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Is there a way to derive second equation from the first one? I mean is there a connection between those two uncertainty relations?

\begin{align} \Delta x \Delta p &\geq \frac{\hbar}{2}\\ \Delta E \Delta t &\geq \frac{\hbar}{2} \end{align}

Abhimanyu Pallavi Sudhir
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2 Answers2

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The uncertainty principle can be seen as a result of space $x$ and momentum $p$ being a Fourier transform pair. The free-particle wave function has, similarly to the exponential $e^{-\frac{i}{\hbar} px}$ an exponential $e^{-\frac{i}{\hbar} E t}$. Thus one could expect a similar uncertainty relation for the variable pair $(E, t)$. An immediate result is that the solutions with a perfectly defined energy, solutions of $\hat H \psi = E \psi$ are stationary, i.e. their physical content does not change with time.

This is unprecise, though. The (minimal) theoretical uncertainty of any two variables can be expressed through their commutator (see Wiki) $$ \sigma_A \sigma_B \geq \big \langle [ \hat A, \hat B ] \big\rangle $$ The hard thing to do is to find an operator that represents time, as non-relativistic quantum mechanics deals with time only as a parameter.

For unstable states, there is a way to derive the uncertainty you give where $t$ is not time in general but the lifetime of the state, thus giving an explanation for the natural width of spectral lines. The derivation (in brief) can be found on the above-linked wiki page.

John Rennie
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Neuneck
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  • Also, there is a quantum information theoretical interpretation. Imagine you have a quantum state undergoing unitary time evolution, psi(t) = U(t) psi(0). You may ask yourself what is the best estimation you can make of the time parameter by performing quantum measurement on the state. The answer is obtained using the Quantum Cramer Rao bound, that in this particular case become the time-energy inequality. See for example http://arxiv.org/abs/0804.2981. – giulio bullsaver Nov 27 '15 at 13:45
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Special relativity has four-vectors $(\Delta t,\Delta \mathbf{r})$ and $(E,\mathbf{p})$, so we would like there to be a direct analogy between these two uncertainty relations. In fact the analogy fails, because position is an operator in quantum mechanics, but time isn't. Peierls has a nice discussion of this in Surprises in Theoretical Physics, pp. 36-37:

...time is not an observable. A measurement of time in itself does not convey any information about a physical system, and a statement about any other physical quantity usually implies that we are talking about its value at some particular time. In the case of a conserved quantity, such as the energy of an isolated system, the result then also gives the energy at any time. Landau was fond of making this point by saying, "There is evidently no such limitation -- I can measure the energy, and look at my watch; then I know both energy and time!"

So the energy-time uncertainty relation has a fundamentally different interpretation from that of the momentum-position one. There are actually multiple ways of interpreting it. One interpretation is given in this answer. Another is that the uncertainty relation holds if $E$ is the amount of energy transferred to or from a system, and $t$ is the time at which that transfer happened.

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