# “Derivation” of the Heisenberg Uncertainty Principle

Ok, so I posted this in the mathematics StackExchange, but got no response. The question I outline below is my textbook's "derivation" of the Heisenberg Uncertainty Principle. The "derivation" my textbook uses involves wave packets.

Suppose there are seven waves of slightly different wavelengths and amplitudes and we superimpose them (textbook is talking about wave packets). The wavelengths range from $\lambda _9 = 1/9$ to $\lambda _{15} = 1/15$. Their wavenumbers ($k = 2\pi / \lambda$) ranges from $k_9 = 18\pi$ to $k_{15} = 30\pi$. Note, the waves are of the form

$$y(x,t) = A\sin(kx - wt)$$

The waves are all in phase at $x = 0$ and again at $x = \pm 12, \pm 24$ etc. My question is the last line. How does my textbook (from which I copied what they wrote) know that they are all in phase at $x = \pm 12$ etc. ?

If you can do this in simple terms that would be great (i.e., no fourier transform math since I have yet to learn about it). Is there some rule to know when $n$ number of waves are in phase? (They have all 7 waves graphed, but not on top of each other. Did they do some mathematics or find this from the graph? Note, looking at this graph its hard to tell that all 7 waves are in phase at $x=\pm 12, \pm 24$, etc).

Second question, my textbook goes on to say that the width of the group $\Delta x$ of superposition is just a big larger than 1/12. There's a graph of the superposition (looks like a beat graph) but did they determine this number from the graph or is it somehow related to the numbers given above?

Then it shows a plot of the amplitude of the waves ($y_0$) vs. $k$. It shows'' that the width at $y_0 = 1/2$ is $4\pi$.

Just fyi, this is a physics textbook which goes on to say that $\Delta k \Delta x \sim 1$ (using the numbers above, $4\pi * 1/12 \approx 1$) and $\Delta w \Delta t \sim 1$ (by similar arguments). It then uses these as a basis to state the Heisenberg uncertainty principle.

-
I think you should study Fourier transformations before beginning topics like QM –  innisfree Apr 16 at 18:00
There will be an $x$ where they are all in phase. Take the CGD of the wavelengths to find where this is. Can can than do a change of coordinates to that node and it wont change the equations, nor the outcome. –  ja72 Apr 16 at 18:00
This derivation is not rigorous, it is just illustrative. –  ja72 Apr 16 at 18:02
Crossposted from math.stackexchange.com/q/742188/11127 –  Qmechanic Apr 16 at 18:07
@DWade64 It mist also be quite helpful to tell us which textbook this is so that we may look it up. –  Flint72 Apr 16 at 20:14