# Wrong formulation of Heisenberg's uncertainty principle [closed]

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 $$\Delta x \Delta p$$.

• Yes, I think that your professor made a simple typo. May 28 at 18:12
• @ApoorvPotnis Thank you for the confirmation! May 28 at 18:14
• See physics.stackexchange.com/q/53802/36194 as to the interpretation since $t$ is not an observable in QM. May 28 at 18:38
• 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. May 29 at 15:46
• 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. Jun 1 at 18:16

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 $$\leq$$.