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?
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?