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At my university, in the during lectures and in the equation sheet for our exams, the formula for the Heisenberg Uncertainty Principle is stated as $\Delta x \Delta p_x \geq h$, for example in one of my lecture notes, the following example is illustrated using this formula

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However I know that in my textbook (University Physics by Young and Freedman) and pretty much universally the Heisenberg Uncertainty Principle is stated as $\Delta x \Delta p_x \geq \frac{\hbar}{2}$. Using this formula, we can see the example above is off by a factor of $4\pi$.

Is the formula, $\Delta x \Delta p_x \geq h$, that my university uses a valid formula? If so is just a weaker version of $\Delta x \Delta p_x \geq \frac{\hbar}{2}$?


marked as duplicate by Qmechanic quantum-mechanics Jan 11 '17 at 20:32

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  • 4
    $\begingroup$ if you are asked to estimate something, you can ignore factors of $\pi$ or 2. In any case, the correct formula is $\Delta x\Delta p\ge\hbar/2$, as you can read it in the wikipedia entry $\endgroup$ – AccidentalFourierTransform Jan 11 '17 at 19:20
  • $\begingroup$ I would suggest that, during an exam, use what is given regardless. You might want to draw attention to the issue later ( e.g. in case of a printing error ), but do not trouble yourself during the actual exam. $\endgroup$ – StephenG Jan 11 '17 at 20:31
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/69604/2451 and links therein. $\endgroup$ – Qmechanic Jan 12 at 17:56

The Heisenberg uncertainty principle (HUP)is encoded formally in the theory of quantum mechanics in the commutations relations of the operators defining the observables. There are a number of other observable variable pairs whose commutator is not zero

Commutators use $\hbar$ so it is the "correct" form in the calculations. On the other hand the HUP, when it was first proposed, was an extension of wave mechanics arguments and there a factor of $\pi$ or so was irrelevant. In the formal mathematics, one should use $\hbar$. See also the answer here.


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