron's wave move at half the electron's speed.
16. Your turn: an electron packet is centered on $k_0 = 10$ nm⁻¹. How fast does it move?, step 3
$v_g = 1.16 \times 10^{6}\ \text{m/s}$
The crests move at half this.