28 Jun 2025

Free Gaussian Wave Packet and Momentum Space

wave-packets fourier-transform momentum-space gaussian-spreading

A free-particle momentum eigenfunction extends over all space. A localized free particle must therefore be a wave packet: a continuous superposition of plane waves with different momenta.

Use the symmetric Fourier-transform pair

\[\boxed{ \phi(p,t)=\frac1{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty}\Psi(x,t)e^{-ipx/\hbar}dx },\] \[\boxed{ \Psi(x,t)=\frac1{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty}\phi(p,t)e^{ipx/\hbar}dp }.\]

The opposite signs make the two operations inverse because

\[\frac1{2\pi\hbar}\int_{-\infty}^{\infty} e^{ip(x-x')/\hbar}dp=\delta(x-x').\]

Parseval’s relation then gives

\[\int\lvert\Psi(x,t)\rvert^2dx=\int\lvert\phi(p,t)\rvert^2dp=1.\]

Thus $\Psi$ has units $\mathrm{m^{-1/2}}$, while $\phi$ has units $(\mathrm{kg\,m\,s^{-1}})^{-1/2}$.

Operators in momentum space

Fourier transformation of $-i\hbar\partial_x\Psi$ and integration by parts give

\[\begin{aligned} \mathcal F[-i\hbar\partial_x\Psi] &=\frac1{\sqrt{2\pi\hbar}} \left[-i\hbar\Psi e^{-ipx/\hbar}\right]_{-\infty}^{\infty}\\ &\quad-\frac1{\sqrt{2\pi\hbar}} \int\Psi(-i\hbar)\left(-\frac{ip}{\hbar}\right)e^{-ipx/\hbar}dx\\ &=p\phi(p,t), \end{aligned}\]

where the boundary term vanishes for a localized packet and $(-i)(-i)=-1$ supplies the displayed sign. Also,

\[i\hbar\frac{\partial}{\partial p}e^{-ipx/\hbar} =x e^{-ipx/\hbar}.\]

Therefore

\[\boxed{\hat p=p},\qquad \boxed{\hat x=i\hbar\frac{\partial}{\partial p}}\]

in momentum representation.

For a free particle, $\hat H=p^2/(2m)$, so

\[i\hbar\frac{\partial\phi}{\partial t} =\frac{p^2}{2m}\phi.\]

At fixed $p$ this is a first-order equation with solution

\[\boxed{\phi(p,t)=\phi(p,0)e^{-ip^2t/(2m\hbar)}}.\]

The momentum probability $\lvert\phi(p,t)\rvert^2$ is constant; only its phase evolves.

Initial minimum-uncertainty Gaussian

Take a packet centred at $x_0$ with mean momentum $p_0$ and initial position standard deviation $\sigma_0$:

\[\Psi(x,0)=\frac1{(2\pi\sigma_0^2)^{1/4}} \exp\!\left[-\frac{(x-x_0)^2}{4\sigma_0^2} +\frac{ip_0(x-x_0)}{\hbar}\right].\]

Its density is a normalized Gaussian,

\[\lvert\Psi(x,0)\rvert^2 =\frac1{\sqrt{2\pi}\sigma_0} e^{-(x-x_0)^2/(2\sigma_0^2)}.\]

To transform it, put $y=x-x_0$ and complete the square:

\[-\frac{y^2}{4\sigma_0^2} -\frac{i(p-p_0)y}{\hbar} =-\frac1{4\sigma_0^2} \left[y+\frac{2i\sigma_0^2(p-p_0)}{\hbar}\right]^2 -\frac{\sigma_0^2(p-p_0)^2}{\hbar^2}.\]

Using $\int_{-\infty}^{\infty}e^{-y^2/(4\sigma_0^2)}dy=2\sqrt\pi\sigma_0$ gives

\[\boxed{ \phi(p,0)= \left(\frac{2\sigma_0^2}{\pi\hbar^2}\right)^{1/4} \exp\!\left[-\frac{\sigma_0^2(p-p_0)^2}{\hbar^2} -\frac{ipx_0}{\hbar}\right] }.\]

The momentum density is Gaussian with

\[\langle p\rangle=p_0,\qquad \Delta p=\frac{\hbar}{2\sigma_0}.\]

Since $\Delta x(0)=\sigma_0$,

\[\Delta x(0)\Delta p=\frac\hbar2.\]

Free evolution and spreading

Insert $\phi(p,t)$ into the inverse transform. Completing its complex Gaussian square gives

\[\boxed{ \begin{aligned} \Psi(x,t)={}&\frac{1}{(2\pi\sigma_0^2)^{1/4}} \frac{1}{\sqrt{1+i\tau}}\\ &\times\exp\!\left[ -\frac{(x-x_0-p_0t/m)^2}{4\sigma_0^2(1+i\tau)} +\frac{ip_0}{\hbar} \left(x-x_0-\frac{p_0t}{2m}\right) \right], \end{aligned}}\]

where

\[\tau=\frac{\hbar t}{2m\sigma_0^2}.\]

The parameter $\tau$ is dimensionless: both $\hbar t$ and $m\sigma_0^2$ have units $\mathrm{kg\,m^2}$. Since

\[\operatorname{Re}\!\left(\frac1{1+i\tau}\right) =\frac1{1+\tau^2},\]

taking the modulus squared gives

\[\boxed{ \lvert\Psi(x,t)\rvert^2 =\frac1{\sqrt{2\pi}\,\sigma_t} \exp\!\left[ -\frac{(x-x_0-p_0t/m)^2}{2\sigma_t^2} \right] },\] \[\boxed{ \sigma_t=\sigma_0 \sqrt{1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2} }.\]

The centre moves at $p_0/m$, the group velocity for $E=p^2/(2m)$. The momentum distribution does not broaden, but its components acquire different phases proportional to $p^2$, so the position distribution spreads. At $t=0$, $\sigma_t=\sigma_0$; for $\lvert t\rvert\gg2m\sigma_0^2/\hbar$, the width grows approximately as $\sigma_t\simeq\hbar\lvert t\rvert/(2m\sigma_0)$.

The plotted curves use the derived centre and width. The editable TikZ source generates the figure, and the normalization, mean, variance, width, group-velocity, and free Schrödinger-equation checks return zero residuals in the Maxima worksheet.