My layman understanding of the Uncertainty Principle is that you can't determine the both the position and momentum of a particle at the same point in time, because measuring one variable changes the other, and both cannot be measured at once.

But what happens if I measure a charged particle with any number of detectors over a period of time? Can I use a multitude of measurements to infer these properties for some point in the past? If not, how close can we get? That is, how precise can our estimate be?

  • $\begingroup$ This sounds a lot like Leibniz's & Newton's thought experiments when they were creating calculus. "Can we measure the speed/position of an object during a single instant? If not, how close can we get?" $\endgroup$
    – trysis
    Jun 13, 2014 at 0:56
  • $\begingroup$ Per the uncertainty principle, each of those detectors is going to have to trade off accuracy in measuring position versus speed, and each is going to affect the position or speed in the process of taking the measurement. (You can't detect anything without energy exchange.) A group of measurements each of which has error within known bounds, and each of which introduces error into the system, doesn't necessarily add up to less error. $\endgroup$
    – keshlam
    Jun 13, 2014 at 3:02
  • $\begingroup$ @keshlam, this would be true if the values were all inaccurate as well as imprecise. This is in the scientific terms, not layman's terms as you use, where "accuracy" is the same thing as "precision". When numbers are inaccurate, they don't round to the "correct" number", they don't have a correct average when taken together, etc., while you seem to simply imply we can't measure them out to an acceptable number of digits. Are you implying this to be the case? $\endgroup$
    – trysis
    Jun 13, 2014 at 3:42
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    $\begingroup$ I'm saying that uncertainty plus uncertainty does not equal greater certainty, and that the experiment as described appears to also introduce error such that averaging is probably meaningless. $\endgroup$
    – keshlam
    Jun 13, 2014 at 4:54
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    $\begingroup$ Whatever detector you use will change the state of the particle, making every other measurement show wrong results. It is just as simple as that. $\endgroup$
    – Vercas
    Jun 13, 2014 at 7:37

2 Answers 2


The uncertainty principle should be understood as follows: The position and momentum of a particle are not well-defined at the same time. Quantum mechanically, this is expressed through the fact that the position and momentum operators don't commute: $[x,p]=i\hbar$.

The most intuitive explanation, for me, is to think about it in terms of wave-particle duality. De Broglie introduced the idea that every particle also exhibits the properties of a wave. The wavelength then determines the momentum through $$p=\frac{h}{\lambda}$$ where $\lambda$ is the De Broglie wavelength associated with the particle. However, when one thinks about a wave, it is clear that the object described by it will not be easy to ascribe a position to. In fact, one needs a specific superposition of waves to create a wave that is essentially zero everywhere except at some position $x$. However, if one creates such a wave packet, one loses information about the exact wavelength (since a wave with a single, well-defined wavelength will simply extend throughout space). So, there is an inherent limitation to knowing the wavelength (i.e. momentum) and position of a particle. On a more technical level, one could say that the uncertainty principle is simply a consequence of wave-particle duality combined with properties of the Fourier transform. The uncertainty is made precise by the famous Heisenberg uncertainty principle, $$\sigma_x\sigma_p\geq \frac{\hbar}{2}$$ More generally, for two non-comuting observables $A$ and $B$ (represented by hermitian operators), the generalized uncertainty principle reads $$\sigma_A^2\sigma_B^2\geq \left(\frac{1}{2i}\langle [A,B] \rangle\right)^2\ \implies \ \sigma_A\sigma_B \geq \frac{|\langle [A,B]\rangle| }{2}$$ Here, $\sigma$ denotes the standard deviation and $\langle\dots\rangle$ the expectation value. This holds at any time. Therefore, the measurement occurring right now, having occurred in the past or occurring in the future has nothing to do with it: The uncertainty principle always holds.

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    $\begingroup$ This doesn't really answer the question. The last paragraph starts to, but doesn't go very far. Therefore, I'm not giving it an upvote or a downvote. $\endgroup$
    – trysis
    Jun 13, 2014 at 0:58

Found this great article (https://arxiv.org/abs/0906.1605) researching this topic myself. The short answer is no, the uncertainty principle doesn't apply to the past in the same way it applies to present measurements. It is possible to "update" our best guess for a particle's position and momentum made at time $t_0$ at some later time $t_1>t_0$ as we make new measurements. The new measurements inform the likelihood that a specific set of conditions at $t_0$ were present beyond the normal limit of the uncertainty principle, but it's important to recognize that some uncertainty always remains (makes me think of the Truman show; no matter how real and consistent your version of events appears, can you really be sure it's not all a massive conspiracy or coincidence?). It seems probable that there is some formulation or extension of the uncertainty principle that bounds this historical knowledge, but I've been as of yet unable to find it. Might be worth a paper if there isn't one already.


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