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The time x energy uncertainty principle, that states $$\Delta E\Delta t\geq\frac{\hbar}{2}$$ is derived from the general uncertainty principle for two non-commuting operators: $$\sqrt{\langle\Delta\hat{A}\rangle^2\langle\Delta\hat{B}\rangle^2}\geq\frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|$$ by setting $\hat A=\hat H$ and $\hat B$ as some arbitrary observable. It follows the notion, that for time-independent observables following equality holds: $$\frac{\mathrm d}{\mathrm dt}\langle\hat B\rangle=\frac{1}{i\hbar}\langle[\hat H,\hat B]\rangle$$ and by setting $$\frac{\Delta\langle\hat B\rangle}{\mathrm d\langle\hat B\rangle/\mathrm dt}=:\Delta t$$ Now I have a few problems with this. Firstly, how can we derive something for a special case of time-independent operator and say it holds generally? Secondly, how can we be sure, that the definition of $\Delta t$ used above holds? Why should this expression be equal to $\Delta t$(yes, it has the dimension of time, but that does not imply it is the time difference)?

This derivation can be edited to lower the strength of requirements for $\hat B$, because its time-independece is quite a strong requirement. If we only require the $\hat B$ to follow the continuity equation $$\frac{\partial \hat B}{\partial t}+\vec\nabla\cdot(\hat{\vec u}_B\hat B)=0$$ where $\vec{\hat u}_B$ is the speed of change of observable $\hat B$, we can only require, that $\vec\nabla\times\hat B=0$ or $(\vec\nabla\times\hat B)\cdot\hat{\vec u}_b=0$ and $\hat{\vec p}\cdot\hat{\vec u}_b=0$. Which allows for $\hat B$ to be time-dependent, but is still a quite strong requirement. Can it be weakened further?

And still, we have the problem with non-obvious, non-trivial definition of $\Delta t$.

Qmechanic
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user74200
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OK, so I studied the problem and went into the past and read the paper of I. Tamm and L. Mandeltsam and concluded, that the condition $$\frac{\partial \hat B}{\partial t}=0$$ is wrong and all of the answers and books that rely on it got it completely wrong. The original general time-energy relation is given not for any case, but for a stationary case. In a stationary case, we can write $$\frac{\mathrm d}{\mathrm dt}\langle\hat B\rangle=0$$ for all observables $\hat B$, which is the definition of a stationary state.

Then we can write: $$\left|\left\langle\frac{\partial \hat B}{\partial t}\right\rangle\right|=|\langle[\hat H,\hat B]\rangle|$$ Now we can apply the Minkowski inequality, Fubini's theorem and the Fundamental theorem of calculus to integrate the Heisenberg's relation $$\langle\Delta E\rangle\langle\Delta B\rangle\Delta t=\int_t^{t+\Delta t}\langle\Delta E\rangle\langle\Delta B\rangle\mathrm dt'\geq\frac{\hbar}{2}\left|\left\langle\int_t^{t+\Delta t}\frac{\partial \hat B}{\partial t'}\mathrm dt'\right\rangle\right|=\frac{\hbar}{2}|\langle\hat B(t+\Delta t)- \hat B(t)\rangle|$$ since the state is stationary, the first equation holds.

Now we identify the variation of $\hat B$ with the $\langle\Delta \hat B\rangle$, which is not very rigorous step, so the completely general relation states: $$\langle\Delta E\rangle\Delta t\geq b\frac{\hbar}{2}$$ where constant $b$ depends on the measured observable. The widely accepted relation holds if and only if $b\leq1$.

Mandelstam & Tamm went a little further and denoted $\Delta T$ as the special time $\Delta t$, for which $|\langle\Delta \hat B\rangle|=|\langle\hat B(t+\Delta t)-\hat B(t)\rangle|$.

Source: http://daarb.narod.ru/mandtamm/mt-eng.pdf

user74200
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