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Measuring the position and momentum of a particle is not simultaneously possible according to Heisenberg's uncertainty principle. Heisenberg's uncertainty principle gives us the uncertainty in measuring the position and momentum of a particle simultaneously. We can think of this as the highest possible resolution in measuring distance as $\Delta{x}$, and highest possible resolution in measuring momentum as $\Delta{p}$. So momentum and position can be thought of as varying in units $\Delta{p}$ and $\Delta{x}$ respectively, so they can't take on continuous values.

This is my interpretation of Heisenberg's uncertainty principle. My interpretation is not only we cannot measure position and momentum simultaneously, but that momentum and position take on discrete values. For position, it would be $x+\Delta{x}$, $x-\Delta{x}$, $x+2\Delta{x}$, ..., $x + n\Delta{x}$, where $n$ is an integer. This implies discretization. Is my interpretation right?

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    $\begingroup$ There is nothing in QM saying that position or momentum can get only discrete values at a distance $\Delta x$. That is your interpretation, but it is based on a misunderstanding. $\endgroup$
    – GiorgioP
    Jul 5 at 6:17
  • $\begingroup$ @GiorgioP has not keeping $x$ continuous led to difficulties in QM, and the creation of QFT and Loop Quantum Gravity theory? "When describing space-time as a continuum, certain statistical and quantum mechanical constructions are not well-defined. " - From wikipedia (en.wikipedia.org/wiki/Renormalization) (My interpretation may be wrong again). $\endgroup$ Jul 5 at 6:42
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    $\begingroup$ No, there is no difficulty in QM with a continuous quantity like $x$. Some difficulties may come in QFT, but they are not directly related to the continuous character of $x$. They are rather due to the way interactions depend on position. $\endgroup$
    – GiorgioP
    Jul 5 at 6:49
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    $\begingroup$ In any case, I would suggest you do not rely too much on Wikipedia. If you want to understand QM. It is much better to start with an introduction to the subject. You may find many, tuned for different backgrounds. $\endgroup$
    – GiorgioP
    Jul 5 at 6:52
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No, just because a quantity has a minimum value doesn't mean it can't vary continuously. Δx is continuously variable, as is Δp.

EDIT TO ADRESS YOUR ADDITIONAL CONTENT

The information you've added to you question is entirely beside the point. Space may well be quantised, but that is not a conclusion that follows from the HUP. Your assertion that if Dx has a minimum value then x must be discrete multiple of that minimum value is a non-sequitur.

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$Δx$ , the increment of x because x is a continuous variable , is also a continuous variable. Your taking discrete values of a particular $Δx$ is one instance of the possible discrete values of the continuous $Δx$, it does not bind it to your $Δx$.

You added:

Heisenberg's uncertainty principle necessitating the treatment of fields as quantized, and thus space as discrete.

This argument has little to do with Heisenberg's uncertainty principle, HUP. Loop quantum gravity,is a proposal for a quantized space , but this does not change the HUP. The HUP is an envelope extension of the commutators of the space and momentum variables, operator relations. It makes no difference if the operators operate on discrete space or continuous for the evaluation of the HUP.

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  • $\begingroup$ Yes, but if $\Delta{x}$ is fixed, then discretization applies, does it not? $\endgroup$ Jul 5 at 5:39
  • $\begingroup$ It applies mathematically by your choice, not to the physics variables described by the uncertainty principle, as the continuum is open to them. $\endgroup$
    – anna v
    Jul 5 at 5:41
  • $\begingroup$ $x$ being continuous has led to problems in quantum mechanics, hasn't it? "When describing space-time as a continuum, certain statistical and quantum mechanical constructions are not well-defined. " - From wikipedia (en.wikipedia.org/wiki/Renormalization) . $\endgroup$ Jul 5 at 6:40
  • $\begingroup$ I've edited my question to include more context. $\endgroup$ Jul 5 at 7:10
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Definitely no, uncertainty principle has nothing to do with the possible discreteness of space and time, or more unlikely, discreteness of energy-momentum. A way to see this is by looking at the modification of the uncertainty principle in the presence of a minimal length : see this article, equation (1). It turns out that the uncertainty principle is : \begin{equation} \Delta x \Delta p=\frac{\hbar}{2}\left( 1+\beta (\Delta p)^2\right) \end{equation} With $\beta$ a constant linked to the discreteness of space. Here, space is discretized but not momentum.

The uncertainty principle is not only a principle but also a theorem in mathematics related to Fourier transform and hence has nothing to do with the discreteness of $\mathbb{R}^n$, because it would be a contradiction.

To finish, I will say that at least for the position and momentum (the general case is false as pointed out in the comments), there are two possible forms of this uncertainty principle: one using the standard deviation of the canonical variables used, and one using the commutator. The latter is given by: \begin{equation} [\hat{x}_i;\hat{p}_j]\propto i\delta_{ij} \end{equation} This one is better suited for interpretation because it is straightforward: if you measure firstly the momentum and then the position you get a result $A$, if you do the converse you get a result $B$, and it turns out that $B\neq A$. Thus, it is impossible to measure simultaneously the position and the momentum with arbitrary resolution. Note that under this form, this uncertainty principle can't be interpreted as a discreteness of space and momentum.

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    $\begingroup$ The commutator is not a "form of the uncertainty principle". You can derive the uncertainty relations of two operators from their commutation relations, but you cannot in general derive the commutation relations of two operators from their uncertainty relations. $\endgroup$
    – ACuriousMind
    Jul 5 at 8:31
  • $\begingroup$ I was implicitly talking about the uncertainty relation of position and momentum, not the general case. But you are right and I will edit my answer to make it clearer. $\endgroup$ Jul 5 at 8:39

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