# What is $\Delta t$ in the time-energy uncertainty principle?

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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Nice explanation by Baez – Cosmas Zachos Jun 7 at 19:16

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 $$\Delta\langle\Omega\rangle =\int_t^{t+\Delta t} \left|\frac{d\langle \Omega\rangle}{dt}\right|dt \approx \left|\frac{d\langle \Omega\rangle}{dt}\right|\Delta t = \sigma_\Omega$$

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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
@BenCrowell Since you're active again, perhaps you could write an answer that expounds your interpretation along with some discussion of how it is a valid one? Welcome back btw. – joshphysics Aug 8 '14 at 1:54
@BenCrowell I would also be interested in an answer from your perspective. $\Delta t$ has always seemed quite shady to me. – Danu Sep 27 '14 at 10:37
@BenCrowell I agree with joshphysics' suggestion. – BMS Oct 21 '14 at 20:10
@joshphysics: Hi! I searched for eq. 3.72 but I got the relation in eq. 3.149 in 1995 edition. So, it would be better to mention which edition you were talking about. – MAFIA36790 May 5 at 17:17

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

The time-energy uncertainty relation (and other time-"observable" uncertainty relations that can be constructed) is (considered) not to have same meaning as canonical uncertainty relations. Meaning uncertainty relations costructed from canonical dynamical variables/observables (in the Hamiltonian sense), like position and momentum, since time parameter is not an observable and also not an operator in QM/QFT formalisms.

In fact, there are various approaches and interpretations of time-energy uncertainty. For example:

1. Energy-dispersion ($\Delta E$) of a state and lifetime ($\Delta t$ or $\tau_s$) of the state itself.

2. Energy exchange ($\Delta E$) and time-frame ($\Delta t$) during which this can happen.

3. Energy measurement ($\Delta E$) and time ($\Delta t$) it needs for accuracy (although this is rigorously disputed, see below )

4. ..other similar or specialised formulations of the above

In L. Mandelstam and I. Tamm, "The uncertainty relation between energy and time in nonrelativistic quantum mechanics", J Phys (USSR) 1945, they show how one can derive time-observable uncertainty relations for any observable $A$ with

$$\Delta t = \tau_A = \frac{\Delta A}{d\left<A\right> /dt}$$

Time and time-energy uncertainty is used heavily in (quantum/mixed) statistical mechanics of systems since it relates half-times and life-times of states and transitions (will have to find some references)

An analysis of various formulations of time-energy uncertainty relations can be found in:

Jan Hilgevoord, The uncertainty principle for energy and time I

and

Jan Hilgevoord, The uncertainty principle for energy and time II

Summary:

The uncertainty principle for energy and time is not a canonical uncertainty relation because it is not based/produced by canonical hamiltonian variables, instead it expresses dispersion and lifetime of a state. There is a confusion of a cartesian space-time $x, t$ (used as parameters) and canonical position and momenta ($q, p$) which are functions of these parameters (however simple in some cases, like $q=x$)

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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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sorry, I'm kind of confused by this interpretation. What exactly does it mean by spending more time measuring the experiment? In order to calculate the standard deviation, we need to measure the different possible results using identically prepared systems and figure out the corresponding probabilities. – M. Zeng Mar 29 '15 at 12:50

Here's another interpretation of the relation $\Delta t \, \Delta E \ge \frac{\hbar}{2}$.

You have a classical system described by a lagrangian $L = \dot{q} \, p - H$, where $H$ is the hamiltonian which is supposed to be time independant. The system's action is $$\tag{1} S = \int_{t_1}^{t_2} L \, dt = \int_{q_1}^{q_2} p \, dq - E \, \Delta t = S_p + S_E.$$ Now consider an arbitrary variation of the classical path. The action would then be changed by the following amount (I'm now sure what to do with the first part, which should give the other Heisenberg relation : $\Delta q \; \Delta p \ge \frac{\hbar}{2}$) : $$\tag{2} \delta S_E = -\: \delta E \, \Delta t.$$ It is postulated that any variation that change the action by an amount less than $\frac{\hbar}{2}$ cannot be observable. This is similar to the minimal cell of phase space in statistical mechanics, for which $\Delta q_{\text{min}} \, \Delta p_{\text{min}} \sim h \equiv 2 \pi \hbar$. Thus, for observable processes we have $|\, \delta S_E | \ge \frac{\hbar}{2}$, which implies the relation $$\tag{3} \Delta t \; \delta E \ge \frac{\hbar}{2}.$$ Here, $\Delta t \equiv t_2 - t_1$ is just the time interval that defines the boundary of the action (1) above. This is a classical "ordinary" interval of time. $\delta E$ is the amount of energy variation that you could get relative to the classical value, during that time interval. If $\Delta t$ is large, then $\delta E$ must be low (only small variations from the classical motion are allowed).

This "derivation" is very rough and is certainly not rigorous.

I think that the "reversed" inequality (for the unobservable variations) is more interesting : $$\tag{4} \Delta t \; \delta E < \frac{\hbar}{2}.$$ This says that any "violation" of the energy is allowed during a very short time interval.

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In addition to Joshphysics' precise answer, let's mention another interpretation (the one I think Ben Crowell is referring to in his comment to the same answer).

There's a formula from time dependent perturbation theory which gives the probability of an induced transition from an initial state $\lvert i \rangle$ to a final state $\lvert f \rangle$ with energy difference $\hbar \omega_{if}$. The transition is supposed to be induced from an harmonic perturbation: $$V=\cal Ve^{i\omega t}+\cal V ^\dagger e^{-i\omega t},$$ and the formula reads, for absorption i.e. transition to an higher energy level:$$P_{i\to f}(t;\omega)=\dfrac{\lvert \cal V _{fi} \rvert ^2}{\hbar ^2}\dfrac{\sin ^2(\frac{\omega _{fi}-\omega}{2}t)}{(\frac{\omega _{fi}-\omega}{2})^2}.$$

As a function of $t$ for fixed $\omega$, the probability grows quadratically for small $t$, reaches its maximum at $t$ given by:$$\frac{\lvert \omega _{fi}-\omega \rvert}{2}t=\frac {\pi} {2},$$ that is: $$t\Delta E =\frac {h}{2},$$where $$\Delta E = \lvert E_f -E_i -\hbar \omega \rvert.$$

Suppose that I am trying to cause a transition beetween two energy levels $i,f$ of an atom by sending on it some radiation at frequency $\omega$. Then $\Delta t$ is the order of the required lenght of the interaction to have a consistent probability of a transition (note that the above formula for $P_{i\to f}$ makes sense at $t=t_{\text{max}}$ only if $|V _{fi}|\ll \Delta E$).

Instead of fixing $\omega$, we could imagine to fix the time of interaction $\Delta t$. Again, the above formula for $P_{i\to f}$ says that we have a consistent probability for the transition to occur if $\Delta E \ll \frac{h} {\Delta t}$. Therefore, if we wish to determine $E_f -E_i$ precisely enough by varying $\omega$ and seeing if the transition does or does not occur, we must have a big $\Delta t$.

Here I'm considering transition beetween two distinct levels and I'm assuming that the spectrum is discrete, in the physical sense, that is, $|E_f'-E_i-(E_f-E_i)|$ for every other level $f'$ is much bigger than the experimental uncertainty on $\hbar \omega$. If this was not the case, we should consider transition not to a single final state but to a group $[f]$ of final states. The correct way to do this is by Fermi's golden rule, which is discussed in every good book of quantum mechanics (see e.g. Sakurai or Griffiths, also for the derivation of the above formula).

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