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In class today we were taught about Heisenberg’s equation, $\Delta x\Delta p\ge\frac{h}{4\pi}$.

Why isn’t De Broglie wavelength a factor here - or is it, but it’s represented behind the deltas instead? After all, if we’re dealing with indeterminism, isn’t that where the wave part of particle-wave duality comes into play?

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It was precisely De Broglie who hypothesized that $p=h/\lambda$. So you can also write $\Delta x / \Delta \lambda = 1/4\pi$.

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  • $\begingroup$ facepalm I should have been able to figure this one out. Thanks! $\endgroup$
    – DonielF
    Commented Nov 12, 2019 at 22:08
  • $\begingroup$ Wait, what is $\lambda$ here? I suppose it would be something like $\frac{\Delta x\Delta \lambda}{\langle \lambda \rangle^2}$ instead of $\frac{\Delta x}{\lambda}$. $\endgroup$
    – user87745
    Commented Nov 12, 2019 at 22:32
  • $\begingroup$ @Dvij Where do you get your first fraction from? This answer seems to flow quite nicely from De Broglie’s equation $\lambda=\frac{h}{p}$. $\endgroup$
    – DonielF
    Commented Nov 13, 2019 at 4:45
  • $\begingroup$ I corrected my answer after @DvijMankad 's comment $\endgroup$
    – my2cts
    Commented Nov 13, 2019 at 5:42

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