Wednesday, 8 May 2019

quantum mechanics - Time-Energy Uncertainty Relation when A depends explicitly on t


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


ΔTAΔH2.


My question is: What happens if we allow A/t0?.


EDIT.


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



ΔTA=AdAdt.


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


ΔTAΔH2|1A/tdA/dt|.


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


ΔTA=AdAdtAt,


Then the Heisenberg's principle takes its usual form, but we have to reinterpret the quantity ΔTA.



Answer



Following the argument in Introduction to Quantum Mechanics by Griffiths, ΔHΔQ=12i[H,Q]=2|dQdtQt|


Then, defining ΔE=ΔH, |dQdtQt|ΔtΔQ

Your first equation after the edit is what Δt 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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