---
title: The Free Particle and Wave-Packet Dynamics
module: Wave Mechanics in One Dimension
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  The free particle has no bound states: its stationary solutions are non-normalizable
  plane waves forming a continuum. Physical states are wave packets built by superposing
  them, and the superposition is a Fourier transform. We delta-normalize the plane waves,
  assemble a Gaussian packet, solve for its exact time evolution, and read off the two
  facts that reconcile the wave picture with mechanics: the packet moves at the group
  velocity ħk/m, the classical velocity, and it spreads because its component momenta
  travel at different speeds.
topics: [Wave Mechanics in One Dimension]
draft: false
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 2 — Time-Independent Schrödinger Equation; §2.4 The Free Particle"
  - book: Shankar
    ref: "Ch. 5 — Simple Problems in One Dimension; §5.1 The Free Particle, §5.4 The Gaussian Wave Packet"
---

The [square wells](/quantum-mechanics/wave-mechanics-1d/particle-in-infinite-and-finite-square-wells)
bind a particle and quantize its energy. Removing the walls removes the quantization: with
$V(x) = 0$ everywhere the spectrum is continuous, and the stationary states are not
normalizable. This is not a defect of the theory but a statement about what a definite
momentum means. A state of exactly known momentum is spread uniformly over all space, so it
cannot be a probability distribution. Every physically realizable free state is a
superposition of these ideal plane waves, localized to the extent that its momenta are
spread. The apparatus for building such superpositions, and for propagating them, is the
Fourier transform.[^gs-24]

## The continuum of stationary states

For $V(x) = 0$ the time-independent Schrödinger equation is

$$
-\frac{\hbar^2}{2m}\frac{\d^2\psi}{\d x^2} = E\psi
\quad\Longrightarrow\quad
\frac{\d^2\psi}{\d x^2} = -k^2\psi,
\qquad
k \equiv \frac{\sqrt{2mE}}{\hbar}.
$$

Writing the solution as $\psi_k(x) = e^{ikx}$ and attaching the standard time phase
$e^{-iEt/\hbar}$ gives a travelling wave

$$
\Psi_k(x,t) = e^{i(kx - \omega t)},
\qquad
\omega(k) = \frac{\hbar k^2}{2m},
$$

where $k$ ranges over all reals: $k > 0$ moves right, $k < 0$ moves left, and the magnitude
fixes $E = \hbar^2 k^2/2m \ge 0$. Unlike the well, nothing restricts $k$, so **every**
non-negative energy occurs, each twice (two directions). The energy spectrum is the
continuous half-line $[0,\infty)$.

$$
% caption: A bound potential admits a discrete ladder of normalizable levels; removing the
% walls collapses the ladder into a continuous band of scattering energies starting at zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % discrete panel
  \draw[black] (0,0) -- (0,5.0) node[above, font=\scriptsize, text=black] {$E$};
  \foreach \h/\lab in {0.5/1, 1.9/2, 3.1/3, 3.9/4, 4.5/5} {
    \draw[acc, thick] (0.15,\h) -- (2.1,\h);
  }
  \node[align=center, font=\scriptsize, text=black] at (1.1,-0.5) {bound: discrete};
  % continuum panel
  \begin{scope}[xshift=5.0cm]
    \draw[black] (0,0) -- (0,5.0) node[above, font=\scriptsize, text=black] {$E$};
    \fill[acc!14] (0.15,0) rectangle (2.1,5.0);
    \draw[acc, thick] (0.15,0) rectangle (2.1,5.0);
    \node[acc, font=\scriptsize, anchor=west] at (2.2,0) {$E = 0$};
    \node[align=center, font=\scriptsize, text=black] at (1.1,-0.5) {free: continuous};
  \end{scope}
\end{tikzpicture}
$$

The travelling wave carries a probability current but no localization. Its phase advances at
the **phase velocity**

$$
v_{\text{ph}} = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m},
$$

exactly half the classical particle speed $p/m$. A single plane wave therefore cannot
represent a particle: it is delocalized, and even its phase moves at the wrong speed. Both
defects are cured by superposing a band of wave numbers.

### Delta-function normalization

The plane wave is not square-integrable, $\int_{-\infty}^\infty |e^{ikx}|^2\,\d x = \infty$,
so it has no Born probability interpretation on its own. The useful statement is the
orthogonality of two different wave numbers, which is the Fourier representation of the Dirac
delta:

$$
\int_{-\infty}^{\infty} e^{-ik'x}\,e^{ikx}\,\d x = 2\pi\,\delta(k - k').
$$

Choosing the normalization constant $1/\sqrt{2\pi}$ absorbs the factor and gives a clean
**Dirac-orthonormal** set

$$
\psi_k(x) = \frac{1}{\sqrt{2\pi}}\,e^{ikx},
\qquad
\int_{-\infty}^{\infty}\psi_{k'}^\ast(x)\,\psi_k(x)\,\d x = \delta(k - k').
$$

> **Definition (Dirac normalization).** Continuum eigenstates cannot be normalized to unity;
> they are normalized to a Dirac delta, $\langle \psi_{k'}\mid\psi_k\rangle = \delta(k-k')$.
> The label $k$ is continuous, so sums over states become integrals $\int\d k$ and the
> Kronecker delta of discrete bases is replaced by the Dirac delta. These states are
> idealizations; only integrals of them over a band of $k$ are physical.

## Wave packets and the Fourier transform

Because the Schrödinger equation is linear, any integral over the stationary states with a
weight $\phi(k)$ is again a solution:

$$
\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,
e^{i\left(kx - \frac{\hbar k^2}{2m}t\right)}\,\d k.
$$

> **Definition (Wave packet).** A normalizable free state formed as a continuous
> superposition of plane waves with amplitude $\phi(k)$. The packet is localized in $x$ to
> the degree that $\phi(k)$ is spread in $k$; a single $k$ gives a delocalized plane wave,
> a broad band of $k$ gives a narrow packet.

At $t = 0$ this is an ordinary inverse Fourier transform,

$$
\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,e^{ikx}\,\d k,
$$

so the amplitude $\phi(k)$ is recovered from the initial wave function by the forward
transform

$$
\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,0)\,e^{-ikx}\,\d x.
$$

The function $\phi(k)$ is the state written in the momentum basis: since $p = \hbar k$, it is
(up to a constant) the momentum-space wave function, and $|\phi(k)|^2\,\d k$ is the
probability that a momentum measurement returns $\hbar k$ in $\d k$. Position and momentum
descriptions are the same state in two bases connected by a Fourier transform, a duality made
general in the [formalism module](/quantum-mechanics/formalism/position-momentum-and-continuous-spectra).

The recipe for propagating any initial free state is now fixed:

> **Free-particle propagation.**
> 1. Fourier-transform the initial state $\Psi(x,0)$ to get $\phi(k)$.
> 2. Multiply each component by its phase $e^{-i\hbar k^2 t/2m}$.
> 3. Inverse-transform back to $x$.
>
> The middle step is where the physics lives: each plane wave rotates at its own frequency
> $\omega(k) = \hbar k^2/2m$, and because $\omega$ is not linear in $k$ the components drift
> out of step. That dephasing is the spreading of the packet.

## The Gaussian packet

The one initial profile that transforms into itself, and whose time evolution can be done in
closed form, is the Gaussian.[^sh-54] Take a packet at rest, centered at the origin, of width
set by a real constant $a > 0$:

$$
\Psi(x,0) = \Big(\frac{2a}{\pi}\Big)^{1/4} e^{-a x^2}.
$$

The prefactor normalizes it: $\int|\Psi(x,0)|^2\,\d x = (2a/\pi)^{1/2}\int e^{-2ax^2}\,\d x = 1$.
Its momentum amplitude is a Gaussian integral,

$$
\phi(k) = \frac{1}{\sqrt{2\pi}}\Big(\frac{2a}{\pi}\Big)^{1/4}\!\int_{-\infty}^{\infty}
e^{-a x^2}e^{-ikx}\,\d x
= \frac{1}{(2\pi a)^{1/4}}\,e^{-k^2/4a},
$$

using $\int e^{-ax^2 - ikx}\,\d x = \sqrt{\pi/a}\,e^{-k^2/4a}$. A narrow packet in $x$ (large
$a$) is a broad band in $k$, and conversely: the two Gaussians have reciprocal widths, the
sharpest quantitative form of wave–particle complementarity.

$$
% caption: The position and momentum profiles of a Gaussian packet are a Fourier pair with
% reciprocal widths; squeezing the packet in position necessarily broadens its momentum
% content.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % position-space narrow Gaussian
  \draw[black] (-2.4,0) -- (2.4,0) node[right, font=\scriptsize, text=black] {$x$};
  \draw[acc, very thick] (-2.2,0.03) .. controls (-0.9,0.05) and (-0.5,2.1) .. (0,2.1)
    .. controls (0.5,2.1) and (0.9,0.05) .. (2.2,0.03);
  \node[acc, font=\scriptsize, anchor=south] at (0,2.1) {narrow};
  \node[align=center, font=\scriptsize, text=black] at (0,-0.55) {position density};
  % momentum-space broad Gaussian
  \begin{scope}[xshift=6.2cm]
    \draw[black] (-2.6,0) -- (2.6,0) node[right, font=\scriptsize, text=black] {$k$};
    \draw[acc, very thick] (-2.5,0.05) .. controls (-1.4,0.4) and (-0.9,1.15) .. (0,1.15)
      .. controls (0.9,1.15) and (1.4,0.4) .. (2.5,0.05);
    \node[acc, font=\scriptsize, anchor=south] at (0,1.15) {broad};
    \node[align=center, font=\scriptsize, text=black] at (0,-0.55) {momentum density};
  \end{scope}
\end{tikzpicture}
$$

### Exact time evolution

Insert $\phi(k)$ into the propagation integral and complete the square in $k$. The exponent
is quadratic,

$$
-\frac{k^2}{4a} + ikx - \frac{i\hbar t}{2m}k^2
= -\Big(\frac{1}{4a} + \frac{i\hbar t}{2m}\Big)k^2 + ikx,
$$

so with $\beta \equiv \frac{1}{4a} + \frac{i\hbar t}{2m}$ the Gaussian integral evaluates to

$$
\Psi(x,t) = \Big(\frac{2a}{\pi}\Big)^{1/4}
\frac{1}{\sqrt{1 + 2i\hbar a t/m}}\;
\exp\!\Big[\frac{-a x^2}{1 + 2i\hbar a t/m}\Big].
$$

The probability density is the modulus squared. Writing the dimensionless time factor

$$
\theta(t) \equiv \frac{2\hbar a t}{m},
\qquad
w(t) \equiv \sqrt{1 + \theta(t)^2},
$$

the algebra collapses to a Gaussian of growing width and shrinking height,

$$
|\Psi(x,t)|^2 = \sqrt{\frac{2a}{\pi}}\;\frac{1}{w(t)}\;
\exp\!\Big[-\frac{2a x^2}{w(t)^2}\Big].
$$

> **Theorem (Spreading of the Gaussian packet).** A free Gaussian packet stays Gaussian, its
> center fixed (for a packet at rest), while its position spread grows as
> $$
> \sigma_x(t) = \frac{1}{2\sqrt{a}}\,\sqrt{1 + \Big(\frac{2\hbar a t}{m}\Big)^2}.
> $$
> The width has a minimum at $t = 0$ and increases monotonically, approaching
> $\sigma_x \to (\hbar\sqrt{a}/m)\,t$ at large $t$. The area under $|\Psi|^2$ stays one; the
> peak falls as $1/w(t)$.

The position variance is read from the Gaussian $e^{-x^2/2\sigma_x^2}$ by matching exponents:
$2a/w^2 = 1/2\sigma_x^2$, giving $\sigma_x = w/(2\sqrt a)$. At $t=0$, $\sigma_x(0) = 1/2\sqrt a$;
as $t$ grows the second term under the root dominates and $\sigma_x$ grows linearly.

$$
% caption: The width of the packet is constant to first order near t equals zero, then grows
% linearly; the crossover time m over 2 hbar a marks where the packet has roughly doubled.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (5.8,0) node[right, font=\scriptsize, text=black] {$t$};
  \draw[->, black] (0,0) -- (0,3.6) node[above, font=\scriptsize, text=black] {width};
  % hyperbola-like sqrt(1+t^2), starting flat at y=1
  \draw[acc, very thick] (0,1.0) .. controls (0.9,1.02) and (1.5,1.25) .. (2.2,1.75)
    .. controls (3.1,2.4) and (4.4,3.0) .. (5.4,3.5);
  % linear asymptote
  \draw[black, dashed] (1.0,0.55) -- (5.4,3.55);
  \node[black, font=\scriptsize, anchor=north west] at (3.75,1.45) {linear};
  \draw[black, dashed] (0.35,1.0) -- (5.4,1.0);
  \node[black, font=\scriptsize, anchor=east] at (-0.1,0.85) {$\frac{1}{2\sqrt{a}}$};
\end{tikzpicture}
$$

The physical time scale for spreading is $t_\ast = m/2\hbar a = 2m\sigma_x(0)^2/\hbar$. A
tightly localized packet (small $\sigma_x(0)$) spreads fastest, because sharp localization
demands a wide momentum band and a wide band of speeds. An electron localized to
$1\ \text{nm}$ has $t_\ast \sim 10^{-15}\ \text{s}$; a macroscopic mass localized to any
practical accuracy has $t_\ast$ longer than the age of the universe, which is why classical
objects keep sharp trajectories.

$$
% caption: Three snapshots of a free Gaussian packet at rest: as time advances the density
% widens and flattens while the enclosed area stays fixed at one.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-3.4,0) -- (3.6,0) node[right, font=\scriptsize, text=black] {$x$};
  % t0 tall narrow
  \draw[acc, very thick] (-1.6,0.03) .. controls (-0.6,0.05) and (-0.35,2.6) .. (0,2.6)
    .. controls (0.35,2.6) and (0.6,0.05) .. (1.6,0.03);
  \node[acc, font=\scriptsize, anchor=south west] at (0.15,2.4) {$t_0$};
  % t1 medium
  \draw[acc!70, very thick] (-2.4,0.03) .. controls (-1.0,0.1) and (-0.7,1.5) .. (0,1.5)
    .. controls (0.7,1.5) and (1.0,0.1) .. (2.4,0.03);
  \node[acc!70, font=\scriptsize, anchor=south west] at (0.9,1.35) {$t_1$};
  % t2 wide low
  \draw[black, very thick] (-3.2,0.03) .. controls (-1.5,0.2) and (-1.1,0.95) .. (0,0.95)
    .. controls (1.1,0.95) and (1.5,0.2) .. (3.2,0.03);
  \node[black, font=\scriptsize, anchor=south west] at (1.7,0.8) {$t_2$};
\end{tikzpicture}
$$

## Group velocity and the classical limit

A packet at rest stays put, but its center should move for a packet built around a nonzero
central wave number $k_0$. Take the initial profile

$$
\Psi(x,0) = \Big(\frac{2a}{\pi}\Big)^{1/4} e^{-a x^2}\,e^{ik_0 x},
$$

which shifts the momentum amplitude to $\phi(k) \propto e^{-(k-k_0)^2/4a}$, a Gaussian peaked
at $k_0$. The generic argument does not need the Gaussian: for any $\phi(k)$ sharply peaked at
$k_0$, expand the frequency to first order,

$$
\omega(k) \approx \omega(k_0) + \omega'(k_0)\,(k - k_0),
\qquad
\omega'(k) = \frac{\d\omega}{\d k} = \frac{\hbar k}{m}.
$$

Substituting into the packet integral and factoring the terms that do not depend on the
integration variable,

$$
\Psi(x,t) \approx e^{i\left[\omega'(k_0)k_0 - \omega(k_0)\right]t}\,
\frac{1}{\sqrt{2\pi}}\int \phi(k)\,e^{ik\left[x - \omega'(k_0)t\right]}\,\d k
= e^{i\gamma t}\,\Psi\big(x - v_g t,\,0\big),
$$

where the envelope depends on $x$ only through the combination $x - v_g t$. The packet
therefore translates rigidly (to this order) at the **group velocity**

$$
v_g = \omega'(k_0) = \frac{\hbar k_0}{m} = \frac{p_0}{m},
$$

the classical velocity of a particle with momentum $p_0 = \hbar k_0$.[^gs-24b]

> **Theorem (Group velocity equals classical velocity).** The envelope of a free wave packet
> centered at wave number $k_0$ moves at $v_g = \d\omega/\d k = \hbar k_0/m = p_0/m$, whereas
> the individual phase crests move at $v_{\text{ph}} = \omega/k = \hbar k_0/2m$. For matter
> waves the group velocity is twice the phase velocity, and it is the group velocity that
> carries the probability and reproduces Newtonian motion.

The distinction resolves the earlier puzzle that a single plane wave moved at half the
particle speed. That was the phase velocity, which is not observable for a delocalized wave.
The observable motion is the drift of the localized envelope, and that runs at $p_0/m$. The
same conclusion follows from [Ehrenfest's theorem](/quantum-mechanics/formalism/time-evolution-schrodinger-and-heisenberg-pictures),
$\d\langle x\rangle/\d t = \langle p\rangle/m$, applied to the free Hamiltonian.

$$
% caption: Inside a travelling packet the individual crests slide forward at the phase
% velocity while the envelope that carries the probability advances at the faster group
% velocity; the two speeds differ because the matter-wave frequency is quadratic in k.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-3.6,0) -- (3.8,0) node[right, font=\scriptsize, text=black] {$x$};
  % envelope (dashed) Gaussian
  \draw[black, thick, dashed] (-3.2,0.05) .. controls (-1.3,0.15) and (-0.8,1.7) .. (0,1.7)
    .. controls (0.8,1.7) and (1.3,0.15) .. (3.2,0.05);
  % carrier oscillation modulated (drawn within envelope, symmetric)
  \draw[acc, very thick]
    (-3.0,0.05) sin (-2.6,0.35) cos (-2.2,0.05) sin (-1.8,-0.55) cos (-1.4,0.05)
    sin (-1.0,1.0) cos (-0.6,0.05) sin (-0.2,-1.5) cos (0.2,0.05)
    sin (0.6,1.0) cos (1.0,0.05) sin (1.4,-0.55) cos (1.8,0.05)
    sin (2.2,0.35) cos (2.6,0.05) sin (3.0,-0.15) cos (3.2,0.05);
  \draw[->, black, thick] (0,2.05) -- (1.4,2.05);
  \node[black, font=\scriptsize, anchor=south] at (0.7,2.05) {group};
  \draw[->, acc, thick] (-0.2,-1.75) -- (0.5,-1.75);
  \node[acc, font=\scriptsize, anchor=north] at (0.15,-1.8) {phase};
\end{tikzpicture}
$$

### Dispersion and why the packet spreads

Spreading is the failure of the linear approximation. Keeping the next term,

$$
\omega(k) = \omega(k_0) + \omega'(k_0)(k-k_0) + \tfrac12\omega''(k_0)(k-k_0)^2 + \cdots,
\qquad
\omega''(k) = \frac{\hbar}{m},
$$

the quadratic term is a nonzero constant for the free particle: the medium is **dispersive**,
different wave numbers travel at different speeds $v(k) = \hbar k/m$. The faster (higher-$k$)
components outrun the slower ones and the packet broadens. The rate follows from the momentum
spread: a band of velocities $\Delta v = \Delta p/m = \hbar\sigma_k/m$ smears an initially
sharp packet at rate $\Delta v$, so after time $t$ the width grows by $\sim (\hbar\sigma_k/m)t$,
matching the exact Gaussian asymptote $\sigma_x \to (\hbar\sqrt a/m)t$ since $\sigma_k = \sqrt a$.

$$
% caption: The free-particle dispersion relation is a parabola; the slope of the tangent at
% k naught is the group velocity, and its upward curvature, constant at hbar over m, is what
% spreads the packet.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-3.0,0) -- (3.2,0) node[right, font=\scriptsize, text=black] {$k$};
  \draw[->, black] (0,-0.2) -- (0,3.4) node[above, font=\scriptsize, text=black] {frequency};
  % parabola omega = k^2 (scaled)
  \draw[acc, very thick] (-2.6,3.38) .. controls (-1.3,0.85) and (-0.6,0.2) .. (0,0)
    .. controls (0.6,0.2) and (1.3,0.85) .. (2.6,3.38);
  % point at k0 = 1.4, omega = 0.98 (scaled to 1.0)
  \fill[acc] (1.4,0.98) circle (2pt);
  \node[acc, font=\scriptsize, anchor=north west] at (1.4,0.98) {$k_0$};
  % tangent line slope 2*1.4=2.8 (scaled): through (1.4,0.98)
  \draw[black, thick] (0.5,-1.28) -- (2.5,3.32);
  \node[black, font=\scriptsize, anchor=west] at (2.2,3.0) {slope $= v_g$};
\end{tikzpicture}
$$

## The uncertainty product over time

The Gaussian packet is a **minimum-uncertainty state** at the instant it is prepared. From
$\sigma_x(0) = 1/2\sqrt a$ and the momentum-space width $\sigma_k = \sqrt a$, so
$\sigma_p = \hbar\sigma_k = \hbar\sqrt a$, the product at $t=0$ is

$$
\sigma_x(0)\,\sigma_p = \frac{1}{2\sqrt a}\cdot \hbar\sqrt a = \frac{\hbar}{2},
$$

saturating the Heisenberg bound. The free Hamiltonian does not change the momentum
distribution, so $\sigma_p$ is constant, while $\sigma_x$ grows. The product therefore
increases,

$$
\sigma_x(t)\,\sigma_p = \frac{\hbar}{2}\,w(t) = \frac{\hbar}{2}\sqrt{1 + \Big(\frac{2\hbar a t}{m}\Big)^2}\ \ge\ \frac{\hbar}{2},
$$

with equality only at $t=0$. A free packet is minimum-uncertainty for one instant and
becomes less certain in position thereafter; the generalized bound and its saturation
condition are derived in the [formalism module](/quantum-mechanics/formalism/commutators-and-the-generalized-uncertainty-principle).
This is the wave-mechanical content behind the [uncertainty principle](/quantum-mechanics/matter-waves/the-uncertainty-principle):
localization and definite momentum are complementary, and the trade-off has an exact,
time-dependent form for the free particle.

The three results — Dirac-normalized plane waves, group velocity $p_0/m$, and dispersive
spreading — carry over to any slowly varying potential, where they become the semiclassical
picture. Adding a sharp feature to the potential instead produces bound states or scattering,
the subjects of the [delta potential](/quantum-mechanics/wave-mechanics-1d/the-dirac-delta-potential)
and [barrier tunneling](/quantum-mechanics/wave-mechanics-1d/barrier-penetration-and-quantum-tunneling).

[^gs-24]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ 3rd ed., §2.4 — the free particle, non-normalizable stationary states, Dirac normalization $\int e^{i(k-k')x}\d x = 2\pi\delta(k-k')$, and the general solution as a Fourier superposition.
[^gs-24b]: **Griffiths & Schroeter**, §2.4 — phase velocity $\omega/k = \hbar k/2m$ versus group velocity $\d\omega/\d k = \hbar k/m$, and the identification of the group velocity with the classical particle speed.
[^sh-54]: **Shankar**, _Principles of Quantum Mechanics_ 2nd ed., §5.1 and §5.4 — the free-particle propagator and the Gaussian wave packet worked in closed form, including the spreading width $\sigma_x(t)$ and the minimum-uncertainty property at $t=0$.