Index
Chapter 2 · Item 2.13
Gaussian wave packet and Fourier width
A localized packet and its momentum-space partner
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Guided reading

The Gaussian packet is the best example because the Fourier transform is also Gaussian. This makes the inverse relation between position width and wave-number width completely explicit.

A packet with carrier and envelope
Fig. 2.3, adapted from the original chapter: Gaussian wave packet and its probability density.
Fig. 2.3, adapted from the original chapter: Gaussian wave packet and its probability density. Copyright © 2026 Elsevier Inc.

A useful localized wave packet is a plane-wave oscillation multiplied by a Gaussian envelope. Here it is the initial condition, specified at \(t=0\):

\[\Psi(x,0)=\frac{1}{(a^2\pi)^{1/4}}\,e^{ik_0x}e^{-x^2/2a^2}.\]

The oscillating factor \(e^{ik_0x}\) carries the average wave number \(k_0\). The Gaussian factor localizes the packet around \(x=0\). The next step is to read from this initial state both its position width and its momentum width.

Position probability

The probability density removes the phase \(e^{ik_0x}\):

\[|\Psi(x,0)|^2=\frac{1}{a\sqrt{\pi}}e^{-x^2/a^2}.\]

The parameter \(a\) controls spatial width. Larger \(a\) means a broader packet in \(x\).

Fourier partner in wave-number space

The wave-number distribution is obtained by Fourier transform:

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

For the Gaussian packet, this gives another Gaussian centered at \(k_0\):

\[\phi(k)=\frac{\sqrt{a}}{\pi^{1/4}}e^{-a^2(k-k_0)^2/2}.\]
Width relation

The width is computed from the variance. For any variable \(q\),

\[\mathrm{var}(q)=\langle q^2\rangle-\langle q\rangle^2,\qquad \Delta q=\sqrt{\mathrm{var}(q)}.\]

For the initial Gaussian, \(|\Psi(x,0)|^2\) is centered at zero. The odd integral vanishes, and the second moment gives

\[\langle x\rangle=0,\qquad \langle x^2\rangle=\frac{a^2}{2},\qquad \mathrm{var}(x)=\frac{a^2}{2}.\]

The Fourier partner \(|\phi(k)|^2\) is centered at \(k_0\). For the shifted variable \(k-k_0\),

\[\langle k-k_0\rangle=0,\qquad \left\langle (k-k_0)^2\right\rangle=\frac{1}{2a^2}.\]

Therefore the wave-number variance is simply

\[\mathrm{var}(k)=\frac{1}{2a^2}.\]

Since \(p=\hbar k\), the momentum variance is \(\mathrm{var}(p)=\hbar^2\mathrm{var}(k)\). Therefore

\[\Delta x=\frac{a}{\sqrt2},\qquad \Delta p=\frac{\hbar}{\sqrt2\,a},\qquad \Delta x\,\Delta p=\frac{\hbar}{2}.\]

This initial Gaussian is a minimum-uncertainty packet. Making it broader in position makes its momentum distribution narrower; the product cannot be pushed below \(\hbar/2\).

Widths are the main result

For a Gaussian packet, the algebra is useful because it makes the width tradeoff explicit. If the initial packet is narrow in \(x\), its Fourier transform must be broad in \(k\).

\[\mathrm{var}(x)=\frac{a^2}{2},\qquad \mathrm{var}(k)=\frac{1}{2a^2}.\]

Since \(p=\hbar k\), the momentum variance is \(\mathrm{var}(p)=\hbar^2\mathrm{var}(k)\). Therefore the minimum product is

\[\Delta x\,\Delta p=\frac{\hbar}{2}.\]

Exercises can use these relations directly while leaving the complete Fourier integral details to the book.

Exercise-ready boundary

This page is designed to support short guided exercises on: Fourier transform, Gaussian distributions in x and k, and the inverse relation between widths.

  • Use from this page: the definitions, physical setup, highlighted equations and conceptual links needed to start a first calculation or explanation.
  • Keep in the book: complete derivations, extended historical discussion, worked solutions and the full textbook narrative remain in the original chapter and linked book resources.
  • Good exercise balance: ask the student to identify assumptions, apply one relation, and interpret the result physically, without requiring material not introduced on this page.
Practice anchors

Use these anchors to design compact exercises. The exercise should be answerable from this page plus standard algebra, while longer derivations, full worked examples and broader context should point back to the original book.

  • Focus: Fourier transform, Gaussian distributions in x and k, and the inverse relation between widths.
  • Conceptual check: state what the main result says physically before using it algebraically.
  • Equation: \[\Psi(x,0)=\frac{1}{(a^2\pi)^{1/4}}\,e^{ik_0x}e^{-x^2/2a^2}.\]
  • Equation: \[|\Psi(x,0)|^2=\frac{1}{a\sqrt{\pi}}e^{-x^2/a^2}.\]
  • Equation: \[\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}\,dx.\]
  • Boundary: use this page for setup and first-step reasoning; cite the book for longer derivations, complete experimental history or solved-problem detail.
  • Typical task: derive, interpret, or apply the relation above to a simple case without introducing topics outside this page.
Source note: Original auxiliary summary for this book-app, based on Chapter 2 of Mario Reis, Quantum Mechanics, Elsevier, 2026. Book text and figures are copyright © 2026 Elsevier Inc. Selected figure material is reproduced/adapted from Chapter 2 of the original book and carries a visible copyright caption.