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

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.

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.

• Thank you so much for your response (you cleared up a lot of confusion). Now, I do understand, mathematically, what you mean when you say, "our state is more spread out": what does that mean probabilistically? What does that tell us about our given state? If the SD is smaller, does that make our state more probable? Commented Feb 10, 2023 at 18:56
• @m_1265 probabilistically, a small standard deviation means that if we measure the position of a wavefunction $\psi(x)$ we will often get values centered around the mean. Think of a gaussian distribution with small $\sigma$. Likewise, a large SD means that the measurements of position will yield a wide range of outcomes. Physically this means that when we constrict a particle to be in a small volume the wavefunction in momentum space gets spread out so the momentum can take on a wide range of values. Commented Feb 10, 2023 at 20:19
• ohhh, that makes much more sense! Thank you so much for taking time to help me out with this. Cheers! Commented Feb 10, 2023 at 20:25