ecoanuncios7x

2022-10-02

Factor 2 in Heisenberg Uncertainty Principle: Which formula is correct?
Some websites and textbooks refer to
$\mathrm{\Delta }x\mathrm{\Delta }p\ge \frac{\hslash }{2}$
as the correct formula for the uncertainty principle whereas other sources use the formula
$\mathrm{\Delta }x\mathrm{\Delta }p\ge \hslash .$
Question: Which one is correct and why?
The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.
This is the mentioned derivation:
$\mathrm{sin}\theta =\frac{\lambda }{\mathrm{\Delta }y}$
assuming $\theta$ is small:
$\mathrm{tan}\theta =\frac{\lambda }{\mathrm{\Delta }y}$
de Broglie equation:
$\lambda =\frac{h}{{p}_{x}}$
$⇒\mathrm{tan}\theta \approx \frac{h}{{p}_{x}\cdot \mathrm{\Delta }y}$
but also:
$\mathrm{tan}\theta =\frac{\mathrm{\Delta }{p}_{y}}{{p}_{x}}$
equalize $\mathrm{tan}\theta$:
$\frac{h}{{p}_{x}\cdot \mathrm{\Delta }y}\approx \frac{\mathrm{\Delta }{p}_{y}}{{p}_{x}}$
$⇒\frac{h}{\mathrm{\Delta }y}\approx \mathrm{\Delta }{p}_{y}⇒\mathrm{\Delta }{p}_{y}\mathrm{\Delta }y\approx h$
$⇒\mathrm{\Delta }{p}_{y}\mathrm{\Delta }y\ge \frac{h}{2\pi }$
$⇒\mathrm{\Delta }p\mathrm{\Delta }x\ge \frac{h}{2\pi }$

lascosasdeali3v

The strongest limit without loss of generality is
$\mathrm{\Delta }p\mathrm{\Delta }x\ge \frac{1}{2}\hslash ,$
this is always true. Whilst $\mathrm{\Delta }p\mathrm{\Delta }x\ge \hslash$ might often be true, it is not always true.
The $\frac{1}{2}$ is often omitted, because, as mentioned in the comments, often only the magnitude of the right-hand-side is important, and not its precise value. Also, it might be omitted for brevity/simplicity.
A further reason is historical: Heisenberg's original statement of his uncertainty principle was a rough estimate that omitted $\frac{1}{2}$. Only later was his estimate refined with a formal calculation and the $\frac{1}{2}$ added.

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