# Time $\times$ energy uncertainty. Rigorous derivation

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 $\Delta t$.

• Related: physics.stackexchange.com/q/53802/2451 , physics.stackexchange.com/q/259334/2451 and links therein. – Qmechanic Nov 27 '16 at 14:53
• This gives a good summary of the general outlook to the time-energy uncertainty principle - it's a fair bit shakier than the position-momentum one. (The computational complexity bit is more esoteric, though.) – Emilio Pisanty Nov 27 '16 at 15:00
• What do you mean "how can we be sure that the definition of $\Delta t$ used above holds"? It's a definition, how could it not hold? That this $\Delta t$ might not be what we intuitively want to call "time difference" is of no importance. – ACuriousMind Nov 27 '16 at 16:08
• ACuriousMinfd - in other words, what does $\Delta t$ defined like this even mean? In general, $\Delta t=t_2-t_1$. How can we be sure it has the right properties? The whole problem is, that we are assigning, defining, some unit or entity, that we know nothing about and then work with it as if it was a unit that we know from our everyday life. This is not trivial. Is the $\Delta t$ defined like this well defined?! This is a classic question you should know from mathematics... Every time you define something that should have some properties, you have to prove it is well-defined. – user74200 Nov 27 '16 at 16:33
• Qmechanic - all of these answers rely on $\partial \hat B/\partial t=0$... – user74200 Nov 27 '16 at 16:57

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 $|\langle\Delta \hat B\rangle|=|\langle\hat B(t+\Delta t)-\hat B(t)\rangle|$.