hiesenberg uncertainty principle, h or hbar?

rash92

i've seen both:

ΔxΔp >= h/2

and

ΔxΔp >= hbar/ 2

used, and i'm not sure which is correct. my physics textbook uses h/2, but wiki and other online rescources seem to use hbar/2

do they apply to different situations? (if so, where do you use hbar and where do you use h?) or is one of them outdated?

vanhees71

No, the only generally correct statement is about standard devitians of two observables. For any (pure or mixed) state, one has

[tex]\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.[/tex]

Since for position and momentum components in the same direction, you have

[tex][x,p]=\mathrm{i} \hbar[/tex]

you have

[tex]\Delta x \Delta p \geq \hbar/2.[/tex]

Other uncertainty relations are found in the literature from hand-waving arguments using other uncertainty measures than the standard deviation!

rash92

so if i'm understanding correctly, if you use standard deviation, it's always hbar/2,
but if you use other measures of spread then it could be different?

Demystifier

Your physics textbook is wrong, the correct inequality is the one with hbar.