Classical uncertainty principles

AI Thread Summary
The discussion revolves around deriving the second uncertainty principle, ΔωΔt ∼ 1, from the first uncertainty principle, ΔxΔk ∼ 1. The user demonstrates their work by relating position and time through velocity, leading to the expression VΔtΔk ∼ 1. They correctly apply the group velocity concept, V_group = Δω/Δk, to transition between the two principles. The user expresses initial uncertainty about the legality of their approach but concludes that it is valid since uncertainty principles apply to wave packets. The derivation is affirmed as correct and well-structured.
danoonez
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Please check my work.

Using the first uncertainty principle:

<br /> \Delta x \Delta k \sim 1<br />

derive the second uncertainty principle:

<br /> \Delta \omega \Delta t \sim 1<br />


My work:

<br /> \Delta x \Delta k \sim 1<br />

<br /> \frac {\Delta x} {\Delta t} = V \Rightarrow \Delta x = V \Delta t<br />

<br /> V \Delta t \Delta k \sim 1<br />

<br /> V_\textrm {group} = \frac {\Delta \omega} {\Delta k}<br />

<br /> \frac {\Delta \omega} {\Delta k} \Delta t \Delta k \sim 1<br />

<br /> \frac {\Delta k} {\Delta k} \Delta \omega \Delta t \sim 1<br />

<br /> \Delta \omega \Delta t \sim 1<br />

What do you think?
 
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Looks good to me. Right on man. What were you uncertain about when you wrote this?
 
Ed Quanta said:
Looks good to me. Right on man. What were you uncertain about when you wrote this?


I wasn't sure if my use of

V_\textrm {group} = \frac {\Delta \omega} {\Delta k}<br />

was legal. But then I realized that the uncertainty principles are used for wave packets so it was fine.
 
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