ecoanuncios7x

## Answered question

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 $\mathrm{tan}\theta$:
$\frac{h}{{p}_{x}\cdot \mathrm{\Delta }y}\approx \frac{\mathrm{\Delta }{p}_{y}}{{p}_{x}}$
$⇒\frac{h}{\mathrm{\Delta }y}\approx \mathrm{\Delta }{p}_{y}⇒\mathrm{\Delta }{p}_{y}\mathrm{\Delta }y\approx h$
$⇒\mathrm{\Delta }{p}_{y}\mathrm{\Delta }y\ge \frac{h}{2\pi }$
$⇒\mathrm{\Delta }p\mathrm{\Delta }x\ge \frac{h}{2\pi }$

### Answer & Explanation

lascosasdeali3v

Beginner2022-10-03Added 10 answers

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.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?