What does "the range of reciprocal wavelengths" mean for a wave packet? (QM)

Question

I'm learning about the derivation of the uncertainty principle via the addition of a finite number of sinusoidal waves of varying wavelengths and frequencies (basically, baby steps towards a Fourier series).

However, in the image provided below, there are two quantities - $\Delta x$ and $\Delta \kappa$ - which I'm a bit shaky about. $\Delta x$, as I understood from the textbook, is the distance from max. amplitude to half-max. amplitude for the main group of waves. However, what does $\Delta \kappa$ represent? The book says, and I quote, "$\Delta \kappa$ is defined as the range of reciprocal wavelengths of the components of psi from maximum amplitude to half-maximum amplitude". What does that mean?

Moreover, the textbook goes on to explain that $\Delta x$ and $\Delta \kappa$ are inversely proportional to each other, and as the range of reciprocal; wavelengths is increased, the $\Delta x$ of the group of waves is decreased. Again, what does this even mean? I'm genuinely confused on how to even begin to visualize this.

I'd really appreciate any assistance.

Answer 1

The quantities $\Delta x$ and $\Delta k$ are the standard deviation of $x$ or $k$ for a certain wavefunction and as such, they measure how spread out these variables are. The standard deviation is defined using $$\Delta x^2=\left\langle\ ( x-\langle x\rangle)^2\ \right\rangle=\langle x^2\rangle-\langle x\rangle^2$$ Where $\Delta x$ is just the square root of this.

For example, let's take $$|\psi\rangle=\sqrt{\tfrac 3 4}|x=0\rangle+\sqrt{\tfrac 1 4}|x=1\rangle$$

I define this basis such that all $x$ states are orthonormal $\langle x=i|x=j\rangle=\delta_{ij}$. Usually you would see these states written as just $|i\rangle$ but I added the $x=$ for clarity. First calculate the mean $\langle x\rangle$. \begin{align} \langle\psi|x|\psi\rangle=\frac{3}{4}\times 0+\frac{1}{4}\times1=\frac 1 4. \end{align} Now calculate $\langle x^2\rangle$. \begin{align} \langle\psi|x^2|\psi\rangle=\frac{3}{4}\times 0^2+\frac{1}{4}\times1^2=\frac 1 4. \end{align} So the standard deviation becomes $\Delta x=\sqrt{\frac 1 4-(\frac 1 4)^2}=\sqrt{\frac 3 {16}}\approx 0.43$.

If instead we would have chosen the state

$$|\psi\rangle=\sqrt{\tfrac 3 4}|x=0\rangle+\sqrt{\tfrac 1 4}|x=4\rangle,$$

we would have gotten $\Delta x=\sqrt{3}\approx 1.73$. Our state is more spread out (covers a larger part of x-space) so our standard deviation is larger. Your textbook is telling something very deep. If we transform a not-so-spread-out function to Fourier space it becomes spread out and vice versa. Transforming to Fourier space is the same as writing it in the $k$ basis.