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In non-relativistic QM, the $\Delta E$ in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of $n$ identically prepared systems as $n$ goes to infinity. What does the $\Delta t$ mean, since $t$ is not even an observable?

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up vote 15 down vote accepted

Let a quantum system with Hamiltonian $H$ be given. Suppose the system occupies a pure state $|\psi(t)\rangle$ determined by the Hamiltonian evolution. For any observable $\Omega$ we use the shorthand $$ \langle \Omega \rangle = \langle \psi(t)|\Omega|\psi(t)\rangle. $$ One can show that (see eq. 3.72 in Griffiths QM) $$ \sigma_H\sigma_\Omega\geq\frac{\hbar}{2}\left|\frac{d\langle \Omega\rangle}{dt}\right| $$ where $\sigma_H$ and $\sigma_\Omega$ are standard deviations $$ \sigma_H^2 = \langle H^2\rangle-\langle H\rangle^2, \qquad \sigma_\Omega^2 = \langle \Omega^2\rangle-\langle \Omega\rangle^2 $$ and angled brackets mean expectation in $|\psi(t)\rangle$. It follows that if we define $$ \Delta E = \sigma_H, \qquad \Delta t = \frac{\sigma_\Omega}{|d\langle\Omega\rangle/dt|} $$ then we obtain the desired uncertainty relation $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$ It remains to interpret the quantity $\Delta t$. It tells you the approximate amount of time it takes for the expectation value of an observable to change by a standard deviation provided the system is in a pure state. To see this, note that if $\Delta t$ is small, then in a time $\Delta t$ we have $$ \Delta\langle\Omega\rangle =\int_t^{t+\Delta t} \left|\frac{d\langle \Omega\rangle}{dt}\right|dt \approx \left|\frac{d\langle \Omega\rangle}{dt}\right|\Delta t = \sigma_\Omega $$

Cheers! ;)

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+1, nice answer. However, it could give the impression that this is the only possible way of interpreting $\Delta t$, which I assume you're not claiming. –  Ben Crowell Jul 27 '13 at 15:55
@BenCrowell Thanks. Yeah I'm definitely not making any uniqueness of interpretation claims. Do you have others? I'm actually pretty curious. –  joshphysics Jul 27 '13 at 18:07
Another interpretation 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. –  Ben Crowell Jul 27 '13 at 20:36
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The time-energy uncertaintly relation has a different interpration and derivation than the uncertaintly relation for non-commuting operators. Try John Baez for an explanation, but, roughly speaking $\delta t$ measures the time it takes for the expectation value of some operator to change noticeably.

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The link is useful, but this is basically a link-only answer. Joshphysics' answer has given a self-contained presentation of the content of Baez's page. –  Ben Crowell Jul 27 '13 at 15:59
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The meaning is pretty much the same as for coordinate-momentum uncertainty. In addition to what joshphysics wrote, I'd like to stress that stationary solution of time-dependent Schroedinger equation is $\vert \psi \rangle \sim e^{i \frac{E}{\hbar}t}$. If you want to measure energy, you should somehow follow this wavefunction evolution in time. To measure energy definitely, you should measure it during infinite time. If the time of measurement is limited, the energy is not definite.

Technically it is more complicated as normally $\Delta t$ is not the measurement time, but the time of some process results of which you measure. However, the main idea is that simple.

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Good answers have been given so far. Let us see it from a different perspective:

Think of two eletrons interacting with each other very briefly. This interaction takes place by means of energy exchage, and let us say this is an amount $\Delta E$. The time $\Delta T$ within which this energy must be exchanged between the two electrons has a limit, and is dictated by Heisenberg's uncertainty principle. The higher the amount of energy exchanged, the shorter the time it should take to exchange it. This is taken care by nature, the electrons just do what they have to do; they exchange energy 'folowing the rules.'

Similarly, a free photon carries an amount of energy $E=hf$. This also has the meaning of Heisenberg's uncertainty principle if you write it in the form $E\times T=h$, since $f=1/T$. This amount of energy, will be carried by the photon a distance of one wavelength, $\lambda =c/f$, in no longer or shorter time than the period of its probability wave. This also applies when we interact with nature durng a measurement, as has been mentioned by other respondens. Nature is very keen in optimising her action, she is not wasteful. A good question is: Why is $h$ as small as it is? What determines its value? I am not aware of any facility that will produce this number, other than measured experimentally.

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In addition to what mentioned in @Michael's link, one of the best ways to think about is as follows:

The more time you spending in measuring your experiment (thus standard deviation will become smaller) the more precisely you will measure energy of this system.

P.S This interpretation widely used in Russian text books.

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