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Considering the Heisenberg uncertainty principle, which states $\Delta x \cdot \Delta p \geq h$, I've explored a modified version by computing $(\Delta x + 1)(\Delta p + 1) \geq \Delta x \cdot \Delta p + \Delta x + \Delta p + 1 \geq h + 1$, leading to the conclusion that, after realizing $\Delta x = |x_1 - x_2|$ and $\Delta p = |p_1 - p_2|$, it also implies $\log(|x_1 - x_2| + 1) + \log(|p_1 - p_2| + 1) \geq \log(h + 1)$. This can be further expressed as $d((x_1, p_1), (x_2, p_2)) \geq \log(h + 1)$, where $d((a,b),(c,d)):= \log(|a-c|+1)+\log(|b-d|+1)$ denotes a certain distance measure in the phase space.

My question to the community is: How does this logarithmic interpretation of the Heisenberg uncertainty principle compare with the conventional understanding, particularly in terms of physical implications and mathematical validity? Is there a recognized theoretical framework that accommodates such a modification, especially considering the logarithmic terms added to the uncertainties in position and momentum? Furthermore, does this approach yield any novel insights or implications for quantum mechanics or related mathematical structures?

Mathematical question: How are metric space called where for two distinct points $x,y \in X$ we always have $d(x,y) \ge \epsilon > 0$ for some constant $\epsilon > 0$?

Here is a an example image of two points in the space $(x,p)$ and the boundaries for these points:

enter image description here

Thanks for your help!

Edit: I do not understand the vote to close and the downvotes.

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    $\begingroup$ Planck's constant $h$ is not dimensionless, what do you mean by $\log (h+1)$? $\endgroup$
    – Carlo Beenakker
    Commented Feb 10 at 11:38
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    $\begingroup$ well, position $x$ and momentum $p$ each have dimensions, and their product has dimensions of Joule/second; if I wish to compute $\log(h+1)$, which value should I enter for $h$? $\endgroup$
    – Carlo Beenakker
    Commented Feb 10 at 11:51
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    $\begingroup$ Since a metric with $d(x,y)\ge\varepsilon$ for all distinct $x,y$ and a fixed $\varepsilon$ generates the discrete topology one could call the metric topologically discrete. I don't know whether an established name exists for this property. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 10 at 12:15
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    $\begingroup$ @JochenWengenroth Since $\epsilon$ is fixed, over the whole space, I'd be inclined to call the metric "uniformly discrete". Mere discreteness of the topology would be the weaker statement $\forall x\,\exists\epsilon>0 \,\forall y\neq x\ d(x,y)>\epsilon$. $\endgroup$
    – Andreas Blass
    Commented Feb 10 at 13:32
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    $\begingroup$ Agreed. Nevertheless, I would keep the adjective topologically because metric aspects (e.g., Lipschitz conditions) are not determined by uniform discreteness. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 10 at 13:44

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Let me try to answer the question "How does this logarithmic interpretation of the Heisenberg uncertainty principle compare with the conventional understanding". There are two issues that prevent a meaningful comparison.

  1. $\Delta x$ and $\Delta p$ in the Heisenberg uncertainty relation do not refer to a difference of two quantities (although the letter $\Delta$ is used), they refer to the standard deviation of the probability density function of $x$ and $p$, respectively. So the identifications $\Delta x = |x_1 - x_2|$ or $\Delta p = |p_1 - p_2|$ have no significance in this context.

  2. The parameter $h$ in the Heisenberg uncertainty principle, Planck's constant, has a dimension, it is not a dimensionless number. So you cannot "add 1" to $h$, and you cannot take the logarithm of $h+1$.

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