0
$\begingroup$

I know that Heisenberg's Uncertainty Principle prevents us from even theoretically predicting the exact position and momentum of a particle. There is always an uncertainty in one or the other, given by $$\Delta x\cdot \Delta p\geqslant \frac h{4\pi} $$

This apparently leads to our inability to see the wave nature of electron when we have an observer observing the slit through which it passes (delayed choice quantum eraser, Kim et al.(1999)).

So, we have enough evidence (the electron not falling into the atom, for instance) that one cannot determine position and momentum accurately and simultaneously. But still, Heisenberg's principle does not prevent us from measuring the exact position at all- but for that we must sacrifice any knowledge about the momentum of an electron.

Is it possible, ideally (assuming we have powerful techniques and instruments, and all errors are accounted for or absent), to devise an experiment which in no way gives us an idea about momentum- with which we can measure the exact position of the particle?

$\endgroup$
  • $\begingroup$ I'm not sure what it means to "theoretically devise an experiment". In my "theoretical" view on experiments, all experimental results have non-zero error bars, so you cannot measure "exact position" even classically, and all the quantum stuff in this question is completely irrelevant! $\endgroup$ – ACuriousMind Apr 26 at 9:41
  • $\begingroup$ @ACuriousMind Maybe we can in an 'ideal' case reduce the error to near zero values- or ignore them for the time being. Thanks, now I have changed it to 'ideal'. $\endgroup$ – Krishna Apr 26 at 10:04
2
$\begingroup$

I interpret your question to be: Does the position operator on, say, $L^2({\mathbb R})$, have any eigenvectors?

The answer is no because if $f$ is a square integrable function (or for that matter any function) with $xf=f$ almost everywhere, then $f=0$ almost everywhere.

You can try to get around this by replacing the state space with a rigged Hilbert Space but at that point you've left the formalism of quantum mechanics.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Every direct measurement of length (and thus position) is a rational number of units, in metric system a decimal with finite number of digits. If possible values of the measured coordinate are from the set of real numbers, almost every measurement will be tainted by a finite error. There is no bottom limit for this error; using interferometry, fractions of atom size can be measured.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.