My professor wrote this on the blackboard:

$$\Delta E\Delta t \leq \hbar$$

Isn't the $\leq$ sign wrong? Or is this "another" possible formulation of Heisenberg's uncertainty principle?

I am at the very beginning of quantum mechanics, so I am basically not aware of anything-ish. Except for the fact that every note and book I have read so far, mention the principle with $\geq$, no matter if $\Delta E \Delta t$ or with $\Delta x \Delta p$.

  • 7
    $\begingroup$ Yes, I think that your professor made a simple typo. $\endgroup$ May 28 at 18:12
  • $\begingroup$ @ApoorvPotnis Thank you for the confirmation! $\endgroup$
    – Numb3rs
    May 28 at 18:14
  • $\begingroup$ See physics.stackexchange.com/q/53802/36194 as to the interpretation since $t$ is not an observable in QM. $\endgroup$ May 28 at 18:38
  • 4
    $\begingroup$ I’m voting to close this question because it arises from a simple typo in a particular source and is unlikely to be useful to future users. $\endgroup$ May 29 at 15:46
  • 1
    $\begingroup$ Ask professor what he meant. This inequality is not very meaningful without explaining what $\Delta t$ is; $\Delta E$ is presumably uncertainty in energy $E$, but time coordinate $t$ usually does not have uncertainty in QT, it is treated as independent variable/parameter. $\endgroup$ Jun 1 at 18:16

1 Answer 1


Don't be confused. It is definitely $$\Delta E \Delta t \geq \frac{\hbar}{2}$$.

Your professor typo'd here, as if what he wrote: Then we could measure energy with any arbitrarily small error. We want to break the uncertainty principle and quantum mechanics, so it is $\geq$. We don’t want the error to be smaller than something, so it must be $\geq$, not $\leq$.


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