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Hiesenberg uncertainty principle, h or hbar?

  1. May 31, 2012 #1
    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 (if so, where do you use hbar and where do you use h?) or is one of them outdated?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. jcsd
  3. May 31, 2012 #2

    vanhees71

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    2016 Award

    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!
     
  4. May 31, 2012 #3
    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?
     
  5. May 31, 2012 #4

    Demystifier

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