- #1
ognik
- 643
- 2
Trying to follow some text:
Given a Gaussian-like distribution for a wave packet, ie a symmetric distribution of $k_x$ about a mean value $ \bar{k}_x $. Making the change of variable $u=k_x - \bar{k}_x$, they write the Fourier transform as: $\psi(x,0) = \frac{1}{2 \pi} e^{i\bar{k}_x x}\int_{-\infty}^{\infty} \phi(u+\bar{k}_x) e^{iux} \,du $ - so far I follow.
Then they say "It is easy to see for any number of simple examples that the width $\Delta k_x$ of the amplitude $\phi$ and the width $\Delta x$ of the wave packet $\psi$ stand in a reciprocal relationship: $\Delta x \Delta k_x \approx 1$
Please explain how they get that?
Given a Gaussian-like distribution for a wave packet, ie a symmetric distribution of $k_x$ about a mean value $ \bar{k}_x $. Making the change of variable $u=k_x - \bar{k}_x$, they write the Fourier transform as: $\psi(x,0) = \frac{1}{2 \pi} e^{i\bar{k}_x x}\int_{-\infty}^{\infty} \phi(u+\bar{k}_x) e^{iux} \,du $ - so far I follow.
Then they say "It is easy to see for any number of simple examples that the width $\Delta k_x$ of the amplitude $\phi$ and the width $\Delta x$ of the wave packet $\psi$ stand in a reciprocal relationship: $\Delta x \Delta k_x \approx 1$
Please explain how they get that?