# 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