In "The Road to Reality" (pg. 523), Roger Penrose writes:

Heisenberg's uncertainty relation tells us that the product of these spreads cannot be smaller than the order of Planck's constant, and we have

$$ \Delta p \Delta x \gtrsim \frac{1}{2} \hbar. $$

However, other texts use

$$ \Delta p \Delta x \ge \frac{1}{2} \hbar $$

What is the meaning of $\gtrsim$ in this context? What is the difference between "greater than" and "greater than the order of"?


The $\gtrsim$ symbol is there because of popsci simplification.

The precise statement of the principle is $$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$ where $\sigma_x$ is the standard deviation of the results we would get by a position measurement, and $\sigma_p$ is the standard deviation of the results we would get by a momentum measurement.

However, in a popular-level presentation, the standard deviation would be too technical to introduce, and the technicalities wouldn't be relevant to the conceptual content anyway. So the standard deviations are replaced by "uncertainties" or "spreads" $\Delta x$. The term "uncertainty" in these discussions is not a technical one and it doesn't have a strict definition. Some authors take it to mean the range of the distribution, some take it to be half the range, and so on.

Since $\Delta x$ isn't usually well-defined in the first place, the best you can say is a statement like $$\Delta x \Delta p \gtrsim \hbar$$ where $\gtrsim$ vaguely means "probably greater than, but possibly smaller by up to an order of magnitude or so but no more". That's good enough because people rarely use the uncertainty principle quantitatively; even practicing physicists just invoke it for intuition and don't care about the constants.

Penrose's book is straddling the line between being pop science and a textbook, which is probably why he included the factor of $1/2$, even though "$\gtrsim$" inherently means you're not working to that accuracy in the first place.

| cite | improve this answer | |
  • $\begingroup$ Yep, this is it. With one addition: it's not just "possibly smaller by up to an order of magnitude or so but no more", but also the fact that if, when you make $\Delta x \Delta p$ precise, it falls bellow $\hbar$, then that fraction is constant and does not scale with the details of the problem. $\endgroup$ – Emilio Pisanty Dec 17 '18 at 11:12
  • $\begingroup$ Thank you. Could you please cite a source of this? $\sigma$ seems simple enough, and a frequentist approach requiring repeatability seems particularly inconvenient. $\endgroup$ – MrMartin Dec 17 '18 at 16:10
  • $\begingroup$ @MrMartin A source for what, in particular? $\endgroup$ – knzhou Dec 17 '18 at 16:12
  • $\begingroup$ A source which explains $\sigma_x \sigma_p \geq \frac{\hbar}{2}$, such as a textbook or the original publication $\endgroup$ – MrMartin Dec 17 '18 at 21:22
  • $\begingroup$ Actually, a quick search for "Heisenberg Uncertainty Standard Deviation" yields lots. Thanks again @knzhou $\endgroup$ – MrMartin Dec 17 '18 at 21:27

The correct statement is $ \Delta p \Delta x \ge \frac{1} {2} \hbar $. According to https://en.m.wikipedia.org/wiki/Uncertainty_principle

"The formal inequality relating the standard deviation of position σx and the standard deviation of momentum σp was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928:

$ \sigma_x \sigma_p \geq \frac{\hbar }{2} $

where ħ is the reduced Planck constant, h/(2π). "

Kennard, E. H. (1927), "Zur Quantenmechanik einfacher Bewegungstypen", Zeitschrift für Physik (in German), 44 (4–5): 326–352, Bibcode:1927ZPhy...44..326K, doi:10.1007/BF01391200. Weyl, H. (1928), Gruppentheorie und Quantenmechanik, Leipzig: Hirzel

| cite | improve this answer | |
  • $\begingroup$ This would be a better answer if you explained why your statement is correct, and maybe explained why Penrose might use the greater than or approximately equal to symbol. $\endgroup$ – John Rennie Dec 16 '18 at 11:52
  • $\begingroup$ Kennard did not really derive $\frac{\hbar}{2}$, but only $\hbar$. $\endgroup$ – DanielC Dec 16 '18 at 19:39
  • $\begingroup$ @DanielC perhaps you can update the Wikipedia article. $\endgroup$ – my2cts Dec 16 '18 at 23:22
  • 1
    $\begingroup$ This isn't an answer - the answer is in knzhou's post. $\endgroup$ – Emilio Pisanty Dec 17 '18 at 11:11
  • $\begingroup$ @Emilio Pisanty I disagree. The difference between $\geq$ and $\gtrsim$ in this context is that $\geq$ is correct. Questions about the difference between these symbols in a broader context belong on a different forum. $\endgroup$ – my2cts Dec 17 '18 at 12:59

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.