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Heisenberg's uncertainty principle tells us that there is a lower limit on the accuracy with which we can determine both the momentum and the position of a particle simultaneously.

$\Delta$$x$ $\Delta$$p$ $\ge$ $\frac{\hbar}{2}$

Suppose we decide to determine the position with 100 % accuracy thus remaining completely uncertain about the momentum.After measurement we can represent such a system by a delta function, which represents a state with definite position.At least this works theoretically.

Is this possible practically? I mean is there an experimental set up for which we can make $\Delta$$x$=$0$ and $\Delta$$p$ = $\infty$ rather than having $\Delta$$x$ as small finite value close to $0$ and $\Delta$$p$ as some large but finite huge value.

For more clarity on what I mean,consider Heisenberg's Microscope.We cannot determine the position of the electron with complete. If we try to use high wavelength light we get a fuzzy image and if we use a low wavelength light, we disturb the electron and its exact position cannot be determined. Is there some way in this experiment of some other experiment where we can arrange the experiment setup in such a way that we are forced to get $\Delta$$x$ =$0$ and $\Delta$$p$=$\infty$ ?(or are we forced to have $\Delta$$x$ as some non-zero finite quantity as long we are doing the experiment practically?)

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  • $\begingroup$ No, we cannot actually put a system in a state with no uncertainty in position at all. The measurement process will always be imperfect. $\endgroup$ – By Symmetry Feb 13 '18 at 12:29
  • $\begingroup$ Is that true even if our apparatus is extremely good and can give accurate results? $\endgroup$ – Akaash Srikanth Feb 13 '18 at 15:13

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