The Free Particle and Wave-Packet Dynamics

The 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.¹

The continuum of stationary states

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

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

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

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

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, ${\int }_{-\infty }^{\infty }|e^{ikx}{|}^2\mathbf{d}x=\infty$, 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:

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

Wave packets and the Fourier transform

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

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

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

The function $\varphi (k)$ is the state written in the momentum basis: since $p=\hslash k$, it is (up to a constant) the momentum-space wave function, and $|\varphi (k){|}^2\mathbf{d}k$ is the probability that a momentum measurement returns $\hslash k$ in $\mathbf{d}k$. Position and momentum descriptions are the same state in two bases connected by a Fourier transform, a duality made general in the formalism module.

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

The Gaussian packet

The one initial profile that transforms into itself, and whose time evolution can be done in closed form, is the Gaussian.² Take a packet at rest, centered at the origin, of width set by a real constant $a>0$:

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

using $\int e^{-a{x}^2-ikx}\mathbf{d}x=\sqrt{\pi /a}{e}^{-k^2/4a}$. 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.

Exact time evolution

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

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

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

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

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.

The physical time scale for spreading is $t_{\ast }=m/2\hslash a=2m{\sigma }_x(0{)}^2/\hslash$. 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.

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

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

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

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

the classical velocity of a particle with momentum $p_0=\hslash k_0$.³

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, $\mathbf{d}⟨x⟩/\mathbf{d}t=⟨p⟩/m$, applied to the free Hamiltonian.

Dispersion and why the packet spreads

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

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

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 ${\sigma }_p=\hslash {\sigma }_k=\hslash \sqrt{a}$, the product at $t=0$ is

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,

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. This is the wave-mechanical content behind 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 and barrier tunneling.