# Fourier transform of wave packet

• schniefen
In summary, the speaker is unsure if ##h(x,t)## is a wave packet, but it appears to be one. They want to find ##\hat{h}(k,t=0)## and have attempted to use the Fourier transform, but are unable to remove the real part from the integral. The function ##a(k)## is provided, and it is in the form of a complex number with real constants A, B, and C, as well as a Dirac delta. The speaker suggests using the alternative form of the real part of a complex number involving the trigonometric functions cosine and sine.
schniefen
Homework Statement
Consider ##h(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \Re\{a(k)e^{i(kx-\omega t)}\}\mathrm{d}k.## What is the Fourier transform ##\hat{h}(k,t)## evaluated at ##t=0##, i.e. ##\hat{h}(k,t=0)##? (##a(k)## is given, but I do not think it needs to be specified)
Relevant Equations
The Fourier transform (FT) ##\hat{f}(k,t)=\int_{-\infty}^{\infty} f(x,t)e^{-ikx}\mathrm{d}x## and the inverse Fourier transform ##f(x,t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(k,t)e^{ikx}\mathrm{d}k##.
I am unsure if ##h(x,t)## really is a wave packet, but it looks like one, hence the title. Anyway, so I'd like to determine ##\hat{h}(k,t=0)##. My attempt so far is recognizing that, without the real part in the integral, i.e.

##g(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} a(k)e^{i(kx-\omega t)} \mathrm{d}k,##
then ##a(k)## is just the Fourier transform of ##g(x,t=0)##. However, I can not remove the real part from the integral and I am unsure how to proceed.

schniefen said:
(##a(k)## is given, but I do not think it needs to be specified)
I disagree. What is a(k)?

DrClaude said:
I disagree. What is a(k)?
##a(k)=A2\pi\delta(k)+2B/((k-C)^2+B^2)##, where ##A, B## and ##C## are real constants. ##\delta## is the Dirac delta.

schniefen said:
However, I can not remove the real part from the integral and I am unsure how to proceed.

For complex numbers $z$ and $w$, $$\Re(zw) = \Re(z)\Re(w) - \Im(z)\Im(w).$$ However for $w = e^{i\theta} = \cos \theta + i\sin \theta$ for real $\theta$ you may prefer to write $$\begin{split} \Re(e^{i\theta}) &= \frac{e^{i\theta} + e^{-i\theta}}2,\\ \Im(e^{i\theta}) &= \frac{e^{i\theta} - e^{-i\theta}}{2i}.\end{split}$$

schniefen

## 1. What is a Fourier transform of a wave packet?

A Fourier transform of a wave packet is a mathematical operation that decomposes a wave packet into its constituent frequencies. It is a tool used in signal processing and quantum mechanics to analyze the frequency components of a wave packet.

## 2. How is a Fourier transform of a wave packet different from a regular Fourier transform?

A regular Fourier transform is used to analyze signals that are continuous and infinite in time, while a Fourier transform of a wave packet is used for signals that are localized in time and space. It takes into account the finite duration of the wave packet and provides a more accurate representation of its frequency components.

## 3. What is the physical significance of a Fourier transform of a wave packet?

The physical significance of a Fourier transform of a wave packet lies in its ability to provide information about the energy and momentum of a particle. In quantum mechanics, the wave packet represents the probability amplitude of finding a particle at a particular position and time. The Fourier transform of this wave packet gives the momentum distribution of the particle.

## 4. How is a Fourier transform of a wave packet used in signal processing?

In signal processing, a Fourier transform of a wave packet is used to analyze and filter signals that are localized in time and space. It helps in identifying the frequency components of a signal and can be used to remove noise or unwanted frequencies from the signal.

## 5. Can a Fourier transform of a wave packet be inverted?

Yes, a Fourier transform of a wave packet can be inverted to obtain the original wave packet. This is known as the inverse Fourier transform and is used to reconstruct the original signal from its frequency components.

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