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If in the Heisenberg's uncertainty principle the quantity $$\frac{\hbar}2=\frac{h}{4\pi}$$ is near to $0$, what happen? What is a simple any explanation that could be given to a high school student?

I have read this: Zero Planck constant, but I found lots of different information that still cause confusion in my mind.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Aug 13 '19 at 23:23
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    $\begingroup$ See "arxiv.org/abs/1201.0150" $\endgroup$ – nmasanta Aug 14 '19 at 8:54
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    $\begingroup$ The question is still unclear to me. What specifically are you thinking about for this scenario. Are you wondering what would happen to atoms in our bodies? Are you wondering if the sun would shine differently. Are you wondering if experiments at the LHC would stop working? What are you interested in here, specifically? $\endgroup$ – Aaron Stevens Aug 19 '19 at 23:09
  • $\begingroup$ @AaronStevens Dear, I am asking for something simple that can be of use to high school students. No details that must be difficult to understand. A simple example that I can understand. $\endgroup$ – Sebastiano Aug 20 '19 at 8:57
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    $\begingroup$ Ok then specify what type of simple example you are looking for. You're starting to do that thing again where you are resisting advice that is being given to you too improve your question. There are many examples that a high school student could understand. You need to be more specific. $\endgroup$ – Aaron Stevens Aug 20 '19 at 10:51
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Planck's constant has a fixed value, $h = 6.626 \,070\, 15 \times 10^{−34}\:\rm J \: s$, which does not depend on any aspects of the state or structure of any given system. This implies that the limit "$h\to 0$" is a completely meaningless gesture. Planck's constant cannot be changed, either up or down, so "making it go to zero" is not something that can be done.

In particular, this means that speculating what will happen to X bit of physics (be it the Heisenberg uncertainty principle or anything else) when "$h\to 0$" is pointless, as the premise is unphysical.

The same goes with "$h\simeq 0$" $-$Planck's constant has a value and it is nonzero. Saying that "$h\simeq 0$" is meaningless.


Nevertheless, what you're describing does have a very closely related concept which does make sense, known as the correspondence principle.

The core intuition that expressions like "$h\simeq 0$" are trying to capture is the basic question

what happens when the dynamics of a system happens on scales at which $h$ is so small as to be irrelevant?

but that question is phrased the wrong way round: it's not that $h$ is small, it's that the characteristic scales of the dynamics of the system are large compared to $h$. So what happens then? basically, you recover the classical limit of your quantum system, i.e. the system behaves more and more like a classical system, and quantum behaviour becomes less and less pronounced, until it is no longer visible.

If you want further details, ask a question in a way which is suitably detailed in a way that correspond to the level of detail that you're expecting in the answer. And, of course, before asking or editing, be sure to research appropriately the several hundred questions on the classical limit and correspondence principle already on this site.

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    $\begingroup$ Might be worth pointing out that approximating something at zero is only really valid when it's all nondimensional... In which case the uncertainty could be "zero" when looking at the position of a car driving the road, but wouldn't be "zero" when looking at the distance between two protons or something. $\endgroup$ – tpg2114 Aug 13 '19 at 22:54
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    $\begingroup$ While I agree with the conclusion, I'm not sure I agree with the argument. The electron mass is fixed, $m=.5\ \mathrm{Mev}$, but that does not mean it is meaningless to ask what happens when $m\to 0$. The massless limit of a QM model is a perfectly well defined concept, and it is sometimes useful. For example, massless quarks is an approximation that works fine for some purposes. So "having a fixed value" and "taking it to zero" are not mutually exclusive concepts. But I agree that "$\hbar\to0$" is a very subtle limit that is ill-defined if considered at face value. $\endgroup$ – AccidentalFourierTransform Aug 13 '19 at 23:19
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    $\begingroup$ @AccidentalFourierTransform You are alluding to the Higgs mechanism and the massless Lagrangians, but that is just an explanation as how particles have non-zero mass. A correct counterpart of this theory to the HUP would be: hey, I have a theory which starts off with $\hbar =0$, then uses some principles and other theoretical constants and at the end comes up with a perfectly valid explanation/formula on why $\hbar$ has a non-zero value. $\endgroup$ – DanielC Aug 13 '19 at 23:25
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    $\begingroup$ @DanielC 1) massless approximations are definitely not unrealistic. Massless quarks is a very useful concept. I used QED because it is easier to write. 2) and even if it is an unrealistic model, that has nothing to do with my point. What I said is that $m\to0$ is a meaningful limit; not that it is useful (although it most definitely is). 3) And even if it were unrealistic and useless, what would it have to do with the Higgs mechanism? I still don't understand why you brought it up, it has nothing to do with E.P.'s answer or my remark... $\endgroup$ – AccidentalFourierTransform Aug 13 '19 at 23:32
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    $\begingroup$ @EmilioPisanty the value of h is arbitrary, it is experimental, not theoretical, why do you claim that it does not make sense to imagine a scenario in which its value is different? $\endgroup$ – Wolphram jonny Aug 14 '19 at 1:18

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