Timeline for Prove: $(\Delta x)(\Delta \lambda) \geq \frac{\lambda^2}{4\pi}$

10 events

when toggle format	what		by	license	comment
Apr 15, 2023 at 10:30	comment	added	Andrew		@asmaier If that were the case, you would allow negative values on standard deviations for $k$, since $\lambda > 0$ and $\Delta \lambda > 0$. See en.wikipedia.org/wiki/Propagation_of_uncertainty#Simplification for the general rule.
Apr 14, 2023 at 22:26	comment	added	asmaier		@AndreasMastronikolis Shouldn't $\Delta k = - \frac{2\pi}{\lambda^2} \Delta\lambda$ ? Did you forget the minus sign?
Oct 7, 2020 at 4:12	history	edited	user276504	CC BY-SA 4.0	added 4 characters in body
Oct 7, 2020 at 0:00	history	tweeted			twitter.com/StackPhysics/status/1313630204158447616
Oct 6, 2020 at 23:39	answer	added	John Dumancic		timeline score: 2
Oct 6, 2020 at 22:14	history	edited	Qmechanic♦		edited tags
Oct 6, 2020 at 21:03	comment	added	Andrew		I hesitate to answer the question because of the second part. But here is how I am thinking. The De Broglie hypothesis asserts: $$p = \hbar k \text{ with } k = \frac{2 \pi}{\lambda}$$ Thus, $$\Delta p = \hbar \Delta k \text{ and } \Delta k= \frac{2 \pi}{\lambda^2} \Delta \lambda$$ So, finally we have: $$\Delta x \cdot \Delta p \geq \frac{\hbar}{2} \Longrightarrow \Delta x \Delta k \geq \frac{1}{2} \Longrightarrow \Delta x \cdot \Delta \lambda \geq \frac{\lambda^2}{4 \pi}$$
Oct 6, 2020 at 20:36	review	First posts
Oct 6, 2020 at 22:11
Oct 6, 2020 at 20:33	history	edited	G. Smith	CC BY-SA 4.0	edited title
Oct 6, 2020 at 20:31	history	asked	user276504	CC BY-SA 4.0