How Does the Uncertainty Principle Relate to Wave Packet Spread?

[ognik]

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?

 

[jvicens]

Hello! It's great that you're trying to understand the concept behind the reciprocal relationship between the widths of the amplitude and the wave packet. Let me explain it to you in more detail.

Firstly, let's consider the Gaussian-like distribution for a wave packet, where the wavepacket is symmetric about a mean value $\bar{k}_x$. This means that the wavepacket has a central peak at $\bar{k}_x$ and the values of $k_x$ are distributed symmetrically around it.

Now, when we make the change of variable $u=k_x - \bar{k}_x$, we essentially shift the origin of the distribution from $\bar{k}_x$ to 0. This allows us to write the Fourier transform as $\psi(x,0) = \frac{1}{2 \pi} e^{i\bar{k}_x x}\int_{-\infty}^{\infty} \phi(u+\bar{k}_x) e^{iux} \,du$. This is because the Fourier transform gives us the amplitude of the wave at a given position $x$ in terms of the wave's frequency $k_x$.

Now, let's consider the width of the amplitude $\phi$. This width is given by $\Delta k_x$, which represents the spread of values of $k_x$ in the distribution. Similarly, the width of the wave packet $\psi$ is given by $\Delta x$, which represents the spread of positions $x$ in the distribution.

Here comes the interesting part. As we can see from the Fourier transform, the wavepacket $\psi$ is essentially a superposition of waves with different frequencies $k_x$. And since the amplitude $\phi$ represents the strength of each frequency in the wavepacket, it is directly related to the width of the wavepacket.

Now, if we have a wide spread of frequencies in the wavepacket (i.e. a large $\Delta k_x$), then we will have a narrow spread of positions in the wavepacket (i.e. a small $\Delta x$). This is because the amplitude $\phi$ will have a larger contribution from each frequency, resulting in a more localized wavepacket.

Conversely, if we have a narrow spread of frequencies (i.e. a small $\Delta k_x$), then we will have a wider spread of positions (i.e. a larger $\Delta x$). This is because the amplitude $\phi$ will have