[Physics] Connection between $\Delta x \Delta p \geq \frac{\hbar}{2}$ and $\Delta E \Delta t \geq \frac{\hbar}{2}$

energyheisenberg-uncertainty-principlequantum mechanicsquantum-field-theorytime

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}

Best Answer

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.

Related Question