I have been taking classes in theoretical physics and naturally been studying the heisenbergs uncertanty principle. So far the position-momentum kinda makes sense, but I can’t wrap my head around the energy-time uncertainty relation. Here is my problem:

Suppose you construct a potential and a wave function with only one eigenstate and thus only one defined energy. Then the energy uncertainty would have to be zero, but that would violate the energy-time uncertainty principle. Does that mean that the time uncertainty is infinite, and thus that you can’t measure the energy. Or does that imply that its impossible to construct a system with only one eigenstate?

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    $\begingroup$ Does this answer your question? What is $\Delta t$ in the time-energy uncertainty principle? $\endgroup$ Jul 27, 2022 at 11:59
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    $\begingroup$ if the operator has only one eigenstate and eigenvalue then the operator is a scalar multiplier, there cannot be any evolution of it and the Mandelstam interpretation is not operative. $\endgroup$
    – hyportnex
    Jul 27, 2022 at 12:12
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    $\begingroup$ @hyportnex you should transform this into an answer. $\endgroup$ Jul 27, 2022 at 12:22
  • $\begingroup$ such systems certainly exist: the deuteron for instance has a single bound state. $\endgroup$ Jul 27, 2022 at 12:47

1 Answer 1


Part of the difficulty is that $\Delta t$ is not generally well defined. When writing $\Delta E\Delta t$, both terms are meant to be related to variance of operators. However, time is not an operator in quantum mechanics so, as proposed by Mandelstam and discussed in this post, one often defines $$ \Delta t= \frac{\sigma_A}{\vert d\langle A\rangle/ dt\rangle\vert}\, .\tag{1} $$ If there is only one bound state, then $\langle A\rangle $ cannot depend on $t$ because $$ \langle \Psi(t)\vert \hat A\vert \Psi(t)\rangle= \langle \psi\vert\hat A\vert\psi\rangle $$ for $\vert\Psi(t)\rangle=e^{iEt/\hbar}\vert\psi\rangle$.

Thus, for any observable $d\langle A\rangle/dt=0$; in Eq.(1), one can understand $\Delta t\to\infty$ "in the physics way" (assuming $\sigma_A\ne 0$, whatever that means in such a system). You can then qualitatively salvage $\Delta E\Delta t$ by suggesting that $0\times \infty\ge \hbar/2$.

In practice, in a system with only one bound state, there is no natural time scale. You cannot (for instance) define a unit of time based the energy of some transitions because your one bound state cannot transition to any other (bound) state. Thus, the meaning of $\Delta t$ is poorly defined, and one must take great care in interpretation of $\Delta E\Delta t$.


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