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Just a short question regarding an interpretation of the Heisenberg uncertainty principle $\sigma_x \sigma_p \geq \frac{\hbar}{2}$.

Question:

  • Why is it also sometimes that $\Delta x$ and $\Delta p$ are considered as change in position and momentum when written as $\Delta x \Delta p \geq \frac{\hbar}{2}$?
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    $\begingroup$ It is so because many school teachers don't know Quantum Mechanics beyond the de Broglie hypothesis... $\endgroup$ – AccidentalFourierTransform May 2 '16 at 15:56
  • $\begingroup$ Are you sure that they are simply considering those to be changes, or are they using classically maximum fluctuations of position and momentum as approximations to the uncertainties? For example, in bound-state problems, the width of the bounding potential might approximate, to first order, the uncertainty in position. Later, a more detailed calculation could be done. $\endgroup$ – Bill N May 2 '16 at 16:17
  • $\begingroup$ @BillN Yes sorry they are defining the $\Delta x$ and $\Delta p$ as half-widths corresponding to the half-maximum of $|\psi_0(x)|^2$ and $|\phi(k)|^2$. Why is this a suitable approximation? $\endgroup$ – user100411 May 6 '16 at 12:10
  • $\begingroup$ @BillN If you have a chance please see my other post. $\endgroup$ – user100411 May 6 '16 at 12:19

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