1
$\begingroup$

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?

enter image description here

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – m_1265
    Feb 10, 2023 at 18:56
  • $\begingroup$ @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. $\endgroup$ Feb 10, 2023 at 20:19
  • 1
    $\begingroup$ ohhh, that makes much more sense! Thank you so much for taking time to help me out with this. Cheers! $\endgroup$
    – m_1265
    Feb 10, 2023 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.