There are several questions asking by the meaning of Time-Energy uncertainly relation and also by its derivation.

My problem is about the second. In all documents I have seen, the same is supposed alwas: Let $A$ an observable which does not depend explictly on $t$. (from John Baez's web page to Griffiths (p.113) and also in the original paper, formula 4). Then

$$\Delta T_A \Delta H \geq \frac{\hbar}{2} . $$

My question is: What happens if we allow $\langle \partial A/ \partial t \rangle \neq 0$?.


@ZoltanZimboras demands me to write the definition of $\Delta T_A$. The truth is I'm not sure how to answer him. Usually, $\Delta T_A$ is defined as

$$ \Delta T_A = \frac{\langle A \rangle}{\frac{d\langle A\rangle}{dt}}. $$

If we set the same definition, then the Heisenberg's inequality comes

$$\Delta T_A \Delta H \geq \frac{\hbar}{2}\left| 1-\frac{\langle\partial A/\partial t\rangle}{d\langle A\rangle /dt } \right|. $$

The advantage of this definition is physics understand very well the meaning of the quantity $\Delta T_A$, but then the above equation comes more difficult. On the other hand, if we set

$$\Delta T_A = \frac{\langle A \rangle}{\frac{d\langle A\rangle}{dt}-\left\langle \frac{\partial A}{\partial t}\right\rangle}, $$

Then the Heisenberg's principle takes its usual form, but we have to reinterpret the quantity $\Delta T_A$.

  • $\begingroup$ Just to make the question a bit more self-explanatory, could you write down how $\Delta T_A$ is defined in terms of $A$ (and $H$)? $\endgroup$ – Zoltan Zimboras Mar 31 '18 at 4:26
  • $\begingroup$ @ZoltanZimboras I'm not sure how to define it. Maybe this is the key of the problem. See the edit. $\endgroup$ – Dog_69 Apr 2 '18 at 17:16

Following the argument in Introduction to Quantum Mechanics by Griffiths, $$\Delta H\Delta Q=\frac{1}{2i}\langle [H,Q]\rangle=\frac{\hbar}{2}\left|\frac{d\langle Q\rangle}{dt}-\left\langle\frac{\partial Q}{\partial t}\right\rangle\right|$$

Then, defining $\Delta E=\Delta H$, $$\left|\frac{d\langle Q\rangle}{dt}-\left\langle\frac{\partial Q}{\partial t}\right\rangle\right|\Delta t \equiv \Delta Q$$ Your first equation after the edit is what $\Delta t$ is defined as in the absence of explicit time dependence of $Q$. As for the interpretation of this new definition, it is the amount of time it takes for the expectation value of $Q$ to move a standard deviation away from the expectation value of its change. Informally, it is a measure of how quickly this operator will "misbehave" in the sense of how far its "average" value is from what one would expect just from the operator's changes.

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  • $\begingroup$ Okey. Nice answer. Thanks for your effort. I'm going to wait until the bounty expires for give you the bounty. Anyway, good answer. $\endgroup$ – Dog_69 May 13 '18 at 20:22
  • $\begingroup$ @Dog_69, I'm sorry, I really don't care if I do get the bounty (though it would be nice!), I don't know what you are trying to say. $\endgroup$ – user195162 May 13 '18 at 20:40
  • $\begingroup$ I'm trying to say that since there is an open bounty I'm going to wat some days to receive more answers. Otherwise, I had already accepted your answer. $\endgroup$ – Dog_69 May 13 '18 at 21:54
  • $\begingroup$ @Dog_69, may I have the bounty now? :P $\endgroup$ – user195162 May 19 '18 at 18:55
  • $\begingroup$ Oh, Yes. Of course. $\endgroup$ – Dog_69 May 19 '18 at 18:56

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