It is usually said/done in textbooks and classes that if $\Delta x$ is known then $\Delta p_x$ can be estimated using the uncertainty principle as $\Delta p_x \sim \hbar/\Delta x$.

But the uncertainty principle does not say that, it says $\Delta p_x\ge\hbar/\Delta x$. That means we can only set a lower bound on $\Delta p_x$, i.e. $\Delta p_x$ could be anything between $\hbar/\Delta x$ and $\infty$

Why then the lower bound is chosen for estimation?

Are there certain situations where the products in uncertainties is of the same order as $\hbar$?


1 Answer 1


It assumes a sort of democracy between $x$ and $p$, and is obviously not valid everywhere. The connection between classical and quantum mechanics happens through coherent states, which are states which minimize the product $\Delta p \Delta x$, and in a sense, behave most classically (and have well-defined classical limits as $\hbar \to 0$).

So if you started with a something behaving reasonably classically, e.g. an electron's motion in a uniform electric field, and you wanted to estimate the uncertainty of it's momentum, you could use this as a starting point. There's nothing really rigorous about it, other than the fact that coherent states seem to enjoy a privileged status in nature.

  • 1
    $\begingroup$ I did not ask for rigorous. It is only an estimation. My problem is the estimation is based on an inequality. So I can estimate something to be n or one million times bigger than n. Sorry I am not sure that I understand your answer. I do not know what coherent states are. Are you saying that in certain situations the uncertainty product can be of the same order as $\hbar$? $\endgroup$
    – Revo
    Commented Nov 28, 2013 at 20:31
  • $\begingroup$ Yes. The ground state of the harmonic oscillator is an example of a coherent state, and satisfies $\Delta x \Delta p = \frac{\hbar}{2}$. $\endgroup$ Commented Nov 28, 2013 at 20:33
  • $\begingroup$ But the cases in which the uncertainty principle is used for estimation are so different from the harmonic oscillator. Or is it because that we are assuming that the particle is localized by a gaussian wave packet (which is the harmonic oscillator ground state)? $\endgroup$
    – Revo
    Commented Nov 28, 2013 at 20:38
  • $\begingroup$ Yes, it's the wave-packet assumption. Can you give an example of where you have encountered this? $\endgroup$ Commented Nov 28, 2013 at 20:44
  • $\begingroup$ Introductory Physics/ Modern physics textbooks. How about energy time uncertainty relations? do coherent states hit the lower bound also? $\endgroup$
    – Revo
    Commented Nov 28, 2013 at 21:00

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