# Hiesenberg uncertainty principle, h or hbar?

1. May 31, 2012

### 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 (if so, where do you use hbar and where do you use h?) or is one of them outdated?

2. May 31, 2012

### vanhees71

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

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

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

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

you have

$$\Delta x \Delta p \geq \hbar/2.$$

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

3. May 31, 2012

### 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?

4. May 31, 2012

### Demystifier

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