# Heisenberg Uncertainty principle defined as $\Delta x \Delta p_x \geq h$ versus $\Delta x \Delta p_x \geq \hbar/2$ [duplicate]

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 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 $\Delta x \Delta p_x \geq \frac{\hbar}{2}$?

• 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 – AccidentalFourierTransform Jan 11 '17 at 19:20
• 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. – StephenG Jan 11 '17 at 20:31
• Possible duplicates: physics.stackexchange.com/q/69604/2451 and links therein. – Qmechanic Jan 12 '19 at 17:56

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.