Wave Packets and Spread

A free particle in quantum mechanics is not described by a single, infinitely extended plane wave. Instead it is described by a wave packet: a superposition of plane waves whose amplitudes are concentrated around some central momentum $p_0$ . Understanding wave packets is the clearest route to seeing why position and momentum spread are inseparable.

Plane waves and their limitations

A pure momentum eigenstate with momentum $p$ has the position-space wave function

$\psi_p(x) = A\,e^{ipx/\hbar}.$

This state has a perfectly sharp momentum — $\Delta p = 0$ — but the probability density $|\psi_p(x)|^2 = |A|^2$ is uniform across all of space. The particle is completely delocalized: $\Delta x = \infty$ . Eigenstates of momentum cannot be normalized over all of space, and they carry no information about where the particle is.

Building a wave packet

To localize the particle, we add (superpose) plane waves with slightly different momenta. Choosing coefficients $\phi(p)$ that form a Gaussian of width $\sigma_p$ centered on $p_0$ , the position-space wave function becomes the Fourier transform of $\phi(p)$ :

$\psi(x,0) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \phi(p)\,e^{ipx/\hbar}\,dp.$

A Gaussian in momentum space,

$\phi(p) = \left(\frac{1}{2\pi\sigma_p^2}\right)^{1/4} \exp\!\left(-\frac{(p-p_0)^2}{4\sigma_p^2}\right),$

transforms into a Gaussian in position space:

\exp\!\left(-\frac{\sigma_p^2 x^2}{\hbar^2}\right) e^{ip_0 x/\hbar}.$$ Carrying through the Gaussian integral, the position-space width is $\sigma_x = \hbar/(2\sigma_p)$. Taking standard deviations as the measure of spread, this gives $$\sigma_x\,\sigma_p = \frac{\hbar}{2}.$$ This is the **minimum-uncertainty product**: a Gaussian wave packet saturates the Heisenberg bound $\Delta x\,\Delta p \geq \hbar/2$. <Callout type="tip"> No matter what shape you choose for $\phi(p)$, the product $\sigma_x \sigma_p$ can never be made smaller than $\hbar/2$. The Gaussian is the unique state that achieves this minimum. </Callout> ## Why localization forces momentum spread The argument above is a direct consequence of Fourier analysis, independent of any physical interpretation. A function that is narrow in one domain must be broad in the conjugate domain. Specifically, if you try to confine a wave packet to a position window of width $\Delta x$, you inevitably need to mix in a range of wave vectors $\Delta k \sim 1/\Delta x$. Since $p = \hbar k$, the corresponding momentum range is $$\Delta p \sim \frac{\hbar}{\Delta x}.$$ This is the origin of the uncertainty principle: it is a **mathematical property of waves** first, and a quantum-mechanical constraint second. Any wave — sound, water, electromagnetic — obeys an analogous time–bandwidth relation. What makes the quantum version remarkable is that $p$ and $x$ are not just labels for the wave; they are the actual observable momentum and position of the particle. ## Spreading in time A wave packet does not stay localized. Different momentum components travel at different group velocities $v = p/m$, so the packet **disperses**: higher-momentum parts race ahead, lower-momentum parts fall behind. For the Gaussian packet above, the position-space width at time $t$ grows as $$\sigma_x(t) = \frac{\hbar}{2\sigma_p}\sqrt{1 + \left(\frac{2\sigma_p^2 t}{m\hbar}\right)^2}.$$ At $t = 0$ the packet is as narrow as the uncertainty principle allows. For $t \gg m\hbar/(2\sigma_p^2)$ the width grows linearly in time. Meanwhile, $\sigma_p$ stays constant — free-particle dynamics do not change the momentum distribution, only the position spread grows. <Callout type="tip"> The time scale for significant spreading, $\tau \sim m\hbar/\sigma_p^2$, is tiny for electrons (femtosecond range for thermal energies) and enormous for macroscopic objects. This is why you never see a baseball spreading out like a quantum wave. </Callout> ## Summary A wave packet is a superposition of momentum eigenstates. The narrower the packet in position space, the wider the spread in momentum space must be — this is the Fourier duality that underlies the Heisenberg uncertainty principle. The Gaussian wave packet is special: it is the only state that achieves the minimum product $\sigma_x \sigma_p = \hbar/2$. All other wave functions satisfy $\Delta x\,\Delta p \geq \hbar/2$, with equality only for coherent Gaussian states.