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As the title suggests. Is there a prediction of quantum mechanics that could be construed as representing an "energy-time uncertainty relation?" Does there exist any reference to such a prediction, or is it obviously false on some level?

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marked as duplicate by Sofia, ACuriousMind, Kyle Kanos, Danu, JamalS Mar 10 '15 at 8:19

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    $\begingroup$ One version of Heisenberg's Uncertainty Principle is just that: $\Delta E \Delta t \geq \hbar / 2$. $\endgroup$ – ahemmetter Mar 9 '15 at 21:10
  • $\begingroup$ @andynirox, you wrote "Et≥ℏ/2"... clearly that makes no sense. Even wrapped in some specific context it is likely useless/misleading. $\endgroup$ – hft Mar 9 '15 at 21:12
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    $\begingroup$ It's important to remember that the energy-time uncertainty relation is a little shakier than the position-momentum one, simply because "time" is not an operator in quantum mechanics. $\endgroup$ – zeldredge Mar 9 '15 at 21:16
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    $\begingroup$ Possible duplicate of physics.stackexchange.com/q/53802/50583 $\endgroup$ – ACuriousMind Mar 9 '15 at 21:19
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    $\begingroup$ See physics.stackexchange.com/q/53802 $\endgroup$ – WetSavannaAnimal Mar 9 '15 at 21:40
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There is a 1961 paper by Aharonov and Bohm on this subject, in which there is defined, among other things, a characteristic time for an operator's expectation value to deviate significantly, measured by the initial dispersion in that operator. This result is essentially a theorem we will prove here (Theorem 2).


Let $\mathcal{S}$ be a Hilbert space, $A$ a (bounded) Hermitian operator on $\mathcal{S}$, and $\psi$ a unit vector in $\mathcal{S}$. By the dispersion of $A$ $($relative to the state $\psi$$)$ we mean the number$$\Delta_\psi A = \left( \left\langle \psi\,\left|\,A^2\,\right|\,\psi \right\rangle - \left\langle \psi\,\left|\,A\,\right|\,\psi\right\rangle^2 \right)^{{1\over2}} = \left( \left\langle \psi\,\left|\,\left(A - \langle \psi\,|\,A\,|\,\psi\rangle I\right)^2\,\right|\,\psi \right\rangle \right)^{{1\over2}}.$$So, for example, the dispersion is nonnegative, and vanishes if and only if $\psi$ is an eigenvector of $A$.

Since $\Delta A$ is merely a number – and not a Hermitian operator on the Hilbert space – we cannot of course make an observation via $\Delta A$. But we can do the next best thing. Introduce two ensembles of copies of system $\mathcal{S}$, each copy in initial state $\psi$. Allow an instrument to observe in succession the systems in the first ensemble, via $A$; and a second instrument for the second ensemble, via $A^2$. Finally, introduce a third instrument that observes the first two instruments via an appropriate operator to "compute $\Delta A$. In this sense, then, the number $\Delta A$ has direct physical significance.

Theorem 1. Let $A$ and $B$ be $($bounded$)$ Hermitian operators on Hilbert space $\mathcal{S}$, and $\psi$ a unit vector. Then $$\Delta_\psi A \Delta_\psi B \ge {1\over2}\left|\langle \psi\,\left|\,[A, B]\,\right|\,\psi\rangle\right|.$$

Proof. This is the standard uncertainty relations for noncommuting observables in quantum mechanics.

$\tag*{$\square$}$

But what of the "energy-time uncertainty relation?"

Theorem 2. Let $H$ be a $($bounded$)$ Hermitian operator on Hilbert space $\mathcal{S}$, and $\psi$ a unit vector. Set $P = I - |\psi\rangle\langle\psi|$, the projection operator orthogonal to $\psi$, and $\psi_t = e^{{{-i}\over{\hbar}} Ht\psi}$. Then, for every number $\Delta t \ge 0$, we have$$\Delta_\psi H \Delta t \ge \hbar \left|\langle \psi_{\Delta t}\, |\, P \,|\, \psi_{\Delta t}\rangle\right|^{{1\over2}}.$$

Proof. Set, for each $t \ge 0$, $\alpha(t) = \langle \psi_t \,|\,P\,|\,\psi_t\rangle$. Then we have $$\left|{{d\alpha}\over{dt}}\right| = \left|\left\langle \psi_t\, \left|\, {i\over\hbar} [H, P]\,\right|\, \psi_t\right\rangle\right| = {2\over\hbar} \Delta_{\psi_t} H \Delta_{\psi_t} P,$$where we used Theorem 1 in the second step. but $\Delta_{\psi_t}H = \Delta_\psi H$, and $$\Delta_{\psi_t} P = \left( \left\langle \psi_t\,\left|\,P^2\,\right|\,\psi_t \right\rangle - \left\langle \psi_t\,\left|\,P\,\right|\,\psi_t\right\rangle^2 \right)^{{1\over2}} = \left(\alpha - \alpha^2\right)^{1\over2} \le \alpha^{1\over2}.$$Substituting, we obtain $$\left|{{d\alpha}\over{dt}}\right| \le \left({2\over\hbar}\right) \left(\Delta_\psi H\right) \alpha^{1\over2}.$$Dividing this inequality by $\alpha^{1\over2}$, integrating from $t = 0$ to $t = \Delta t$, and using that $\alpha(0) = 0$, the result follows.

$\tag*{$\square$}$

To apply this theorem to physical problems, we choose for $H$, of course, the Hamiltonian of the system. Then $\psi_t$ is the solution of Schrödinger's equation with initial $(t = 0)$ state $\psi$. Further, $\Delta_{\psi} H$ is the energy dispersion relative to this initial state. We may interpret $\left|\left\langle \psi_t\,\left|\,P\,\right|\,\psi_t\right\rangle\right|^{1\over2} = \|P\psi_t\|$ as a measure of how much the state $\psi_t$ differs from the initial state $\psi$. Thus, this expression vanishes for $t = 0$ $($when $\psi_t = \psi$$)$, and grows from zero only as $\psi_t$ acquires a component orthogonal to $\psi$. So, the theorem states, roughly the following: "In order to obtain an evolved state $(\psi_{\Delta t})$ appreciably different from the initial state $(\psi)$, you must wait sufficiently long $(\Delta t)$, and have sufficient dispersion in the initial state $(\Delta_\psi H)$ that the product of these two is greater than the order of $\hbar$." Of course, the theorem, with these choices, does make a testable prediction of quantum mechanics, using suitable ensembles as described earlier.


One cannot, in time $\Delta t$, measure the energy of a system within error $\Delta E$ unless $\Delta E\Delta t \ge \hbar$.

I do not know what this statement means, for I do not know what experiment is being contemplated. The part "...measure the energy of a system within error..." suggests the idea that quantum systems have some "actual energy." But, according to quantum mechanics, they do not. What they "have" is a ray in their Hilbert space of states, while the energy is an operator on this Hilbert space. (Perhaps this idea is a holdover from classical mechanics, in which systems do "have an energy," for there the energy is a function on the space of classical states.) Further, even if we thought of quantum systems as having some true energy, it is not clear how we are supposed to acquire the information as to the discrepency between this true energy and our measured value.

On an ensemble of systems, all in initial state $\psi$, let there be made a determination, carried out in time $\Delta t$, of the dispersion of the Hamiltonian, $\Delta_\psi H$. Then $\Delta_\psi H \Delta t \ge \hbar$. $($See earlier.$)$

This is intended as a clarified version of the first statement. Now, it is claimed, the experiment is more or less clear. But, unfortunately, this statement is false. For fixed $\psi$ and $H$, I see no obstacle to making this determination in an arbitrarily small time $\Delta t$. We merely turn the interaction on and then off quickly.

Consider an ensemble of systems, all in initial state $\psi$, and an ensemble of instruments, each of which will make an observation on a corresponding system via the Hamiltonian $H$, in time $\Delta t$. Introduce observable $E$ on the instrument Hilbert space corresponding to "the reading of the instrument." Determine the dispersion of $E$, $\Delta E$. Then $\Delta E \Delta t \ge \hbar$.

This is intended as a second possible version of the first statement. It is perhaps more in the spirit of that statement, for now we determine the dispersion of the "energy readings of the instruments," and not of the system's energy. But this statement is also false in quantum mechanics.

A measurement of the energy of a system, made in time $\Delta t$, will disturb the energy of that system by at least amount $\Delta E$, where $\Delta E \Delta t \ge \hbar$.

I do not know what this system means, for I do not know what experiment is contemplated. Again, the part about "...disturb the energy of that system..." suggests that systems in quantum mechanics "have a true energy." But even if they did, it is not clear how we are supposed to acquire the knowledge of by how much that energy was disturbed.

On an ensemble of systems, all in initial state $\psi$, let there be made an observation via the Hamiltonian of each system, in time $\Delta t$. Subsequent to this, let there be made on this ensemble a determination of the dispersion in the Hamiltonian, $\Delta H$. $($See earlier.$)$ Then $\Delta H \Delta t \ge \hbar$.

This is intended as a clarified version of the statement above. Now, we make an energy observation on each system in the ensemble, and then, for the ensemble taken as a whole, we determine its Hamiltonian-dispersion. The idea is that this "subsequent Hamiltonian-dispersion" will be at least so large, by virtue of the earlier observation, via the Hamiltonian, at time $\Delta t$. But this last statement is false. For instance, suppose that the initial state $\psi$ were a Hamiltonian eigenstate. Then the result of the first observation would leave each system in that eigenstate; whence the determination of $\Delta H$ would yield zero. Clearly, then, we would in this case violate the assertion above.


I do not know whether there are any statements that are both meaningful and true, along the lines above. But there does seem to be at least one statement that does appear to reflect "energy-time uncertainty," i.e. Theorem 2.

Another paper that is relevant is one by Sorkin $($Foundations of Phys, 9, 123 $($1979$)$$)$. While it is true that one can time-Fourier-transform the wave function, and thereby derive an "uncertainty relation" between $\Delta t$ and $\Delta \omega$ $($frequency$)$, it is not clear what that means physically. I could see someone quoting the old Einstein-Bohr argument, and in particular the example of the box hanging by a spring, emitting a photon under the control of a shutter, but I have never really understood what that debate was about.

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Ok, so the Energy-Time "Uncertainty Principle" is often way misrepresented, but it does mean something.

In $\Delta t \Delta E \geq \frac{\hbar}{2}$ the $\Delta t$ represents the amount of time is takes for the energy expectation value to change by one standard deviation. It does not represent a measurement.

Note: This is explained in Griffith's "Quantum Mechanics" text. I also just looked at my notes for this particular topic and can edit in the derivation later. It's just derived from the generic uncertainty principle.

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    $\begingroup$ See Joshphysics's answer physics.stackexchange.com/q/53802 $\endgroup$ – WetSavannaAnimal Mar 9 '15 at 21:41
  • $\begingroup$ @Sean I disagree. See my answer. $\endgroup$ – Sofia Mar 9 '15 at 21:57
  • $\begingroup$ @Sean I am sorry, you say $\Delta t$ represents the amount of time is takes for the energy expectation value to change by one standard deviation. No, the energy doesn't change. The energy is undefined. $\endgroup$ – Sofia Mar 9 '15 at 22:31
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Time is not an operator, s.t. this type of uncertainty relation differs from the usual form, in which participate operators. (There are though attempts to define time as an operator in time-arrival problems). And though, in nuclear physics we are well acquainted with unstable (radioactive) nuclei. The unstable nuclei are so-called resonant states, and the energy of such states is not well-defined, it has a width $\Gamma$. As these states decay, they also have a so-called half-life. Indeed these two quantities obey the uncertainty principle about time and energy, $\Gamma \cdot \tau_{1/2}$ is of the order of $\hbar$. It is not exactly $\hbar/2$ unless we adjust $\tau_{1/2}$ by some multiplicative constant to obtain a suitable $\Delta t$ so as to obtain $\hbar/2$.

Now, $\Delta E$ does not represent variation in time of the energy, when the energy is undefined, that doesn't mean that it varies in time, but that the system is not in an eigenstate of the Hamiltonian.

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  • $\begingroup$ But $\Delta t$ is not an observable , hence the confusion over where the energy time uncertainty principle comes from $\endgroup$ – Sean Mar 9 '15 at 22:00
  • $\begingroup$ @Sean no delta is an observable, $\Delta E$ is also not an observable. As to the $\tau_{1/2}$ it is a central issue in the nuclear physics, with models that predict, with an arsenal of formulas. There are even people that speak of operator time, but I am not among them. $\endgroup$ – Sofia Mar 9 '15 at 22:01
  • $\begingroup$ @Sean in fact, see the question Is there a prediction of quantum mechanics that could be construed as representing an energy-time uncertainty relation?. Of course, nuclear models are, among other things, for predicting $tau_{1/2}$. $\endgroup$ – Sofia Mar 9 '15 at 22:05
  • $\begingroup$ Yeah, that's this question... I answered it... $\endgroup$ – Sean Mar 9 '15 at 22:05

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