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Edited:-

$\hat{H} = -\frac {\hbar^2}{2m} \frac {\partial^2}{\partial^2 x}~+~V$

$\hat{U}(t) = e^{-iHt/\hbar}$

The Robertson Uncertainty Principle states:-

$\sigma^2_A\sigma^2_B \ge (\frac {1}{2i}[\hat{A},\hat{B}])^2$

The Energy-Time Uncertainty Principle states:-

$\sigma_H\sigma_{U(t)} \ge \frac{\hbar}{2}$

Also written as:- $∆E∆t \ge \frac{\hbar}{2}$

$\Rightarrow [\hat{H},\hat{U(t)}] = i\hbar$

$[\hat{H},\hat{U(t)}] = 0$ as stated by some sources.

This is contradictory? Why is it so?

2 Answers2

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First off - looks like you are using two different variables for the same thing. Usually, $H$ is the Hamiltonian operator (not a number, an operator). And $E$ is the energy of the particle (this is a real number). Basically, replace every $E$ with an $H$ in your question. Your first equation $H=i \hbar \frac{\partial}{\partial t}$ is also incorrect. (Edit: OP fixed).

Now, also note that one of the commutators is incorrect, because $[H, U(t)]=0$ and not $i \hbar$. How do we know this? As you wrote, we know that

$$U(t)=e^{-itH/\hbar}=\sum_{n=0}^{\infty}\frac{(-itH/\hbar)^n}{n!}$$

The last expression there is the series expression for $U(t)$, basically a taylor series but for operators. This is actually how it is defined. We also know that $H$ commutes with $H^n$, and of course it commutes with any real number. So from the series expression for $U(t)$ it is clear that $H$ must commute with $U(t)$.

Edit: In an edit, OP mentioned the uncertainty principle between operators $H$ and $U(t)$. The uncertainty principle using these operators just gives $0$ because $[H, U(t)]=0$ as demonstrated above. So wherever the equation $\sigma_H \sigma_{U(t)}\geq i \hbar$ came from, it is not correct.

doublefelix
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  • The question asks about $\hat{E} = i\hbar \partial_t$. $\hat H$ is a different operator, usually acting on spatial coordinates, not on time coordinate. – Ján Lalinský Jun 01 '23 at 22:20
  • @doublefelix the uncertainty relation between the energy and time is very well known:- $∆E∆t \ge \frac{\hbar}{2}$ –  Jun 02 '23 at 05:20
  • @AgnibhoDutta Yes, but this is completely different from $\Delta E \Delta U(t)$. There is a different relation for every pair of operators. In the case of $\Delta E \Delta t$ it is actually somewhat special, because there is no time operator. In that case the interpretation of the uncertainty relation is a bit more context-dependent and there are a few different but similar energy-time uncertainty relations that can be derived in different ways. The formula using commutators generally is not how time-energy uncertainty relations are derived. – doublefelix Jun 02 '23 at 06:58
  • @doublefelix U(t) is not equal to t. Am I correct now? –  Jun 02 '23 at 07:54
  • Yes, $U(t)=e^{-itH/\hbar}$ while simply $t=t$. They are not equal. – doublefelix Jun 02 '23 at 14:40
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Actually, the fundamental confusion here has nothing to do with the uncertainty principle or the commutators. It is just bad notation.

The commutators are in terms of operators. Operators should be written with little hats on them. In my notation, the momentum operator that returns a vector is written as $$\vec{\hat p}$$ whereas a unit vector is written as $$\hat{\vec n}$$ so that you can visually see the difference.

The fundamental commutator is $$\left[\hat x,\hat p\right]=i\hslash\,\hat{\mathbb I}$$ and this is the thing that makes $\hat p=-i\hslash\hat\partial_x$ and the following equation will only make sense as an eigenvalue equation if you are very careful with how things are written: $$\hat p\left|p\right>=p\left|p\right>$$ Note that this corresponds to $$\hat H\left|E\right>=E\left|E\right>$$

If you want to use the unitary operator to move a state from origin to $x$, the operator is $$e^{ix\hat p/\hslash}$$ whereas the plane wave is $$e^{ipx/\hslash}$$

You should now see that the following are correct: $$[p,e^{ix\hat p/\hslash}]=0=[p,e^{ipx/\hslash}]$$ $$[\hat p,e^{ipx/\hslash}]=p\,e^{ipx/\hslash}$$ If you translate these into $E$ and $\hat H$ and $t$ and so forth, you will see where the mistake is.

There is actually no operator shorthand for $i\hslash\hat\partial_t$, even though Schrödinger's equation asserted that it is equivalent to the Hamiltonian operator $\hat H$, because the Hamiltonian operator is in terms of $\hat x$ and $\hat p$, not $\hat\partial_t$. This complicates this problem because you have both confusions working at once.

naturallyInconsistent
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